1

PHYSICS

SECTION-A (MULTIPLE CHOICE QUESTIONS)

1. A particle is making simple harmonic motion along the X-axis. If at a distances x1 and x2 from the mean position the velocities of the particle are v1 and v2 respectively. The time period of its oscillation is given as

(a) Tx x

v v=

−

+2 2

212

12

22

π (b) Tx x

v v=

−

−2 2

212

12

22

π

(c) Tx x

v v=

+

+2 2

212

12

22

π (d) Tx x

v v=

+

−2 2

212

12

22

π

2. Consider a binary star system of star A and star B with masses mA and mB revolving in a circular orbit of radii rA and rB, respectively. If TA and TB are the time period of star A and star B, respectively, then

(a) TT

rr

A

B

A

B=

32 (b) TA > TB(if rA > rB)

(c) TA = TB (d) TA > TB (if mA > mB)

3. In an electromagnetic wave the electric field vector

and magnetic field vector are given as E E i= 0

^ and

B B k= 0

^ respectively. The direction of propagation of

electromagnetic wave is along

(a) ( )^k (b) ( )

^− j (c) ( )

^−k (d)

4. If the kinetic energy of a moving body becomes four times its initial kinetic energy, then the percentage change in its momentum will be(a) 100% (b) 200% (c) 400% (d) 300%

5. The length of a metal wire is l1, when the tension in it is T1 and is l2 when the tension is T2. The natural length of the wire is

(a) l l1 2

2+

(b) l T l T

T T1 2 2 1

2 1

−−

(c) l T l T

T T1 2 2 1

2 1

++

(d) l l1 2

j^

6. For a series LCR circuit with R = 100 W, L = 0.5 mH and C = 0.1 pF connected across 220 V – 50 Hz AC supply, the phase angle between current and supplied voltage and the nature of the circuit is(a) 0°, resonance circuit(b) 0°, resistive circuit(c) ≈ 90°, predominantly inductive circuit(d) ≈ 90°, predominantly capacitive circuit

7. A satellite is launched into a circular orbit of radius R around earth, while a second satellite is launched into a circular orbit of radius 1.02R. The percentage difference in the time period of the two satellites is(a) 1.5 (b) 3.0 (c) 0.7 (d) 2.0

8. Two small drops of mercury each of radius R coalesce to form a single large drop. The ratio of total surface energy before and after the change is

(a) 1 213: (b) 2 : 1 (c) 2 1

13 : (d) 1 : 2

9. A body at rest is moved along a horizontal straight line by a machine delivering a constant power. The distance moved by the body in time t is proportional to

(a) t32 (b) t

12 (c) t

34 (d) t

14

10. An electron having de-Broglie wavelength l is incident on a target in a X-ray tube. Cut-off wavelength of emitted X-ray is

(a) 0 (b) hcmc

(c) 2 2 2 2

2m c

hλ

(d) 2 2mch

λ

11. At an angle of 30° to the magnetic meridian, the apparent dip is 45°. Find the true dip.

(a) tan−1 13

(b) tan−1 3

(c) tan−1 32

(d) tan−1 23

12. For a certain radioactive process the graph between In R and t(sec) is obtained as shown in the figure. Then the value of half life for the unknown radioactive material is approximately

JEE MAIN 2021 ONLINE20th July 2nd Shift

2

10 20Time, t(sec)

30 40 50 60

2ln

R4

6

8R = decay rate

(a) 6.93 sec (b) 4.62 sec (c) 9.15 sec (d) 2.62 sec

13. A boy reaches the airport and finds that the escalator is not working. He walks up the stationary escalator in time t1. If he remains stationary on a moving escalator then the escalator takes him up in time t2. The time taken by him to walk up on the moving escalator will be

(a) t2 – t1 (b) t t1 2

2+

(c) t t

t t1 2

2 1− (d)

t tt t

1 2

2 1+

14. A body rolls down an inclined plane without slipping. The kinetic energy of rotation is 50% of its translational kinetic energy. The body is(a) solid sphere (b) hollow cylinder(c) ring (d) solid cylinder

15. The magnetic susceptibility of a material of a rod is 499. Permeability in vacuum is 4p × 10–7 H/m. Absolute permeability of the material of the rod is (a) p × 10–4 H/m (b) 2p × 10–4 H/m(c) 3p × 10–4 H/m (d) 4p × 10–4 H/m

16. The correct relation between the degrees of freedom f and the ratio of specific heat g is

(a) f =+2

1γ (b) f =

+1

1γ

(c) f =−2

1γ (d) f =

+γ 12

17. With what speed should a galaxy move outward, with respect to earth so that the sodium -D line at wavelength 5890 Å is observed at 5896 Å?(a) 296 km/sec (b) 322 km/sec(c) 336 km/sec (d) 306 km/sec

18. Which of the following graphs represent the behavior of an ideal gas? Symbols have their usual meaning.

(a)

PV

T

(b)

PV

T

(c)

PV

T

(d)

PV

T

19. If time (t), velocity (v), and angular momentum (l) are taken as the fundamental units. Then the dimension of mass (m) in terms of t, v, and l is(a) [t–2v–1l1] (b) [t–1v1l–2](c) [t1v2l–1] (d) [t–1v–2l1]

20. Two vectors P Qand have equal magnitudes. If the

magnitude of P Q+ is n times the magnitude of

P Q− ,then angle between

P Qand is

(a) cos− −+

1 11

nn

(b) sin− −+

1 11

nn

(c) cos− −+

12

211

nn

(d) sin− −+

12

211

nn

SECTION-B (NUMERICAL VALUE TYPE)

Attempt any 5 questions out of 10.

21. A radioactive substance decays to 116

thof its initial

activity in 80 days. The half life of the radioactive substance expressed in days is ____.

22. A series LCR circuit of R = 5 W, L = 20 mH and C = 0.5 mF is connected across and AC supply of 250 V, having variable frequency. The power dissipated at resonance condition is ______ × 102 W.

23. In the given figure switches S1 and S2 are in open condition. The resistance across ab when the switches S1 and S2 are closed is ______ W.

a bS2S1

12 W

12 W6 W 4 W

4 W 6 W

24. A certain metallic surface is illuminated by monochromatic radiation of wavelength l. The stopping potential for photoelectric current for this radiation is 3V0. If the same surface is illuminated with a radiation of wavelength 2l, the stopping potential is V0. The threshold wavelength of this surface for photoelectric effect is ______ l.

25. One mole of an ideal gas at 27°C is taken from A to B as shown in the given PV indicator diagram. The work done by the system will be _____ × 10–1 J.[Given : R = 8.3 J/mole K, ln2 = 0.6931] (Round off to nearest integer)

A(P1V1)

B(P2V2)

V(m3)

P(N

/m2 )

200

100

2 4

3

26. A body rotating with an angular speed of 600 rpm is uniformly accelerated to 1800 rpm in 10 sec. The number of rotations made in the process is _____.

27. For the forward biased diode characteristics shown in the figure, the dynamic resistance at ID = 3 mA will be ______ W.

VD(V)

I D(m

A)

1

0.1 0.2 0.30.4 0.6 0.7 0.8 0.90.5

2345678

28. A body of mass ‘m’ is launched up on a rough inclined plane making an angle of 30 ° with the horizontal. The coefficient of friction between the body and plane is

x5

if the time of ascent is half of the time of descent.

The value of x is ______.29. A zener diode having zener votage 8 V and power

dissipation rating of 0.5 W is connected across a potential divider arranged with maximum potential drop across zener diode is as shown in the diagram. The value of protective resistance RP is _____ W.

RP

pn

Vz = 8 V

+

20 V–

30. Two bodies a ring and a soild cylinder of same material are rolling down without slipping an inclined plane. The radii of the bodies are same. The ratio of velocity of the centre of mass at the bottom of the inclined plane of

the ring to that of the cylinder is x

2. Then, the value

of x is ______.

CHEMISTRY

SECTION - A (MULTIPLE CHOICE QUESTIONS)

31. The single largest industrial application of dihydrogen is (a) in the synthesis of nitric acid(b) manufacture of metal hydrides(c) rocket fuel in space research(d) in the synthesis of ammonia.

32. Spin only magnetic moment of an octahedral complex of Fe2+ in the presence of a strong field ligand in BM is(a) 0 (b) 4.89 (c) 2.82 (d) 3.46

33. A B

NH2

SO3H

(majorproduct)

(majorproduct)

273 K

NaNO2, HCl

273-278 K

CH3

CH3N

Consider the above reaction, compound B is

(a)

N

HO3S

CH3

CH3N N

(b) NHO3SCH3

CH3N N

(c) NCH3

CH3N N

(d) HO3S NCH3

CH3

34. Cu2+ salt reacts with potassium iodide to give(a) Cu2I2 (b) CuI (c) Cu(I3)2 (d) Cu2I3

35. A solution is 0.1 M in Cl– and 0.001 M in CrO42− .

Solid AgNO3 is gradually added to it. Assuming that the addition does not change in volume and Ksp(AgCl) = 1.7 × 10–10 M2 and Ksp(Ag2CrO4) = 1.9 × 10–12 M3.Select correct statment from the following.(a) AgCl will precipitate first as the amount of Ag+

needed to precipitate is low.(b) Ag2CrO4 precipitates first because the amount of

Ag+ needed is low.(c) AgCl precipitates first because its Ksp is high.(d) Ag2CrO4 precipitates first as its Ksp is low.

36. Bakelite is a cross-linked polymer of formaldehyde and (a) dacron (b) buna-S(c) novolac (d) PHBV.

37. Benzene on nitration gives nitrobenzene in presence of HNO3 and H2SO4 mixture, where(a) both H2SO4 and HNO3 act as acids(b) both H2SO4 and HNO3 act as bases(c) HNO3 acts as an acid and H2SO4 acts as a base(d) HNO3 acts as a base and H2SO4 acts as an acid.

38. PBr (major product)

(i) (C6H5CO)2O2, HBr(ii) CoF2

Major product P of above reaction is

(a) Br

F

4

(b) Br F

(c) F

F

(d) Br

F

39. Which one of the following species doesn’t have a magnetic moment of 1.73 BM, (spin only value)?(a) CuI (b) [Cu(NH3)4]Cl2(c) O2

+ (d) O2−

40. In Carius method, halogen containing organic compound is heated with fuming nitric acid in the presence of(a) CuSO4 (b) HNO3 (c) BaSO4 (d) AgNO3

41. The hybridisations of the atomic orbitals of nitrogen in NO2

− , NO2+ , and NH4

+ , respectively are(a) sp3, sp and sp2 (b) sp2, sp and sp3

(c) sp3, sp2 and sp (d) sp, sp2 and sp3.

42. Which one of the following pairs of isomers is an example of metamerism?

(a)

O O

and

(b) OH

OH

H5C6H5C6

and

(c)

O

O

C6H5H5C6 Hand

(d) CH3CH2CH2CH2CH3 and H3C C CH3

CH3

CH3

43. (A)

O

ClR (B)

O O

OR R

(C)

O

ORR

(D)

O

NRH

HThe correct order of their reactivity towards hydrolysis at room temperature is(a) (D) > (A) > (B) > (C)(b) (A) > (B) > (C) > (D)(c) (A) > (C) > (B) > (D)(d) (D) > (B) > (A) > (C)

44. Which one of the following statements is not true about enzymes? (a) Almost all enzymes are proteins.(b) Enzymes work as catalysts by lowering the

activation energy of a biochemical reaction.(c) Enzymes are non-specific for a reaction and

substrate.(d) The action of enzymes is temperature and pH

specific.

45. The major product (P) in the following reaction is

(i) KOH (alc.)(ii) H+, D (major product)

O

P

O

CHO

(a)

O

O (b)

OHC

O

(c)

O

O (d)

CHO

O

46. Metallic sodium does not react normally with(a) but-2-yne (b) tert-butyl alcohol(c) gaseous ammonia (d) ethyne.

47. Outermost electronic configuration of a group 13 element, E, is 4s2, 4p1. The electronic configuration of an element of p-block period-five placed diagonally to element, E is(a) [Xe] 5d10 6s2 6p2 (b) [Ar]3d10 4s2 4p2

(c) [Kr] 3d10 4s2 4p2 (d) [Kr] 4d10 5s2 5p2

48. NH2

Br

(major product)KOBr A

O

NH2

Br

(major product)

LiAlH4

H3O+ B

O

In the above reactions, product A and product B respectively are

(a) NH2 NH2

,

5

(b)

Br Br

NH2 NH2,

(c)

Br

NH2 NH2,

(d)

Br

NH2 NH2,

49. Which one of the following gases is reported to retard photosynthesis?(a) CFCs (b) CO (c) NO2 (d) CO2

50. Consider two chemical reactions (A) and (B) that take place during metallurgical process.(A) ZnCO3(s)

∆ → ZnO(s) + CO2(g)

(B) 2ZnS(s) + 3O2(g)∆ → 2ZnO(s) + 2SO2(g)

The correct option of names given to them respectively is(a) both (A) and (B) are producing same product so

both are calcination(b) (A) is roasting and (B) is calcination(c) (A) is calcination and (B) is roasting(d) both (A) and (B) are producing same product so

both are roasting.

SECTION - B (NUMERICAL VALUE TYPE)

Attempt any 5 questions out of 10.

51. 100 mL of 0.0018% (w/v) solution of Cl– ion was the minimum concentration of Cl– required to precipitate a negative sol in one h. The coagulating value of Cl– ion is _______. (Nearest integer)

52. The vapour pressures of A and B at 25°C are 90 mm Hg and 15 mm Hg respectively. If A and B are mixed such that the mole fraction of A in the mixture is 0.6, then the mole fraction of B in the vapour phase is x × 10–1. The value of x is ______. (Nearest integer)

53. Diamond has a three dimensional structure of C atoms formed by covalent bonds. The structure of diamond has face centred cubic lattice where 50% of the tetrahedral voids are also occupied by carbon atoms. The number of carbon atoms present per unit cell of diamond is ____.

54. 4 g equimolar mixture of NaOH and Na2CO3 contains x g of NaOH and y g of Na2CO3. The value of x is ________ g.(Nearest integer)

55. For a given chemical reaction A → B at 300 K the free energy change is –49.4 kJ mol–1 and the enthalpy of reaction is 51.4 kJ mol–1. The entropy change of the reactions is _______ J K–1 mol–1.

56. PCl5(g) → PCl3(g) + Cl2(g)In the above first order reaction the concentration of PCl5 reduces from initial concentration 50 mol L–1 to 10 mol L–1 in 120 minutes at 300 K. The rate constant for the reaction at 300 K is x × 10–2 min–1. The value of x is _______. [Given log 5 = 0.6989]

57. Potassium chlorate is prepared by electrolysis of KCl in basic solution as shown by following equation.6 3 63OH Cl ClO H O2

− − − −+ → + + eA current of x A has to be passed for 10 h to produce 10.0 g of potassium chlorate. The value of x is ____. (Nearest integer)(Molar mass of KClO3 = 122.6 g mol–1, F = 96500 C)

58. The wavelength of electrons accelerated from rest through a potential difference of 40 kV is x × 10–12 m. The value of x is ______. (Nearest integer)Given : Mass of electron = 9.1 × 10–31 kg Charge on an electron = 1.6 × 10–19 CPlanck’s constant = 6.63 × 10–34 Js

59. When 0.15 g of an organic compound was analyzed using Carius method for estimation of bromine, 0.2397 g of AgBr was obtained. The percentage of bromine in the organic compound is _______.(Nearest integer)[Atomic mass : Silver = 108, Bromine = 80]

60. An aqueous solution of NiCl2 was heated with excess sodium cyanide in presence of strong oxidizing agent to form [Ni(CN)6]2–. The total change in number of unpaired electrons on metal centre is _______.

MATHEMATICS

SECTION - A (MULTIPLE CHOICE QUESTIONS)

61. For the natural numbers m, n, if (1 – y)m (1 + y)n = 1 + a1y + a2y2 + .... + am + n ym + n and a1 = a2 = 10, then the value of (m + n) is equal to(a) 100 (b) 80 (c) 88 (d) 64

62. Let y = y(x) satisfies the equation dydx

A− =| | ,0

for all x > 0, where Ay x

x= −

sin

/

10 1 12 0 1

. If y(p) = p +

2, then the value of y(p/2) is

(a) 32

1ππ

− (b) ππ24

+

(c) π

π21

− (d) ππ24

−

63. Let r1 and r2 be the radii of the largest and smallest circles, respectively, which pass through the point (–4, 1) and having their centres on the circumference of

the circle x2 + y2 + 2x + 4y – 4 = 0. If rr

a b1

22= + , then

a + b is equal to(a) 5 (b) 7 (c) 3 (d) 11

6

64. Let f R R: −

→α6

be defined by f xxx

( ) .=+−

5 36 α

Then the value of a for which (fof)(x) = x, for all

x R∈ −

α6

, is

(a) 6 (b) 8(c) 5 (d) No such a exists.

65. If f : R → R is given by f(x) = x + 1, then the value of

lim ( ) ...( )

n nf f

nf

nf

nn→∞

+

+

+ +−

10

5 10 5 1, is

(a) 1/2 (b) 3/2 (c) 7/2 (d) 5/266. Let in a right angled triangle, the smallest angle be q. If

a triangle formed by taking the reciprocal of its sides is also a right angled triangle, then sinq is equal to

(a) 5 14+ (b) 2 1

2− (c) 5 1

2− (d)

5 14−

67. Let g t t f x dx( ) cos ( )/

/= +

−∫

π

π

π

42

2

, where

f x x xe( ) log ( )= + +2 1 , x ∈ R. Then which one of the following is correct?(a) g(1) = g(0) (b) g(1) + g(0) = 0(c) g g( ) ( )1 2 0= (d) 2 1 0g g( ) ( )=

68. If sum of the first 21 terms of the series log91/2 x + log91/3 x + log91/4 x + ...., where x > 0 is 504, then x is equal to(a) 9 (b) 243 (c) 7 (d) 81

69. The value of k ∈ R, for which the following system of linear equations3x – y + 4z = 3,x + 2y – 3z = –2,6x + 5y + kz = –3,has infinitely many solutions, is(a) 5 (b) 3 (c) –5 (d) –3

70. Let A, B and C be three events such that the probability that exactly one of A and B occurs is (1 – k), the probability that exactly one of B and C occurs is (1 – 2k), the probability that exactly one of C and A occurs is (1 – k) and the probability of all A, B and C occur simultaneously is k2, where 0 < k < 1. Then the probability that at least one of A, B and C occur is(a) exactly equal to 1/2(b) greater than 1/4 but less than 1/2(c) greater than 1/8 but less than 1/4(d) greater than 1/2

71. Let P be a variable point on the parabola y = 4x2 + 1. Then, the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line y = x is(a) (3x – y)2 + (x – 3y) + 2 = 0

(b) 2(3x – y)2 + (x – 3y) + 2 = 0(c) 2(x – 3y)2 + (3x – y) + 2 = 0(d) (3x – y)2 + 2(x – 3y) + 2 = 0

72. If the mean and variance of six observations

7, 10, 11, 15, a, b are 10 and 203

, respectively, then the

value of |a – b| is equal to(a) 7 (b) 11 (c) 1 (d) 9

73. The sum of all the local minimum values of the twice differentiable function f : R → R defined by

f x x xf

x f( )( )

( )= − −′′

+ ′′3 233 2

21 is :

(a) 0 (b) 5 (c) –22 (d) –27

74. If [x] denotes the greatest integer less than or equal to x,

then the value of the integral [[ ] sin ]/

/

x x dx−−∫π

π

2

2

is

equal to(a) –p (b) 0 (c) p (d) 1

75. Consider the following three statements :(A) If 3 + 3 = 7, then 4 + 3 = 8(B) If 5 + 3 = 8, then earth is flat.(C) If both (A) and (B) are true, then 5 + 6 = 17.Then, which of the following statements is correct?(a) (A) is true while (B) and (C) are false.(b) (A) and (C) are true while (B) is false.(c) (A) and (B) are false while (C) is true.(d) (A) is false, but (B) and (C) are true.

76. The lines x = ay – 1 = z – 2 and x = 3y – 2 = bz – 2, (ab ≠ 0) are coplanar, if(a) b = 1, a ∈ R – 0 (b) a = 2, b = 2(c) a = 1, b ∈ R – 0 (d) a = 2, b = 3

77. Consider the line L given by the equationx y z−

=−

=−3

21

12

1. Let Q be the mirror image of the

point (2, 3, –1) with respect to L. Let a plane P such that it passes through Q, and the line L is perpendicular to P. Then which of the following points is on the plane P?(a) (–1, 1, 2) (b) (1, 1, 2)(c) (1, 1, 1) (d) (1, 2, 2)

78. If the real part of the complex number (1 – cosq + 2isinq)–1 is 1/5, for q ∈ (0, p), then the value of the integral

sin x dx0

θ

∫ is equal to

(a) 1 (b) 0 (c) –1 (d) 2

79. The value of tan tan sin235

513

1 1− −

+

is equal to

(a) 15163

(b) 22021

(c) −181

69 (d)

−29176

7

80. In a triangle ABC, if | | , | | , | | ,BC CA BA

= = =3 5 7and then the projection of the vector BA BC

on is equal to

(a) 192

(b) 132

(c) 112

(d) 152

SECTION-B (NUMERICAL VALUE TYPE)

Attempt any 5 questions out of 10.81. The number of solutions of the equation log(x + 1)(2x2 + 7x + 5) + log(2x + 5)(x + 1)2 – 4 = 0,

x > 0, is _______.82. If the point on the curve y2 = 6x, nearest to the point

(3, 3/2) is (a, b), then 2(a + b) is equal to _______.

83. If limlog ( )

sin,

x

xe

xxe x x e

x x→

−− + +=

0

2

21

10α β γ

a, b, g ∈ R,

then the value of a + b + g is _______.

84. For p > 0, a vector v i p j2 2 1= + +^ ^( ) is obtained by rotating the vector v p i j1 3= +^ ^ by an angle q about origin in counter clockwise direction.

If tan( )( )

θα

=−+

3 24 3 3

, then the value of a is equal

to _______ .85. Let a curve y = y(x) be given by the solution of the

differential equation cos cos ( ) .12

11 2− −

= −e dx e dyx x

If it intersects y-axis at y = –1, and the intersection point of the curve with x-axis is (a, 0), then ea is equal to ________.

86. Let a function g : [0, 4] → R be defined as

g xt t t x

x xt x( )

max ,

,=

− + − ≤ ≤

− < ≤

≤ ≤0

3 26 9 3 0 3

4 3 4, then

the number of points in the interval (0, 4), where g(x) is NOT differentiable, is ________.

87. For k ∈ N, let 1

1 2 20 0

20

α α α α α( )( ) ... ( ),

+ + +=

+=∑ A

kk

k

where a > 0. Then the value of 100 14 15

13

2A A

A+

is

equal to ________.

88. Let an n =∞

1 be a sequence such that a1 = 1, a2 = 1

and an + 2 = 2an + 1 + an for all n ≥ 1. Then the value of

4723

1

ann

n=

∞

∑ is equal to ________.

89. Let A = aij be a 3 × 3 matrix, where

ai ji j

i jij

j i

i j

=− <

=

− >

−

+

( ) ,, ,

( ) ,

12

1

ifif

if

then det(3 Adj (2A–1)) is equal to ________.

90. Consider a triangle having vertices A(–2, 3), B(1, 9) and C(3, 8). If a line L passing through the circum-center of triangle ABC, bisects line BC, and intersects y-axis at point (0, a/2), then the value of real number a is ________.

1. (b) : From mean positions, let at = x1 velocity of the particle is v1 and at x = x2, the velocity is v2.The formula of velocity of a body executing SHM is given

by v A x= −ω 2 2

where, w is angular frequency, x is distance from mean position and A is amplitude.

Also, ω π= 2T

v A x12

12= −ω ...(i)

v A x22

22= −ω ...(ii)

From equation (i) and (ii) v1

2 – v22 = w2 (A2 – x1

2 – A2 + x22)

v v

x x T12

22

22

12

22−

−=

π

T

x x

v v=

−

−2 2

212

12

22

π

2. (c) : Mass of star A = mA, mass of star B = mBRadius of orbit for star A = rARadius of orbit for star B = rBFrom center of mas, we getmArA = mBrA ...(i)Force acting on both stars i.e., gravitational force and centripetal force are balanced.

∴ = =Gm m

rm v

rm v

rA B A A

A

B B

B2

2 2

or m vr

m vr

A A

A

B B

B

2 2= ...(ii)

Also, v rT

v rTA

A

AB

B

B= =2 2π π and

Putting vA and vB in equation (ii), we get

8

m rT

m rT

A A

A

B B

B( ) ( )2 2= ...(iii)

From equation (i) and (iii), we haveTA – TB

3. (b) : The direction of propagation of wave is

n E B E i B k E B j^ ^ ^ ^= × = × = −

0 0 0 0

So, EM wave propagates along − j^.4. (a) : Let the initial kinetic energy, Ki = KFinal kinetic energy, Kf = 4 KThe relation between kinetic energy and momentum is

K pm

=2

2

So, K K pmi = =2

2 ...(i)

Let the final momentum is p′.

42

2K p

m=

′ ...(ii)

% change in momentum =′ −

×p p

p100

=

′−

×p

p1 100

=

××

−

×

4 22

1 100K mK m

= (2 – 1) × 100 = 100%5. (b) : Let the actual length of wire is L and area of crossection is A. Let Y is the Young’s modulus. Use the formula of Young’s Modulus

YT L

A l=

⋅∆where Dl is change in length.

So, TA

Y l LL

1 1=−( ) ...(i)

TA

Y l LL

2 2=−( )

...(ii)

From equation (i) and (ii) we get

⇒ TT

l Ll L

1

2

1

2=

−−

⇒ Ll T l T

T T=

−−

1 2 2 1

2 1

6. (d) : R = 100 W, L = 0.5 mH, C = 0.1 pFVrms = 220 V, Frequency, f = 50 HzInductive reactance, XL = 2pfL = 2 × p × 50 × 0.5 × 103 = 0.157 WCapacitive reactance,

XfCC = =

×× ×

= ×12

1 102 50 0 1

3 18 1012

10π π .

. Ω

tan ( )φ phase =−

= ∞X X

RL C

So, XC > XL; f = 90°Predominantly capacitive circuit.7. (b) : The time period of a satellite is given by

T R

M= 2

4

3π

⇒ T ∝ R3/2

For first satellite T1 ∝ R3/2

For second satellite T2 ∝ (1.02R)3/2

So, ∆ ∆TT

RR

RR

= = ×−3

232

1 02 1( . )

= × =32

0 02 0 03. .

∆TT

×100% = 0.03 × 100% = 3%8. (c) : Number of drops = 2, Radius of each drop = R.Let the radius of bigger drop is R′.So, volume of 2 small drops = volume of 1 large drop

⇒ 2 43

43

3 3× = ′π πR R

⇒ 2R3 = R′3

⇒ R′ = 21/3RThe surface energy is E = T × Awhere T is surface tension and A is area.

So,EE

AA

R

R1

2

1

2

2

22 4

4= =

×

′

π

π

EE

RR

1

21 3

2

2 31 32

22

22 1=

= =/ // :

9. (a) : KE = 12

mv2 = Power × time

where m is mass of body, and v is speed.

So, v t∝

dsdt

K t=

where K is a constant.

ds K t dt∫ ∫=

S K t C= +2

3

3 2/

\ S ∝ t3/2

10. (d) : de-Broglie wavelength = l

Formula of de-Broglie wavelength is, λ = hp

where p is momentum and h is Planck’s constant.

9

p h=λ

...(i)

Kinetic energy of electron

E p

mhm

= =2 2

22 2 λ(from (i))

For l′ to be cut-off wavelength, E hc hm

=′

=λ λ

2

22∴ ′ =λ λ2 2mc

h11. (c) : Let BV and BH be the vertical and horizontal components of earth’s magnetic field.

So, tanθ =BB

V

H ....(i)

Now BH = BHcos30°; BV = BV

So, tancos

4530

° =°

BB

V

H

BB

V

H= ° ° =tan cos tan45 30 θ (from (i))

tanθ = 3

2

θ =

−tan 1 3

212. (b) : Let T1/2 is the half life of the unknown radioactive substance.By the formula of radioactivity R = R0e–lt

where, R0 is initial activity, R at time t, l is decay constant.Take log on both sides, lnR = lnR0 – ltSlope is = –l

Slope by graph, λ = =640

320

The half life is given by

T1 22 20 2

34 62/

ln ln .= = =λ

s

13. (d) : Let the length of escalator is L.

Speed of boy with respect to escalator (stationary) = Lt1

Speed of escalator = Lt2

Speed of boy with respect to the ground when escalator is moving

= +Lt

Lt1 2

So, the time taken is, t LLt

Lt

t tt t

=+

=+

1 2

1 2

1 2

14. (d) : KE of rotation = 50% of KE of translation

Rotational KE = 12

2Iω

qTranslational KE = 1

22mv

where, I is moment of inertia of body, m is mass, w is angular speed and v is linear speed.

So, 12

12

12

2 2I mvω =

I vr

mv⋅

=2

212

I mr= 12

2

This is moment of inertia of solid cylinder. 15. (b) : Magnetic susceptibility, cm = 499.m0 = 4p × 10–7 H/mLet m is the absolute permeability.m = m0(1 + cm) = 4p × 10–7 (1 + 499) = 2p × 10–4 H/m16. (c) : The ratio of two specific heat is

γ = +1 2

fwhere f is degree of freedom.

∴ − =γ 1 2f

or f =

−2

1γ

17. (d) : The speed of planet is

vc

= ∆λλ

where, c is speed of light, Dl is change in wavelength, l is original wavelength. l = 5890 Å, l′ = 5896 Å

v = × × = ×3 10 6

5890305 6 108 3. m/s

v = 306 km/s18. (d) : For ideal gas, the relation between pressure (P), volume (V), number of moles (n), absolute temperature (T) is PV = nRTSo, the graph for PV ∝ T is a straight line with positive slope.19. (d) : Mass m ∝ ta vb lc[t] = [T], [v] = [LT–1], [l]= [ML2T–1][M] = [Ta][LT–1]b [ML2T–1]c

[M] = [Mc Lb + 2c Ta – b – c]By comparing, c = 1, b + 2c = 0 ⇒ b = –2and a – b – c = 0 a + 2 – 1 = 0 a = –1\ [m] = [t–1 v–2 l1]

10

20. (c) : | | | | P Q x= =

Let the angle between P Qand is q.

Given, | | | | P Q n P Q+ = −

On squaring both the sides, we get⇒ P2 + Q2 + 2PQcosq = n2(P2 + Q2 – 2PQcosq)⇒ x2 + x2 + 2x2cosq = n2x2 + n2x2 – 2n2x2cosq 2 + 2cosq = 2n2 – 2n2cosq 2(1 + n2)cosq = 2(n2 – 1)

cos ;θ =−

+

n

n

2

21

1 θ =

−

+

−cos 12

21

1

n

n

21. (20) : Initial amount = NRemaining amount = N0/16Time, t = 80 daysLet the half life is T and n is the number of half lives.By the equation of radioactivity,

N Nn

=

0

12

NN n

n0

01612

4=

⇒ =

n tT

=1 2/

⇒ T1 2804

20/ = = days

22. (125) : R = 5 W, L = 20 mH, C = 0.5 mFVrms = 250 V

At resonance, IV

Rrmsrms= as R = Z

Power Wrms= =×

= ×V

R

22250 2 50

5125 10

23. (10) : When the switches are closed, 12 W and 6 W are in parallel.

∴ ′ =×+

= =Rp12 612 6

7218

4 Ω

12 W

12 W

4 W

4 W

6 W

6 Wa b

4 W and 4 W in parallel

′′ =

×+

=Rp4 44 4

2 Ω

6 W and 12 W in parallel

′′′=

×+

=Rp12 612 6

4 Ω

Now Req = R′p + R′′p + R′′′p = 4 + 2 + 4 = 10 W

24. (4) : Incident wavelength = lStopping potential = 3V0So, by the equation of photoelectric effect,

hc W eVλ

= + 3 0 ...(i)

where, W is work function.

Now, hc W eV2 0λ

= + ...(ii)

From equation (i) and (ii),

hc eV hc eVλ λ

− = −320 0

hc eV2

2 0λ= ...(iii)

Using equation (iii) in equation (ii),⇒ 2eV0 – eV0 = W⇒ W = eV0

So, hc eVλ0

0= (l0 = threshold wavelength)

∴ = =× ×

λλ

00

2 2hceV

hchc

(Using (iii))

l0 = 4l25. (17258) : Temperature, T = 27°C = 300 K, n = 1Work done in isothermal process

W nRTVV

= ln 2

1

W = × ×1 8 3 300 42

. ln

W = 8.3 × 300 × 0.6931 = 1725.8 J = 17258 × 10–1 J26. (200) : Initial angular velocity, w0 = 600 rpm

= ×600

602π rad/s

Final angular velocity, w = 1800 rpm

= ×180060

2π rad/sTime, t = 10 secLet the number of rotations is n. a is angular acceleration.Using first equation of rotational motion, w = w0 + at

1800 60060

210

−

× =π α

∴ =×

=απ

π1200 2

6004 2rad/s

Using third equation of rotational motion, w2 = w02 + 2aq

1800 260

600 260

2 42 2×

−

×

= × ×

π ππ θ

(60p)2 – (20p)2 = 8p × 2pn (602 – 202)p2 = 16p2 n\ n = 200

27. (25) : Dynamic resistance, r dVdid = = −

− × −0 75 0 655 1 10 3

. .( )

= 25 W

11

28. (3) : m x=5

Time of ascent = 12

time of descent

t ta d= 1

2 q = 30°

Let L is the length of the incline, aa is acceleration of ascent and ad is acceleration of descent.

So, 2 12

2La

Laa d

=

2 14

2La

Laa d

= ×

4ad = aa ...(i)Acceleration of ascent, aa = g(sinq + mcosq)Acceleration of descent, ad = g(sinq – mcosq) Using values of aa and ad in equation (i), we get⇒ 4⋅g(sinq – mcosq) = g(sinq + mcosq)

⇒ 4 12

32

12

32

−

= +

µ µ

⇒ 4 4 3 1 3− = +µ µ

3 5 3= µ ⇒ µ = 35

29. (192) : Zener voltage, VZ = 8 V Power, P = 0.5 WLet the current is I. P = I⋅VZ ⇒ 0.5 = 8I

∴ =I 116

A

Now, 20 = 8 + IRP

⇒ 12 16

1×

= RP

⇒ RP = 192 W30. (3) : MI of ring, IR = mR2, About contact point, I′R = mR2 + mR2 = 2 mR2

MI of solid cylinder, IC = 12

mR2

About contact point, ′ = + =I mR mR mRC12

32

2 2 2

where, m is mass and R is radius of both ring and cylinder.Let the height is h.For ring :

hLet velocity is vR.Using conservation of energy,

mgh IR R= ′12

2ω

mgh mRv

RR= ⋅ ×1

22 2

2

2 ⇒ gh vR= 2

∴ =v ghR ...(i)For solid cylinder :Let vC is the speed.Using conservation of energy,

mgh IC C= ′12

2ω

mgh mRv

RC= × ⋅1

232

22

2

43

2gh vC=

⇒ v ghC = 43

...(ii)

From equation (i) and (ii),

vv

ghgh

R

C=

×=

34

32

31. (d) : The largest single use of dihydrogen is in the synthesis of ammonia which is used in the manufacturing of nitric acid and nitrogenous fertilizers.32. (a) : Fe2+ = [Ar]3d6

In presence of strong field ligand, low spin complex is formed. Hence all electrons get paired with the configuration. Fe+2 = t6

2g e0g

Number of unpaired electrons = 0µ = + =n n( )2 0 BM33. (b) :

NNN

NH2

NN2

+Cl–

SO3H SO3H

SO3H

NaNO2 + HCl

(Azo compound)

(A)

(B)

0°–5°C 0° C

The first reaction is called diazotisation reaction.34. (a) : 2Cu2+ + 4KI Cu2I2(s) + I2 + 4K+

35. (a) : (i) Concentration of Ag+ required for precipitation of AgCl. Ksp(AgCl) = [Ag+][Cl–]

1.7 × 10–10 = [Ag+](0.1) [Ag+] = 1.7 × 10–9M(ii) Concentration of Ag+ required for precipitation of Ag2CrO4 Ksp(Ag2CrO4) = [Ag+]2 [CrO4

2–]

1.9 × 10–12 = [Ag+]2(0.001)

[Ag+] = 1 9 10 19 109 5. × = ×− −

12

Amount of Ag+ required to precipitate is low for Cl–.So, AgCl gets precipitated first.36. (c) : Novolac on heat ing with formaldehyde undergoes cross linking to form an infusible mass called bakelite.

s

Bakelite

37. (d) :

O

O

O O

H

N H N

H H

NSO3 H2O

OH

OO

O O

+

+

++

++

Base Acid Nitronium ion

38. (d) :

Br Br

Br

Br

F

(C6H5CO)2O2, HBr

CoF2

HBr in presence of organic peroxide gives anti – Markovnikov addition product.39. (a) : m = 1.73 BM, means number of unpaired electron = 1

Species Unpaired electron

O2– 1

O2+ 1

Cu+ 0

Cu2+ 1

Magnetic moment of Cu+ = 040. (d) : Carius method is used for determining the amount of sulphur and halogens in organic compounds by heating the compound in sealed tube with AgNO3 in conc. HNO3. The compound is decomposed and silver sulphide or halides are precipitated.

41. (b) : NO2– O O–N sp2

NO2+ O ON

+ sp

NH4+ N

H

HH H

+ sp3

42. (a) : Metamers are compounds which have different alkyl groups attached to same polyvalent functional group or atom.

O Oand are metamers.

43. (b) : Rate of hydrolysis is directly proportional to positive charge present on carbon of C O group. Rate of hydrolysis : Acid chloride > Acid anhydride > ester > amide.44. (c) : Enzymes are very specific for a reaction and substrate.45. (c) : This is an intramolecular aldol condensation

OO

OO

O–

HH Oalc. KOH

–H+

O–OH

OOH

OOH+

–H2O D

O

O46. (a) : But-2-yne does not have any acidic hydrogen and hence does not react with metallic sodium.47. (d) : 13th 14th 15th

2nd period 2s22p1 B C N3rd period 3s23p1 Al Si P4th period 4s24p1 Ga Ge As5th period 5s25p1 In Sn SbThe element is 50Sn ⇒ [Kr]4d10 5s2 5p2

48. (b) : O

NH2

Br(A)LiAlH4

Br

NH2(Hoffmann bromamide reaction)

KOBr

NH2

(B)Br

(reduction)

13

49. (c) : Higher concentrations of NO2 damage the leaves of plants and retard the rate of photosynthesis.

50. (c) : Calcination is done in absence of oxygen while roasting is done in presence of oxygen.

51. (1) : Coagulation value : The minimum concentration of electrolyte in millimoles required to cause coagulation of 1 L of colloidal solution.Given : 0.0018 gram Cl– present in 100 mL solution.

Coagulation value of Cl– =

0 001835 5

10

0 10 507 1

3..

..

×= ≈

52. (1) : p°A = 90 and p°B = 15Ptotal = 90 × 0.6 + 15 × 0.4 = 54 + 6 = 60 mm of Hg

YB = pP

B

T=

660

= 0.1 = 1 × 10–1

53. (8) : Number of carbon atoms in fcc in diamond = 4.Total number of tetrahedral voids = 4 × 2 = 8Number of tetrahedral voids occupied by carbon in diamond

= 8 × 12

= 4

Total number of carbon atoms = 4 + 4 = 8

54. (1) : Given that solution is equimolar so,

x yy x

40 10610640

= ⇒ =

Also x + y = 4

x + 10640

x = 4 ⇒ = ⇒ =×146

404

4 40146

x x

x = 1⋅095 ≈ 1

55. (336) : DG° = DH° – TDS°–49.4 = 51.4 – TDS°

–DS° = –49 4 51 4

300. .+

DS° = 0.336 kJ mol–1/K = 336 J K–1 mol–1

56. (1) : For first order reaction,

k = 1t

aa x

ln−

=

=

×2 303120

5010

2 303 0 6989120

.log

. .

= 0.0134 min–1 = 1.34 × 10–2 min–1

Value of x ≈ 1.

57. (1) : W = Z × I × t

Z of KClO3 = 122 66 96500

.×

10122 6

6 9650010 3600=

×× × ×

.x

x = 1.31 ≈ 1A

58. (6) : λ = =hp

h2meV

=×

× × × × × ×

−

− −

6 6 10

2 9 1 10 1 6 10 40 10

34

31 19 3

.

. .m

= 0.614 × 10–11 m = 6.14 × 10–12 m ⇒ x = 6

59. (68) : Organic compound AgNO3 → AgBr 0.15 g 0.2397 g

Mole of AgBr = 0 2397

188.

Weight of Br– = 0 2397

188.

× 80 = 0.102 gram

% of Bromine in organic compound

= 0 1020 15..

× 100 = 68%

60. (2) : NiCl2 ⇒ Ni2+ ⇒ 3d8 ⇒ Unpaired electrons, n = 2[Ni(CN)6]2– ⇒ Ni4+ ⇒ 3d6 ⇒ t2g

2,2,2 eg0,0

Unpaired electron = 0Difference in unpaired electrons = 2

61. (b) : Given, (1 – y)m(1 + y)n = 1 + a1y + a2y2 + .... + am+n ym+n ...(i)

and a1 = a2 = 10Differentiating (i) both sides with respect to y, we get –m(1 – y)m–1 (1 + y)n + (1 – y)m ⋅ n(y + 1)n–1 = a1 + 2a2 y + 3a3 y2 + ...+ am+n(m+n)ym+n–1

⇒ n(y + 1)n–1 (1 – y)m – m(1 – y)m–1 (1 + y)n = a1 + 2a2 y + 3a3 y2 + 4a4 y3 +.... ....(ii)Again, differentiating (ii) w.r.t. y, we getn(y + 1)n–1 m(1 – y)m–1 (–1) + (1 – y)m n(n –1)(y + 1)n–2 – [m(1 – y)m–1 n(1 + y)n–1 – (1 + y)n m(m –1)(1 – y)m–2] = 2a2 + 6a3y + 12a4y2 +.... ...(iii)Putting y = 0 in (ii) and (iii), we getn – m = 10 ...(iv)and m2 + n2 – (m + n) – 2mn = 20⇒ (n – m)2 – (m + n) = 20 ...(v)By (iv) and (v), we get⇒ (10)2 – (m + n) = 20 ⇒ m + n = 80.

62. (b) : We have, A =

y x

x

sin 10 1 1

2 01

−

\ |A| = y − −

10

x + 2(sinx + 1) = −

yx

+ 2 sinx + 2

Now, dydx

Adydx

yx

− = ⇒ +0 – 2 sinx – 2 = 0

⇒ dydx

y x+ / = 2sinx + 2, which is a linear differential

equation.

14

So, integrating factor (I.F.) = e xdx1

∫= x

Now, solution is given by, y⋅x = ∫x(2sinx + 2)dx⇒ y⋅x = x2 – 2x cosx + 2sinx + c ...(i)Applying the given condition, x = p, y = p + 2, we get(p + 2)p = p2 + 2p + c ⇒ c = 0\ Solution becomes, y⋅x = x2 – 2x cosx + 2sinx ...(ii)

Now, put x = π2

in (ii), we getπ π2 4

22

y = + ⇒ = +yπ

π24

63. (a) : Let C be the centre of the

BC

A

P(–4, 1)given circle and P be the fixed point (–4, 1), through which two circles of radii r1 and r2 and centres B and A, respectively, passes.Since, x2 + y2 + 2x + 4y – 4 = 0\ (x + 1)2 + (y + 2)2 = 32

⇒ C ≡ (–1, –2) andradius (R) = CA = 3 = BC\ r1 = |BC + CP| = 3 3 2+ = +3 3 2

Q PC = − + + + =

( ) ( )4 1 1 2 3 22 2

and r PC AC2 3 2 3 3 2 3= − = − = −

\ rr1

2

3 1 2

3 2 1

2 1

2 1

2 1

2 13 2 2=

+( )−( ) =

+

−×

+

+= +

⇒ a b+ = +2 3 2 2⇒ a = 3, b = 2 ⇒a + b = 5

64. (c) : We have, f(x) = 5 36

xx

+− α

∴ = +−

f f x f xx

( ( )) 5 36 α

⇒ x

xxxx

=

+−

+

+−

−

5 5 36

3

6 5 36

α

αα

[Q ( fof )(x) = x]

⇒ x x xx x

= + + −+ − +

25 15 18 330 18 6 2

αα α

⇒ (30 – 6a)x2 + (a2 – 25)x + 3a – 15 = 0⇒ 6(5 – a)x2 + (a – 5)(a + 5)x + 3(a – 5) = 0⇒ (a – 5) [–6x2 + (a + 5)x + 3] = 0So, required value of a is given by a – 5 = 0 ⇒ a = 565. (c) : We have, f(x) = x + 1

\ lim ( ) ...( )

n nf f

nf

nf

nn→∞

+

+

+ +

−

10

5 10 5 1

=

→∞ =

−∑lim

n r

n

nf r

n1 5

0

1= f x dx( )5

0

1

∫ = ( )5 10

1x dx+∫

= 52

2

0

1x x+

= 72

66. (c) : Let DABC be the right angled triangle with ∠C = 90° and ∠A = q(say) be the smallest angle. ⇒ ∠B = 90° – q

∠A is the smallest angle⇒ a is the smallest.

⇒ 1a

is the largest ...(i)

Now, let DA′B′C′ be the triangle formed by taking the reciprocal of sides of DABC; which is also a right angled triangle.

From (i), 1a

is largest ⇒ 1a

is the hypotenuse of DA′B′C′.

⇒ 1 1 12 2 2a b c

= + ⇒ 12

2

2

2= +ab

ac

...(ii)

Using DABC, we have

sin ; sin( ) cosθ θ θ= ° − = ⇒ =ac

bc

bc

90

\ From (ii), we have 1 = tan2q + sin2q⇒ cos2q = tan2q ⇒ cos4q = sin2q⇒ (1 – sin2q)2 – sin2q = 0⇒ (1 – sin2q)2 – (sinq)2 = 0⇒ 1 – sin2q – sinq = 0 or 1 – sin2q + sinq = 0⇒ sin2q + sinq – 1 = 0 or sin2q – sinq – 1 = 0

⇒ sinθ θ=− ± + ± +1 1 4

21 1 4

2or sin =

⇒ sinθ =− ± ±1 5

21 5

2or

⇒ sinθ =−5 1

2 [ 0 < q < 90° ⇒ 0 < sinq < 1]

67. (d) : We have, g(t) = cos ( )/

/ ππ

π

422

t f x dx+

−∫

=

−

−∫ cos cos ( ) sin sin ( )

// π ππ

π

4 422

t f x t f x dx

=

−−∫cos cos ( )

//ππ

π

40

22

t f x dx

[Q sinf(x) is an odd function]

⇒ g t t x x dxe( ) cos cos log/

/=

+ +( )

−∫

ππ

π

412

2

2

g x x dxe( ) cos log/

/1

12

122

2= + +( )

−∫ π

π ...(i)

and g(0) = cos log/

/e x x dx+ +( )

−∫

22

21

π

π ...(ii)

⇒ g(1) = 12

g(0) [Using (i) & (ii)]

⇒ 2 g(1) = g(0)

A

b

q

Ca B

c

90°– q

15

68. (d) : We have, log91/2 x + log91/3 x + log91/4

x + .... + log91/22 x = 504⇒ 2log9x + 3log9x + 4log9x + .....+ 22log9x = 504⇒ (2 + 3 + 4 + ....... + 22)log9x = 504

⇒ 22 23

21

×−

log9x = 504

⇒ 252 log9x = 504 ⇒ log9x = 2 ⇒ x = 81.69. (c) : The necessary condition for the infinitely many solutions is the determinant of the coefficient matrix is zero.

∴−

− =3 1 41 2 36 5

0k

⇒ 3(2k + 15) + k + 18 – 28 = 0 ⇒ k = –570. (d) : Given, probability that exactly one of A and B occurs is (1 – k).⇒ P(A ∩ B) + P(A ∩ B) = 1 – k ...(i)Probability that exactly one of B and C occurs is (1 – 2k).⇒ P(B ∩ C) + P(B ∩ C) = 1 – 2k ....(ii)Probability that exactly one of C and A occurs is (1 – k).⇒ P(A ∩ C) + P(A ∩ C) = 1 – k ...(iii)and probability of all A, B and C occur simultaneously is k2

⇒ P(A ∩ B ∩ C) = k2

Now, as P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A∩B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C)

\ P(A ∪ B ∪ C) = 12

[P(A) – P(A∩B) + P(B) – P(A∩B)

+ P(B) –P(B∩C) +P(C) – P(B∩C) +P(A) – P(A∩C) + P(C) – P(A∩C)] + P(A∩B∩C).

⇒ P(A∪B∪C) = 1 1 2 1

2− + − + −k k k

+ k2

=− +

+4 3

22k

k =− +

=− +

= − +2 4 3

22 1 1

21

12

2 22k k k

k( )

( )

\ P(A∪B∪C) > 12

71. (b) : Let R(h, k) be the mid point of the point P and the foot of the perpendicular drawn from P to the line y = x.Let Q(t, t) is the foot of perpendicular.

Then, k th t

−−

= −1

⇒ t = h k+

2Also, R h k

x t y t( , ) ,=

+ +

2 2

⇒ R h kx h k y h k

( , ) ,= + + + +

2 4 4 2 4 4

∴ = − = −xh k

yk h3

2 232 2

and

P(x, y)

Q(t, t)

O

R

y

x

Now, put x & y in y = 4x2 + 1, we get

32

43

21

2k h h k−=

−

+

⇒−

=−

+3

24 3

41

2k h h k( )

⇒−

= − +3

23 12k hh k( )

⇒ (3k – h) = 2(3h – k)2 + 2⇒ 2(3h – k)2 – (3k – h) + 2 = 0⇒ 2(3h – k)2 + (h – 3k) + 2 = 0\ The required locus is 2(3x – y)2 + (x – 3y) + 2 = 0.72. (c) : Mean = 10 (Given)

⇒ + + + + +7 10 11 156

a b = 10

⇒ a + b = 17 ...(i)

Also we have, variance = 7 10 11 156

2 2 2 2 2 2+ + + + +a b

–(10)2 =203

⇒ 49 + 100 + 121 + 225 + a2 + b2 – 600 = 40⇒ a2 + b2 = 145 ...(ii)Now, (a + b)2 = a2 + b2 + 2ab⇒ (17)2 = 145 + 2ab (Using (i) & (ii))⇒ ab = 72Since, (a – b)2 = (a + b)2 – 4ab\ (a – b)2 = 289 – 4 × 72 = 1⇒ |a – b| = 1

73. (d) : We have, f(x) = x3 – 3x2 – 32

f ″(2)x +f ″(1)

⇒ f ′(x) = 3x2 – 6x – 32

f ″(2)

⇒ f ″(x) = 6x – 6 ⇒ f ″(2) = 6 and f ″(1) = 0For maxima and minima, f ′(x) = 0

⇒ 3x2 – 6x – 32

f ″(2) = 0

⇒ 3x2 – 6x – 9 = 0 ⇒ x2 – 2x – 3 = 0⇒ x = 3, – 1Now, f ″(x) = 6x – 6f ″(3) = 12 (+ve)f ″(–1) = –12 (–ve)⇒ f(x) has minimum value at x = 3.

∵ f(3) = 27 – 27 – 32

× 6 × 3 + 0 ⇒ f(3) = –27

\ Local minimum value is – 27.

74. (a) : Let I x x dx= −−∫ [[ ] sin ]

/

/

π

π

2

2

⇒ I x x dx= + −( )−∫ [ ] [ sin ]

/

/

π

π

2

2 ...(i)

⇒ I x x dx= − +( )−∫ [ ] [sin ]

/

/

π

π

2

2 ...(ii)

16

Adding (i) and (ii), we get

22

2I x x x x dx= − + + + −( )

−∫ ([ ] [ ]) ([sin ] [ sin ])/

/

π

π

⇒ 2 1 12

2I dx= − + −

−∫ (( ) ( ))/

/

π

π

= − = −− −∫ 2 2

2

22

2dx xπ

ππ

π/

//

/[ ]

= − +

= −22 2

2π π

π

⇒ I = –p75. (b) : In (A), 3 + 3 = 7 is false statement and 4 + 3 = 8 is also a false statement. So, the given implication is true i.e., (A) is true.In (B), 5 + 3 = 8 is true while earth is flat is false. So, (B) is false.In (C), both (A) and (B) are true is a false statement and 5 + 6 = 17 is also false. So, (C) is true.76. (a) : Given lines are : x = ay – 1 = z – 2 and x = 3y – 2 = bz – 2.

We can write them as xa

yz

a+

= =−1 1 and

xy

z

b

+= =

23 3

.

∵ The lines are coplanar.

\

a a

b

1

3 1 3

1 0 1

0

− −

=

⇒ a(–1 –0) –1 − +

3

3b

+ a(0 + 1) = 0

⇒ –a + 3 –3b

+ a = 0 ⇒ 3 = 3b

⇒ b = 1 and a ∈ R and a ≠ 0, i.e., a ∈ R –0 [∵ ab ≠ 0]

77. (d) : Given equation of line L is x y z−=

−=

−32

11

21

Let the coordinates of Q be (x, y, z), which is the mirror image of the point M (2, 3, –1) w.r.t. L.∵ Plane P is passing through point Q and perpendicular to line L. \ 2(x – 2) + 1(y – 3) + 1(z + 1) = 0 [∵ Sum of product of corresponding Dr’s of QM and L will be 0.]⇒ Equation of plane is, 2x + y + z – 6 = 0Now, from options only point (1, 2, 2) lie on the plane.78. (a) : Let z = (1 – cosq + 2i sinq)–1, then

zi

ii

=− +

×− −− −

11 2

1 21 2( cos sin )

cos sincos sinθ θ

θ θθ θ

=− −

− +1 2

1 42 2cos sin

( cos ) sinθ θ

θ θi

=− ⋅

− +

22

42 2

1 4

2

2 2

sin sin cos

( cos ) sin

θ θ θ

θ θ

i

=−

+

sin cos

sin sin cos

θ θ

θ θ θ2

22

22 2

42

2 2

i

⇒ Re( )sin cos

z =+

=1

22

42

152 2θ θ

⇒ sin2 θ2

+ 4cos2 θ2

= 52

(Given)

⇒ 1 – cos2 θ2

+ 4cos2 θ2

= 52

⇒ 3cos2 θ2

= 32

⇒ cos2 θ2

= 12

⇒ cos θ2

12

= [ q ∈(0, p) ⇒ 0 < q/2 < p/2]

⇒ θ π2

4= / ⇒ q = p/2

∴ ∫ sin/

x dx0

2π= −[ ]cos /x 0

2π = –(0 – 1) = 1.

79. (b) : Let x = 235

513

1 1tan sin− −

+

⇒ x =−

−tan 1

65

1 925

+

−tan 1 512

⇒ x =

+− −tan tan1 115

85

12=

+

− ×

−tan 1

158

512

1 158

512

=

−tan 1 22021

∴ =

=−tan( ) tan tan .x 1 220

2122021

80. (c) : Given, in DABC, BC CA BA

= = =3 5 7, and

Projection of the vector BA BC

on =⋅

= ⋅BA BC

BCBA B

cos

Now, cos Ba c b

ac=

+ −2 2 2

2

=+ −

× ×( ) ( ) ( )3 7 5

2 3 7

2 2 2

= =3342

1114

\ Projection of BA BC

on = × =71114

112

.

81. (1) : We have,log(x+1)(2x2 + 7x + 5) + log(2x + 5)(x + 1)2 – 4 = 0⇒ log(x +1)(2x + 5)(x + 1) + log(2x + 5)(x + 1)2 = 4⇒ log(x+1)(2x + 5) + log(x +1)(x + 1) + 2log(2x + 5)(x + 1) = 4

⇒ 1 + log(x+1)(2x + 5) + 2

2 51log ( )( )x x+ + = 4

Let log(x +1)(2x + 5) = u

A

B

7 5

3 C

17

⇒ 1 + u + 2u

= 4

⇒ u2 – 3u + 2 = 0 ⇒ u = 1, 2Now, u = 1⇒ log(x+1)(2x + 5) = 1 ⇒ 2x + 5 = x + 1 ⇒ x = –4 (not possible, as x > 0).and u = 2 ⇒ 2x + 5 = (x + 1)2 ⇒ x2 = 4⇒ x = 2 or x = –2. But x = –2 is not possible.\ x = 2 is the only solution.82. (9) : We have, y2 = 6xMinimum distance is along the normal at point

P ≡ 32

32t t,

Normal at point P is tx + y = 3t + 32

t3, which passes through

the point 332

,

.

\ 3t + 32

= 3t + 32

t3

⇒ t3 = 1 ⇒ t = 1

\ P ≡

32

3,

So, a + b = 32

+ 3 = 92

⇒ 2(a + b) = 9.

83. (3) : Given, limlog ( )

sinx

xe

xxe x x ex x→

−− + +=

0

2

2

110

α β γ

⇒ lim ...x

x xx x

x x x→

+ + + +

− − +0

2 32 31

2 312

13

α β

+ + − + − +

=....) ...

sin

γx x x x

x x

2 2 3

2

1 12

13

10

⇒+ + +

− − + −

+ − +

→lim

... ... ( ..

x

x x x x x x x x

0

2 2 321

2 2 31α β γ ..)

x3 10=

⇒ lim( )

x

x x x

x→

− + + +

+ − −

0

2 3

32 2 3

α β α β γ α β γ

+ Higher powers of x = 10For existence of limit

a – b = 0, a + β2

+ g = 0

and α βγ

2 310− − = ...(i)

⇒ b = a, g = −3

2α

.

Put it in (i), we getα α α2 3

32

10− + = ⇒ 53

10α

= ⇒ a = 6

\ a = 6, b = 6 and g = –9

\ a + b + g = 6 + 6 – 9 = 3.84. (6) : For p > 0, v2 is obtained by rotating v1 by an angle q.\ | | | | v v1 2=⇒ 3p2 + 1 = 4 + (p + 1)2

⇒ 3p2 + 1 = 4 + p2 + 1 + 2p⇒ p2 – p – 2 = 0 ⇒ p = 2 (Q p ≠ –1 as p > 0)

Now, angle between v v1 2& is given by, cos| || |

θ =⋅

v vv v

1 2

1 2

=+

⋅=

+4 3 313 13

4 3 313

⇒ tan( )

θ =− +

+=

−+

13 4 3 34 3 3

112 24 34 3 3

2 2

=−+

=−+

6 3 24 3 3

3 24 3 3α

(Given)

On comparing both sides, we get a = 6.85. (2) : The given differential equation is

cos cos12

11 2− −( )

= −e dx e dyx x

⇒ dy edx

e

xx

= ( )

⋅−

− −cos cos12 1

12

...(i)

Put ex = sec q ⇒ exdx = sec q.tan q dq⇒ dx = tanq dq\ Equation (i) becomes

dyd

= ( )( )

⋅−

−cos cos costan

sec

12 1

12

θθ θ

θ

⇒ dy d=∫ ∫ cosθ

θ2

⇒ y c y c= + ⇒ =−

+22

21

2sin

cosθ θ

⇒ y e cx= − +−2 1( )Now, y(0) = – 1⇒ –1 = 0 + c ⇒ c = – 1\ Solution becomes, y e x= − −−2 1 1( ) ...(ii)

Put x = a and y = 0 in (ii), we get

0 2 1 1= − −−( )e α

⇒ − = ⇒ − =− −2 1 1 1 12

( )e eα α

⇒ = ⇒ =−e eα α12

2

86. (1) : We have,

g xt t t x

x xt x( )

max ,

,=

− + − ≤ ≤

− < ≤

≤ ≤0

3 26 9 3 0 3

4 3 4Let f(x) = x3 – 6x2 + 9x – 3

18

⇒ f ′(x) = 3x2 – 12x + 9 = 3(x – 1)(x – 3)Put f ′(x) = 0 ⇒ x = 1, 3Now, the value of f(x) at x = 1 and at x = 3 isf(1) = 1 and f(3) = –3

∴ =≤ <≤ ≤

− < ≤

g xf x x

xx x

( )( ),

,,

0 11 1 3

4 3 4

Clearly, g(x) is continuous.

Now, ′ =− − < <

< <− < <

g xx x x

xx

( )( )( ),

,,

3 1 3 0 10 1 31 3 4

⇒ g(x) is not differentiable at x = 3\ The number of points in (0, 4), where g(x) is not differentiable, is 1.87. (9) : We have,

11 2 20 0

20

α α α α α( )( )....( ),

+ + +=

+∈

=∑ A

kk Nk

k

⇒+ + +

= ++

++

+11 2 20 1 2

0 1 2α α α α α α α( )( )...( )

...A A A

++

A2020α

⇒ 1 = A0 (a + 1) (a + 2)...(a + 20) + A1 (a)(a + 2)... (a + 20) +... + A20(a) (a + 1)...(a + 19)Now, putting a = –13 in above relation, we get

A131

13 12 11 13 12 13 14 13 20=

− − − − + − + − +( )( )( )...( )( )...( )

= −⋅

113 7

Similarly,

A141

14 13 14 13 14 15 14 201

14 6=

− − − + − + − +=

⋅( )( )...( )( )...( )

and A151

15 14 15 201

15 5=

− − − += −

⋅( )( )....( )

Now, AA

14

13

13 714 6

714

12

=−

= − = − ; AA

15

13

13 715 5

15

= ⋅⋅

=

Value of 100A A

AAA

AA

14 15

13

214

13

15

13

2

100+

= +

= − +

100

12

15

2

= ×−

=100

310

92

.

88. (7) : We have, a1 = 1, a2 = 1 and an + 2 = 2an + 1 + an for all n ≥ 1 ...(i)

Now, a a

Snn

nn

nn 2 8311

= ==

∞

=

∞

∑∑ (say)

Dividing both sides of (i) by 8n, we geta a an

nnn

nn

+ += +2 1

82

8 8 ⇒ = ++

++

+648

168 8

22

11

a a ann

nn

nn

⇒ 648

168 8

22

1

11

11

a a ann

n

nn

nn

nn

++

=

∞++

=

∞

=

∞

∑ ∑∑= +

⇒ 648 8

168

1 22

1Sa a

Sa

S− −

= −

+

⇒ − −

= −

+64

18

164

1618

S S S

⇒ 64 S – 9 = 16S – 2 + S ⇒ 47S = 7

⇒ ==

∞

∑472

731

ann

n

89. (108) : We have,

aij =− <

=

− >

−

+

( ) ,,

( ) ,

12

1

j i

i j

i ji j

i j

ifif

if

∴ =−

− −−

A2 1 11 2 1

1 1 2⇒ |A| = 2(4 – 1) + 1(–2 + 1) + 1 (1 – 2) = 6 – 1 – 1 = 4Now, det (3Adj(2A–1)) = |3⋅22 Adj(A–1)|

= (12)3 |Adj(A–1)| = (12)3 |A–1|2 = = =( )| |

( )( )

12 124

1083

2

3

2A90. (9) : Given, A(–2, 3), B(1, 9) and C(3, 8).\ AB = ( ) ( )− − + − =2 1 3 9 452 2

BC = − + − =( ) ( )3 1 8 9 52 2

AC = + + − =( ) ( )3 2 8 3 502 2

Q (AC)2 = (AB)2 + (BC)2

\ ∠ABC = 90°

Clearly, mid point of BC = 2172

,

Q DABC is right angled triangle.\ Circumcentre is the mid-point of hypotenuse.

\ Circumcentre of DABC = 12

112

,

So, the equation of line L passing through 12

112

,

and

2172

,

is y x−

= −

112

212

⇒ y – 2x = 92

Now, as line passing through the point 02

, .α

∴ − =α2

092

⇒ a = 9.