Aieee 2012 Solution (12th may)

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Aieee Online 2012 12th May(paper 3) phy & maths Solution

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Detailed solution given below

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By

Saurav gupta

Electronics amp telecomm engg(2nd year)

Jadavpur university

Q89

1)

2)

3)

4)

Ans 2

Q90

1)

2)

3)

4)

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

Q3

1)

2)

3)

4)

Ans 2

Q4

1)

2)

3)

4)

Ans 2

Q5

1)

2)

3)

4)

Ans 4

Q6

1)

2)

3)

4)

Ans 2

Q7

1)

2)

3)

4)

Ans 4

Q8

1)

2)

3)

4)

Ans 3

Q9

1)

2)

3)

4)

Ans 3

Q10

1)

2)

3)

4)

Ans 4

Q11

1)

2)

3)

4)

Ans 1

Q12

1)

2)

3)

4)

Ans 1

Q13

1)

2)

3)

4)

Ans 4

Q14

1)

2)

3)

4)

Ans 4

Q15

1)

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3)

4)

Ans 3

Q16

1)

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3)

4)

Ans 4

Q17

1)

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3)

4)

Ans 1

Q18

1)

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3)

4)

Ans 4

Q19

1)

2)

3)

4)

Ans 1

Q20

1)

2)

3)

4)

Ans 2

Q21

1)

2)

3)

4)

Ans 3

Q22

1)

2)

3)

4)

Ans 4

Q23

1)

2)

3)

4)

Ans 4

Q24

1)

2)

3)

4)

Ans 2

Q25

1)

2)

3)

4)

Ans 3

Q26

1)

2)

3)

4)

Ans 4

Q27

1)

2)

3)

4)

Ans 4

Q28

1)

2)

3)

4)

Ans 1

Q29

1)

2)

3)

4)

Ans 1

Q30

1)

2)

3)

4)

Ans 3

Q61

1)

2)

3)

4)

Ans 4

Q62

1)

2)

3)

4)

Ans 2

Q63

1)

2)

3)

4)

Ans 1

Q64

1)

2)

3)

4)

Ans 4

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 3

Q67

1)

2)

3)

4)

Ans 2

Q68

1)

2)

3)

4)

Ans 2

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 1

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 2

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 3

Q75

1)

2)

3)

4)

Ans 3

Q76

1)

2)

3)

4)

Ans 2

Q77

1)

2)

3)

4)

Ans 4

Q78

1)

2)

3)

4)

Ans 1

Q79

1)

2)

3)

4)

Ans 1

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 4

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 3

Q86

1)

2)

3)

4)

Ans 2

Q87

1)

2)

3)

4)

Ans 1

Q88

1)

2)

3)

4)

Ans 3

By

Saurav gupta

Electronics amp telecomm engg(2nd year)

Jadavpur university

Q89

1)

2)

3)

4)

Ans 2

Q90

1)

2)

3)

4)

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

Q5

1)

2)

3)

4)

Ans 4

Q6

1)

2)

3)

4)

Ans 2

Q7

1)

2)

3)

4)

Ans 4

Q8

1)

2)

3)

4)

Ans 3

Q9

1)

2)

3)

4)

Ans 3

Q10

1)

2)

3)

4)

Ans 4

Q11

1)

2)

3)

4)

Ans 1

Q12

1)

2)

3)

4)

Ans 1

Q13

1)

2)

3)

4)

Ans 4

Q14

1)

2)

3)

4)

Ans 4

Q15

1)

2)

3)

4)

Ans 3

Q16

1)

2)

3)

4)

Ans 4

Q17

1)

2)

3)

4)

Ans 1

Q18

1)

2)

3)

4)

Ans 4

Q19

1)

2)

3)

4)

Ans 1

Q20

1)

2)

3)

4)

Ans 2

Q21

1)

2)

3)

4)

Ans 3

Q22

1)

2)

3)

4)

Ans 4

Q23

1)

2)

3)

4)

Ans 4

Q24

1)

2)

3)

4)

Ans 2

Q25

1)

2)

3)

4)

Ans 3

Q26

1)

2)

3)

4)

Ans 4

Q27

1)

2)

3)

4)

Ans 4

Q28

1)

2)

3)

4)

Ans 1

Q29

1)

2)

3)

4)

Ans 1

Q30

1)

2)

3)

4)

Ans 3

Q61

1)

2)

3)

4)

Ans 4

Q62

1)

2)

3)

4)

Ans 2

Q63

1)

2)

3)

4)

Ans 1

Q64

1)

2)

3)

4)

Ans 4

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 3

Q67

1)

2)

3)

4)

Ans 2

Q68

1)

2)

3)

4)

Ans 2

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 1

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 2

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 3

Q75

1)

2)

3)

4)

Ans 3

Q76

1)

2)

3)

4)

Ans 2

Q77

1)

2)

3)

4)

Ans 4

Q78

1)

2)

3)

4)

Ans 1

Q79

1)

2)

3)

4)

Ans 1

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 4

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 3

Q86

1)

2)

3)

4)

Ans 2

Q87

1)

2)

3)

4)

Ans 1

Q88

1)

2)

3)

4)

Ans 3

By

Saurav gupta

Electronics amp telecomm engg(2nd year)

Jadavpur university

Q89

1)

2)

3)

4)

Ans 2

Q90

1)

2)

3)

4)

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

Q7

1)

2)

3)

4)

Ans 4

Q8

1)

2)

3)

4)

Ans 3

Q9

1)

2)

3)

4)

Ans 3

Q10

1)

2)

3)

4)

Ans 4

Q11

1)

2)

3)

4)

Ans 1

Q12

1)

2)

3)

4)

Ans 1

Q13

1)

2)

3)

4)

Ans 4

Q14

1)

2)

3)

4)

Ans 4

Q15

1)

2)

3)

4)

Ans 3

Q16

1)

2)

3)

4)

Ans 4

Q17

1)

2)

3)

4)

Ans 1

Q18

1)

2)

3)

4)

Ans 4

Q19

1)

2)

3)

4)

Ans 1

Q20

1)

2)

3)

4)

Ans 2

Q21

1)

2)

3)

4)

Ans 3

Q22

1)

2)

3)

4)

Ans 4

Q23

1)

2)

3)

4)

Ans 4

Q24

1)

2)

3)

4)

Ans 2

Q25

1)

2)

3)

4)

Ans 3

Q26

1)

2)

3)

4)

Ans 4

Q27

1)

2)

3)

4)

Ans 4

Q28

1)

2)

3)

4)

Ans 1

Q29

1)

2)

3)

4)

Ans 1

Q30

1)

2)

3)

4)

Ans 3

Q61

1)

2)

3)

4)

Ans 4

Q62

1)

2)

3)

4)

Ans 2

Q63

1)

2)

3)

4)

Ans 1

Q64

1)

2)

3)

4)

Ans 4

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 3

Q67

1)

2)

3)

4)

Ans 2

Q68

1)

2)

3)

4)

Ans 2

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 1

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 2

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 3

Q75

1)

2)

3)

4)

Ans 3

Q76

1)

2)

3)

4)

Ans 2

Q77

1)

2)

3)

4)

Ans 4

Q78

1)

2)

3)

4)

Ans 1

Q79

1)

2)

3)

4)

Ans 1

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 4

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 3

Q86

1)

2)

3)

4)

Ans 2

Q87

1)

2)

3)

4)

Ans 1

Q88

1)

2)

3)

4)

Ans 3

By

Saurav gupta

Electronics amp telecomm engg(2nd year)

Jadavpur university

Q89

1)

2)

3)

4)

Ans 2

Q90

1)

2)

3)

4)

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

Q10

1)

2)

3)

4)

Ans 4

Q11

1)

2)

3)

4)

Ans 1

Q12

1)

2)

3)

4)

Ans 1

Q13

1)

2)

3)

4)

Ans 4

Q14

1)

2)

3)

4)

Ans 4

Q15

1)

2)

3)

4)

Ans 3

Q16

1)

2)

3)

4)

Ans 4

Q17

1)

2)

3)

4)

Ans 1

Q18

1)

2)

3)

4)

Ans 4

Q19

1)

2)

3)

4)

Ans 1

Q20

1)

2)

3)

4)

Ans 2

Q21

1)

2)

3)

4)

Ans 3

Q22

1)

2)

3)

4)

Ans 4

Q23

1)

2)

3)

4)

Ans 4

Q24

1)

2)

3)

4)

Ans 2

Q25

1)

2)

3)

4)

Ans 3

Q26

1)

2)

3)

4)

Ans 4

Q27

1)

2)

3)

4)

Ans 4

Q28

1)

2)

3)

4)

Ans 1

Q29

1)

2)

3)

4)

Ans 1

Q30

1)

2)

3)

4)

Ans 3

Q61

1)

2)

3)

4)

Ans 4

Q62

1)

2)

3)

4)

Ans 2

Q63

1)

2)

3)

4)

Ans 1

Q64

1)

2)

3)

4)

Ans 4

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 3

Q67

1)

2)

3)

4)

Ans 2

Q68

1)

2)

3)

4)

Ans 2

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 1

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 2

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 3

Q75

1)

2)

3)

4)

Ans 3

Q76

1)

2)

3)

4)

Ans 2

Q77

1)

2)

3)

4)

Ans 4

Q78

1)

2)

3)

4)

Ans 1

Q79

1)

2)

3)

4)

Ans 1

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 4

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 3

Q86

1)

2)

3)

4)

Ans 2

Q87

1)

2)

3)

4)

Ans 1

Q88

1)

2)

3)

4)

Ans 3

By

Saurav gupta

Electronics amp telecomm engg(2nd year)

Jadavpur university

Q89

1)

2)

3)

4)

Ans 2

Q90

1)

2)

3)

4)

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

Q12

1)

2)

3)

4)

Ans 1

Q13

1)

2)

3)

4)

Ans 4

Q14

1)

2)

3)

4)

Ans 4

Q15

1)

2)

3)

4)

Ans 3

Q16

1)

2)

3)

4)

Ans 4

Q17

1)

2)

3)

4)

Ans 1

Q18

1)

2)

3)

4)

Ans 4

Q19

1)

2)

3)

4)

Ans 1

Q20

1)

2)

3)

4)

Ans 2

Q21

1)

2)

3)

4)

Ans 3

Q22

1)

2)

3)

4)

Ans 4

Q23

1)

2)

3)

4)

Ans 4

Q24

1)

2)

3)

4)

Ans 2

Q25

1)

2)

3)

4)

Ans 3

Q26

1)

2)

3)

4)

Ans 4

Q27

1)

2)

3)

4)

Ans 4

Q28

1)

2)

3)

4)

Ans 1

Q29

1)

2)

3)

4)

Ans 1

Q30

1)

2)

3)

4)

Ans 3

Q61

1)

2)

3)

4)

Ans 4

Q62

1)

2)

3)

4)

Ans 2

Q63

1)

2)

3)

4)

Ans 1

Q64

1)

2)

3)

4)

Ans 4

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 3

Q67

1)

2)

3)

4)

Ans 2

Q68

1)

2)

3)

4)

Ans 2

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 1

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 2

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 3

Q75

1)

2)

3)

4)

Ans 3

Q76

1)

2)

3)

4)

Ans 2

Q77

1)

2)

3)

4)

Ans 4

Q78

1)

2)

3)

4)

Ans 1

Q79

1)

2)

3)

4)

Ans 1

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 4

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 3

Q86

1)

2)

3)

4)

Ans 2

Q87

1)

2)

3)

4)

Ans 1

Q88

1)

2)

3)

4)

Ans 3

By

Saurav gupta

Electronics amp telecomm engg(2nd year)

Jadavpur university

Q89

1)

2)

3)

4)

Ans 2

Q90

1)

2)

3)

4)

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

Q14

1)

2)

3)

4)

Ans 4

Q15

1)

2)

3)

4)

Ans 3

Q16

1)

2)

3)

4)

Ans 4

Q17

1)

2)

3)

4)

Ans 1

Q18

1)

2)

3)

4)

Ans 4

Q19

1)

2)

3)

4)

Ans 1

Q20

1)

2)

3)

4)

Ans 2

Q21

1)

2)

3)

4)

Ans 3

Q22

1)

2)

3)

4)

Ans 4

Q23

1)

2)

3)

4)

Ans 4

Q24

1)

2)

3)

4)

Ans 2

Q25

1)

2)

3)

4)

Ans 3

Q26

1)

2)

3)

4)

Ans 4

Q27

1)

2)

3)

4)

Ans 4

Q28

1)

2)

3)

4)

Ans 1

Q29

1)

2)

3)

4)

Ans 1

Q30

1)

2)

3)

4)

Ans 3

Q61

1)

2)

3)

4)

Ans 4

Q62

1)

2)

3)

4)

Ans 2

Q63

1)

2)

3)

4)

Ans 1

Q64

1)

2)

3)

4)

Ans 4

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 3

Q67

1)

2)

3)

4)

Ans 2

Q68

1)

2)

3)

4)

Ans 2

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 1

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 2

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 3

Q75

1)

2)

3)

4)

Ans 3

Q76

1)

2)

3)

4)

Ans 2

Q77

1)

2)

3)

4)

Ans 4

Q78

1)

2)

3)

4)

Ans 1

Q79

1)

2)

3)

4)

Ans 1

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 4

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 3

Q86

1)

2)

3)

4)

Ans 2

Q87

1)

2)

3)

4)

Ans 1

Q88

1)

2)

3)

4)

Ans 3

By

Saurav gupta

Electronics amp telecomm engg(2nd year)

Jadavpur university

Q89

1)

2)

3)

4)

Ans 2

Q90

1)

2)

3)

4)

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

Q17

1)

2)

3)

4)

Ans 1

Q18

1)

2)

3)

4)

Ans 4

Q19

1)

2)

3)

4)

Ans 1

Q20

1)

2)

3)

4)

Ans 2

Q21

1)

2)

3)

4)

Ans 3

Q22

1)

2)

3)

4)

Ans 4

Q23

1)

2)

3)

4)

Ans 4

Q24

1)

2)

3)

4)

Ans 2

Q25

1)

2)

3)

4)

Ans 3

Q26

1)

2)

3)

4)

Ans 4

Q27

1)

2)

3)

4)

Ans 4

Q28

1)

2)

3)

4)

Ans 1

Q29

1)

2)

3)

4)

Ans 1

Q30

1)

2)

3)

4)

Ans 3

Q61

1)

2)

3)

4)

Ans 4

Q62

1)

2)

3)

4)

Ans 2

Q63

1)

2)

3)

4)

Ans 1

Q64

1)

2)

3)

4)

Ans 4

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 3

Q67

1)

2)

3)

4)

Ans 2

Q68

1)

2)

3)

4)

Ans 2

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 1

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 2

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 3

Q75

1)

2)

3)

4)

Ans 3

Q76

1)

2)

3)

4)

Ans 2

Q77

1)

2)

3)

4)

Ans 4

Q78

1)

2)

3)

4)

Ans 1

Q79

1)

2)

3)

4)

Ans 1

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 4

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 3

Q86

1)

2)

3)

4)

Ans 2

Q87

1)

2)

3)

4)

Ans 1

Q88

1)

2)

3)

4)

Ans 3

By

Saurav gupta

Electronics amp telecomm engg(2nd year)

Jadavpur university

Q89

1)

2)

3)

4)

Ans 2

Q90

1)

2)

3)

4)

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

Q20

1)

2)

3)

4)

Ans 2

Q21

1)

2)

3)

4)

Ans 3

Q22

1)

2)

3)

4)

Ans 4

Q23

1)

2)

3)

4)

Ans 4

Q24

1)

2)

3)

4)

Ans 2

Q25

1)

2)

3)

4)

Ans 3

Q26

1)

2)

3)

4)

Ans 4

Q27

1)

2)

3)

4)

Ans 4

Q28

1)

2)

3)

4)

Ans 1

Q29

1)

2)

3)

4)

Ans 1

Q30

1)

2)

3)

4)

Ans 3

Q61

1)

2)

3)

4)

Ans 4

Q62

1)

2)

3)

4)

Ans 2

Q63

1)

2)

3)

4)

Ans 1

Q64

1)

2)

3)

4)

Ans 4

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 3

Q67

1)

2)

3)

4)

Ans 2

Q68

1)

2)

3)

4)

Ans 2

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 1

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 2

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 3

Q75

1)

2)

3)

4)

Ans 3

Q76

1)

2)

3)

4)

Ans 2

Q77

1)

2)

3)

4)

Ans 4

Q78

1)

2)

3)

4)

Ans 1

Q79

1)

2)

3)

4)

Ans 1

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 4

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 3

Q86

1)

2)

3)

4)

Ans 2

Q87

1)

2)

3)

4)

Ans 1

Q88

1)

2)

3)

4)

Ans 3

By

Saurav gupta

Electronics amp telecomm engg(2nd year)

Jadavpur university

Q89

1)

2)

3)

4)

Ans 2

Q90

1)

2)

3)

4)

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

Q22

1)

2)

3)

4)

Ans 4

Q23

1)

2)

3)

4)

Ans 4

Q24

1)

2)

3)

4)

Ans 2

Q25

1)

2)

3)

4)

Ans 3

Q26

1)

2)

3)

4)

Ans 4

Q27

1)

2)

3)

4)

Ans 4

Q28

1)

2)

3)

4)

Ans 1

Q29

1)

2)

3)

4)

Ans 1

Q30

1)

2)

3)

4)

Ans 3

Q61

1)

2)

3)

4)

Ans 4

Q62

1)

2)

3)

4)

Ans 2

Q63

1)

2)

3)

4)

Ans 1

Q64

1)

2)

3)

4)

Ans 4

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 3

Q67

1)

2)

3)

4)

Ans 2

Q68

1)

2)

3)

4)

Ans 2

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 1

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 2

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 3

Q75

1)

2)

3)

4)

Ans 3

Q76

1)

2)

3)

4)

Ans 2

Q77

1)

2)

3)

4)

Ans 4

Q78

1)

2)

3)

4)

Ans 1

Q79

1)

2)

3)

4)

Ans 1

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 4

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 3

Q86

1)

2)

3)

4)

Ans 2

Q87

1)

2)

3)

4)

Ans 1

Q88

1)

2)

3)

4)

Ans 3

By

Saurav gupta

Electronics amp telecomm engg(2nd year)

Jadavpur university

Q89

1)

2)

3)

4)

Ans 2

Q90

1)

2)

3)

4)

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

Q24

1)

2)

3)

4)

Ans 2

Q25

1)

2)

3)

4)

Ans 3

Q26

1)

2)

3)

4)

Ans 4

Q27

1)

2)

3)

4)

Ans 4

Q28

1)

2)

3)

4)

Ans 1

Q29

1)

2)

3)

4)

Ans 1

Q30

1)

2)

3)

4)

Ans 3

Q61

1)

2)

3)

4)

Ans 4

Q62

1)

2)

3)

4)

Ans 2

Q63

1)

2)

3)

4)

Ans 1

Q64

1)

2)

3)

4)

Ans 4

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 3

Q67

1)

2)

3)

4)

Ans 2

Q68

1)

2)

3)

4)

Ans 2

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 1

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 2

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 3

Q75

1)

2)

3)

4)

Ans 3

Q76

1)

2)

3)

4)

Ans 2

Q77

1)

2)

3)

4)

Ans 4

Q78

1)

2)

3)

4)

Ans 1

Q79

1)

2)

3)

4)

Ans 1

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 4

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 3

Q86

1)

2)

3)

4)

Ans 2

Q87

1)

2)

3)

4)

Ans 1

Q88

1)

2)

3)

4)

Ans 3

By

Saurav gupta

Electronics amp telecomm engg(2nd year)

Jadavpur university

Q89

1)

2)

3)

4)

Ans 2

Q90

1)

2)

3)

4)

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

Q27

1)

2)

3)

4)

Ans 4

Q28

1)

2)

3)

4)

Ans 1

Q29

1)

2)

3)

4)

Ans 1

Q30

1)

2)

3)

4)

Ans 3

Q61

1)

2)

3)

4)

Ans 4

Q62

1)

2)

3)

4)

Ans 2

Q63

1)

2)

3)

4)

Ans 1

Q64

1)

2)

3)

4)

Ans 4

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 3

Q67

1)

2)

3)

4)

Ans 2

Q68

1)

2)

3)

4)

Ans 2

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 1

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 2

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 3

Q75

1)

2)

3)

4)

Ans 3

Q76

1)

2)

3)

4)

Ans 2

Q77

1)

2)

3)

4)

Ans 4

Q78

1)

2)

3)

4)

Ans 1

Q79

1)

2)

3)

4)

Ans 1

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 4

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 3

Q86

1)

2)

3)

4)

Ans 2

Q87

1)

2)

3)

4)

Ans 1

Q88

1)

2)

3)

4)

Ans 3

By

Saurav gupta

Electronics amp telecomm engg(2nd year)

Jadavpur university

Q89

1)

2)

3)

4)

Ans 2

Q90

1)

2)

3)

4)

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

Q29

1)

2)

3)

4)

Ans 1

Q30

1)

2)

3)

4)

Ans 3

Q61

1)

2)

3)

4)

Ans 4

Q62

1)

2)

3)

4)

Ans 2

Q63

1)

2)

3)

4)

Ans 1

Q64

1)

2)

3)

4)

Ans 4

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 3

Q67

1)

2)

3)

4)

Ans 2

Q68

1)

2)

3)

4)

Ans 2

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 1

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 2

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 3

Q75

1)

2)

3)

4)

Ans 3

Q76

1)

2)

3)

4)

Ans 2

Q77

1)

2)

3)

4)

Ans 4

Q78

1)

2)

3)

4)

Ans 1

Q79

1)

2)

3)

4)

Ans 1

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 4

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 3

Q86

1)

2)

3)

4)

Ans 2

Q87

1)

2)

3)

4)

Ans 1

Q88

1)

2)

3)

4)

Ans 3

By

Saurav gupta

Electronics amp telecomm engg(2nd year)

Jadavpur university

Q89

1)

2)

3)

4)

Ans 2

Q90

1)

2)

3)

4)

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

Q61

1)

2)

3)

4)

Ans 4

Q62

1)

2)

3)

4)

Ans 2

Q63

1)

2)

3)

4)

Ans 1

Q64

1)

2)

3)

4)

Ans 4

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 3

Q67

1)

2)

3)

4)

Ans 2

Q68

1)

2)

3)

4)

Ans 2

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 1

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 2

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 3

Q75

1)

2)

3)

4)

Ans 3

Q76

1)

2)

3)

4)

Ans 2

Q77

1)

2)

3)

4)

Ans 4

Q78

1)

2)

3)

4)

Ans 1

Q79

1)

2)

3)

4)

Ans 1

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 4

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 3

Q86

1)

2)

3)

4)

Ans 2

Q87

1)

2)

3)

4)

Ans 1

Q88

1)

2)

3)

4)

Ans 3

By

Saurav gupta

Electronics amp telecomm engg(2nd year)

Jadavpur university

Q89

1)

2)

3)

4)

Ans 2

Q90

1)

2)

3)

4)

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

Q63

1)

2)

3)

4)

Ans 1

Q64

1)

2)

3)

4)

Ans 4

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 3

Q67

1)

2)

3)

4)

Ans 2

Q68

1)

2)

3)

4)

Ans 2

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 1

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 2

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 3

Q75

1)

2)

3)

4)

Ans 3

Q76

1)

2)

3)

4)

Ans 2

Q77

1)

2)

3)

4)

Ans 4

Q78

1)

2)

3)

4)

Ans 1

Q79

1)

2)

3)

4)

Ans 1

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 4

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 3

Q86

1)

2)

3)

4)

Ans 2

Q87

1)

2)

3)

4)

Ans 1

Q88

1)

2)

3)

4)

Ans 3

By

Saurav gupta

Electronics amp telecomm engg(2nd year)

Jadavpur university

Q89

1)

2)

3)

4)

Ans 2

Q90

1)

2)

3)

4)

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

Q66

1)

2)

3)

4)

Ans 3

Q67

1)

2)

3)

4)

Ans 2

Q68

1)

2)

3)

4)

Ans 2

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 1

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 2

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 3

Q75

1)

2)

3)

4)

Ans 3

Q76

1)

2)

3)

4)

Ans 2

Q77

1)

2)

3)

4)

Ans 4

Q78

1)

2)

3)

4)

Ans 1

Q79

1)

2)

3)

4)

Ans 1

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 4

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 3

Q86

1)

2)

3)

4)

Ans 2

Q87

1)

2)

3)

4)

Ans 1

Q88

1)

2)

3)

4)

Ans 3

By

Saurav gupta

Electronics amp telecomm engg(2nd year)

Jadavpur university

Q89

1)

2)

3)

4)

Ans 2

Q90

1)

2)

3)

4)

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 1

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 2

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 3

Q75

1)

2)

3)

4)

Ans 3

Q76

1)

2)

3)

4)

Ans 2

Q77

1)

2)

3)

4)

Ans 4

Q78

1)

2)

3)

4)

Ans 1

Q79

1)

2)

3)

4)

Ans 1

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 4

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 3

Q86

1)

2)

3)

4)

Ans 2

Q87

1)

2)

3)

4)

Ans 1

Q88

1)

2)

3)

4)

Ans 3

By

Saurav gupta

Electronics amp telecomm engg(2nd year)

Jadavpur university

Q89

1)

2)

3)

4)

Ans 2

Q90

1)

2)

3)

4)

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

Q72

1)

2)

3)

4)

Ans 2

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 3

Q75

1)

2)

3)

4)

Ans 3

Q76

1)

2)

3)

4)

Ans 2

Q77

1)

2)

3)

4)

Ans 4

Q78

1)

2)

3)

4)

Ans 1

Q79

1)

2)

3)

4)

Ans 1

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 4

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 3

Q86

1)

2)

3)

4)

Ans 2

Q87

1)

2)

3)

4)

Ans 1

Q88

1)

2)

3)

4)

Ans 3

By

Saurav gupta

Electronics amp telecomm engg(2nd year)

Jadavpur university

Q89

1)

2)

3)

4)

Ans 2

Q90

1)

2)

3)

4)

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

Q75

1)

2)

3)

4)

Ans 3

Q76

1)

2)

3)

4)

Ans 2

Q77

1)

2)

3)

4)

Ans 4

Q78

1)

2)

3)

4)

Ans 1

Q79

1)

2)

3)

4)

Ans 1

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 4

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 3

Q86

1)

2)

3)

4)

Ans 2

Q87

1)

2)

3)

4)

Ans 1

Q88

1)

2)

3)

4)

Ans 3

By

Saurav gupta

Electronics amp telecomm engg(2nd year)

Jadavpur university

Q89

1)

2)

3)

4)

Ans 2

Q90

1)

2)

3)

4)

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

Q78

1)

2)

3)

4)

Ans 1

Q79

1)

2)

3)

4)

Ans 1

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 4

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 3

Q86

1)

2)

3)

4)

Ans 2

Q87

1)

2)

3)

4)

Ans 1

Q88

1)

2)

3)

4)

Ans 3

By

Saurav gupta

Electronics amp telecomm engg(2nd year)

Jadavpur university

Q89

1)

2)

3)

4)

Ans 2

Q90

1)

2)

3)

4)

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 4

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 3

Q86

1)

2)

3)

4)

Ans 2

Q87

1)

2)

3)

4)

Ans 1

Q88

1)

2)

3)

4)

Ans 3

By

Saurav gupta

Electronics amp telecomm engg(2nd year)

Jadavpur university

Q89

1)

2)

3)

4)

Ans 2

Q90

1)

2)

3)

4)

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 3

Q86

1)

2)

3)

4)

Ans 2

Q87

1)

2)

3)

4)

Ans 1

Q88

1)

2)

3)

4)

Ans 3

By

Saurav gupta

Electronics amp telecomm engg(2nd year)

Jadavpur university

Q89

1)

2)

3)

4)

Ans 2

Q90

1)

2)

3)

4)

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

Q86

1)

2)

3)

4)

Ans 2

Q87

1)

2)

3)

4)

Ans 1

Q88

1)

2)

3)

4)

Ans 3

By

Saurav gupta

Electronics amp telecomm engg(2nd year)

Jadavpur university

Q89

1)

2)

3)

4)

Ans 2

Q90

1)

2)

3)

4)

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

By

Saurav gupta

Electronics amp telecomm engg(2nd year)

Jadavpur university

Q89

1)

2)

3)

4)

Ans 2

Q90

1)

2)

3)

4)

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q1 Case 1

PV = Pᵢ x 2V

Pᵢ = P2

Case 2

PV⁵ sup3 = Pₐ (2V)⁵ sup3

Pₐ = P 2ˉ⁵ sup3

PₐPᵢ = 2ˉsup2 sup3

Ans 3

Q2 Since force is perpendicular to the motion of the particle it doesnrsquot do any work hence KE remains

unchanged

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q3 dMdt =αv

now

F = 0

dPdt=0

Mdvdt + vdMdt = 0

Ma + v x αv = 0

a = -vsup2αM

Ans 2

Q4 Case 1

N = Nᵢ( frac12 ) ⁰⁵

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

100 = 1600 ( frac12 ) ⁸ ⁰⁵

T frac12 = 2 sec

Case 2

N = 1600 frac12 )⁶ sup2

= 200

Ans 2

Q5 R = 2radich1 (H-h1)

For same range

h1 = h2

Ans 4

Q6

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

RP = 1dθ = 122λD

dθ = 122 x 55 x 10ˉ⁷3 x 10ˉsup2

= 223 x 10ˉ⁵

X(min) 80 = 223 x 10ˉ⁵

X(min) = 179 x 10ˉsup3

Hence minimum spacing should be 179 x 10ˉsup3 to be resolved

Ans 2

Q7 fact

Ans 4

Q8

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

I (max)I(min) = [(a+b)(a-b)]sup2

W1w2 = (ab)sup2

125 = (ab)sup2

ab = 15

I (max)I(min) = 94

Ans 3

Q9 The maximum energy of photoelectrons is

hυ-ф

Ans 3

Q10 L = constant

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Mωrsup2 = c

Now

T =mωsup2r

= (mrsup2ω)sup2mrsup3

= Lsup2mrsup3

= Arˉsup3

Where A = Lsup2m

Ans 4

Q11 r= (llrsquo - 1)R

given r =R

llrsquo ndash 1 = 1

llrsquo = 2

lrsquo = l2

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q12 V dvdx = -Gmxsup2

˳int V dv = -Gm intʳdxxsup2

Solving we get

Vsup2 = 2Gmr[1 ndash 1radic2]

Ans 1

Q13 Sinθ = 1μ

For total internal reflection

θ ge 45⁰

radic2r

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

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ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

sinθ ge 1radic2

1μ ge 1radic2

μ le radic2

μ le 1414

hence only green amp blue colours can be seen

Ans 4

Q14 V =2 rsup2(ρ-σ)g9ᶯ

V proprsup2

Ans 4

Q15 ωprop1radicc

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

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SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

ω1ω2=radicc2c1

now

c propϵ

ω1ω2=radicϵ2ϵ1

υ1υ2=radicϵ2ϵ1

10⁴(10⁴-50) = radicϵ21

ϵ2 =101

Ans 3

Q16 Repulsive force

Also no two electric field lines can intersect

Ans 4

Q17

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

E= cB

= 3x10⁸x482x10ˉsup1sup1

= 001446 NC

λ=cυ

= 360m

Ans 1

Q18

h d

case 1

d-h = vsup22g

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

h = d ndash vsup22g---(1)

case 2

usup2 = k x 2gd klt1

vsup2 = 2kgd ndash 2gh

h = kd ndash vsup22g---(2)

Ans 4

Q19

R prop lA

RrsquoR = (lrsquol)(AArsquo)

= frac12 x frac12

= frac14

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q20 Capacitanceofearth=4πϵ˳R

=4πx885x10ˉsup1sup2 x 64 x 10⁶

=711μF

Statement 2---false

Ans 2

Q21 Y = 1[1 + (x - t2)sup2]

Eqᶯ of wave

partsup2ypartsup2x = 1csup2 (partsup2ypartsup2t)

Solving we get

C= frac12

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q22 Force applied by you + frictional force = force

applied by horse

Statement 2---false

Ans 4

Q23 Forceunitlength=μ˳IIrsquo2πd

= 4 x 10ˉ⁵

Total force = 8 x 10ˉ⁵N

Ans 4

Q24 Case 1

V˳=[2gh(1+ksup2Rsup2)]⁰⁵---(1)

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Case 2

Vrsquo = radic2gh

5v˳4=radic2gh

Substituting this value in eqᶯ (1)

1 + ksup2Rsup2 = 2516

K =34 R

Ans 2

Q25 λ=4l(2n-1)

υ=330(2n-1)4l

l= 5(2n-1)16 m

put n=2

l = 9375cm

Ans 3

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q26 Temp will increase

Ans 4

Q27 Vprop1n

Ans 4

Q28 Y=(Arsquo+Brsquo)rsquo

= (Arsquo)rsquo(Brsquo)rsquo----de morganrsquos law

=AB

Ans 1

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY PHYSICS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q29 KE(final) = KE(initial) + work done

frac12 x 2 x 5sup2 + area of F-x diagram

=25 + 2 ndash 2 + 45

= 295

Ans 1

Q30 σ=Mˉsup1Lˉsup3Tsup3Asup2

Ans 3

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q61 a(b x c) = 0

=gt a | b x c

ie a is || to the plane of b and c

Ans 4

Q62 ax+2by+3b = 0

bx-2ay-3a=0

solving we get

x=0 y = -32

Ans 2

Q63

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

σsup2= Σxᵢsup2n ndash xsup2

x = [1 + 3 +5 +helliphellipupto n terms]n

= n

Σxᵢsup2 = 1sup2+3sup2+5sup2+helliphellipnterms

= n[4nsup2 - 1]3

Statement 2---false

σsup2= (nsup2-1)3

Ans 1

Q64 1 k 3

3 k -2 = 0

2 3 -4

K = 332

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 4

Q65 Point of intersection= (12) amp (44)

Area = int⁴ [2radicx ndash (2x+4)3] dx

=13

Ans 2

Q66 Let th eqᶯ of plane be

ax+by+cz=d

it passes through (07-7)

ax+b(y-7)+c(z+7)=0-----(1)

it also passes through (-13-2)

1

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+4b+5c = 0----(2)

now

dr of normal to the plane will be perpendicular to the line

-3a+2b+c = 0-----(3)

Solving (2) amp(3)

a1 = b1 = c1

substituting this value in eqᶯ (1)

x+y+z = 0

Ans 3

Q67 C1=(41)

C2= (-3-4)

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

R1= 4

R2= 5

|C1C2| = radic74

R1+R2 = 9

R1+R2 gt |C1C2|

Ans 2

Q68 Lim sin(πcossup2x)

X -gt 0 xsup2

C1 C2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Differentiating num amp den (00 form Lrsquo Hospital rule)

U have to differentiate 2 times

= π

Ans 2

Q69

F(x) = intdxsin⁶x

Put tanx= t

Secsup2x dx = dt

1 + tsup2 dx = dt

dx = dt1+tsup2

Sinx = tradictsup2+1

Substituting these values

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F = int (1+tsup2)sup2t⁶ dt

= -[1t + 23tsup3 + 15t⁵] + c

= - [cotx + 23 cotsup3x + 15 cot⁵x] + crsquo

Ans 2

Q70 Cos(255⁰) + sin (195⁰)= -2sin(15⁰)

= -2 [(radic3-1)2radic2]

= (1-radic3)radic2

Ans 1

Q71 Ysup2 = 3(1- xsup216)

dxdy|sup2 sup3 sup2 = -4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

(y-32) = 4(x-2)

Y = 4x ndash 132

am = -132

a4 = -132

a =-26

ysup2= -104x

Ans 3

Q72 int⁰⁹ [xsup2] + log[(2-x)(2+x)] dx

log[(2-x)(2+x)] is an odd funcᶯ

[xsup2]= 0 for -1ltxlt0 amp 0ltxlt1

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q73 Q(42)

P(-42)

Eqᶯ of PQ x = 4

Ysup2 = x

dydx = frac14

Slope of normal = -4

Ans 4

Q74

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

F(x) = xsup3 + 1

Frsquo(x) = 3xsup2

Frsquorsquo(x)= 6x

Now

Frsquo(x) = 0

3xsup2 = 0

X = 0

Frsquorsquo(0)=0

Not local extremum point

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Statement 1---false

Ans 3

Q75 dydx + 2x(xsup2-1)= x(xsup2-1)

IF = exp int 2x(xsup2-1) dx

= xsup2-1

Ans 3

Q76 1sup2+22sup2+3sup2+24sup2+helliphellipn terms

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

When n is even

S = n(n+1)sup22---given

Let n = 2m

S (even)= 2m(2m+1)sup22

S (odd) = 2m(2m+1)sup22 + (2m+1)th term

= 2m(2m+1)sup22 + (2m+1)sup2

=(2m+1)sup2[m+1]

Now we put 2m+1 = n

= nsup2(n+1)2

Ans 2

Q77 axsup2 + bx + c=0

1 is a root

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

a+b+c=0

Y = 4axsup2+3bx+2c

4axsup2 + 3bx + 2(-a-b) = 0

D = 9bsup2 + 32(a+b) ne0

Hence two roots

Ans 4

Q78 P amp R are parallel as direction cosines are same

Ans 1

Q79 White-2 black-1

P=23

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q80 AB = I

Ans 2

Q81 [(x-1)x]ᶯ(1-x)ᶯ

= (-1x)ᶯ[1-x]sup2ᶯ

= (-1x)ᶯ[ sup2ᶯCn (-x)ᶯ]

sup2ᶯCn

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q82 Men and women has to sit alternately

67

Ans 4

Q83 ~(Pq)

= ~(~P v q)

= p Ʌ~q

Can also check it by truth table

Ans 4

Q84 L1||L4 L2||L3 L1||L2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

L1 | L3 L2 | L4

Ans 2

Q85 F(x)= a|sinx| + b exp|x| + c|x|sup3

For x gt 0

F(x) = asinx + beˣ + cxsup3

Frsquo(x) = acosx + beˣ + 3cxsup2

For x lt 0

F(x) = -asinx +beˉˣ - cxsup3

Frsquo(x) = -acosx ndashbeˉˣ -3cxsup2

Lim x0+ f(x) = a + b

Lim x0ˉ f(x) = -a-b

For f(x) to be differentiable a=0 b=0

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 3

Q86 AM1 = AM2

(tp + tq)2 = (tr + ts)2

a + (p-1)d + a + (q-1)d = a + (r-1)d + a + (s-1)d

p + q = r + s

Ans 2

Q87 Eqᶯ of the line is (x-2)6 = (y-3)3 = (z+4)-4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

P(-126)

Q(6λ+23λ+3-4λ-4)

dr of PQ =(6λ+33λ+1-4λ-10)

since these 2 lines are |

(6λ+3)6 + (3λ+1)3 + (-4λ-10)-4 = 0

λ = -1

Q = (-400)

|PQ| = 7

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Ans 1

Q88 Statement 1---true

Statement 2----true amp implies f is right invertible but not sufficient explanation of statement 1

Ans 3

Q89 |Z1 + Z2|sup2 + |Z1 - Z2|sup2

= (Z1+Z2)(Z1 + Z2) + (Z1 ndash Z2)(Z1 ndash Z2)

= 2(|Z1|sup2+|Z2|sup2)

Ans 2

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

AIEEE 2012 ONLINE 12TH MAY MATHS SOLUTION

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG (2ND YEAR)

JADAVPUR UNIVERSITY

Q90 drdt = 01 cmhr

A = πrsup2

dAdt = 2πr drdt

= 10π cmsup2hr

Ans 4

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