SOLUTIONS & ANSWERS FOR JEE MAINS-2021

22nd July Shift 2

[PHYSICS, CHEMISTRY & MATHEMATICS]

PART – A – PHYSICS

Section A

Ans: 2.77 10−8 T

Sol: 0

20

av2

CB

=

==

20

03

0

20

C

1C

2

B

T1077.2C

2B 8

30

0−=

=

Ans: ( )ji +

Sol: A cos BB

BAB

AB

BAAB

=

=

ji2

ji

2

2+=

+=

Ans: 2 eV

Sol: We have band gap = 0

hc

eV2621

1242=

Ans: 0T4

1

Sol: g

L2T0 =

g

L

4

2

g

16L

2T'0

==

4

T

g

L

2

0=

=

Ans:

Sol: In a capacitor, current leads voltage by 2

. Hence option (3) is the correct answer.

Ans: True dip is less than he apparent dip Sol:

H

V

B

Btan = (true dip circle)

=cosB

B'tan

H

V (apparent dip circle)

The apparent dip is greater than true dip

BH

BV

’ BH cos

BV

As cos < 1

Ans: The sphere has the greatest and the ring has the least velocity of the centre of mass at the bottom of the inclined plane

Sol:

2mR1

singa

+

=

spheresolid

cy lindersoildring

aring < asolid cylinder < asolid sphere

Vring < Vsolid cylinder < Vsolid sphere

Ans:

−

+ 1

R

h1

g

R2

3

1 23

e

e

Sol:

e

1R

GM2V = = Escape Velocity

According to energy conservation

rR

GMmmV

2

1

R

GMm

R

GM2m

2

1 2

ee +−=−

rR

GMmmV

2

1 2

+=

V2 = V

2

r

1

dt

dr

rR

GM2V =

+=

+

+=

hR

R

t

0

e

e

drrRdtGM2

( )hR

R

23 e

e

rR3

2tGM2

+

+=

−

+= 1

R

h1

GM2

R

3

2t

23

e

3e

But 2

eR

GMg =

i.e,

−

+= 1

R

h1

g

R2

3

1t

23

e

e

Ans: 0.2 kg ms-1, 0.05 ms-1 Sol: By the conservation of momentum 4 × 10-3 (50 – V) – 4V = 0

1

3

3

ms05.01044

50104V −

−

−

=+

=

i.e., Impulse J = mV = 4 × 0.05 = 0.2 kg ms-1

Ans: e

p

m

m

Sol: KE = e V

( )Vem2

h

e

e

=

( )Vem2

h

p

p

=

e

p

p

e

m

m=

Ans: -627.2 J

Sol: Wporter + Wmg = KE = 0 Wporter = -Wmg = -mgh i.e., = -80 × 9.8 × 0.80 = -627.2 J

Ans: 5.38 MeV Sol: Rest

( ) 5.5180

V4m180

2

1Vm4

2

12

2 =

+

5.5180

4451mV4

2

12

2 =

+

MeV38.5

180

4451

5.5KE

2=

+

=

184 m 180 m 4 m

V

180

V4

Ans: 0.858 m

Sol: 21

21

21

21

ARR

RRR

+

=

+=

=

=

−

−

−

−

m858.0103.4

106.27.1

103

1025R

8

16

6

2

Ans: Statement I is true but statement II is false Sol: As the T increases, ferromagnetism decreases and above curie temperature, it will become

paramagnet.

Ans: TK2

3B

Sol: In monoatomic gas f = 3

Average Energy = 3 × TK2

1B

TK2

3B=

Ans: 1( )

( ) ( )( )

=+

= −

0x

0Vtan,0x

0VC 12

2

2

Sol: x = A sin t + B cos t

tsinbwtcosadt

dxV −==

At t = 0, x (0) = B

V(0) = A

x = A sin t + B sin (t + 90)

22net BAA +=

B

Acot

A

Btan ==

( )++= tsinBAx 22

( )( )−−+= 90tcosBAx 22

x = C cos (t - )

22 BAC +=

Anet

A t

B

( )

( ) 22

2

0x0V

C +

=

= 90 -

tan = cos = ( )

( )=

0x

0Vtan

B

A

( )

( )

= −

0x

0Vtan 1

Ans: sin-1 (tan r)

Sol: r + r’ + 90 = 180 r’ = 90 – r = 90 – i n1 sin i = n2 sin r’ = n2 sin (90 – i) n1 sin i = n2 cos i

tan i = 1

2

n

n

sin C = 1

2

n

n= tan i

C = sin-1 (tan i) = sin-1 (tan r)

Ans: (a) – (ii), (b) – (i), (c) – (iv), (d) – (iii) Sol: (a) – (ii) (b) – (i) (c) – (iv) (d) – (iii)

Ans: Height = 1731 m Population covered = 1413 × 105

Sol: RH2r,Radius =

H105.62km150 6 =

H = 1731 m

Population covered = r2 × 2000/ km2 = 3.14 × (150)2 × 2000 = 1413 × 105

Ans: 4.44 C Sol: Let r = 10 mm, x = 2

qr

k2Fq

=

qxr

k2F q

+

=−

( )xrr

qxk2

xr

qk2

r

qk2Fnet

+

=

+

−=

mm12mm10

mm2q10310924

69

=

−

q = 4.4 C

Section B

Ans: 2.00

Sol: Center of mass of semi circular ring is at

R2

Distance from the centre x = 2

Ans: 2.00

CM

R2

Sol:

For no slipping, V0 = R

VA = VB = ( ) 022

0 V2RV =+

x = 2.00

Ans: 60.00 Sol: For minimum deviation

r1 = r2 = 2

A

given i = 2r1 = A

1 sin i = 3 sin r1

1 sin A = 3 sin 2

A

2 sin 2

Acos

2

A = 3 sin

2

A

cos == 302

A

2

3

2

A

A = 60

Ans: 15.00 Sol:

rR

RRV

+

==

1st case 1.25 = )1(r5

5−−−−−

+

2nd case 1 = )2(r2

2−−−−−

+

R

V0 C V0

A

B

I R

V

, r

From (1) and (2)

r = 1 Ω, V10

15V

2

3==

i.e., x = 15

Ans: 4.00

Sol: Electric flux density =

=

20

02 r4

rQr

r4

Q

0

xx

0

kz2jycoseiysineDE

+−=

=

−−

By Gauss’s law

Ekrz

rj

ry

ri

rx

r

0

++=

0

Dk

rz

rj

ry

ri

rx

r

++=

( ) ( ) ( )z2rz

rycose

ry

rysine

rx

r xx +−+= −−

= -e-x sin y + e-x sin y + 2

At origin = -e-0 sin 0 + e-0 sin 0 + 2

= 2 C/m3

Charge = × V = 2 × 2 × 10-9 = 4 × 10-9 = 4 nC

Ans: 57.00 Sol: Tp – Ts = (Tf – Ts) e-Q 65 – 25 = (75 – 25) e-5C

e-5C = 5

4

T - 25 = (65 - 25) e-5C

325

440 =

i.e., T = 57C

Ans: 3.00

Sol: Direction of P BA

BAV1

=

3

kji +−=

Direction of Q CA

CAV2

=

k2

k2==

Angle between 21 VandV

3

1

11

31

|V||V|

VV

21

21 =

=•

i.e., x = 3

Ans: 1.00

Sol: 2

2

T

4g

g2T

==

T

T2

g

g +

=

nT

T2

g

g 0+

=

As ℓ and T0 are same for all observations so g

g is minimum for highest value of ℓ, n and T

Minimum percentage error in g is for student number -1

Ans: 500.00 Sol:

L

2R

5=

V across R = 50 – 5 = 45 volt

V = IR 21

45V

+=

LR

5mA90

45R

+

=

When RL → , current in the zener diode will be maximum i.e., I2 →0, I1 → 1

So R = = 50090

45

Ans: 5.00

Sol: ( )2strain2

yE = × A × L

( ) Astrain2

y

L

E 2=

= ( ) 22411

10102

10 −−

= 5 J/m

PART – B – CHEMISTRY

Section A

Ans: Vitamin B1 and Vitamin B6

I R

Vz

I1

I2

RL z V1 = 50 V

Sol: Thiamine − Vitamin B1

Pyridoxine − Vitamin B6

Ans: 3-Ethylhex-3-ene

Sol: CH3−CH2 C=C−CH2−CH3

CH2CH3

3-Ethylhex-3-ene does not show geometrical or optical isomerism.

Ans: (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) Sol: (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)

Ans: CH3−CH2CH=CHCH2NH2

Sol: CH3−CH2−CH=CH2−NH2 has no conjugation between -electrons and lone pair on nitrogen. Hence three will be no resonance

Ans: Cu2+, Cr3+, Sc+ Sol: Cu2+, Sc+, Cr3+ has 1, 2, 3 unpaired electron respectively. Hence it is paramagnetic and coloured.

Ans: BiH3

Sol: On moving down the group the E−H bond weakness and hence reducing property increases.

Ans: (a)-(ii), (b)-(iii), (c)-(i) and (d)-(iv) Sol: (a)-(ii), (b)-(iii), (c)-(i) and (d)-(iv)

Ans: (a) and (c) only Sol: The softer acid (Ag+) and border line acid (Pb2+) can be precipitated as sulphides from their

aq.solutions. Since S2− ion is a softer base

Black

22 PbSPbS →+ +−

black

22 SAgAgS →+ +−

Ans: AgI / I− Sol: When silver nitrate solution is added to KI solution, the precipitated silver iodide absorbs iodide ions

from the dispersión médium and negatively charged coloidal sol results

Ans: A =

OH

, B =

OH

CHO

, C =

OH

CH2OH

Sol:

OH

CHO

OH

(A)

CHCl3 / KOH

(B)

OH

CH2−OH

(C)

PCC

Ans:

O CH3C

O

Sol:

O CH3C

O

CH3MgBrCH3−C−CH3 +

(ketone)

O OH

CH3−C=O

CH3 CH3

OH

(3 alcohol)

(i) CH3MgBr

(ii) H3O+ CH3−C−CH3

Ans:

NH2

+ AlCl3 + CH3Cl

CH3

NH2

Sol: Aniline does not undergo Friedel-Crafts reaction (alkylation and acetylation) due to salt formation with

AlCl3, the Lewis acid, which is used as a catalyst.

Ans: At the time, he proposed Periodic Table of elements structure of atom was known Sol: At the time, be prepared periodic table of elements structure of atom was known

Ans: water at 4C Sol: Solubility of oxygen increases with decrease in temperature

Ans: Al2(SO4)3

Sol: Tf = i Kf.m, Tf I, Tf =Tf − Tf Greater the i value lower will be the freezing point

Ans: (a)-(iii), (b)-(i), (c)-(v) and (d)-(iv)

Sol: SF4 No. of hybrid orbitals dsp52

46 3=+

=

IF4 No. of hybrid orbitals 23dsp62

57=

+=

+2NO No. of hybrid orbitals 2

2

15=

−= sp

+4NH No. of hybrid orbitals 3sp4

2

145=

−+=

Ans: Tritium

Sol: Only tritium is radioactive and emits low energy − particles with t1/2 value 12.33 yrs

Ans: (a) only

Sol: 262ether

24 HNaI2HBINaBH2 ++⎯⎯⎯ →⎯+

Ans: H3PO2 and CH3CH2Cl

Sol: (A)

N2Cl−

+ H3PO2 + H2OCH3CH2Cl (B)

Anhy.AlCl3

CH2−CH3

+

Ans: Reduced pressure distillation

Sol: This process is used if the liquid has a tendency to decompose near its boiling point

Section B

Ans: 12

Sol: 222622

23s4d3s3p2s2s1V →

No. of electrons in p-orbitals = 6 + 6 = 12

Ans: 4

Sol: 3

)L(B

B 104004.01180

72.0

VM

WMolarity −==

==

Ans: 7

Sol: 2

20

106.1

104.2log

60

303.2

]R[

]R[log

t

303.2K

−

−

==

]3010.0477.0[60

303.2

2

3log

60

303.2−==

31075.6006755.0176.060

303.2 −===

Ans: 78

Sol: )Toluene(356

nMethy latio

)Benzene(66 CHHCHC ⎯⎯⎯⎯⎯ →⎯

78 g → 92 g

10 g → x

(Theoretical wt), g79.1178

1092x =

=

Actual weight of Toluene = 9.2 g

03.7810079.11

2.9100

W

Wyield%

ltheoretica

actual ===

Ans: 3 Sol: [Co(NH3)6]Cl2 Co2+ : [Ar] 3d7 4s0 4p0 No pairing of electrons takes place Hence, no. of unpaired electrons = 3 [Co(NH3)6]Cl3 Pairing of electrons takes place

Hence no. of unpaired electrons = 0

Total no. of unpaired electrons = 3 + 0 = 3

Ans: 6

Sol: CH3−CH2−CH2−CH=CH2 ⎯ (1)

CH3−CH2−CH=CH−CH3 ⎯ (2) [1 cis and 1 trans]

CH3−CH2−C−CH2

CH3

⎯ (1)

CH3−CH−CH=CH2

CH3

⎯ (1)

CH3−C=CH−CH3

CH3

⎯ (1)

Total = 6

Ans: 2

Sol: 112n)RT(KK gn

cpg =−==

47.9 = Kc (0.083 288)1

033.2288083.0

9.47Kc =

=

Ans: 3

Sol: 2

2

cellcell]Ag[

]Cu[log

2

06.0EE

+

+

−=

23 )101(

250.0log

2

06.097.2

−−=

610

250.0log03.097.2

−−=

= 2.97 −0.03 log 0.250 106

= 2.97 − 0.03 log 2.5 105

= 2.97 − 0.03 [log 2.5 + 5 log 10]

= 2.97 − 0.03 5.3979

= 2.97 − 0.1619 = 2.808

Ans: 106

Sol: 7.1054

10023.6)104518.0(62.7

Z

NadM,

Na

ZMd

2337A

3

A3

=

===−

Ans: 21

Sol: mol/kg1048.2HCOOC 22

)graphite(2 −=→+

12 g → 2.48 102 kg

1 g → x

66.20102066.012

1048.2x 2

2

==−

=

PART – C – MATHEMATICS

Section A

Ans: 3

Sol: A matrix which satisfies the given condition is ,

A=

100

010

001

=

===

100

010

001

A 33

Sum of all elements =3

Ans: 100 < C < 156

Sol: [C is denoted by k to avoid confusion]

36

kC,

3

10f2,3g2 ==−=

Centre: (-g , -f) =

−

3

5,

2

3

Clearly r <2

3

4

9cfg 22 −+

4

9

36

k

9

25

4

9−+

100k36

k

9

25 —(1)

Point of intersection of x-2y=4 and 2x-y=5 is (2 , -1) [After solving the equations ]

0S )1,2( − (Point (2, -1) is inside the circle S=0)

036

k)1(

3

10)2(3)1(2 22 +−+−−+

036

k

3

1065 +−−

036

k

3

13+−

156k3

13

36

k —(2)

• 23 3

5

From (1) and (2) , 100<k<156 ie;100<C<156

Ans: 1862

Sol: 106d9a2530)d9a2(2

10530S10 =+=+= —(1)

56d4a2140)d4a2(2

5140S5 =+=+= —(2)

(1)—(2) 10d50d5 == and

2a+90=106 8a16a2 ==

)d5a2(2

6)d19a2(

2

20SS 620 +−+=− = 10(16+190)-3(16+50) = 1862

Ans: 1

Sol: )xe1log(2)xe21log(.)xsin2(

xlim)x(flim xx2

22

3

0x0x

−−

→→−−+=

−−+

=

−−

→ x

)xe1log(2)xe21log(

xsin

xlim

4

1 xx2

4

4

0x—(1)

Where 0x

lim→

xsin

x4

4

=0x

lim→

−−+

−−

→ x

)xe1log(2)xe21log(lim.

4

1)1(

xx2

0x

=

+−

−

−−+−

+

−−

−

−−

−→

xx

x

x2x2

x20xe)1(xe

)xe1(

)1(2e)2.(xe2.

)xe21(

1lim

4

1

144

1== (Using L’Hospitals rule)

Ans:

−

2

3,1

Sol: f(x) =

Case 1 (x>0)

y = x3x2x3

4 22 ++−

03x4x4dx

dy 2 =++−= (say)

03x4x4 2 =−−

8

84

8

48164x

=

+=

8

12x = or

2

3x

8

4=− or

2

1−

Hence if x>0 , f(x) is increasing on

2

3,0 — (1)

Case 2 (x<0)

y = xxe3

xx exe3dx

dy+= )1x(e3 x +=

Hence if x<0, f(x) is increasing on [-1 , 0) — (2) Hence from (1) and (2),

−

2

3,1x

− • •

2

1− −

2

3

+ −

− •

1− −

+

Ans: −42

Sol:

a

is parallel to c)b.d(b)c.d()cb(d

−=

c14b7

c)626(b)623(

−=

++−+−=

( ) )kji(2kji27 +−−++=

kj3ā −=

coplanararecandb,asince0cba

=

dcbadcadba

+=+

636)6(1)618(3

623

203

130

−−=−−−=

−

= = −42

Ans: 51c2ba32

=−+

)kj3i0(7 −+=

a

b

c

•

d

Sol: ccba

= is perpendicular to both a

and b

aacb

= is perpendicular to both b

and c

2cand

1b2b2.b

)2(2cb

a90sincb

acb

)1(cb2

c90sinba

cba

=

==

−=

=

=

−=

=

=

5316136

)2(4)1()2(9

c4ba9c2-ba3

true.is)1(

82122

cba2]cba[2

]cba[]cba[]bac[]cba[

222

2222

=++=

++=

++=+

==

==

+=+

Ans: 3

4

Sol: z3z2 −=

Ry,x;)iyx(3)iyx( 2 −−=+

Case 1: y=0 x2

– y2 +3x=0 So, x =0 and x=-3 Therefore z=0 and z=-3

Case 2:

x2 – y2 +3x=0

a

b

c

=

+++=

=

0k210k

to.............n

1

n

1

n

1

n

1

2n

3

4

Ans: )e1(200 1−−

Sol: Both sin2x and

−

xx

e are periodic functions with period π

−

−

==

0xx

2100

0xx

2

dx

e

xsin100dx

e

xsin

If 0x

1x

0thenx0 =

( )

( )

−

−

−

−

−

−

−

−

−

−

−

−

+

−

+

−

−−=

−−=

−−−=

−

−=

−=

−==

0

x

22

1

0

x

1

0

x

01

0

x

0

x

0

x1

0

x

0

x1

0xx

2

x2sin2x2cos1

e

21

50e150

xdx2cos.e.50e150

xdx2cos.eee50

xdx2cose1

e50xdx2cosedxe50

dx2

x2cos1.e100dx

e

xsin100

( )

++

+= cbxsinbbxcosa

ba

ebxdxcose

22

axax

( )

( )( )

( )

( )

2

13

2

21

2

11

2

21

1

2

1

41

)e1(200

14

4.e150

14

11e150)e1(

1

14

50e150

110

1e

41

50e150

+

−=

+

−=

+−−=−

+

−−=

−−

+

−

+

−−=

−

−

−−−

−−

)e1(200 1−−=

Ans: 1e−

Sol: dx.xeccos)x2cosy1(dx2xdyeccos 22 +=+

dxxsin

)xsin21(y

xsin

1dx2dy

xsin

12

2

22

−+=+

)xsin21)(y1(dx

dy

)y1(xsin2)y1(

xsin2xsiny2y1dx

dy

.xsiny2y1xsin2dx

dy

dxxsiny2y1xdxsin2dy

2

2

22

22

22

−+=

+−+=

−−+=

−+=+

−+=+

=+

+=

dx.x2cosy1

dy

x2cos).y1(dx

dy

( )

1

2

2

1

22

0

2

1

2

1

2

1

x2sin2

1

C2

x2sin

eek1)0(y

1k1ke)0(y

ek1e.k

1e.k004

y

1e.ky

e.ey1

C2

x2sin)y1log(

−−

−

=

==+

−=−=

==

−==

−=

=+

+=+

Ans: 5

Sol:

5solutionofNumber

4,2,0,2

2,2

xx

1xcosor1xsin

1xcosxsin

xcosxsinxcosxsin

xcosxcosandxsinxsin

77

2277

2727

=

+=

==

+

++

Ans: = 2, 10

Sol: )1(6zyx =++

)3(zy2x

)2(26z5y5x3

=++

=++

Ans: (~ p q) (~ q p)

Sol: we know that p q'porq'pq += (in Boolean algebra)

(1) p)''q()'q'p( +++ )Tq~q(Tautology'pp)'qq(pq'q'p =+++=+++=

(2) Tautologyqq)p'p(pqq'p]p)''q[()q'p( +++=+++=+++

(3) .Tautologypqp'q]p)''q[()p'q( +++=+++

(4). )qp()qp(pqqp]p)''q[(]q)''p[( +++=+++=+++

qpqp =+= tautologyaNot

Ans: 3

Sol: 3

1z

1

2y

2

1x

21

1z

61

2y

31

)1x(:L1

−=

−=

−

−=

−=

−

4

3z

2

y

1

2x

1

3z

21

y

41

2x:L2

−=

−=

−

−=

−=

−

1a

:(1,2,1) ; 1b

:(2,1,3) ; 2a

:(2, ,3) ; 2b

:(1,2,4)

21 bb

= k3j5i2

421

312

kji

+−−=

2a

- 1a

= k2j)2(i +−+

S.D38

1

38

514

9254

61052

bb

)aa).(bb(

21

1221 =−

=++

++−−=

−=

13or15511451514 ===−

Integral value of 3=

Ans: )2log,0[ e

Sol: 03]1e[]e[ x2x =−++

2]e[or1]e[

0)2]e)([1]e([

02]e[]e[

031]e[]e[

xx

xx

x2x

x2x

−==

=+−

=−+

=−++

Since ex >0 , ]e[ x=-2 is not possible.

2log)elog(1log2e11]e[ xxx =

)2log,0[x2logx0 ee

Ans: 2

3

Sol: f(x) =

−

+−

−

−

2

1x2sin

)1xx(cos

1

21

For denominator,

12

1x20

−

)1(2

3x

2

1

3x21

21x20

−

−

For Numerator,

11xxo 2 +−

( )rootpositiverepresents""

11xx0 2 +−

Since D = b2-4ac<0 and a>0, x2-x+1≥0

)2(]1,0[x0)1x(x

0xx11xx 22

−−

−+−

Intersection of (1) and (2) is

1,

2

1

2

3=+

Ans: 24−

Sol: y=-2x+k —(1)

kcand2m =−=

Hyperbola: 13

y

3

x 23

=− —(2)

3b,3a 22 ==

Since (1) is tangent to (2),

2222 bmac −=

)0k(3k

93)2(3k 22

=

=−−=

3kc ==

Parabola: x4

4y2

=

2m,3c,4

a −==

=

m

ac = (Condition for tangency)

82

43

−

=

−

=

24−=

Ans: 2

51+−

Sol: For E1

2

2

1a

b1e −=

For E2

Let E2 : 12

2

2

2

=+A

y

B

x

Where B= a —(1) Focii: (0,±c)=(0,±b)

222222

22

baAbBA

bBAbc

+==−

=−=

b •

−a

y

x 0 • −b

+a

E1

E2

Ans: 119

Sol:

The vector normal to p2 is )n(kji2 2

=−+

DR of L are given by 21 nn

21 nn

= k3j5i

112

211

kji

++−=

−

−

The vector along L is kjib ˆ3ˆ5ˆ ++−=

),,( satisfy the equation of the planes

)1(22

)1(22

−=−+

−=+−

0Ap.b =

11941453335

41

7

9

35

3335)(35

35

33

35

3722

35

37

235

82452

35

82

7

9)1(

35

41

7

41

7

36115115

7

36)A(

7

997)B()A(

)B(253)1(2)2(

)A(1154)3()1(

)3(935

031051

0)3)(()5)(2()1)(1(

=++=

++=++

=−==+

=+−

+=+−

==−==+

==+

−−=−−

−=++

−=++−

=+−++−

=+−+−−

P1

P2

•

• A (1, 2, 0)

x − y + 2z = 2

p(, , )

2x + y − z = 2

Ans: 162

43

Sol: Let

=

dc

baA where 0bcadanddcba −

Ie; ad ≠ bc Let us consider the cases in which ad = bc and a≠b≠c≠d (1 , 6 ; 2 , 3) 4 Cases.

(2 , 3 ; 1 , 6) 4 Cases.

(2 , 6 ; 4 , 3) 4 Cases.

(3 , 4 ; 2 , 6) 4 Cases.

Total = 16 Cases a≠b≠c≠d 6×5×4×3 = 360 Cases

162

43

6666

344

6666

16360P =

=

−=

Section B

Ans: 8.00

Sol: 102_r )xx2( +

)k2krr10(.k10

k10

k2krr10k10k

10k2k10rk

101k

x2.C

x.x.2C)x.()x2(CT

−−−

−−−−−+

=

==

For constant term, 0k2krr10 =−−

k10k

102 2.C245180

10k,k where2r

r10kr10)2r(k

−==

+

==+

Clearly k=8

Ans: 2.00

Sol:

−

−=

−−=+−=

2

1y

2

14

2

1y21y2x2

)1x)(1(44x4y

)1x)(1(4)1x(44x4y

2

2

++=

−−=−−=+−=

Area

−−+−==

1

0

2

22

)x1(4x42A2

23

6

3

2_

3

8

3

11]01[

3

8

3

xx

)x1(4

dx)x1(2

12dxx122

1

0

31

0

)1(2

3

23

1

0

21

0

===

−−−=

−−

−

=

−−−=

−

Ans: 96

Sol: 11n>10n+9n —-(1)

)2()9.0(1)1.1(

10

91

10

11

nn

nn

−+

+

will not be satisfied for n=1,2,3,4.

For n=5,(1) is true.

As n increases from 5,(2) holds good.

5n

Possible values of n are 5,6,7,8,……,100

Number of values = 96

Ans: 96.00

Sol: Given digits : 0,2,4,6,8

Answer = 4 4 3 2 1=96

4 4 3 2 1

x2 = −2y + 1

• 2

1

1 −1

y2 = −4x + 4 y2 = −4x + 4

Ans: 4

Sol: dy).2x(dx)1y(e)2x( 2x

1y

+=

+++

+

+

2x

1ye

dx

dy 2x

1y

+

++=

+

+

Let y+1=Y and x+2=X

+

++−=+

+

+=

+

+−

+

+=+++−=

−−=+=−=

++=−

+=−

+=−

=

=

+=+

+==

+=

=

=+=

−

−

−−

+

+−

−−

+

+−

−

−

−

32

32

32

32

2x

1y

32

32

2x

1y

XY

V

V

V

V

XY

e2x

3loglog).2x(1y

e2x

3loglog

2x

1y

e2x

3loge3log)2xlog(e

.e3logCC3loge1)1(y

C)2xlog(e

CXloge

CXloge

X

dXdVe

edX

dVX

VedX

dVXV

dX

dVXV

dX

dYVXY

X

Ye

dX

dY

dX

dY

dx

dy

X2xanddYdy

Clearly x>-2 and |𝛼 + 𝛽| =4

Ans: 3804

Sol: 2040=23x3x5x17 n does not have factors 2,3,5 and 17 n(1)=100

S(2k)=2+4+6+…….+100=2(1+2+…..+50)=2

250(50+1)=2550

S(3k)=3(1+2+…..+33)= 168334332

3=

S(5k)=5(1+2+…..+20)= 105021202

5=

S(17k)=17(1+2+….5)= 255652

17=

S(6k)=6(1+2+…..+16)= 81617162

6= 2,3,5,17.

S(10k)=10(1+2+…..+10)= 55011102

10=

S(34k)= 102)21(34 =+

S(15k)=15(1+2+…..+6)= 315762

15=

S(51k)=51

S(85k)=85

S(30k)=30(1+2+3)=180

Ans=(2550+1683+1050+255)-(816+550+102+315+51+85)+(180)=3804

Ans: 3125

Sol: B=

ihg

fed

cba

AB = BA b=d,a=e,c=f and g=h

Number of ways =55=3125

Ans: 4.00

A

5K

C

D

17K

B

3K 2K

Sol:

−

−

−

+

=

2xor2xif0

2x0if2

x13

0x2if2

x13

)x(f

+−+

+

+−

+−

++

=+

22xor22xif0

22x0if2

2x13

02x2if2

2x13

)2x(f

−

−−

−−

+

=+

0xor4xif0

0x2if2

x3

2x4if2

4x3

)2x(f

−−−

−

−−

−−

−+

=−

22xor22xif0

22x0if2

2x13

02x2if2

2x13

)2x(f

−

=−

4xor0xif0

4x2if2

)x4(3

2x0if2

x3

)2x(f

−

−

−

−−

−−+

−

=

4xif0

4x2if2

)x4(3

2x0ifx2

3

0x2ifx2

3

2x4if)4x(2

3

4xif0

)x(g

−

−−

−−

−

=

4xif0

4x2if2

3

2x0if2

3

0x2if2

3

2x4if2

3

4xif0

)x('g

Clearly g(x) is continuous for all Rx

g’(x) is not defined at x = −4,x = −2,x = 2, x = 4

n + m = 4

Ans: 4.00

Sol:

Class fi xi fixi

0-6 A 3 3a

6-12 B 9 9b

12-18 12 15 180

18-24 9 21 189

24-30 5 27 135

)1(1018b37a813054b111a243

11088b198a66b309a3098034

22

309

ba26

504b9a3

f

xfx

i

ii

−=+=+

++=++

=++

++=

=

If b=10 , a=8 (a possible value) We can see that in this case median =14

(Using the formula )h.f

c2

N

lm

−

+=

4)ba( 2 =−

Ans: 720

Sol: A=0, 1, 2, 3, 4, 5, 6, 7 f(1) + f(2) + f(3) = 3 Case 1 f(1) = 0, f(2) = 1, f(3) = 2 Case 2 f(1) = 1,f(2) = 0, f(3) = 2 etc. (6 Cases) Case 1

Numbers of bijective function (8 − 3)!=5!

Total= 5! 6 = 6! = 720