iit - jee 2014 (advanced) - vidyalankar classes · 2021. 1. 20. · iit jee 2014 advanced :...

IIT - JEE 2014 (Advanced)

CODE : 1

(2) Vidyalankar : IIT JEE 2014 − Advanced : Question Paper & Solution

2

IIT JEE 2014 Advanced : Question Paper & Solution (Paper – II) (3)

3

PART I : PHYSICS

SECTION − 1 : (Only One Option Correct Type)

This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE option is correct.

1. A tennis ball is dropped on a horizontal smooth surface. It bounces back to its original position after hitting the surface. The force on the ball during the collision is proportional to the length of compression of the ball. Which one of the following sketches describes the variation of its kinetic energy K with time t most appropriately? The figures are only illustrative and not to the scale. (A) (B) (C) (D)

1. (B)

K = 21m

2υ = 2 21

mg t2

2. A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire is fixed vertically on ground as shown in the figure. The bead is released from near the top of the wire and it slides along the wire without friction. As the bead moves from A to B, the force it applies on the wire is (A) always radially outwards. (B) always radially inwards. (C) radially outwards initially and radially inwards later. (D) radially inwards initially and radially outwards later.

2. (D)

mgR(1 − cos θ) = 21mv

2

∴ 2mv

R = 2mg(1 cos ) − θ

∴ N = mg cos θ − 2mv

R

= mg cos θ − 2mg (1 − cos θ) = mg cos θ + 2 mg cos θ − 2 mg

= 3 mg cos θ − 2 mg = 3 mg 2

cos3

⎛ ⎞ θ − ⎜ ⎟⎝ ⎠

if cos θ > 2

3 N is positive

∴ wire will exert force radially outward. so bead will exert force radially inward initially & radially outward later.

θ R cos θ

N

R


4

3. During an experiment with a metre bridge, the galvanometer shows a null point when the jockey is pressed at 40.0 cm using a standard resistance of 90 Ω, as shown in the figure. The least count of the scale used in the metre bridge is 1 mm. The unknown resistance is (A) 60 ± 0.15 Ω (B) 135 ± 0.56 Ω (C) 60 ± 0.25 Ω (D) 135 ± 0.23 Ω

3. (C)

R

40 =

90

60

∴ R = 60 Ω R

R

Δ = 1 2

1 2

Δ Δ + � �

� � =

0.1 0.1

40 60 + =

5

1200 =

1

240

ΔR = R

240 =

60

240=

1

4 = 0.25

∴ 60 ± 0.25 Ω

4. Charges Q, 2Q and 4Q are uniformly distributed in three dielectric solid spheres 1, 2 and 3 of radii R/2, R and 2R respectively, as shown in figure. If magnitudes of the electric fields at point P at a distance R from the centre of spheres 1, 2 and 3 are E1, E2 and E3 respectively, then (A) E1 > E2 > E3 (B) E3 > E1 > E2 (C) E2 > E1 > E3 (D) E3 > E2 > E1

4. (C)

E1 = 2

KQ

R

E2 = 2

K2Q

R

E3 = 2

KQ

2R

E2 > E1 > E3


5

5. A point source S is placed at the bottom of a transparent block of height 10 mm and refractive index 2.72. It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter 11.54 mm on the top of the block. The refractive index of the liquid is (A) 1.21 (B) 1.30 (C) 1.36 (D) 1.42

5. (C)

tan θ = 11.54

2 10×

sin θ = ( )2

11.54

11.54 (400)+

2.72 sin θ = μ × sin 90°

( )2

11.542.72

11.54 400 ×

+ = μ

μ = 1.36 6. Parallel rays of light of intensity I = 912 Wm−2 are incident on a spherical black body kept

in surroundings of temperature 300 K. Take Stefan-Boltzmann constant σ = 5.7 × 10−8 Wm−2K−4 and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close to (A) 330 K (B) 660 K (C) 990 K (D) 1550 K

6. (A)

σ × T4 = σ (300)4 + 912

4

T4 = 48

912(300)

(4 5.7 10 )−+× ×

= 4 8912(300) 10

22.8+ ×

= (300)4 + (40 × 108) = (81 + 40) × 108 = 121 × 108 ∴ T = 330 K

7. A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are u1 and u2, respectively. If the ratio u1 : u2 = 2 : 1 and hc = 1240 eV nm, the work function of the metal is nearly (A) 3.7 eV (B) 3.2 eV (C) 2.8 eV (D) 2.5 eV

7. (A)

1

2

k

k =

21

2

u

u

⎛ ⎞⎜ ⎟⎝ ⎠

= 4

k1 = hc

λ − φ =

1240

248 − φ

k1 = (5 − φ)

k2 = 1240

310 − φ

θ

2.72

r = 11.54

2 μ

10 mm


6

k2 = (4 − φ)

1

2

k

k =

5

4

⎛ ⎞− φ⎜ ⎟ − φ⎝ ⎠

= 4 ⇒ 5 − φ = 4(4 − φ)

⇒ 5 − φ = 16 − 4φ

⇒ 3 φ = 11 ⇒ φ = 11

3 = 3.7ev

8. If λCu is the wavelength of Kα X-ray line of copper (atomic number 29) and λMo is the wavelength of the Kα X-ray line of molybdenum (atomic number 42), then the ratio λCu/λMo is close to (A) 1.99 (B) 2.14 (C) 0.50 (D) 0.48

8. (B) ν = a (z − b)

2

1

νν

= 2

1

z b

z b

− −

= 41

28⎛ ⎞⎜ ⎟⎝ ⎠

(b = 1)

1

1

λλ

= 241

28⎛ ⎞⎜ ⎟⎝ ⎠

= 2.14

9. A planet of radius R = 1

X10

(radius of Earth) has the same mass density as Earth.

Scientists dig a well of depth R

5 on it and lower a wire of the same length and of linear

mass density 10−3 kgm−1 into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of Earth = 6 × 106 m and the acceleration due to gravity on Earth is 10 ms−2) (A) 96 N (B) 108 N (C) 120 N (D) 150 N

9. (B) fnet = dm g ′∫

= R

o

4R5

g rdr

Rλ∫

= R

0

4R5

grdr

R

λ∫ =

R2o

4R5

g r

R 2

⎛ ⎞λ⎜ ⎟⎝ ⎠

= 2

2og 16RR

2R 25

⎛ ⎞λ −⎜ ⎟⎝ ⎠

= 2

og 9R

2R 25

⎛ ⎞λ⎜ ⎟⎝ ⎠

= o9 g R

50

λ

Re = 6 × 105

go = 2

GMR

1000100

=×

1 m/s2

⇒ fnet =

3 59 10 (1) 6 10

50

−× × × =

5400

50 = 108 N


7

10. A glass capillary tube is of the shape of a truncated cone with an apex angle α so that its two ends have cross sections of different radii. When dipped in water vertically, water rises in it to a height h, where the radius of its cross section is b. If the surface tension of water is S, its density is ρ, and its contact angle with glass is θ, the value of h will be (g is the acceleration due to gravity)

(A) 2 S

cos( )b g

θ − αρ

(B) 2 S

cos( )b g

θ + αρ

(C) 2 S

cos( 2)b g

θ − αρ

(D) 2 S

cos( 2)b g

θ + αρ

10. (D)

2π b × s cos 2

α⎛ ⎞θ +⎜ ⎟⎝ ⎠

= π b2 h νg

2S

b gυcos

2

α⎛ ⎞θ +⎜ ⎟⎝ ⎠

= h

SECTION − 2 : Comprehension Type (Only One Option Correct)

This section contains 3 paragraphs, each describing theory, experiments, data etc. Six questions relate to the three paragraphs with two questions on each paragraph. Each question has only one correct answer among the four given options (A), (B), (C) and (D).

Paragraph for Questions 11 & 12

In the figure a container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled with 2 moles of an ideal monatomic gas at 700 K and the upper compartment is filled with 2 moles of an ideal diatomic gas at 400 K.

The heat capacities per mole of an ideal monatomic gas are V P3 5

C R, C R2 2

= = , and those

for an ideal diatomic gas are V P5 7

C R, C R2 2

= = .

11. Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final temperature of the gases will be (A) 550 K (B) 525 K (C) 513 K (D) 490 K 11. (D) Heat loss = heat gain n1 CP1 (T − 400) = n2 CV2 (700 − T) 7 (T − 400) = 3 (700 − T) 10 T = 4900 T = 490K

12. Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be (A) 250 R (B) 200 R (C) 100 R (D) −100 R


8

12. (D) n1 CP1 (T − 400) = n2 CP2 (700 − T) 7 (T − 400) = 5 (700 − T) 12 T = 6300

T = 6300 2100

=12 4

= 525 K

w1 = n1 R ΔT1 = 2(R) 125 = 250 R w2 = n2 R ΔT2 = 2 (R) (−175) = − 350 R wnet = − 100 R

Paragraph For Questions 13 & 14

A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniform cross section is connected to the nozzle. The other end of the tube is in a small liquid container. As the piston pushes air through the nozzle, the liquid from the container rises into the nozzle and is sprayed out. For the spray gun shown, the radii of the piston and the nozzle are 20 mm and 1 mm respectively. The upper end of the container is open to the atmosphere. 13. If the piston is pushed at a speed of 5 mms−1, the air comes out of the nozzle with a speed of (A) 0.1 ms−1 (B) 1 ms−1 (C) 2 ms−1 (D) 8 ms−1 13. (C) Av = constant ⇒ 5 mms1 × (20 mm)2 = ν × (1mm)2 ⇒ ν = 2ms−1 14. If the density of air is ρa and that of the liquid ρ

�, then for a given piston speed the rate

(volume per unit time) at which the liquid is sprayed will be proportional to

(A) aρρ�

(B) aρ ρ�

(C) a

ρρ� (D) ρ

�

14. (A) Speed of ejection of liquid = speed of outgoing air. In the limiting case ρL → ∞; volume flow rate = 0

Paragraph For Questions 15 & 16

The figure shows a circular loop of radius a with two long parallel wires (numbered 1 and 2) all in the plane of the paper. The distance of each wire from the centre of the loop is d. The loop and the wires are carrying the same current I. The current in the loop is in the counterclockwise direction if seen from above.


9

15. When d ≈ a but wires are not touching the loop, it is found that the net magnetic field on the axis of the loop is zero at a height h above the loop. In that case (A) current in wire 1 and wire 2 is the direction PQ and RS, respectively and h ≈ a (B) current in wire 1 and wire 2 is the direction PQ and SR, respectively and h ≈ a (C) current in wire 1 and wire 2 is the direction PQ and SR, respectively and h ≈ 1.2 a (D) current in wire 1 and wire 2 is the direction PQ and RS, respectively and h ≈ 1.2 a 15. (C)

B1oop = 0

2

μ

2

2 2 3/2

ia

(h a )+

0

2 2

i

2 d h

μ

π +

∴ 2

2 2 3/2

ia

2 (h a )

μ ⋅+

= 0i

2a

μ

2 2

d

d h+

⇒ 2

2 2 3/2

a

2(h a )+ =

2 2

a

(a h )π + ⇒ π a = 2 2 2h a+

⇒ a = 2

2h

( 4)π −

16. Consider d >> a, and the loop is rotated about its diameter parallel to the wires by 30° from the position shown in the figure. If the currents in the sires are in the opposite directions, the torque on the loop at its new position will be (assume that the net field due to the wires is constant over the loop)

(A) 2 2

0I a

d

μ (B)

2 20I a

2d

μ (C)

2 203 I a

d

μ (D)

2 203 I a

2d

μ

16. (B)

B = 2 0i

2 d

μ⎛ ⎞⎜ ⎟π⎝ ⎠

∴ τ = iπa2 0

d

μπ

sin (30°)

SECTION − 3 : Matching List Type (Only One Option Correct)

This section contains four questions, each having two matching lists. Choices for the correct combination of elements from List-I and List-II are given as options (A), (B), (C) and (D), out of which one is correct.

4

d

O


10

17. Four charges Q1, Q2, Q3 and Q4 of same magnitude are fixed along the x axis at x = −2a, −a, +a and +2a, respectively. A positive charge q is placed on the positive y axis at a distance b > 0. Four options of the signs of these charges are given in List I. The direction of the forces on the charge q is given in List II. Match List I with List II and select the correct answer using the code given below the lists.

List I List II P. Q1, Q2, Q3, Q4 all positive 1. +x Q. Q1, Q2 positive; Q3, Q4 negative 2. −x R. Q1, Q4 positive; Q2, Q3 negative 3. +y S. Q1, Q3 positive; Q2, Q4 negative 4. −y

Code : (A) P-3, Q-1, R-4, S-2 (B) P-4, Q-2, R-3, S-1 (C) P-3, Q-1, R-2, S-4 (D) P-4, Q-2, R-1, S-3 17. (A) (P) → (3), (Q) → (1), (R) → (4), (S) → (2) (P) → (3)

All x component will be cancelled only y will be exist i.e. along +ve.

(Q) → (1)

All y component will be cancelled and +ve x will exist.

(R) → (4) Each pair will cancel x of each other and y of +ve pair will be smaller than −ve so net will

be along −ve y. (S) → (2) Each dipole pair will cancelled y of each other and x of Q2 − Q3 pair along −ve x is

stronger than that of Q1 − Q4 along +ve x.

q (0, b)

Q1 Q2 Q3 Q4

+ + − −

+ − − +

+ − + − Q1 Q2 Q3 Q4


11

18. Four combinations of two thin lenses are given in List I. The radius of curvature of all curved surfaces is r and the refractive index of all the lenses is 1.5. Match lens combinations in List I with their focal length in List II and select the correct answer using the code given below the lists.

List I List II

P.

1. 2r

Q.

2. r/2

R.

3. −r

S.

4. r

Code : (A) P-1, Q-2, R-3, S-4 (B) P-2, Q-4, R-3, S-1 (C) P-4, Q-1, R-2, S-3 (D) P-2, Q-1, R-3, S-4 18. (B) (P) → (2), (Q) → (4), (R) → (3), (S) → (1)

( )biconvexR

f = = R2 µ 1−

( )planoconvexR

f = = 2Rµ 1−

( )plano convexR

f = = 2Rµ 1

− −−

(P) → (2) eqeq 1 2

1 1 1 1 1 R= + = + f =

f f f R R 2⇒

(Q) → (4) eqeq

1 1 1 1= + = f = R

f 2R 2R R⇒

(R) → (3) eqeq

1 1 1 1= = f = R

f 2R 2R R

−− − ⇒ −

(S) → (1) eq1 1 1 1

= = f = 2Rf R 2R 2R

− ⇒

19. A block of mass m1 = 1 kg another mass m2 = 2 kg, are placed together (see figure) on an

inclined plane with angle of inclination θ. Various values of θ are given in List I. The coefficient of friction between the block m1 and the plane is always zero. The coefficient of static and dynamic friction between the block m2 and the plane are equal to μ = 0.3. In List II expressions for the friction on block m2 are given. Match the correct expression of the friction in List II with the angles given in List I, and choose the correct option. The acceleration due to gravity is denoted by g.

[Useful information : tan (5.5°) ≈ 0.1; tan(11.5°) ≈ 0.2; tan(16.5°) ≈ 0.3]


12

List I List II P. θ = 5° 1. m2g sin θ Q. θ = 10° 2. (m1 + m2) g sin θ R. θ = 15° 3. μm2 g cos θ S. θ = 20° 4. μ (m1 + m2) g cos θ

Code : (A) P-1, Q-1, R-1, S-3 (B) P-2, Q-2, R-2, S-3 (C) P-2, Q-2, R-2, S-4 (D) P-2, Q-2, R-3, S-3 19. (D) (P) → (2), (Q) → (2), (R) → (3), (S) → (3) for not to slip μ m2g cosθ = (m1 + m2) g sinθ μ 2 cos θ = 3 sin θ

tan θ = 23

μ = 0.63

= 0.2

in case (P) (Q) it will not slip ⇒ friction is (m1 + m2) g sinθ for (R) (S) of they must slip so friction should be kinetic i.e. μ m2gcos θ

20. A person in a lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance d of 1.2 m from the person. In the following, state of the lift’s motion is given in List I and the distance where the water jet hits the floor of the lift is given in List II. Match the statements from List I with those in List II and select the correct answer using the code given below the lists.

List I List II P. Lift is accelerating vertically up. 1. d = 1.2 m Q. Lift is accelerating vertically down with an

acceleration less than the gravitational acceleration.

2. d > 1.2 m

R. Lift is moving vertically up with constant speed. 3. d < 1.2 m S. Lift is falling freely. 4. No water leaks out of the jar

Code : (A) P-2, Q-3, R-2, S-4 (B) P-2, Q-3, R-1, S-4 (C) P-1, Q-1, R-1, S-4 (D) P-2, Q-3, R-1, S-1 20. (C) V = 2gh

d = 2h 2H

V 2ghg g

= = 2 h(H)

for other value of geff it is independent of value of geff except geff = 0

where we cannot cancel g in equation 0

0⎛ ⎞⎜ ⎟⎝ ⎠

in that case

it will not come out as there is no pressure at all (P) (Q) (R) → (I) (S) → (4)

θ

m1 m2 μ = 0

μ = 0.3

H

h


13

PART II : CHEMISTRY

SECTION − 1 : (Only One Option Correct Type)

This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE option is correct.

21. The acidic hydrolysis of ether (X) shown below is fastest when [Figure]

(A) one phenyl group is replaced by a methyl group. (B) one phenyl group is replaced by a para−methoxyphenyl group.

(C) two phenyl groups are replaced by two para−methoxyphenyl groups. (D) no structural change is made to X. 21. (C) Para −methoxy phenyl group is electrodonating which can stabilize carbocation.

22. Isomers of hexane, based on their branching, can be divided into three distinct classes as shown in the figure.

[Figure]

The correct order of their boiling point is (A) I > II > III (B) III > II > I (C) II > III > I (D) III > I > II 22. (B) With increase in branching boiling point decreases.

23. The major product in the following reaction is [Figure]

(A) (B)

(C) (D)


14

23. (D)

O

ClCH3

3(1) CH MgBr, dry ether ⎯⎯⎯⎯⎯⎯⎯⎯→ Cl

O

CH3

O CH3

CH3

�

24. Hydrogen peroxide in its reaction with KIO4 and NH2OH respectively, is acting as a (A) reducing agent, oxidising agent (B) reducing agent, reducing agent (C) oxidising agent, oxidising agent (D) oxidising agent, reducing agent 24. (A) 25. The product formed in the reaction of SOCl2 with white phosphorous is (A) PCl3 (B) SO2Cl2

(C) SCl2 (D) POCl3 25. (A)

26. Under ambient conditions, the total number of gases released as products in the final step of the reaction scheme shown below is

(A) 0 (B) 1 (C) 2 (D) 3 26. (C)

6 2 3 2 2XeF 3H O XeO F + ⎯⎯→ + 3Η

3 4XeO OH HXeO− − + ⎯⎯→

44 6 2 22HXeO 2OH XeO Xe 2H O O− − − + ⎯⎯→ + + +

27. For the identification of β−naphthol using dye test, it is necessary to use

(A) dichloromethane solution of β−naphthol.

(B) acidic solution of β−naphthol.

(C) neutral solution of β−naphthol.

(D) alkaline solution of β−naphthol. 27. (D)

28. For the elementary reaction M → N, the rate of disappearance of M increases by a factor of 8 upon doubling the concentration of M. The order of the reaction with respect to M is

(A) 4 (B) 3 (C) 2 (D) 1


15

28. (B) r1 = K[M]x …(1) 8r1 = K[2M]x …(2)

⇒ 1

8 =

x1

2⎛ ⎞⎜ ⎟⎝ ⎠

⇒ 31

2⎛ ⎞⎜ ⎟⎝ ⎠

= n1

2⎛ ⎞⎜ ⎟⎝ ⎠

⇒ x = 3

29. For the process

2 2H O( ) H O(g) → �

at T = 100 °C and 1 atmosphere pressure, the correct choice is

(A) ΔSsystem > 0 and ΔSsurroundings > 0 (B) ΔSsystem > 0 and ΔSsurroundings < 0

(C) ΔSsystem < 0 and ΔSsurroundings > 0 (D) ΔSsystem < 0 and ΔSsurroundings < 0 29. (B)

2 2H O( ) H O (g) (100 C) ⎯⎯→ � °

ΔS > 0 system ΔS surrounding < 0 30. Assuming 2s−2p mixing is NOT operative, the paramagnetic species among the following is (A) Be2 (B) B2 (C) C2 (D) N2 30. (B)

SECTION − 2 : Comprehension Type (Only one Option Correct)


Paragraph For Questions 31 and 32

Schemes 1 and 2 describe sequential transformation of alkynes M and N. Consider only the major products formed in each step for both the schemes.

N H

M OH


16

31. The product X is

31. (C)

2NaNH

2 2HO CH CH C CH− − − ≡

2 5C H r(leq)Β

2 2Na O CH CH C C Na− ≡ −ⓛ

3CH I

5 2 2 2H C O CH CH C C Na− ≡ −

2 4H | Pd BaSO−

5 2 2 2 3H C O CH CH C C CH− ≡ −

5 2 2 2 3H C O CH CH C C CH− = −

H H 32. The correct statement with respect to product Y is (A) It gives a positive Tollens test and is a functional isomer of X. (B) It gives a positive Tollens test and is a geometrical isomer of X. (C) It gives a positive iodoform test and is a functional isomer of X. (D) It gives a positive iodoform test and is a geometrical isomer of X. 32. (C)

3 2CH CH C CH− − ≡

2NaNH

C C C C Na− − ≡ −

BrOH

3H O+

2C C C C CH CH OH− − ≡ − − −

CH3

(B)

(D) (C)

3 2CH CH O

H H

3H CO

H H

(A)


17

2C C C C CH CH OH− − − − − −

O

CH3

2C C C C CH CH OH− − − − − −

CH32H | Pd | C

2C C C C CH C O− − − − − =

CH32 3Cr O

Paragraph For Questions 33 and 34 An aqueous solution of metal ion M1 reacts separately with reagents Q and R in excess to give tetrahedral and square planar complexes, respectively. An aqueous solution of another metal ion M2 always forms tetrahedral complexes with these reagents. Aqueous solution of M2 on reaction with reagent S gives white precipitate which dissolves in excess of S. The reactions are summarized in the scheme given below:

SCHEME:

33. M1, Q and R, respectively are (A) Zn2+, KCN and HCI (B) Ni2+, HCI and KCN (C) Cd2+, KCN and HCI (D) Co2+, HCI and KCN 33. (B)

HCl KCN2 24 4

tetra hedral Sq. Planar[NiCl ] Ni Ni(CN)− +

←⎯⎯⎯ ←⎯⎯⎯

34. Reagent S is (A) K4[Fe(CN)6] (B) Na2HPO4 (C) K2CrO4 (D) KOH 34. (D)

[ ] 2 Cl KCN24 2 4

tetratetra

ZnCl Zn K Zn(CN)− −− + ⎯⎯⎯→ ⎯⎯⎯→

It gives a white precipitate with excess of NaOH.


18

Paragraph for Questions 35 and 36

X and Y are two volatile liquids with molar weights of 10 g mol−1 and 40 g mol−1 respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L = 24 cm, as shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature of 300 K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and Y to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours. 35. The value of d in cm (shown in the figure), as estimated from Graham′s law, is (A) 8 (B) 12 (C) 16 (D) 20 35. X M1 = 10, Y M2 = 40

x 40

24 x 10=

−

x 2

24 x 1=

−

x = 48 − 2x 3x = 48

x = 48

163

=

36. The experimental value of d is found to be smaller than the estimate obtained using

Graham′s law. This is due to (A) larger mean free path for X as compared to that of Y. (B) larger mean free path for Y as compared to that of X. (C) increased collision frequency of Y with the inert gas as compared to that of X with the

inert gas. (D) increased collision frequency of X with the inert gas as compared to that of Y with the

inert gas. 36. (D) Larger the size of the molecules, smaller the mean free path.

Collision frequency (z) = 2 2

avn

2

π σ μ

Mean free path (λ) = 2

1

2 n σ

where, n = No. of molecules per unit molar volume. σ = Collision diameter.


19

SECTION − 3 : Matching List Type (Only one Option Correct) This section contains four questions, each having two matching lists. Choices for the correct combination of elements from List-I and List-II are given as options (A), (B), (C) and (D), out of which one is correct. 37. Different possible thermal decomposition pathways for peroxyesters are shown below.

Match each pathway from List I with an appropriate structure from List II and select the correct answer using the code given below the lists.

List-I List-II

P. Pathway P 1.

Q. Pathway Q

2.

R. Pathway R

3.

S. Pathway S

4.

Codes :

P Q R S (A) 1 3 4 2 (B) 2 4 3 1 (C) 4 1 2 3 (D) 3 2 1 4

37. (A)


20

3 H C C H − ≡ − Red hotiron, 873 K

⎯⎯⎯⎯⎯→ 3 2 4HNO H SOheat + ⎯⎯⎯⎯⎯⎯→ 2 3H S NH ⋅ ⎯⎯⎯⎯⎯→

NO2

NO2

NO2

NH2

2Na NO

2 4SO+ Η

NO2

OH

Hydrolysis←⎯⎯⎯⎯⎯

38. Match the four starting materials (P,Q,R,S) given in List I with the corresponding reaction schemes (I, II, III, IV) provided in List II and select the correct answer using the code given below the lists.

List I List II

P. 1. Scheme I

(i) KMnO4 , H O�

, heat (ii) H⊕ , H2O (iii) SOCl2 (iv) NH3 ? C7H6N2O3

Q. 2. Scheme II (i) Sn/HCl (ii) CH3COCl (iii) conc. H2SO4

(iv) HNO3 (v) dil.H2SO4 , heat (vi) H O�

? C6H6N2O2

R. 3. Scheme III (i) red hot iron, 873 K (ii) fuming HNO3 , H2SO4, heat (iii) H2S.NH3 (iv) NaNO2, H2SO4 (v) hydrolysis ? C6H5NO3

S.

4. Scheme IV

(i) conc. H2SO4 , 60°C (ii) conc. HNO3, conc. H2SO4

(iii) dil. H2SO4, heat

? C6H5NO4

Code : P Q R S (A) 1 4 2 3 (B) 3 1 4 2 (C) 3 4 2 1 (D) 4 1 3 2 38. (C)

(P → 3)


21

NO2

COOH

⎯⎯→ ⎯⎯→

NO2

C Cl

O

NO2

C NH2

O

NO2

CH3

⎯⎯→

(S → 1)

39. Match each coordination compound in List−I with an appropriate pair of characteristics from List − II and select the correct answer using the code given below the lists.

{en = H2NCH2CH2NH2 ; atomic numbers : Ti = 22 ; Cr = 24, C = 27, Pt = 78}

List I List II P. [Cr(NH3)4Cl2]Cl 1. Paramagnetic and exhibits ionisation isomerism

Q. [Ti(H2O)5Cl](NO3)2 2. Diamagnetic and exhibits cis−trans isomerism

R. [Pt(en) (NH3)Cl]NO3 3. Paramagnetic and exhibits cis−trans isomerism

S. [Co(NH3)4(NO3)2]NO3 4. Diamagnetic and exhibits ionisation isomerism

Code : P Q R S (A) 4 2 3 1 (B) 3 1 4 2 (C) 2 1 3 4 (D) 1 3 4 2 39. (B)

(P) 3 33 4 2 18

Paramagnetic[Cr(NH ) Cl ] Cl i.e. Cr i.e.[Ar] 3d+ ⎯⎯→

Ma4b2 shows geometrical isomerism. P → 3.

40. Match the orbital overlap figures shown in List−I with the description given in List− II and select the correct answer using the code given below the lists

List I List II

P.

1. p − d π antibonding

Q.

2. d − d σ bonding

R.

3. p − d π bonding

S.

4. d − d σ antibonding


22

Code : P Q R S (A) 2 1 3 4 (B) 4 3 1 2 (C) 2 3 1 4 (D) 4 1 3 2 40. (D) P → 4 (LCAO of d − d σ anti-bonding) Q → 1 (LCAO of p − d π anti-bonding orbitals) R → 3 (LCAO of p − d π bonding orbitals) S → 2 (LCAO of d − d σ bonding orbitals)

PART III - MATHEMATICS

SECTION − 1 : (Only One Option Correct Type) This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE option is correct.

41. The function y = f(x) is the solution of the differential equation

2

dy xy

dx x 1 +

− =

4

2

x 2x

1 x

+

−

in (−1, 1) satisfying f(0) = 0. Then 3

2

32

f (x) dx

−

∫ is

(A) 3

3 2

π − (B) 3

3 4

π − (C) 3

6 4

π − (D) 3

6 2

π −

41. (B)

( )2

2 2

x 2xdy xy

dx x 1 1 x

++ =

− −

I.F. = 2

e2x 1dx log 1 x 2x 1 2e e 1 x

−−

∫= = −

∴ Solution is 4

2 2

2

x 2xy 1 x . 1 x dx

1 x

+− = −−

∫

5 2 5

2 2x 2x xy 1 x c x c

5 2 5⇒ − = + + = + +

5 2

2

x 5xy

5 1 x

+⇒ =

− ( )( )f 0 0=∵

Now, 3 2 3 25 2

2 23 2 3 2

x x 3I dx

3 45 1 x 1 x− −

π= + = −− −

∫ ∫


23

42. The following integral

217

4

(2 cosec x) dx

π

π ∫ is equal to

(A) ( )log(1 2)

16u u

0

2 e e du+

− + ∫ (B) ( )( )log 1 2

17u u

0

e e du +

− + ∫

(C) ( )( )log 1 2

17u u

0

e e du +

− − ∫ (D) ( )( )log 1 2

16u u

0

2 e e du +

− − ∫

42. (A)

/2

17

/4

(2cos ec(x)) dxπ

π∫

Let 2 cosec (x) = eu + e−u ⇒ cosec (x) = u ue e

2

−+

⇒ − cosec x cot x dx = u ue e

du2

−⎛ ⎞−⎜ ⎟⎜ ⎟⎝ ⎠

⇒ dx = ( )u u

2u u u u

e e du

e e e e2 1

2 2

−

− −

−

⎛ ⎞ ⎛ ⎞+ + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⇒ dx = −u u

2du

e e−+

∴ I = ( )( )e

0u u

u ulog 1 2

2e e du

e e−

−+

− ++∫

I = ( )( )elog 1 2

16u u

0

2 e e du

+−+∫

43. Coefficient of x11 in the expansion of (1 + x2)4 (1 + x3)7 (1 + x4)12 is

(A) 1051 (B) 1106 (C) 1113 (D) 1120 43. (C) Coefficient of x11 in (1 + x2)4 (1 + x3)7 (1 + x4)12

The number of ways in variation of x, x0 x3 x8 x8 x3 x0 x2 x9 x0 x4 x3 x4

Coefficient of x11 = (4C0 × 7C1 × 12C1) + (4C4 × 7C1 × 12C0) + (4C1 × 7C3 × 12C0) + (4C2 × 7C1 × 12C1) = (1 × 7 × 66) + (1 × 7 × 1) + (4 × 35 × 1) + (6 × 7 × 12) = 462 + 7 + 140 + 504 = 1113


24

44. Let f: [0, 2] → � be a function which is continuous on [0, 2] and is differentiable on (0, 2) with f(0) = 1. Let

F(x) = ( )2x

0

f t dt∫

for x ∈ [0, 2]. If F′(x) = f (x)′ for all x ∈ (0, 2), then F(2) equals

(A) e2 − 1 (B) e4 − 1 (C) e − 1 (D) e4 44. (B) f : [0, 2] → R

F(x) =

2x

0

f ( t )dt∫

F′(x) = {f(x)} 2x Given F′(x) = f′(x)

f (x)

f (x)

′∫ = 2x∫

ln f(x) = x2 + K

f(x) = 2x Ke +

f(0) = 1 eK = 1

f(x) = 2x Ke e

Therefore,

f(x) = 2xe

F(x) =

2xt

0

e dt∫

F(2) = 4

t

0

e dt∫

F(2) = 4t0[e ]

F(2) = e4 − 1 45. The common tangents to the circle x2 + y2 = 2 and the parabola y2 = 8x touch the circle at

the points P, Q and the parabola at the points R, S. Then the area of the quadrilateral PQRS is (A) 3 (B) 6 (C) 9 (D) 15

45. (D) Let the slope of common tangent is m. The eqn. w.r.t. circle is

y = mx + 22 1 m+ … (1) and w.r.t parabola is

y = mx + 2

m … (2)

(1) & (2) are same

Therefore, 22 1 m+ = 2

m

2 (1 + m2) = 2

4

m

R

S

x x′

y′

y

P

Q


25

Let m2 = t t (1 + t)= 2 t2 + t − 2 = 0 t = 1, −2 When m2 = 1, then m = ±1 and m2 = −2 is not possible. Now eqn. of tangent are, y = x + 2 and y = − x − 2. By solving with circle P and Q are (−1, 1) and (−1, −1) And by solving with parabola R and S are (2, 4) and (2, −4) By knowing the co-ordinates area is 15. 46. for x ∈ (0, π), the equation sin x + 2 sin 2x − sin 3x = 3 has

(A) infinitely many solutions (B) three solutions (C) one solution (D) no solution

46. (D) sinx + 2 sinx − sin 3x = 3 (2 sin 2x) − (sin 3x − sin x) = 3 (2 sin 2x) − (2 cos 2x . sin x) = 3 2(2 sin x . cos x) − 2 sin x(cos 2x) = 3 2 sin x(2 cos x − cos 2x) = 3 2 sin x{2 cos x − (2 cos2x − 1)} = 3 2 sin x{−2 cos2 x + 2 cos x + 1} = 3 {−2 cos2 x + 2 cos x + 1} will attain maximum value 3/2 when cos x = 1/2 and at the same time sin x must be equal to 1. But it is not possible. Hence no solution.

47. In a triangle the sum of two sides is x and the product of the same two sides is y. If

x2 − c2 = y, where c is the third side of the triangle, then the ratio of the in-radius to the circum-radius of the triangle is

(A) 3y

2x(x c) + (B)

3y

2c(x c)+ (C)

( )3y

4x x c + (D) ( )

3y

4c x c+

47. (B) a + b = x ab = y a + b + c = x + c 2s = x + c

x c

s2

+ =

Given x2 − c2 = y (a + b)2 − c2 = y a2 + b2 + 2ab − c2 = ab ⇒ a2 + b2 − c2 = −ab

cos C = 2 2 2a b c 1

2ab 2

+ − = −

C = 120°

So, r =

1absin C ysin C y sin120 y. 32

a b cs x c x c 2(x c)2

Δ ⋅ °= = = =+ + + + +

c b

a


26

R = c c c

2sin C 3 32

2

= =⋅

y 3r 3y2(x c)

cR 2c(x c)3

+= =+

48. Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in

envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover the card numbered 1 is always placed in envelope numbered 2. Then the number of ways it can be done is (A) 264 (B) 265 (C) 53 (D) 67

48. (C) Total Dearrangement of 6

1 1 1 1 1 1

6! 11! 2! 3! 4! 5! 6!

⎛ ⎞− + − + − +⎜ ⎟⎝ ⎠

= 360 − 120 + 30 − 6 + 1 = 265

In these dearrangements, there are 5 ways in which card numbered 1 is going wrong.

So when it is going in envelop numbered 2 is 265

5= 53 ways.

49. Three boys and two girls stand in a queue. The probability, that the number of boys ahead

of every girl is at least one more than the number of girls ahead of her, is

(A) 1

2 (B)

1

3 (C)

2

3 (D)

3

4

49. (A) n(S) = 5! n(E) = 3 ! × 2 ! × 5

p(E) = n(E) 3! 2! 5

n(S) 5!

× ×= = 1

.2

50. The quadratic equation p(x) = 0 with real coefficients has purely imaginary roots. Then

the equation p(p(x)) = 0 has (A) only purely imaginary roots (B) all real roots (C) two real and two purely imaginary roots (D) neither real nor purely imaginary roots

50. (D) When roots are purely imaginary. Then the form of equation is x2 + K = 0 where K is positive no. Let, p(x) = x2 + K p(p(x)) = (p(x))2 + K p(p(x)) = (x2 + K)2 + K p(p(x)) = x4 + 2Kx2 + K p(p(x)) = 0 x4 + 2Kx2 + K = 0 All coefficients are positive and no odd degree of x are present.


27

SECTION − 2 : Comprehension Type (Only One Option Correct)


Paragraph for Questions 51 and 52 Let a, r, s, t be nonzero real numbers. Let P(at2, 2at), Q, R(ar2, 2ar) and S(as2, 2as) be distinct points on the parabola y2 = 4ax. Suppose that PQ is the focal chord and lines QR and PK are parallel, where K is the point (2a, 0).

51. The value of r is

(A) − 1

t (B)

2t 1

t

+ (C)

1

t (D)

2t 1

t

−

51. (D) P ≡ (at2, 2at), F ≡ (a, 0)

Applying condition of collinearity, we get Q ≡ 2

a 2a,

tt

⎛ ⎞−⎜ ⎟⎝ ⎠

Since QR || PK,

2

22

12a r

2att1 at 2art

⎛ ⎞+⎜ ⎟⎝ ⎠ =

⎛ ⎞ −−⎜ ⎟⎝ ⎠

⇒ r = t − 21 t 1

t t

−=

52. If st = 1, then the tangent at P and the normal at S to the parabola meet at a point whose ordinate is

(A) 2 2

3

(t 1)

2t

+ (B)

2 2

3

a(t 1)

2t

+ (C)

2 2

3

a(t 1)

t

+ (D)

2 2

3

a(t 2)

t

+

52. (B)

S = 1

t

Equation of normal at S is 3

x 2a ay

t t t+ = + …………. (1)

Equation of tangent at P is ty = x + at2 …………. (2) Solving Eq. (1) and (2) simultaneously, we get,

y = ( )22

3

a t 1

2t

+

(a, 0) k

Q

R

P

F


28


Given that for each a ∈ (0, 1) 1 h

a a 1

h 0 h

lim t (1 t) dt+

−− −

→ − ∫ , exists. Let this limit be g(a). In addition, it is given

that the function g(a) is differentiable on (0, 1).

53. The value of g1

2⎛ ⎞⎜ ⎟⎝ ⎠

is

(A) π (B) 2π (C) 2

π (D)

4

π

53. (A)

g1 h

1/2 1/2

h 0 h

1Lt t (1 t) dt

2 +

−− −

→

⎛ ⎞ = −⎜ ⎟⎝ ⎠

∫ = 1 h

2h 0 h

1Lt dt

t t+

−

→ −∫

= 1 h

1

2 2h 0 h 0 h

dt 2t 1Lt Lt sin

11 1t

2 2

+ +

−−

→ →

−⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

∫

= 1 1

h 0Lt sin (1 2h) sin (2h 1)

+− −

→− − − = π

54. The value of g′ 1

2⎛ ⎞⎜ ⎟⎝ ⎠

is

(A) 2

π (B) π (C) −

2

π (D) 0

54. (D)

gh 0

1 1g g

1 2 2Lt

2 →

⎛ ⎞ ⎛ ⎞+ δ −⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎝ ⎠ ⎝ ⎠=⎜ ⎟ δ⎝ ⎠

=

( )1 1 1 11 h 1 12 2 2 2

h

h 0 0

t (1 t) t (1 t) dt

Lt Lt

− − −δ − +δ− − −

→ δ→

− − −

δ

∫

= ( )1 1

2 2

h 0 0

t (1 t) 1Lt Lt t (1 t) dt

−δ δ− −

→ δ→

⎛ ⎞− −⎜ ⎟−⎜ ⎟δ⎝ ⎠

∫

= 1 1 e2 2

h 0 0

1 t 1 tlog

t tLt t (1 t) Lt dt

1

δ

− −

→ δ→

⎛ ⎞− −⎛ ⎞ ⎛ ⎞⋅⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎜ ⎟− ⎜ ⎟

⎝ ⎠∫

= ( )1 11 h2 2

e eh 0

h

Lt t (1 t) log (1 t) log (t) dt− − −

→− − −∫

∴ Also, ( )1 11 h2 2

e eh 0

h

1g ' Lt t (1 t) log (1 t) log (t) dt

2

− − −

→

⎛ ⎞ = − − −⎜ ⎟⎝ ⎠

∫ b b

a a

, f (x)dx f (a b x)dx⎛ ⎞

= + −⎜ ⎟⎜ ⎟⎝ ⎠∫ ∫∵

∴ 1

g ' 02

⎛ ⎞ =⎜ ⎟⎝ ⎠


29


Box 1 contains three cards bearing numbers 1, 2, 3; box 2 contains five cards bearing numbers 1, 2, 3, 4, 5; and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A cards is drawn from each of the boxes. Let xi be the number on the card drawn from the ith box, i = 1, 2, 3.

55. The probability that x1 + x2 + x3 is odd, is

(A) 29

105 (B)

53

105 (C)

57

105 (D)

1

2

55. (B) 1) If x1 = 1, x2 and x3 can be selected in 4 + 3 + 4 + 3 + 4 = 18 ways 2) If x1 = 2, (x2 , x3) can be selected in 3 + 4 + 3 + 4 + 3 = 17 ways 3) In x1 = 3, (x2 , x3) can be selected in 18 ways

∴ P = 18 17 18 53

3 5 7 105

+ + =× ×

56. The probability that x1, x2, x3 are in an arithmetic progression, is

(A) 9

105 (B)

10

105 (C)

11

105 (D)

7

105

56. (C) The favourable cases are (1, 1, 1), (1, 2, 3), (1, 3, 5), (1, 4, 6), (2, 2, 2), (2, 3, 4), (2, 4, 6), (3, 3, 3), (3, 4, 5), (3, 5, 7), (3, 2, 1).

Hence, answer is 11

105.

SECTION − 3 : Matching List Type (Only One Option Correct)

This section contains four questions, each having two matching lists. Choices for the correct combination of elements from List-I and List-II are given as options (A), (B), (C) and (D), out of which one is correct.

57. Let zk = cos 2k

10

π⎛ ⎞⎜ ⎟⎝ ⎠

+ i sin 2k

10

π⎛ ⎞⎜ ⎟⎝ ⎠

; k = 1, 2, …, 9.

List I List II P. For each zk there exists a zj such that zk ⋅ zj = 1 1. True Q. There exists a k ∈ {1, 2, …, 9} such that z1 ⋅ z = zk has no solution z in the set of complex numbers.

2. False

R. 1 2 91 z 1 z 1 z

10

− − ... − equals

3. 1

S. 1 − 9k 1

2kcos

10=π⎛ ⎞

⎜ ⎟⎝ ⎠

∑ equals 4. 2

P Q R S (A) 1 2 4 3 (B) 2 1 3 4

(C) 1 2 3 4 (D) 2 1 4 3


30

57. (C)

π

=2k

i10

Kz e , K = 1, 2, …. 9

z = α, α2 …… α9, where π

α =2k

i10e

so, all are roots ( )1

10z 1=

P → 1 For Option Q z1.z = zk α.z = zk No z satisfy equation ⇒ Q → 2 For Option C z10 − 1 = (z − 1) (z − z1) (z − z2) …… (z − z9) z9 + z8 + …… + z + 1 = (z − z1 (z − z2) …… (z − z9) Put z = 1 10 = (1 − z1) (1 − z2) …… (1 − z9)

∴ 1 210 1 z 1 z= − − …… − 9z z

C → 3 For Option D Sum of roots = 0 1 + z1 + z2 + …… + zq = 0

2 2 4 4 18 181 cos isin cos i sin .... cos i sin 0

10 10 10 10 10 10

π π π π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + + + + + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 4 18equate real part 1 cos cos ...... cos 0

10 10 10

π π π⇒ ⇒ + + + + =

So, a

k 1

2k1 cos

10=

π⎛ ⎞− ⎜ ⎟⎝ ⎠

∑

= 1 − (−1) = 2 S → 4

58. List I List II (P) The number of polynomials f(x) with non-negative integer coefficients

of degree ≤ 2, satisfying f(0) = 0 and 1

0

f (x)dx 1,=∫ is 1 8

(Q) The number of points in the interval 13, 13⎡ ⎤−⎣ ⎦

at which f(x) = sin (x2) + cos (x2) attains its maximum value, is 2 2

(R)

( )2 2

x2

3xdx

1 e− +∫ equals 3 4

(S) 12

12

12

0

1 xcos 2x log dx

1 x

1 xcos 2x log dx

1 x

−

⎛ ⎞⎜ ⎟+⎛ ⎞⎜ ⎟⎜ ⎟−⎝ ⎠⎜ ⎟⎝ ⎠⎛ ⎞⎜ ⎟+⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠⎝ ⎠

∫

∫

equals 4 0


31

P Q R S (A) 3 2 4 1 (B) 2 3 4 1 (C) 3 2 1 4 (D) 2 3 1 4

58. (D) (P) → (2), (Q) → (3), (R) → (1), (S) → (4) (P) → (2)

f(x) = ax2 + bx + c f(0) = 0 ⇒ c = 0 a, b, c (non−negative integer coefficient)

( )1

0

f x dx 1=∫

1 13 2

0 0

x xa b. 1

3 2+ =

a b

13 2

⇒ + =

2a 3b 6⇒ + = (a & b non−negative integer coeff.) (Q) → (3)

( ) 2 2f ' x 2x cos x sin x⎡ ⎤= −⎣ ⎦ …… (1)

( ) ( ) { }( )2 2 2 2f ' x 2 cos x sin x x sin x cos x 2x⎡ ⎤= − + − −⎣ ⎦

= ( )2 2 22 cos x sin x 2x⎡ ⎤− −⎣ ⎦

…… (2)

( )f ' x 0 x 0= ⇒ = or cosx2 − sinx2 = 0

⇒ tanx2 = 1

2x n4

π= π +

2 5 9 13x , , ,

4 4 4 4

π π π π=

But ( )f '' x 0< at 2 5 9 13x , , ,

4 4 4 4

π π π π=

(R) → (1)

2 2

x2

3xI dx

1 e−

=+∫ …… (1)

=2 2

2x

3xdx

11

e− +∫

2 2 x

x2

3x eI dx

e 1−

=+∫ …… (2)

(1) + (2) 2

2

2

2I 3x dx−

= ∫ =23

2x

− = 8 − (−8) = 16


32

I = 8 (S) → (4)

12

12

1 xcos 2x.log dx 0

1 x−

+⎛ ⎞ =⎜ ⎟−⎝ ⎠∫ as odd function.

59. List I List II (P)

Let y(x) = cos (3 cos−1x), x ∈ [−1, 1], x ≠ ± 3

2.

Then ( )2

22

1 d y(x) dy(x)x 1 x

y(x) dxdx

⎧ ⎫⎪ ⎪− +⎨ ⎬⎪ ⎪⎩ ⎭

equals 1 1

(Q) Let A1, A2, …., An (n > 2) be the vertices of a regular polygon of n

sides with its centre at the origin. Let ka��

be the position vector of the

point Ak, k = 1, 2, …., n. If ( ) ( )n 1 n 1k 1 k k 1 k 1 k k 1a a a a ,− −

= + = +Σ × = Σ ⋅��

then

the minimum value of n is

2 2

(R) If the normal from the point P(h, 1) on the ellipse

2 2x y1

6 3+ = is

perpendicular to the line x + y = 8, then the value of h is

3 8

(S) Number of positive solutions satisfying the equation

1 1 12

1 1 2tan tan tan

2x 1 4x 1 x− − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ ⎝ ⎠

4 9

P Q R S (A) 4 3 2 1 (B) 2 4 3 1 (C) 4 3 1 2 (D) 2 4 1 3

59. (A) (P) → (4), (Q) → (3), (R) → (2), (S) → (1) (P) → (4) y = 4x3 − 3x

dy

dx = 12x2 − 3

2

2

d y

dx = 24x

So, 2

22

1 d y dy(x 1) x

y(x) dxdx

⎧ ⎫⎪ ⎪− +⎨ ⎬⎪ ⎪⎩ ⎭

= 2 23

1{(x 1)24x x(12x 3)}

4x 3x− + −

−

= 9 (Q) → (3) (R) → (2)


33

Slope of tangent to ellipse = − 1

1

x

2y

Given, 1

1

x

2y− = −1

x1 = 2y1 So, point P(h, 1) So, h = 2 × 1 h = 2 (S) → (1)

1

1 12x 1 4x 1tan

1 11

2x 1 4x 1

−+

+ +−

+ +

= 12

2tan

x−

⇒ 2

6x 2

8x 6x

++

= 2

2

x

⇒ 2 (3x 1)

x (8x 6)

++

= 2

2

x

⇒ 3x2 + x = 8x + 6 ⇒ 3x2 − 7x − 6 = 0

x = 27 7 4.3.( 6)

2 3

± − −×

= 7 11

6

± = 3,

2

3−

No. of +ve solution 1 (Q) → (3) 60. Let 1 2 3 4f : , f :[0, ] , f : and f : [0, )→ ∞ → → → ∞� � � � � � be defined by

1 x

| x | if x 0,f (x)

e if x 0;

<⎧⎪= ⎨≥⎪⎩

22f (x) x ;=

3

sin x if x 0,f (x)

x if x 0

<⎧= ⎨ ≥⎩

and

( )( )

2 14

2 2

f f (x) if x 0,f (x)

f f (x) 1 if x 0.

<⎧⎪= ⎨− ≥⎪⎩

List I List II (P) f4 is 1 onto but not one−one (Q) f3 is 2 neither continuous nor one−one (R) f2of1 is 3 differentiable but not one-one (S) f2 is 4 continuous and one-one


34

P Q R S (A) 3 1 4 2 (B) 1 3 4 2 (C) 3 1 2 4 (D) 1 3 2 4

60. (D) (P) → (1), (Q) → (3), (R) → (2), (S) → (4) (P) → (1)

f1(x) = x

x if x 0

e if x 0

− <⎧⎪⎨

≥⎪⎩

22f (x) x=

3

sin x if x 0f (x)

x if x 0

<⎧= ⎨ ≥⎩

2 2

24 x x 2 2x

2

f ( x) if x 0 x if x 0 x if x 0f (x)

f (e ) if x 0 (e ) 1 if x 0 e 1 if x 0

⎧ ⎧− <⎧ < <⎪ ⎪ ⎪= = =⎨ ⎨ ⎨≥ − ≥ − ≥⎪ ⎪ ⎪⎩ ⎩ ⎩

f4(x) is onto & not one-one. (Q) → (3) f3(x) → 3 (differentiable but not one-one) (R) → (2)

f2 � f1 (x) = f2 (f1(x)) = ( )22

x 2x2

f ( x) if x 0 x if x 0

f e if x 0 e if x 0

− <⎧ ⎧ <⎪ ⎪=⎨ ⎨≥ ≥⎪⎪ ⎩⎩

(S) → (4) f2(x) = x2 is domain [0, ∞) ⇒ continuous & one-one.


35

iit - jee 2014 (advanced) - vidyalankar classes · 2021. 1. 20. · iit jee 2014 advanced :...

Documents