1

dgV

2dgV

While inside the fluid

SOLUTIONS TO IRP FULL TEST-1 ONLINE (HELD ON 3RD, 4TH, 5TH MAY 2018) PHYSICS (PAPER-1)

SECTION 1:

1. A block is given some

horizontal speed at point

A of frictionless heavy

track as shown in figure.

Block breaks off the track at B and strikes horizontal track at point C.

Height of point A is 15m and speeds at points

A, B and C are in ratio 1:2:2. If height of B above ground is h and speed at C is u, then

value of

2u

2his-

(A) 8 m/s2 (B) 20 m/s2

(C) 8 m/s2 (D) 8 m/s2

Solution (B):

A B Cv : v : v :: 1: 2 : 2

A B CKE :KE :KE ::1: 2 : 4

A BKE mg 15 KE mgh

A B2KE KE AKE mg(15 h)

A CKE mg15 KE

A AKE 15mg 4KE AKE 5mg

A C4KE KE 20mg 21 mu 20mg

2

BKE 10mg mg.h =10mg h 10

2

2u 20m/s2h

2. Twelve rods of mass m and length are rigidly joined as edges (sides) of a cube.

Moment of inertia of the cube about a face

diagonal is-

(A) 213

.m5

(B) 223

.m3

(C) 241

.m5

(D) 231

.m3

Solution (B):

22 2 2 2mL 1 mL mL mL

I 4 2 23 3 12 22

222mL 1 4 4mL

3 2

223m

3

3. A conductor having cross section A = 1mm2 is

carrying I=1.0A current. The linear

momentum of electrons in unit length of

conductor is 9×10-N. Value of N is- (A) 3.2×10-17 N-s (B) 5.7×10-19 N-s (C) 9.1×10-25 N-s (D) 2.9×10-27 N-s

Solution (B):

dI neAv

p= n×volume× me×vd= n.(A.1) vd.m

mp .I

e

3112

19

9.1 10 15.68 10

1.6 10

4. A spherical ball of density d is released at

depth h in a non-viscous fluid of density 2d

kept in a large tank. The ball returns to initial

position after time-

(A) 2h

g (B)

h2.

g

(C) 2h

2.g

(D) h

4.g

Solution (C): While inside the fluid

2dgV dgV Vd a

a g

2ht

g

The ball rises up height h in air up and comes

down then returns to same depth.

5/2total

2h ht 4 2

g g

5. A block pushed gently from the edge of a 15m

high tower, immediately breaks into two pieces of equal mass. One piece reaches

ground after one second, then time in second

when other piece reaches ground is- (A) 2s (B) 3s

(C) 4s (D) 5s

Solution (B):

2115 u 1 10 1

2

u 10m/s

Other peace moves at 10 m/s upwards

2115 10 t 10t2

t 3

SECTION 2:

1. A charged particle of mass m carrying charge

q is performing uniform circular motion at the

center of a large ring of radius R carrying current I. Particle obeys Bohr's principle of

angular momentum quantization. When the

particle is in nth orbit- (A) its speed is proportional to n1/3

(B) radius is proportional to n1/2

(C) ratio of speed and radius is independent of n

(D) time period is proportional to n3/2

Solution (BC):

BCentre 0I

2R

and

nhmvr

2

A

A

2

0I

mv q.B.r q .r2R

20r q I nh

2R 2

1/2r n

mv

rqB

0

0

r 2mRn

v qr I (independent of n)

2. After absorbing a slow moving alpha particle a

nucleus of mass number 87 breaks into two

smaller nuclei. The product nuclei move away with speeds in ratio approximately 3:7. Then

the correct statement is-

(A) mass numbers are in ratio7:3

(B) ratio of de Broglie wavelengths is 7:3 (C) atomic radii are in ratio 4:3

(D) mass decreases in the process

Solution (ACD): Since the alpha particle is slow moving, its

linear momentum is neglected. 1 1 2 2m .v m .v

1 1

2 2

m v 7

m v 3

11 1

hand

m v 2

2 2

h

m v

1

2

1

11 3 33

2 2

m 7 64 4r

r m 3 27 3 (C)

(D) KE to both parts is result of nuclear

energy released due to conversion of mass

into energy.

3. Three dielectric slabs P, Q and R are inserted between the metal

plates of a capacitor. Area of

plates is A and distance between

them is d. Dielectric slab P has

area of cross section A and thickness d/2 and dielectric constant K, slab Q and R have area

A/2 and thickness d/2 each and dielectric

constants 2K and 3K respectively. When

capacitor is charged, electric fields at points

A, B, C and D as shown in figure, are EA, EB,

EC and ED respectively. The correct relationship is- (A) C AE E (B) A DE E

(C) C BE E (D) D BE E

Solution (ABC): Conceptual

4. Electric field in a region is E (a.i b.j) N/C

(a b) . There are four points A at (a, b), B at

(b, -a), C at (-a, -b) and D at (-b, -a). Electric potential at these points are V1, V2, V3 and V4

respectively, then- (A) V2

3

R

d

2 2

GM G9Mx 2R

x (8R x)

21 GM.m G9Mmu2 R 7R

GM.m G9M.m

2R 6R

21 GM.m 9 1 3 2GM.mmu 12 R 7 2 2 7

4GMu

7R

2

1 4GM GM.m G9M.mm.

2 7R R 7R

GM.m G9M.m 1mv

7R R 2

21 GM.m 1 2 9 62mv 9 1 GM.m2 R 7 7 7 7

124GMv

7

7. Two students A and B conduct experiment to find the resistance of same wire. Voltmeter

and ammeter readings by student A are

(6.000.02) volt and (2.500.05) amp respectively. Corresponding values by student

B are (6.000.05) volt and (2.50.02) amp respectively. (A) A measures potential diff. more precisely

than B (B) B measures current more precisely than A

(C) error in resistance is less in measurement

by A

(D) error in resistance is less in measurement

by B

Solution (ABD): V by A is more precise (A)

I by B is more precise (B)

Student A: r v I .02 .05

.023r v I 6 2.5

Student B: r v I .05 .02

.0163r v I 6 2.5

B has more precise resistance than A

8. A particle is moving in x-y plane. At certain

instant, the components of its velocity and

acceleration are as follows Vx=3m/s, Vy=

4m/s, ax= 2m/s2 and ay= 1m/s

2.

(A) angle between velocity and acceleration is

1 2cos5

(B) tangential acceleration is 2 m/s2

(C) radial acceleration is 1 m/s2

(D) The radius of curvature of trajectory at the

instant is 20m

Solution (ABC):

Angle between ˆ ˆ ˆ ˆv 3i 4j & a 2i j is

v.a 6 4 2cos

v . a 5. 5 5

1 2cos (A)5

2radial

v.aa 2m/s

v (C)

2 2 2tangential radiala a a 5 4 1m/s (B)

SECTION 4: 1. There is uniform magnetic field of intensity B

in an equilateral triangular region, directed

vertically downwards. Magnetic field of same

magnitude but directed vertically upwards

exists outside the triangle. A charged particle of mass m carrying charge q horizontally

enters the triangle perpendicular to one side

at midpoint. If the side of triangular region is

2mv

q.B, the time period of particle is

.mn.

q.B

,

value of n is?

Solution (6):

Particle makes 3 complete

revolutions before coming back

to same state.

Time taken= 3T=6 m

qB

2. A point charge +q is placed at distance d from a center of a disc of radius R. Flux through

the disc is o

q

3

. Ratio R/d is 2n/2, value of

n is-

Solution (5):

Rtan

d

0

q 4

3 3

4 12 (1 cos ) cos

3 3

3/2R tan 8 2d n 3

3. When a 4 resistance is connected in parallel with an ammeter, its range increases four

times. If 2 resistance is also connected in parallel, how many times the range will

increase compared to initial range? Solution (10):

AI

4

AA

nI

B

2 I2

I

3I

4

For the ammeter reading to be 1

th4

of the

incoming current

1I 3I

23I 4 I 2 2I 6I

6I 3I I 10I

range of ammeter is increased to 10 fold.

4. A body at temperature 40ºC is kept in a

surrounding of constant temperature 20ºC. It

is observed that its temperature falls to 35ºC

in 10 minutes. Approximate time taken for the body to attain temperature of 30ºC is 4.X

minutes, then value of x is-

Solution (3):

1 2

2

1

2

40-35 35-30=-k.t and =-k.t

37.5 32.5

t 37.5=

t 32.5

t 12minute

5. Glycerin is filled in 25 mm wide space

between two large plane horizontal surfaces.

Calculate the force required to drag a very

thin plate 0.75 m2 in area between the

surfaces at a constant speed of 0.5m/s if it is

at a distance of 10 mm from one surfaces and

15 mm from other surface in horizontal

position? Take coefficient of viscosity = 0.5

N-s/m2. Find the value of X where X=

20×force required to drag (in newton).

Solution (625):

F=dv

Adr

= 0.5×0.75. 3 30.5 0.5

10 10 15 10

= 0.5×0.75×0.5×103 .1 1

10 15

= 31.25 N

AI4I

A B

4

I1

10mm

15mm

0.5m/s

Fixed surface

5

5

2NO

2NO2 2NH NH

O

NH

NH

O

+ 2H N

2NO

2NO

NO2

NO2

NH2

NO2

NH2

F

NO2

NO2

NH2

NO2

NO2NH2

CHEMISTRY (PAPER-1)

SECTION 1: 1. How many times does Cannizzaro reaction take place in the given conversion?

H

O

H

O

H

OH- OH

OH

OHHCOO

-+

(excess)

(A) 0 (B) 1 (C) 2 (D) 3

Solution (B):

After undergoing two times aldol reaction further there is no -hydrogen so it undergoes one time cross cannizzaro reaction in which formaldehyde is oxidised to give formate ion and corresponding

alcohol is formed.

2. Predict the product(Y) of the following reaction:

N-K

+

O

O

+

NO2

NO2

F

NH2NH

2

RefluxProduct(X) Product(Y)

(A) (B) (C) (D)

Solution (B):

3. 3FeCl dissolved in ether and water forming respectively

(A) 3 3

2 2 6Et O Fe and Fe(H O)

(B) 3

2 3 2 6Et O FeC and Fe(H O)

(C) 2 3 2 4 2Et O FeC and Fe(H O) C

(D) 3 3

2 2 66Et O Fe and Fe(H O)

Solution (B):

3 2 2 6 3FeCl CH O [Fe(H O) ]Cl

N-K

+

O

O

+

NO2

NO2

F

NH2NH

2

RefluxProduct(X) Product(Y)

N-K

+

O

O

+

NO2

NO2

F

NH2NH

2

RefluxProduct(X) Product(Y)

2NO

2NO2 2NH NH

O

NH

NH

O

+ 2H N

2NO

2NO

3 2 3 2Co ordinatebond

FeCl OEt FeCl OEt

6 6

OH

and

O

Br Br

H H

and

Br

Br

H

H

and

4. An equilibrium mixture at 300K contains 2 4N O and 2NO at 0.28 and 1.1 atmosphere respectively. If

the volume of container is doubled. Calculate the new equilibrium pressure of two gases (in atm)

2 4N O 2NO

(A) 0.095 0.64

(B) 0.64 0.095

(C) 0.17 0.32

(D) 0.32 0.17

Solution (A):

5. Identify the incorrect statement.

(A) Elemental silicon is known only in diamond structure and no form of silicon possess graphite

structure.

(B) BN has hexagonal and cubic arrangement.

(C) Carbon can form d p bond.

(D) Silicon exists in crystallite and amorphous forms. Solution (C):

SECTION 2: 1. CORRECT statement is: (A) Ethanol and Ethanal can be distinguished by using aqueous sodium bisulphite solution.

(B) Carbolic acid and Aspirin can be distinguished by using aqueous sodium bicarbonate solution.

(C) Acetone and Acetyl acetone can be distinguished by using neutral FeCl3 solution.

(D) 1.0 equivalents Acetyl acetone on reaction with excess of NaOI forms more than one equivalents of

yellow precipitate. Solution (A,B,C):

3 2 3

O O|| ||

CH C CH C CH NaOI

3 3

O||

2CH C O CHI

So, 1 equivalents of acetylacetone will give 1equivalent of 3CHI .

2. Which of the following represents a pair of constitutional isomers?

(A)

(B) 3 2 2 3 3CH CH CH CH and CH CH CH CH

(C)

(D)

Solution (B,C,D):

Molecular formula in (A) is not same so not isomers, whereas in (B) (C) & (D) they have same molecular

formula but different structural formula.

3. Which of the following statement is/are true?

(A) In a metal carbonyl c od larger compared to that of c od of free CO.

(B) The pair of compounds 2 6 3Cr(H O) Cl and 2 3 3 2Cr(H O) Cl 3H O are hydrate isomers.

(C) The „d‟ orbital of central metal atom/ion involved in 2dsp hybridisation is

2dz .

(D) The geometrical isomers of 3 3(Ma b ) complex are optically inactive.

Solution (A,D):

Metal carbonyls CO bond order is less than free carbonyls lesser the bond order greater c 0d

3 3Ma b both the geometrical isomers have plane of symmetry

7

7

optically inactive

4. Correct statement about the structures of 2 24 2 7

CrO &Cr O is/are

(A) The structure of 24CrO

is tetrahedral

(B) 2

2 7Cr O predominantly exist in acid medium and 24CrO

predominantly exist in basic medium

(C) 6Cr-O bonds are similar in 2

2 7Cr O ion

(D) Bridged Cr–O bond length is greater to terminal Cr–O bond length in 2

2 7Cr O

Solution (A,B,C,D):

5. Identify the incorrect statements:

(A) In the metallurgy of Al, purified Al2O

3 is mixed with Na

3AlF

6 (or) CaF

2 which lowers the melting

point and conductivity

(B) Alkali metals are extracted by the electrolytic reduction of their aqueous salt solutions (C) Leaching process is used in the concentration of red bauxite

(D) The gases liberated at anode during electrolysis of Al2O

3 + Na

3AlF

6 in Hall-Heroult process is

2F gas

Solution (A,B,D): Fact.

6. In the decay process

A B C D

(A) A and B are isobars (B) A and D are isotopes

(C) B,C and D are isobars (D) A and C are isotones Solution (B,C):

7. If adsorption of a gas on a solid is limited to monolayer formation then which of the following

statements are true?

(A) At low pressures, x

m varies proportionately with p

(B) At moderate pressures x

m varies less than proportionately with p

(C) At high pressures, x

m becomes independent of p

(D) At high pressures, x

m varies more than proportionately with p

Solution (A,B,C):

8. Select the correct statements

(A) A NaCl type AB crystal lattice can be interpreted to be made up of two individual fcc type unit lattice

of A and B fused together in such a manner that the corner of one unit lattice becomes the edge

centre of the other. (B) In a FCC unit lattice the body centre is an octahedral void.

(C) In a SC lattice there can be no octahedral void.

(D) In a SC lattice the body centre is the octahedral void.

Solution (A,B,C):

O Cr O

O

O

Cr Cr

O

O O

O

O

O

O

, According to structure

8 8

3H C3CH

( )

3H C

* *

4 isomer (2 enantiomeric pair)

( )

2 3CH CH

CH3

C H3( )

CH3CH3

CH3

CH3

CH3

CH3

Cl

+ 3CCl C H

O

2 4H SO

3CCl CH Cl

Cl(DDT)

SECTION 4: 1. How many of them will form atleast one enantiomeric pair during monochlorination ?

(i) 2-methylbutane (ii) isobutane

(iii) neopentane (iv) (Optically pure)

(v) 2,2-di methyl butane (vi)

(vii) (viii)

Solution (6):

(i)

( )

3 2 3

3

H C CH CH CH|CH

(ii) 3 3

3

H C CH CH|CH

No enantiomeric pair

(iii)

3

3 3

3

CH|

H C CH CH|CH

No enantiomeric pair (iv)

(v)

( )3

3 2 3

3

CH|

H C C CH CH|CH

(vi)

(vii) (viii)

2. Number of chlorine atoms present in a molecule of organic product of following reaction

Solution (5):

Cl2 4H SO

3CCl C H

O

Chlorobenzene Trichloroacetaldehyde

Cl2 4H SO

3CCl C H

O

Chlorobenzene Trichloroacetaldehyde

9

9

3. Among following the number of reactions with correct products is

(i) HCl

2 2 4SnCl I SnCl HI

(ii) 3 3 4 2 4 2 2NaIO NaHSO NaHSO Na SO H O I

(iii) 2 2 3 2FeBr Cl FeCl Br

(iv) 2 2 4 2I N H HI N

Solution (4): Factual

4. Among 2 2 2 2 3Cu ,Mn ,Cd ,Zn ,Bi the number of cations which forms sulphide precipitate with

4NH Cl and 4NH OH and 2H S is ____________

Solution (5):

Both cations of group- II and IV are precipitated

5. A compound contains 28% N and 72% of a metal by weight 3 atoms of metal combine with two atoms of

nitrogen. If atomic weight of metal is „M‟ find M

?6

Solution (4):

3 M 2 14 100 M 24

M 244

6 6

10

MATHEMATICS (PAPER-1) SECTION 1:

1. If the value of the definite integral /6

0

1 sin xdx

cos x

2 ln A B C 1

then A B C is (where A, B, C are different natural numbers)

(A) 0 (B) 5 (C) 6 (D) 7

Solution (B): /6

0

dxI

x xcos sin

2 2

/6

0

x x2 ln sec tan

2 4 2 4

2 ln 2 3 2 1

A 4,B 3,C 2

2. If A is non-singular square matrix where TB A &

2A B I . Such that 3A I KA

then K is equal to

(A) 0 (B) 1

(C) 2 (D) 3 Solution (C):

2

TA A I …(i)

Take transpose on both side: T 2A A I …(ii)

From equation (i) & (ii):

2

2A I A I 4 2A A 2A O

1 1 1 1 2A A A A 2A A 0

3A I 2A K 2

3. Number of values of x satisfying the equation

7x 10 x is (A) 0 (B) 1

(C) 2 (D) 3

Solution (D):

7x 10 x x

3x

x10

4. Let ˆ ˆ ˆ ˆ ˆ ˆA 2i 3j 5k ,B i 3j 2k and

ˆ ˆ ˆC i 5j k be the vertices of triangle ABC and the median through A is equally inclined

to the positive directions of the axes then the

value of 2 is equal to (A) 4 (B) 3 (C) 2 (D) 1

Solution (A):

Midpoint of 1/2 2

BC D i 4j k2 2

5 8AD i j k

2 2

As AD is equally inclined with axes.

AD.i AD. j AD.k

5 8

12 2

7, 10

2 4

5. If

n1

2 2 3r 1

2r 1tan tan

r r 1 r r 1 2r

961 , then the sum of digits of n is equal to

(A) 10 (B) 19 (C) 8 (D) 12

Solution (A):

1r 2 4 3

2r 1T tan

1 r r 2r

1

22

2r 1tan

1 r r

1

22

2r 1tan

1 r r 1

221

22

r r 1tan

1 r r 1

21 2 1

rT tan r tan r 1

n1 2

r n 0

r 1

T V V tan n

n1 2

r

r 1

tan T tantan n 361

2n 316 n 19

SECTION 2:

1. Let

2x 4x , 3 x 0

f x sinx ,0 x2

cos x 1 , x2

, then:

(A) x 2 is the point of global minima (B) x is the point of global maxima

(C) f x is non-differentiable at x2

(D) f x is discontinuous at x 0

11

Solution (ABC):

From above figure clearly option (A), (B) and (C) are correct.

2. Let f : a,b a,b (where a b and are real numbers) be a differentiable non-constant

function for which f a b & f b a . Then

(A) at least one c a,b such that f c c

(B) 1 2 1 2c ,c a,b ,c c such that

1 2f c .f c 1

(C) c a,b such that f c 1

(D) c a,b such that f c 1

Solution (ABCD):

Let g x f x x

g x is continuous function

g a b a 0

g b a b 0

g a g b 0

g c 0 for some c a,b

f x x for some c a,b

Use LMVT on f x in a,c

1

f c f af c

c a

1c a,c

1c b

f cc a

Use LMVT on f x in c,b

2

f b f cf c

b c

2c c,b

2a c

f cb c

1 2f c f c 1

1f c 1 , 2f c 1

3. Contents of the two urns is as given in this

table. A fair die is tossed. If the face 1, 2, 4 or

5 comes, a marble is drawn from the urn A

otherwise a marble is chosen from the urn B.

Urn Red Marbles

White marbles

Blue marbles

A 5 3 8

B 3 5 0

Let 1E : Denote the event that a red marble is

chosen.

2E : Denotes the events that a white marble is

chosen

3E : Denote the event that a blue is chosen.

Then

(A) Events 1 2 3E ,E & E are equiprobable.

(B) 1 2 3P E ,P E ,P E are in AP. (C) If the marble drawn is red, the probability

that it came from the urn A is 1

2.

(D) If the marble drawn is white, the probability that the face 5 appeared on the die

is 3

32.

Solution (ABD):

1A : urn A is selected 14 2

P A6 3

.

2A : urn B is selected 22 1

P A6 3

1 1 1 1 2 1 2P E P A P E /A P A P E /A

4 5 2 6 1

6 16 6 16 3

2 1 2 1 2 2 2P E P A P E /A P A P E /A

4 3 2 10 1

6 16 6 16 3

3 1 3 1 2 3 2P E P A P E /A P A P E /A

4 8 1

6 16 3

(C)

1 1 1

1 11

20P E /A P A 20 596P A /E

1P E 32 8

3

(D)

2 1

1 22

P E P A1 1P A /E

4 4P E

121 396

32 4 32

96

4. The line x 6 y 10 z 14

5 3 8

is the

hypotenuse of an isosceles right angled

triangle whose opposite vertex is 7,2,4 . Then which of the following is/are sides of the triangle?

(A) x 7 y 2 z 4

2 3 6

(B) x 7 y 2 z 4

3 6 2

(C) x 5 y 5 z 2

2 3 6

(D) x 4 y 4 z 2

3 6 2

Solution (ABCD):

Let M 5 6,3 10,8 14

BM 5 13 i 3 12 j 8 18 k

12

BM. 5i 3j 8k 0 5

2

Coordinate of 13 5 12

M , ,2 2 2

and

7BM AM CM

2

Let A 5 6,3 10,8 14

25 15

AM 5 i 3 j 8 20 k2 2

7AM

2

2 2

225 155 3 8 20

2 2

49

2

2 5 6 0 2,3

A 4, 4,2 & C 9, 1,10 Lines in options A & C passes through points

B & C lies in option B & D passes through

point A & B.

5. At a point P on the parabola 2y 4x , tangent

and normal are drawn. Tangent intersects the

x-axis at Q and normal intersects the curve at

R such that chord PR subtends an angle of o90 at its vertex, then

(A) PQ 2 6

(B) PR 6 3

(C) Area of PQR 18 2

(D) PQ 3 2

Solution (ABC):

Let 2 21 1 2 2P t 2t & R t ,2t

2 1 1 21

2t t & t t 4

t

1t 2 , P 2, 2 2

2t 2 2 , R 8, 4 2 Equation of tangent at P:

21 1x yt t 0

x 2y 2 0

Coordinates of Q 2,0

PQ 16 8 2 6

PR 36 72 108 6 3

1PQR 2 6 6 3 18 2

2

6. Identify the correct statements:

(A) An ellipse passes through the point 4, 1

and touches the line x 4y 10 0 . If its

axes coincide with the coordinate axes,

then its equation is 2 2x y

180 5/4

(B) In the expansion of

203

4

14

6

, there

are 19 irrational terms

(C) The set S 1,2,3,..,12 is to be partitioned into three sets A, B, C of equal

size such that A B C S ,

A B B C A C . The number of

ways to partition S is

3

12!

4!

(D) If the normal at 11

cct ,

t

on 2xy c meets

it again in the point 22

cct ,

t

, then

31 2t t 1

Solution (ABC):

(A) Let equation of ellipse 2 2

2 2

x y1

a b

Passes through 4, 1

2 2

16 11

a b …(i)

Touches the line 1 5

y x4 2

2225 a b

4 16 …(ii)

From (i) & (ii): 2a 80,20

2 5b ,54

Equation of ellipse 2 2x y

180 5/4

or 2 2x y

120 5

(B)

r20 r420 3

r r1

T C 46

For rational terms r 8,20

Number of irrational terms

21 2 19

(C)

3

12! 3! 12!

4! 4! 4! 3! 4!

(D) 2c

yx

2

2

dy c

dx x

13

Slope of normal at 21 11

cct t

t

2 2 112 1

c 1

t tt

ct ct

1 2

1

t t

31 2t t 1

7. If x

1

ln tf x dt

1 t

, then

(A)

x

1

1 lntf dt

x t 1 t

(B)

x

1

1 lntf dt

x t 1 t

(C) 1 1f e f8e

(D) 21 1

f x f lnxx 2

Solution (BCD):

x

1

ln tf x dt

1 t

1/x

1

1 lntf dt

x 1 t

, Put

1t

z ,

2

1dt dz

z

x

21

1ln

1 1zf dz1 zx z

z

, when t 1,z 1

1t ,z x

x

x x

1 1

lnz lntdz

z 1 z 1 t

x

1

1 lnt lntf x f dt

x 1 t t 1 t

x 2

1

ln xlntdt

t 2

1 1f e f8e

8. If the cubic equation 3 2z az bz c 0 , a,

b, c R , c 0 has a purely imaginary root then

(A) c ab

(B) b ac

(C) number of purely imaginary root is 2

(D) a bc

Solution (AC):

Let be purely imaginary root 3 2a b c 0 …(i)

Take conjugate 3 2a b c 0

3 2a b c 0 …(ii)

(i) + (ii) 2c

a

cb

a

(i) – (ii) 2 b c ab

Non real root occur in pair

2 purely imaginary roots

SECTION 4:

1.

n

n

1lim n e 1 a

n

and

n 1

n

1lim n 1 e b

n

, then 2a 2b

Solution (5):

n 1 n

n

1 12a 2b 2 lim n 1 1

n n

n

n

1 12 lim n 1 1 1

n n

n

n

12 lim 1 2e

n

2a 2b 5 2. Let F : , be anti-derivative of the

function 2

f : ,3 3

,

xf x ;

sin x and

2F F lnc

3 3 2

. Then c

Solution (3):

x

x

/3/3

xf x dx

sin x

x

/3

xF x F

3 sinx

2 /3

/3

2 xf F dx

3 3 sinx

/2

/3

xnC

2 sin x

2 /3

/3

xnc dx

2 sin x

2 /3

/3

nc cosecx

2 /3

/3nc n cosecx cot x

14

2 /3

/3

xntan

2

1n 3 n

3 2 n 3

c 3

3. Let a, b be arbitrary real numbers. Find the

smallest natural number b for which the

equation 2x 2 a b x a b 8 0 has

unequal real roots for all x R Solution (5):

For unequal real roots: D 0

2

4 a b 4 a b 8 0

2 2a b 2ab a b 8 0

2 2a 2b 1 a b b 8 0

D 0 2 22b 1 4 b b 8 0

8b 33 0 33

b8

as b N

Smallest b 5 .

4. If , are two distinct real roots of the

equation 3ax x 1 a 0 , a 1,0 , none of which is equal to unity, then the value of

3 2

1 x1x

1 a x x alim

e 1 x 1

is

am n

, where

mn … m,n N Solution (1):

3ax x 1 a 0

2x 1 a x x 1 1 0

2a x x 1 1 a x x

3 2

1 x1x

1 a x x alim

e 1 x 1

, Put

1x

t

3

t 1t

3

1 a t atlim

e 1 1 t1 t

t t1

t

t

a t 1 t tlim

1 t t t 1

a

m 1,n 1 m n 1

5. Through P 2,2 is drawn a chord of the

ellipse 2 2x y

125 16

such that it intersects the

ellipse at A and B. If maximum value of

PA.PB is m then the value of m/2 is (where [.] is greatest integer function)

Solution (7):

Let A 2 rcos ,2 rsin lie on the ellipse.

2 2

2 rcos 2 rsin1

25 16

2 22cos sin r

25 16

4cos 4sin 236r 0

25 16 400

Roots PA & PB

2 2 2

236 236PA PB

16cos 25sin 16 qsin

max

236PA PB m

16

m 236

2 32

m7

2

15

PHYSICS (PAPER-2) SECTION 1:

1. A metallic rod of length 1 m is rigidly clamped

at its mid-point. Longitudinal stationary waves are set up in the rod in such a way that

there are two nodes on either side of the

midpoint. The amplitude of an antinode is

2×106m. [Young‟s modulus = 2 × 1011 Nm

2,

density = 8000 Kg/m3). Origin is at O which

is mid-point of the rod. P is a point on the rod

such that OP=10 cm. Magnitude of maximum

velocity of the point P is-

(A) 20

m/s (B)

10

m/s

(C) 5

m/s (D)

4

m/s

Solution (A):

V 5000 m/s, =0.4m

4f 1.25 10 Hz

Maximum Velocity = a. =20

m/s

because x= 10 cm is antinode

2. Figure shows

two, identical

narrow slits S1

and S2. A very

small completely

absorbing strip

is placed at

distance „y‟ from

the point C. „C‟ is the point on the screen

equidistant from S1 and S2. Assume

16

=5 3 2 2

a(10 P 10 m 5 10).(50 10 )

1 R

T = 75

RK.

PARAGRAPH FOR NEXT TWO QUESTIONS:

A rectangular block of area of cross section A, length L and

density d is submerged and

floating at the interface of two

liquids. Densities of liquids are d1 and d2

(d1

17

Electrons are accelerated through 5V, will

reach the anode with maximum energy (5–+ 5) eV

10 – = 8 eV

= 2eV Current is less than saturation current

because if slowest electron also reached the plate it would have 5eV energy at the anode,

but there it is given that the minimum energy

is 6eV.

2. A sample of hydrogen atom gas contains 100 atoms. All the atoms are excited to the same

nth excited state. The total energy released by

all the atoms is 4800

49Rch (where Rch = 13.6

eV), as they come to the ground state through

various types of transitions. (A) Maximum energy of the emitted photon

may be 48

49Rch

(B) Initially atoms are in 6th excited state.

(C) Initially atoms are in 7th excited state.

(D) Maximum total number of photons that

can be emitted by this sample is 100. Solution (AB):

Maximum energy of one photon=

4800Rch

49

100

=48

49Rch=Rch

22

1 1

1 n

Rch22

1 1

1 n

=

48

49Rch

n=7 n=6 there are 100 atoms, maximum number of

photons that can be emitted=600.

3. Radionuclide of 56Mn is being produced in a

cyclotron at a constant rate P by bombarding

a maganese target with deutrons. 56Mn has a

half life of 2.5 h and the target contains large

number of only the stable maganese isotope 55Mn. The reaction that produces 56Mn is:

55Mn + 21H 56Mn + 11H . After being

bombarded for a long time, the activity of the

target due to 56Mn becomes constant equal to

13.86×1010 s–1. (Use ln2 = 0.693; Avogadro No = 6 × 1023;

atomic weight 56Mn = 56 gm/mole)

(A) 56Mn nuclei are being produced in the

cyclotron at rate 13.86×1010 nuclei/sec

(B) After a long time, number of 56Mn nuclei present in the target, is 1.8×1015

(C) maximum number of 56Mn nuclei depends

on number of 55Mn nuclei

(D) decay constant is 1.8 × 10–5 per second

Solution (AB):

In equilibrium - rate of production = rate of

decay= 13.86 × 1010 nuclei/s

P= N N=P

=

1/2Pt

n2=1.8×1015

4. A partition divides a

container having

insulated walls into

two compartments

and . The same gas fills the two compartments whose initial

parameters are given. The partition is a

conducting wall which can move freely

without friction. Which of the following statements is/are correct, with reference to

the final equilibrium position? (A) Pressure in the two compartments are

equal

(B) Volume of compartment is3V

5

(C) Volume of compartment is12V

5

(D) Final pressure is5P

3

Solution (ABCD):

In the equilibrium position the net force on

the partition will be zero. Hence pressure on

both sides is same. Hence (A) is correct.

Initially, PV=nRT

n1=1 1

1

P V

RT=

PV

RT & n2=

(2P)(2V)

RT=4

PV

RT

n2 = 4n1

1 1

2 2

V n

V n =

1

4 V2=4V1

V1+4V1=3V V1=3

5V

and V2=12

5V

13V

P5

=P V

RT

RT P1'=5PV

3V=

5

3P

5. Uniform rod AB is hinged at the end A in a

horizontal position as shown in the figure. The other end is connected to a block through

a massless string as shown. Pulley is smooth

and massless. Masses of the block and the

rod are same and equal to 'm'. Choose correct

(A) acceleration of block is 3g

8

(B) tension in string is 5mg

8

(C) string force at A is zero

(D) magnitude of initial acceleration of centre

of mass of system is approx g

10

18

Solution (AD):

T( )–mg2

=2m a

3 and mg–T=ma

a=3g

8

5mg

T mg ma8

rod block

cmrod block

am . m .a

3g g2am m 32 10

6. When some energy is given to an electron its

de Broglie wavelength decreases from

9 92 10 m to 0.5×10 m choose correct

(A) momentum becomes four times of its

initial value

(B) energy becomes four times its initial value

(C) energy required is 5.6 eV (D) final energy is 5.6 eV

Solution (AC):

1 2 11 2 1

h h hp , p 4. 4p

=h

p=

h

2mE E=

2

2

h

2m

E=2h

2m.

2 21 2

1 1

= 5.6 eV

7. A 4kg block and a 2kg block are kept on a

rough horizontal surface. Coefficient of static

friction is s 0.4 and coefficient of kinetic

friction is d 0.2 . Blocks are connected by

massless spring of force constant K = 20 N/m.

Spring is neither extended nor compressed

initially. 4 kg block is given some speed towards 2kg block. 2kg block starts moving

just before 4kg block comes to rest.

(A) Maximum compression in spring is 0.4 m

(B) Initial speed of 4kg block is 3

2 m/s5

(C) Finally spring is compressed by 0.2 m

(D) Finally there is no compression or

extension in spring

Solution (ABD):

When 2kg block starts motion

2m g kx x 0.4m

2 2 k 11 1

4 v k.x m .g.x2 2

2 21 1

4.V 20 (0.4) 0.2 4 10 0.42 2

3

V 2 m/s5

Finally- 2

k 2

1kx .m .g.x x 0.4m

2

8. There is a stream of neutrons with a kinetic

energy of 0.0327eV. If the half life of

neutrons is 700 seconds. Choose correct (A) Speed of neutrons is approx 2500 m/s (B) neutrons decay into proton, electron

(C) neutrons decay into proton and electron

and anti neutrino

(D) nearly 0.0004% neutrons decay after

travelling 10 m.

Solution (ACD):

19

27

2KE 2 0.0327 1.6 10V 2500m/s

m 1.67 10

1 1 00 1 1n H e E

10m

t 0.004s2500m/s

dN dN

.N .dtdt N

Fraction decay dN ln2

. t 0.004N 700

0.0004%

19

19

2 gN one mole

1 atm

2 gH one mole

1 atm

V L V L

Temperature 298 K

P 1 atm

O CH3

O2N

O2N

O CH3

O

CH3 NO2

O

NO2

CH3

isC,C42SOH

H

OHOH

NO2CH3

CHEMISTRY (PAPER-2)

SECTION 1:

1.

(A) (B)

(C) (D)

Solution (C):

2. Assume that 6 6 6 5 3C H andC H CH form an ideal solution then 0G, Hand S at 25 C for the

addition of 1 mole of 6 6C H to an infinitely large sample of solution with with a mole fraction of 0.35 for

6 6C H is

(A) G H S 0 (B) H 0, G S 4582.57J

(C) H 0, G 4582.57J, S 15.38 j/k

(D) H 15.38J/k G 4582.57J, S 15.38J/k

Solution (C):

3.

If we remove the wall in between the two container then change in entropy of mixing is:

(A) 1 111.54 J mol K (B) 1 111.54 J mol K

(C) 1 123.08 J mol K (D) 1 15.77 J mol K

Solution (A):

2

1

V2.303 nR log

V

4. 2 2BaC N A

2 2CaC N B

3CH2NO

OH 2OH

5

4

32 1

H

HO2O N

+3CH

H

O2O N 3CH

20 20

O

O

O

O

CH3O

O

CH3

O

O

CH3

A and B are respectively

(A) 2 2BaCN ,CaCN (B) 2 2Ba CN ,Ca CN

(C) 22Ba CN ,CaCN (D) BaCN,CaCN Solution (C):

2 2 2

2 2 2

BaC N Ba CN

CaC N CaCN

PARAGRAPH FOR NEXT TWO QUESTIONS

In organic chemistry the functional groups plays major roll. Each functional group has it‟s own

characteristic property. Few functional groups interact with specific reagents and sometimes the reactivity

also depends on the structural connectivity of the functional groups in a molecule. Answer the following questions based on their conditions:

5. The structure formula for “vitamin-C” shown below contains four different hydroxyl groups, in which,

which hydroxy group that is most readily methylated with CH2N2

(a)

(b)

OHCH

2

C

OHH O

O

OH OH

H

(c) (d)

(A) a (B) b (C) c (D) d

Solution (C):

Protonation of 2 2CH N takes place first, so more is the acidity of O H bond, faster will be

protonation and faster will be the reaction.

6. Predict the end product in the given reaction:

CH3

42.NaBH

3. /H

1. O3; H2O/H2O2

(A) (B) (C) (D)

Solution (A):

22 3H C N N CH N3Ph OH Ph O Ph OCH

3CH

3 2 2 21. O H O/H OO

O

OH

3CH

42.NaBH

OOH

OH

3CH

3.H /

Esterification

3CH

O

O

21 21

H /NiH O 233heat heat

A CH CHNC B

3CH

PARAGRAPH FOR NEXT TWO QUESTIONS

Lead acid storage battery may be recharged after its usage. It consists of Pb anode with coating of spongy lead, Pb with a coating of

2PbO , serves as cathode. 2 4H SO is used as an electrolytic solution.

2

4 4Pb SO PbSO 2e 0E 0.31 V

2

2 4 4 2PbO 4H SO 2e PbSO 2H O

0E 1.70 V

7. During discharge of battery, which of the following statements are TRUE?

(P) At anode and cathode respectively 2Pb,PbO are involved in the redox reaction

(Q) Density and concentration of 2 4H SO decreases during discharging process

(R) It needs no salt bridge since the oxidizing and reducing agents in both discharge and recharge

processes are solids.

(S) Equivalent weight of 2 4H SO is 49 and that of 2H O is 18.

(A) PQ (B) PS (C) PQR (D) PQS

Solution (C):

(P) At anode and cathode respectively 2Pb,PbO are involved in the redox reaction

(Q) Density and concentration of 2 4H SO decreases during discharging process

(R) It needs no salt bridge since the oxidizing and reducing agents in both discharge and recharge

processes are solids.

8. Two moles of 2 4H SO consumed during the discharge of lead storage battery. With this how many

moles of gases will be liberated during the electrolysis of acidified water (A) 4 (B) 3 (C) 1.5 (D) 1

Solution (C):

SECTION 2: 1. Some statements are given about following compound

3 2 3 2

O||

CH C CH CH (OCH )

Identify correct statements

(A) It gives yellow precipitate with 2I in aq NaOH Solution

(B) It does‟t form silver mirror with Tollens reagent

(C) One of Hydrolysis products of given compound gives red precipitate with Benedicts solution

(D) Both, the given compound and its one of hydrolysis products gives positive test with 2, 4 – DNP

Solution (A,B,C,D):

2.

Products „A‟ and „B‟ can be distinguished by (A and B are Nitrogen containing products)

(A) the treatment of 3CHCl ,OH

gives a foul smelling compound

(B) the action of 2HNO A liberates 2N gas whilst B does not

(C) the actions of 2 2CS ,HgCl . B gives odour of mustard oil whilst A doesn‟t

(D) the treatment of p-toluene sulphonyl chloiride; „A‟ gives alkali soluble product Solution (A,B,D):

H /NiH O 23

3 2 3 3 3

3 3 3

CH CH NH CH CHNC CH CH NH CH| | |CH CH CH

O

2H /H O3 2 3 2 3 2Hydrolysis

CH C CH CH(OCH ) CH C CH C H

O O

22 22

Cl

C

N N

CC

NCl Cl

O O

O

O

OO

S

S S

OO

O

O O

O

OO

OO

OO

P

P P

3CH 3CH

Si

O O

Si Si

O

3CH

3CH

3CH

3CH

3. Select compound which is / are isoelectronic

(A) (B)

(C) (D)

Solution (A,B,C):

(A,B,C) are isoelectronic 4. Which of the following reactions give the same nitrogen containing gaseous product?

(A) Heating of 4 2NH NO (B) By passing 3NH over heated CuO

(C) Heating of 3 2Ba(N ) (D) Cu cold and dilute 3HNO

Solution (A,B,C):

4 2 2 2NH NO N 2H O

3 2 22NH 3Cuo N 3H O 3Cu

3 2 3 2 23Ba(N ) Ba N 8N

5. Which of the following statements are true

(A) All crystalline solids are isotropic

(B) 6SF molecule has no net dipole moment.

(C) Amorphous solids are super cooled liquids with high viscosity.

(D) In crystals, short range order exists

Solution (B,C):

6. Dry air is slowly passed through three solutions of different concentrations, 1 2 3c , c and c ; each

containing (non volatile) NaCI as solute and water as solvent, as shown in the Fig. If the vessel 2 gains weight

and the vessel 3 loses weight, then

(A) 1 2c c (B) 1 2c c (C) 2 3c c (D) 2 3c c Solution (B, D):

1 2c c , 2 3c c

7. 0.1mol of 4MnO

in acidic medium can oxidize:

(A) 2 41

mol of FeC O6

(B) 20.5 mol of Fe

(C) 2

2 40.25 mol of C O

(D) 2

2 70.6 mol of Cr O

23 23

T

S

Solution (A,B,C):

2 41

mol of FeC O6

, 20.5 mol of Fe , 22 40.25 mol of C O

8. The Variation of solubility (S) of a salt s

A B in water with temperature (T) is given by the

following graph then the correct statements is/are (A) The solubility process is endothermic

(B) The lattice energy of salt is less than the hydration energy of ions.

(C ) Solubility decreases with increase of temperature.

(D) The salt may be 4 sNH Cl

Solution (B,C): Conceptual

24

MATHEMATICS (PAPER-2) SECTION 1:

1. If 2 22 2 2

2 21 0

x 1 sin x xsin xdx dx

x 2x 3 x 2x 3

2 2

21

xsin x p2 dx

qx 2x 3

(p and q are relatively

prime). Then p q is equal to

(A) 0 (B) 1

(C) 2 (D) 3

Solution (D):

2 22 2 22 2

1 1

x 1 sin x xsin xI 2 dx

x 2x 3 x 2x 3

2 2

20

xsin xdx

x 2x 3

2 22

121

x 2x 1 sin xI dx I

x 2x 3

2 2 22

121 1

sin xI sin x dx 2 I

x 2x 3

2

1

11 cos2 x dx 0

2

1 p

2 q

p q 3

2. Three distinct points

and are collinear and equation

,

has roots u, v and

w, then which of the following is true?

(A)

(B)

(C)

(D) cannot say anything

Solution (B):

2 3

2 3

2 3

3u 2u 1

3u 2v 1 0

3w 2w 1

u v v w w u uv vw uw 0

uv vw uw 0 c

0a c 0

3. If a pair of variable straight lines

(where is a real

parameter) cut the ellipse at two

points A and B, (where A, B and origin are

non collinear), then the point of intersection of

tangents at A and B, satisfies

(A)

(B)

(C)

(D)

Solution (D):

Let point of intersection of tangent at A & B is

P h,k .

Equation of chord of contact AB:

hx 4ky 4 0 …(iii)

22 2 hx 4kyx 4y 4 0

4

(homogenize equation (ii) with the help of (iii))

2 2 2 2 2 24x 16y h x 16k y 8hkxy 0

2 2 2 24 x 16 16k y 8hk xy 0 ..(iv) Comparing equation (i) & (iv):

2 24 h 16 16k 8hk

1 4

2 24 4 14k 2 2x 4y

4. If the eccentricity of

is and e be the

eccentricity of then is

equal to

(A) (B)

(C) (D)

Solution (A):

2 2 1

nb na 12

2nb 1

na 2

2

2na1 e 1nb

23

e2

2 3 2 3P 3u ,2u ,Q 3v ,2v

2 3R 3w ,2w3 2ax bx cx d 0

a 0 & a,b,c,d,u,v,w R

b 0

c 0

d 0

2 2x 4y xy 0

2 2x 4y 4

2 2x 4y 8y 0

2x y 2x y 0 2 2x 4y 4xy 0

x 2y x 2y 0

2 2

2 2

x y1

na nb

a b 0,a,b 1 1

2

22

2b

xy 1

log a 2e

3

2

1

2

2

3

5

4

25

PARAGRAPH FOR NEXT TWO QUESTIONS:

Let O be an interior point of such that

. Points D and E are the

midpoints of the sides AC and BC respectively.

5.

ar DOE

ar ABC

(A) (B)

(C) (D) none of these

Solution (D):

6.

ar ABC

ar AOC

(A) 2 (B) 3

(C) 4 (D) 8

Solution (B):

SOLUTION FOR QUESTION NUMBER (5 & 6):

Let O be the origin

PV of a c

D2

, PV of

b cE

2

, a 2b 3c

1 1ABC a c b c a b b c c a2 2

1

2b 3c b b c c 2b 3c2

3 b c

1 a c b cDOE

2 2 2

1 2b 2c b c 1b c 0

2 2 2 2

1 1AOC a c 2b 3c c2 2

b c

ABC3

AOC

DOE0

ABC

PARAGRAPH FOR NEXT TWO QUESTIONS:

Let be a sequence of

numbers satisfying and

.

7.

(A) (B)

(C) (D)

Solution (C):

8.

8

ii 0

1

a

(A) (B)

(C) 100 (D) none of these

Solution (A):

SOLUTION FOR QUESTION NUMBER (7 & 8):

n 1 0n

18a 3 ,a 3

6 a

118

a 3 19

218 3

a 37 7

318 18 7

a 3 33 45 5

67

418 3

a 31 31

65

518 6 31 1

a 3 36 3/31 63 21

b 1 2S a a a ………. n 1a

1 2 3 nT T T .....T

n n

3T

2 1

8 9 9 n

i ni 0 n 1 n 1

1 1 2 1 1013

a T 3 3

SECTION 2:

1. Identify the correct statements:

(A) If angle between the line

x 1 y 1 z 2

1 2 2

and the plane

2x y z 4 0 is such that 1

sin3

,

then the value of is 5

3

(B) A tetrahedron has vertices O 0,0,0 ,

A 1,2,1 ,B 2,1,3 and C 1,1,2 , then angle between faces OAB and ABC will be

1 19cos35

(C) Number of divisors of 2 3 3 52 .3 .5 .7 of the

form 4n 1,n N is 48

ABC

OA 2OB 3OC 0

1

3

1

6

1

9

0 1 2 na ,a ,a ,......,a ,...

n 1 n3 a 6 a 18

0a 3

5a

1

3

1

7

1

21

1

63

1013

3

1010

3

26

(D) If the graph of

3 2f x 2x ax bx a,b N with the x-

axis at three distinct points, then the

minimum value of a b equals 4

Solution (ABCD):

(A) Angle between line and plane:

2 2 2 1sin

33 5

4 5 5

3

(B) Normal to the face OAB

1

i j k

n 1 2 1 5i j 3k

2 1 3

Normal to the face ABC

2

i j k

n 1 1 2 i 5 j 3k

2 1 1

Angle between plane 1 2

1 2

n .n 19cos

n n 35

(C) Number of division 4 2 3 3 2 48

(D) For 3 distinct roots 2a 8b 0

Minimum value of a b 4

2. A player tosses a coin. He sets one point for

head and two points for tail. He plays till he

gets sum of points equal to n. If np be the

probability that his score becomes n, then

(A) 31

p2

(B) n n 1 n 21 1

p p p2 4

(C) n n 1 n 21

p p p2

(D) 411

p16

Solution (CD):

n n 1 n 21 1

P P P2 2

31 1 1 1 1

P 22 2 2 2 2

1 1 5

8 2 8

4 2

41 1 1 1 1

P 32 2 2 2 2

1 1 3

16 4 8

1 4 6 11

16 16

3. Vectors a,b,c are three unit vectors and c is

equally inclined to both a & b . Let

2 2a b c b c a 4 x b 4xcos a ,

then which of the following case be correct

(a & b are non-collinear vectors, x 0 )

(A) x 2 (B) 0

(C) (D) x 4

Solution (ABC):

As c is equally inclined with a & b

a.c b.c

a.c b a.b c a.b c b.c a

2 24 x b 4xcos a 2

2

a.c 4 x

b.c 4x cos

2 2x 2 4xsin

x 2& x 0 sin 0

4. In ABC with usual notations 2 2 2a 4 b 4c 4a 4bc then

(A) Length of internal angle bisector of A is

b Acos

3 2

(B) sinB

2sinC

(C) sinB 1

sinC 2

(D) Length of internal angle bisector of angle

2b AA cos

3 2

Solution (AD): 2 2 2a 4a 4 b 4c 4bc 0

2 2

a 2 b 2c 0

a 2,b 2c sinB

2sinC

Length of angle bisector of A

A2bccos

2

b c

2 Abcos

3 2

5. Let xe

f xx

then f x k has

(A) 2 real solution when k e,

(B) one real solution when k ,0

(C) no real solution when k 0,e

(D) 4 real solution when k 0,e Solution (ABC):

xey

x x R 0

x

x

elim 0

x ,

x

x

elim

x

x x

x 0 x 0

e elim , lim

x x

27

x

2

e x 1dy

dx x

6. Consider twice differentiable function f x

such that f 1 3 , f 2 6 , f 3 9 then which of the following is/are true

(A) f x 8 atleast one in 1,3

(B) f x 3 atleast twice in 1,3

(C) f x 0 atleast once in 1,3

(D) f x 0 nowhere in 1,3 Solution (ABC):

f x is continuous and f 1 3 , f 2 6 ,

f 3 9

f x must take all values between 3,9

f x 8 for some x 1,3

Let g x f x 3x

g x is continuous and differentiable.

g 1 g 2 g 3 0

Rolles theorem holds in 1,2 and in 2,3

1 2g c g c 0 where 1 2c 1,2 & c 2,3

Now use Rolle‟s theorem on g x in 1 2c ,c

g x 0 for atleast one 1 2x c ,c

i.e., x 1,3

7. If 2 2 2sin 2x cos 3y tan 4z sin2xcos3y

cos3y tan4z tan4zsin2x 0 , where

x,y,z 0,2

then possible values of

x y z is/are

(A) 11

12

(B)

7

6

(C) 5

4

(D)

3

2

Solution (ABCD):

2 2

sin2x cos3y cos3y tan4z

2

tan4z sin2x 0

sin2x cos3y 0,cos3y tan4z 0

& tan4z sin2x 0

sin2x cos3y tan4z 0

2x n

2m 1n kx ,y ,z

2 6 4

k m,n, I

8. The value of 2

2 2

sec xdx

a sec x btan x is

(A) 11 b a

sin tanx Cab a

if b a 0

(B) e1

log tanx b ab a

2 2cosec x btan x C

if b a 0

(C) 11 a b

sin tanx Caa b

if a b 0

(D) e1

log tanx a ba b

2 2asec x btan x

C if a b 0

(where C is integration coust) Solution (AD):

2

2

sec xI

a a b tan x

(i) if a b 0

1

n a b tanxa b

2 2asec x b tan x

(ii) if b a 0

11 b aI sin tanxab a

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