test - 2 (code-c) (answers) all india aakash test series ......6. answer (4) hint : 110 60 6 a...

Test - 2 (Code-C) (Answers) All India Aakash Test Series for JEE (Main)-2020

1/16

1. (2)

2. (1)

3. (4)

4. (3)

5. (3)

6. (4)

7. (3)

8. (1)

9. (3)

10. (4)

11. (3)

12. (4)

13. (4)

14. (3)

15. (2)

16. (3)

17. (1)

18. (2)

19. (3)

20. (1)

21. (2)

22. (4)

23. (4)

24. (2)

25. (2)

26. (1)

27. (3)

28. (1)

29. (2)

30. (4)

PHYSICS CHEMISTRY MATHEMATICS

31. (1)

32. (1)

33. (3)

34. (2)

35. (4)

36. (4)

37. (3)

38. (4)

39. (3)

40. (2)

41. (1)

42. (1)

43. (1)

44. (2)

45. (4)

46. (2)

47. (1)

48. (1)

49. (2)

50. (3)

51. (3)

52. (2)

53. (4)

54. (4)

55. (4)

56. (1)

57. (2)

58. (3)

59. (3)

60. (4)

61. (3)

62. (2)

63. (2)

64. (3)

65. (1)

66. (1)

67. (3)

68. (4)

69. (2)

70. (1)

71. (4)

72. (4)

73. (2)

74. (3)

75. (1)

76. (2)

77. (2)

78. (4)

79. (2)

80. (3)

81. (4)

82. (1)

83. (3)

84. (1)

85. (4)

86. (2)

87. (1)

88. (1)

89. (2)

90. (4)

Test Date : 11/11/2018

ANSWERS

TEST - 2 - Code-C

All India Aakash Test Series for JEE (Main)-2020

All India Aakash Test Series for JEE (Main)-2020 Test - 2 (Code-C) (Hints & Solutions)

2/16

1. Answer (2)

Hint :

( )

| | | |

A B B A

A B B A

v v r r

v v r r

� ��

� ��

Solution :

( )

| | | |

A B B A

A B B A

v v r r

v v r r

� ��

� ��

2 2

ˆ ˆ ˆ ˆ ˆ3 4 ( 1) ( 2)

5 1 ( 1) ( 2)

i j i y j z k

y z

2

3 12 and

5 1 ( 1) 0z

y

2 251 ( 1)

9y

2. Answer (1)

Hint : ˆ( ) ( | | cos )R P Q P Q P � ��

Solution : We have,

ˆ( ) ( | | cos )R P Q P Q P � ��

ˆ | | cosR Q P Q � ��

3. Answer (4)

Hint : Since both collides, then we have

0

0

0

4 1005 2

5

vv

g v

⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠

Solution : Since both collides, then we have

0

0

0

4 1005 2

5

vv

g v

⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠

02 4

(5 2) (3) 120 m10

vR

⎡ ⎤ ⎢ ⎥⎣ ⎦

4. Answer (3)

Hint : 1 cos cos

24 2

tg g

⎛ ⎞ ⎜ ⎟⎝ ⎠

� �

Sol. : 1 cos cos

24 2

tg g

⎛ ⎞ ⎜ ⎟⎝ ⎠

� �

5. Answer (3)

Hint : T = 20

2m v

l

Solution : T = 20

2m v

l

PART - A (PHYSICS)

6. Answer (4)

Hint : 110 60

6a

Solution : 250

m/s6

ma

7. Answer (3)

Hint : F = mg

Solution : F = mg

8. Answer (1)

Hint : x = 5 tan

Solution : x = 5 tan

2(5 sec )dx d

vdt dt

2

52

4

5

⎛ ⎞⎜ ⎟⎝ ⎠

1252

16

125m/s

8

9. Answer (3)

Hint : (arel

) along incline plane = 0

Solution :

310

4 52 10 12 m

4510

5

PQ

⎛ ⎞ ⎜ ⎟⎝ ⎠

10. Answer (4)

Hint : = 21

2t

Solution : = 21

2t

= 21

2

at

R t =

2 R

a

v = at = 2

2R

a Raa

anet

=

2

2 2 Raa

R

⎛ ⎞ ⎜ ⎟⎝ ⎠

= 2 2 24a a

= 21 4a

Test - 2 (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020

3/16

11. Answer (3)

Hint : 2

2

2tan (1 tan )

2

gxy x

u

Solution : 2

2

2tan (1 tan )

2

gxy x

u

tan = 2, 5 m/su

y = 2x – 5x2

12. Answer (4)

Hint : dy a

dx g

Solution : dy a

dx g

2dy

kxdx

2 1

2 kg 2 5 10 50

ax

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠

13. Answer (4)

Hint : 0T mg PF F F � � �

Solution : 0T mg PF F F � � �

T = mg cos

410 8

5

T

m

14. Answer (3)

Hint :

53°T T

T

4mg

Solution :

53°T T

T

4mg

2T cos53° = 4mg

3(2 ) 4

5T mg

10

3T mg

cos37 sin37

ˆ ˆT T

a i jm m

�

10 4 10 3ˆ ˆ( )3 5 3 5

g i g j⎛ ⎞ ⎜ ⎟⎝ ⎠

80ˆ ˆ20

3i j

15. Answer (2)

Hint : 10 sin37° = 4 sin30°

Solution : 3 1

105 2

u

u = 12 m/s

16. Answer (3)

Hint : 2a BC b � ��

b BC c � ��

Solution :

2a BC b � ��

...(i)

b BC c � ��

...(ii)

From equation (i) and (ii),

2 2a b b c � � � �

1( 2 )

3b a c � � �

AB b a ��

2

3 3

ac a

�

� �

2( )

3c a � �

17. Answer (1)

Hint : Distance = speed × time

Solution :

Take x direction along the flow direction and y along

width

Let swimmer starts with speed u w.r.t

water and makes an angle with flow velocity

ˆ ˆcos sinSR

v u i u j �

0

ˆ ˆcos sinSg

v v u i u j �


4/16

sin

at

u

0cos ·

sin

ab v u

u

0

sin cos

avu

b a

f() = bsin – acos will have maximum value of

2 2a b

0

min2 2

3 2012 km/h

5

avu

a b

18. Answer (2)

Hint : AB = Range of projectile

Solution : AB = (10) × 2 = 20 m

19. Answer (3)

Hint : T2 – T

1 = m2l

Solution : T2 = (m)�2+ T

1

T1 = (2m)(2�)2

T2 = 5m�2

20. Answer (1)

Hint :

22

2 max(1)v

gR

⎛ ⎞ ⎜ ⎟⎜ ⎟

⎝ ⎠

Solution :

22

2 max(1)v

gR

⎛ ⎞ ⎜ ⎟⎜ ⎟

⎝ ⎠

2

2

max4 1

10

v⎛ ⎞ ⎜ ⎟

⎝ ⎠

1/2max

10 3 m/sv

21. Answer (2)

Hint : f1 – f

2 = 4a

max

Solution : 2

max

8 6 1m/s

4 2a

22. Answer (4)

Hint : 10 m/s

v

37°

53°

53°

53°

Solution :

10 m/s

v

37°

53°

53°

53°

3 4( ) (10)

5 5v

40m/s

3v

23. Answer (4)

Hint : 10 cos37° = v cos53°

Solution : We have,

1 2

42 10

5v v

v1 – 2v

2 = 8

24. Answer (2)

Hint :

a

ay

Solution : F = 4 + 4a

a

ay

= 4

4 (4)3

g⎛ ⎞ ⎜ ⎟⎝ ⎠

3

4

ya

a

= 12 160

3

4

3

ya

a

⎛ ⎞ ⎜ ⎟⎝ ⎠

= 172

3

⎛ ⎞⎜ ⎟⎝ ⎠

N

25. Answer (2)

Hint : u

R

Solution : u

R

2

2

2 cos sin

cos

uR

g


5/16

PART - B (CHEMISTRY)

31. Answer (1)

Hint: 1 1 2 2

1 2

P V P V

T T

Solution:

2

1 6 250V 12.5 L

0.4 300

32. Answer (1)

Hint: At constant volume mean free path remain

constant.

Solution: T

P

at constant volume

= constant

33. Answer (3)

Hint: Covalent character increases if polarising

power of cation increases.

Solution: Order of covalent character

ZnCl2 < CdCl

2 < HgCl

2

As we move Zn2+

to Hg2+

size of cation increases but

due poor shielding of inner electrons Zeff

increases so

polarizing power increases.

Zn2+

< Cd2+

< Hg2+

34. Answer (2)

Hint:

C

A

10 cm

Solution: PB = P

C = P

A + gh

3

5

10 10 9.8 0.1 760gh 74.48 torr

10

PA + gh = 760

PA

= 760 – gh

= 760 – 74.48 = 685.52

35. Answer (4)

Hint: real

ideal

VZ

V

Solution: Vreal

< Videal

means force of attraction is dominant

Z = 0.88

m

ZRTV 1.204 L

P

26. Answer (1)

Hint : f = mg sin

Solution : f = mg sin

= 3

2 105

= 12 N

27. Answer (3)

Hint : � = constant

Solution : 4a2 + a

3 + a

1 = 0

28. Answer (1)

Hint : f2 < (f

2)max

Solution : f1 = 2 N

f2 = 2 N

T = 2 N

29. Answer (2)

Hint : x x x

v u a t

Solution : 2 (10 2) cos45

Tg

t = T = 2 s

30. Answer (4)

Hint : T1 = 20 + T

2

T2 = 40 + 20

Solution : T1

= 40 + 20 + 20

= 80 N

T2

= 40 + 20

= 60 N


6/16

36. Answer (4)

Hint: TiC > ScN > MgO > NaF

Solution: Lattice energy Charge

1

Size

Na+, F

– = NaF

Mg2+

, O2–

= MgO

Sc3+

, N3–

= ScN

Ti4 + C

4 = TiC

So order of lattice energy is

NaF < MgO < ScN < TiC

37. Answer (3)

Hint: PV = nRT

PVn

RT

Solution: If V, P and T are same then number of

molecules of all gases must be same.

1 1 1 2

2 2 2 1

n P V T

n P V T

P1 = P

2, V

1 = V

2, T

1 = T

2

38. Answer (4)

Hint: If temperature remain same, then R.M.S.

speed does not change.

Solution: Due to mixing,

T = constant so individual pressure is

P1V

1 = P

2V

2

V2 = 12.5 L

He f

2 2.5P 0.4 atm

12.5

∵ T = constant.

K.E. remains same and rms speed also

remains same.

Due to mixing pressure of both gases will decrease

due to increase in volume.

39. Answer (3)

Hint: Bent's rule

Solution:

H

C

Cl

ClCl

x

H

C

H

ClH

y

I II

s-character of hybrid orbital involved in bonding with

H-atom is more in CHCl3 as compared to that in

CH3Cl.

C–H bond length in CHCl3 is less as compared

to C–H bond length in CH3Cl.

x < y

40. Answer (2)

Hint: Strength of -bond decreases as internuclear

distance increases.

Solution: Strength of -bond depend on internuclear

distance.

2p – 2p > 2p – 3d > 2p – 3p > 3p – 3p

Due to inclined nature of d-orbitals 2p – 3d overlap

is more so 2p – 3d is more stronger than 2p – 3p.

41. Answer (1)

Hint:3 3

C(CH ) is pyramidal

3C(CN) is planar

Solution: 3

C(CN) is planar due to back bonding.

42. Answer (1)

Hint: Hybridisation of 'B' in BF3 and BF

4

– is sp

2 and

sp3 respectively

Solution: In BF3, due to back bonding bond length

of B – F bond reduces.

43. Answer (1)

Hints: 2 >

1

4 >

3

Solution:

3

4

O

FF

O

HH

Due to high electronegativity of F bond angle

reduces

O

SiH3

H3Si

Due to back bonding bond angle increases

44. Answer (2)

Hint: Orbital is bonding.

Solution:

+

–

+

+

– +

– +

–+

–

The orbital is bonding.


7/16

45. Answer (4)

Hint: Al2Br

6 has 3c – 4e

– bond

Solution: (AlH3)n 3c – 2e

– bond + 2c – 2e

–

B2H

6 3c – 2e

– bond + 2c – 2e

–

Al2(CH

3)6 3c – 2e

– bond + 2c – 2e

–

Al2Br

6 3c – 4e

– bond + 2c – 2e

–

Al

Br

Br

Br

Br

Al

Br

Br

The bridge bond of Al2Br

6 are 3c – 4e

– bond.

46. Answer (2)

Hints: Bent's rule

Solution: H—O—O—H F—O—O—F

Due to high electronegativity of F, p-character in OF

bond increases so s-character increases in O—O

bond.

As percentage of p-character increases, bond length

increases. If percentage of s-character increases,

bond length decreases.

y y

H—O—O—H F—O—O—F

y > x

47. Answer (1)

Hint: Bond order depends on resonance also.

Solution:

NO2

–1.5

NO3

–1.33

O3

1.5

CO3

2–1.33

O2

+2.5

N2

+2.5

48. Answer (1)

Hint: Law of diffusion.

Solution:

I II

4

4 4

CHHe He

CH CH He

Mr P 2 16 1= = =

r P M 4 4 1

1

1 Becomes the mole ratio in container (II)

II III

4

He

CH

r 1 16 2=

r 1 4 1

III IV

4

He

CH

r 2 16 4=

r 1 4 1

IV Container

4

He

CH

r 4 16 8=

r 1 4 1

49. Answer (2)

Hint:

A

F

F

F

F

Solution: 2 lone pair and 4 bond pair so A belongs

to 18th

group.

From the shape we can say that A has 8 valence

electron.

So AF3

+ has T-shape with axial and equatorial bond

but AF5

+ has octahedral geometry with one lone pair

so all bonds are identical.

50. Answer (3)

Hint: sp3 and sp

3

Solution:

O O

P P

OH HOH OHH H,

(H PO )3 3

(H PO )3 2

51. Answer (3)

Hint: I2Cl

6 has 8 atom in the plane.

Solution: I2Cl

6 is a planar molecule.

I

Cl

Cl

Cl

Cl

I

Cl

Cl

52. Answer (2)

Hint: HF > HCl > HBr > HI

Solution: As we move from HF to HI bond length

increases but polarity decrease so order of dipole

moment is

HF > HCl > HBr > HI


8/16

53. Answer (4)

Hint:C

1b V

3

a = PC 27b

2

Solution: b = 4Vm

V volume of 1 mol of molecules

C

8aT

27Rb

VC = 3b

3 1

C

1b V 49.3 cm mol

3

= 0.0493 dm3 mol

–1

3

A

4b 4 r N

3

⎛ ⎞ ⎜ ⎟⎝ ⎠

r = 1.69 × 10–8

cm = 1.69 Å

a = 27Pcb

2

= 27 (48.20) (0.0493)

= 3.16 dm6 atm mol

–2

C

8(3.16)T 231K

27(0.082) 0.0493

54. Answer (4)

Hint: H-bonding

Solution: M.P. : H2O > NH

3 > HF

B.P. : H2O > HF > NH

3

55. Answer (4)

Hint: ClO4

– has the greatest number of resonating

structures.

Solution: As resonance increases, stability

increases.

56. Answer (1)

Hint: Equation is virial equation for the van der

Waal's gas. So,

B

aT

Rb

Solution: dZ

0 assume P 0dP

2 3

m m m

a 1 b cZ 1 b

RT V V V

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

as P 0

PVm

= RT

m

P 1

RT V

2

2

a P bPZ 1 b

RT RT (RT)

⎛ ⎞ ⎜ ⎟⎝ ⎠

�

B

dZ a 1b 0 (at T T )

dP RT RT

⎛ ⎞ ⎜ ⎟⎝ ⎠

B

aT

Rb

57. Answer (2)

Hint:C C

8aT V 3b

27Rb

C 2

aP

27b

Solution:

2

C

C

T 8a 27b 8b

P 27Rb a R

Higher the value of C

C

T

P

⎛ ⎞⎜ ⎟⎝ ⎠

, higher will be the value of b,

so higher will be size of molecule and VC.

C

C

T 8b

P R

For NO = 177 8b

2.7364.85 R

For CCl4 =

556 8b12.06

45.6 R

So, b of CCl4 is more than NO.

58. Answer (3)

Hint: x = 1

3RT

M

28RT

yM

32RT

zM

Solution: For same value of x = y = z

31 22RT3RT 8RT

M M M

2

1 3

8T3T 2T

3T

1 = 2.547T

2 = 2T

3

So, T1 < T

2 < T

3


9/16

59. Answer (3)

Hint: Pressure mole of gas

Solution: Let mol of C4H

8 = x

mol of O2 = y

4 1x y 0.1218

0.0821 400

C4H

8 + 6O

2 4CO

2(g) + 4H

2O(g)

x y – –

1x y

6 4

y6

4y

6

Total mol = 4 4 1

x y y y6 6 6

Final total moles = x + y + y

6

= 4.5 1

0.0821 400

1

3

x

y

60. Answer (4)

Hint: mol of H2 = x

mol of O2 = y

mol of He = z

x + y + z = 5

Due to combustion H2 and O

2 consumed so moles

decreased at constant volume and temperature

decreases the pressure.

Solution: x + y + z = 5

2 2 2

1H O H O

2

x y 0

x 2y 0

x – 2y + z

3y = 3, y = 1

Add amount of O2: 3 mol

2 2 2

1H O H O

2

x 2y 3 mol

x 2yx 2y 1.5

2

2x – 4y + x – 2y = 3

3x – 6y = 3

3x – 6 = 3

3x = 9

x = 3

z = 1

PART - C (MATHEMATICS)

61. Answer (3)

Hint: Use cot(A + B) = cot3

4

Solution: Given A + B = 3

4

cot(A + B) = cot3

4

cot cot 11

cot cot

A B

B A

⇒

cot A cot B – 1 + cot B + cot A = 0

cot B(cot A + 1) + (cot A + 1) = 2

(cot A + 1)(cot B + 1) = 2 = n (given)

2 3sin sin

1 3 2

n

n

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

62. Answer (2)

Hint: Use sum and product of roots.

Solution: 2p = 1 + q and + = –p, = q

= 2p – 1 = –2( + ) – 1

2 + 2 + = –1 ( + 2)( + 2) = 3

Either + 2 = ±1, + 2 = ±3

or + 2 = ±3, + 2 = ±1

So, the polynomials are x2 – 1 = 0, x2 + 8x + 15 = 0.

63. Answer (2)

Hint: Solve the expression on L.H.S. of equation and

then find + and .

Solution: From the given equation,

3x2 + 12x – 1 = 0

+ = –4, = –1

3


10/16

1 1 1

3( 2)( 2) 3( 3)( 3) 3( 5)( 5)

1 1

3 6( ) 12 3 9( ) 27

1

3 15( ) 75

1 1 1

1 24 12 1 36 27 1 60 75

1 1 1 48

13 10 14 455

64. Answer (3)

Hint: Use condition for both common roots.

Solution: 2x2 – 5x + 8 = 0 has both roots non-real,

since discriminant = (–5)2 – 4 × 2 × 8 < 0

Both roots of given two equations are common.

, say2 5 8

a b c

a = 2, b = –5, c = 8

(2 8)1

2 10

a c

b

65. Answer (1)

Hint: Use sin =

2

2 2

2tan /2 1 tan /2, cos

1 tan /2 1 tan /2

Solution: Given equation is sin + cos = 2

3

2

2 2

2tan 1 tan22 2

31 tan 1 tan

2 2

25 tan 6 tan 1 0

2 2

6 36 4 5 ( 1)tan , 0

2 10

0 tan 02 2 2

⇒

6 56 3 14tan

2 10 5

66. Answer (1)

Hint: Use, cosx – cosy = 2 sin 2

x y· sin

2

y x

and sinx + siny = 2 sin 2

x y· cos

2

x y = 0

Solution: cosx – cosy = 2 sin sin2 2

x y y x = 0

sin x + sin y = 2 sin cos2 2

x y x y = 0

Here, sin2

y x and cos

2

x y are not simultaneously

zero.

Hence, sin2

x y = 0 x + y = 2n, n I

sin2020x + sin2020y

= 2 sin1010(x + y) · cos1010(x – y) = 0

67. Answer (3)

Hint:

f(x, ) = cos2x + cos(x + )[cos(x + ) – 2 cos cosx]

= cos2x – cos(x + ) · cos( – x)

Solution: f(x, ) = cos2x + cos2(x + ) –

2 cos · cosx · cos( + x)

= cos2x – cos( + x) · cos( – x)

= cos2x – (cos2x – sin2)

= sin2

2

23 1 1, sin

4 4 4 22f

⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

68. Answer (4)

Hint:

Put n = 1, 2 and then check it for n = 3

Solution:

Given, pn + qn = rn + sn ...(i)

Put n = 1, p + q = r + s ...(ii)

Put n = 2, p2 + q2 = r2 + s2 ...(iii)

From (ii),

(p + q)2 = (r + s)2

p2 + q2 + 2pq = r2 + s2 + 2rs

pq = rs ...(iv) [From (iii)]


11/16

Also from (ii),

(p + q)3 = (r + s)3 ...(iv)

p3 + q3 + 3pq(p + q) = r3 + s3 + 3rs(r + s)

p3 + q3 = r3 + s3 [From (ii) and (iv)]

(i) is true for n = 3 also.

Again from (iii),

p2 – s2 = r2 – q2

(p – s)(p + s) = (r – q)(r + q)

[∵ From (ii), p – s = r – q]

p + s = r + q

p – q = r – s ...(v)

Multiply (ii) and (v),

(p + q)(p – q) = (r + s)(r – s)

p2 – q2 = r2 – s2

69. Answer (2)

Hint: z is purely imaginary, then use 0z z and

find value of sin.

Solution: Given z = 3 2 sin

1 2 sin

i

i

∵ z is purely imaginary

0z z

3 2 sin 3 2 sin0

1 2 sin 1 2 sin

i i

i i

6 – 8 sin2 = 0

sin2 = 3

4, cos2 =

3 11

4 4 , tan2 = 3

cosec2 + tan2 = 4

3 + 3 =

13

3

70. Answer (1)

Hint: Find critical points and they are 1, 2.

Now solve the equation for x < 1, 1 x < 2 and x 2

Solution: Here, critical points are given by x – 1 = 0

and x – 2 = 0 x = 1 and x = 2

+ve

Put = 3x

+ve +ve–ve

–ve–ve– +1 2

Sign for – 2x

Sign for – 1x

Case I : When – < x < 1

x – 1 < 0 and x – 2 < 0

From given inequation,

1 – x + 2 – x > 5

3 – 5 > 2x

x < –1

Common values of x are given by – < x < –1.

Case II : When 1 x < 2

x – 1 0 and x – 2 < 0


x – 1 + 2 – x > 5 1 > 5 (absurd)

Case III : When 2 x <

Then from given inequation,

x – 1 + x – 2 > 5

2x > 8 x > 4

Common values are given by 4 < x <

Required solution set is (–, –1] [4, )

71. Answer (4)

Hint: p N, cot x = ([cot2x] – p) I

p = cot x (cot x – 1)

= product of two consecutive integers.

Solution: Here, [cot2x] = I (an integer) and p N

cot x is also an integer.

From given equation,

cot2x – cot x – p = 0 p = cot x(cot x – 1)

Here, p can be the product of two consecutive

integers.

p = 2 × 1, 3 × 2, 4 × 3, 5 × 4, 6 × 5, 7 × 6,

8 × 7, 9 × 8, 10 × 9

Total number of elements in set S = 9.

72. Answer (4)

Hint: From given inequation,

(x2 + 1) < 3x2 – 7x + 8 < 2(x2 + 1)

Solving both inequations x [1, 6]

Solution: Here, x2 + 1 < 3x2 – 7x + 8 < 2(x2 + 1)

2x2 – 7x + 7 > 0 x R

and x2 – 7x + 6 < 0 x [1, 6]

Required solution set is

x [1, 6] R x [1, 6]


12/16

73. Answer (2)

Hint: Find cos2x = a + 3 or –1, by solving equation

and then 0 < a + 3 < 1.

Solution: From given equation,

cos2x =

2( 2) 4 4 4 12

2

a a a a

( 2) ( 4)

2

a a

cos2x = a + 3 or –1

But cos2x 0 and 0 < cos2x < 1

0 < a + 3 < 1 a [–3, –2]

74. Answer (3)

Hint: Use transformation formula.

Solution:

cot70° + 4 cos70°cos70 4cos70 sin70

sin70

cos70 2sin140

sin(90 20 )

sin20 2sin40

cos20

2cos10 sin30 sin40

cos20

cos10 cos50

cos20

2 cos30 cos20

cos20

3 cos203

cos20

75. Answer (1)

Hint: From given equation,

sinx = 1 x = (4n + 1)2


sinx = 1 x = (4n + 1)2

cos cos odd multiple of 02

⎛ ⎞ ⎜ ⎟⎝ ⎠

x

tan2x = tan 2 (4 1) 02

n⎛ ⎞ ⎜ ⎟

⎝ ⎠

cot3x = cot 3 (4 1)2

n⎛ ⎞ ⎜ ⎟

⎝ ⎠

cot odd multiple of 02

⎛ ⎞ ⎜ ⎟⎝ ⎠

sin4x = sin 4 (4 1)2

n⎡ ⎤ ⎢ ⎥

⎣ ⎦ = 0

Required sum = 0 + 0 + 0 + 0 = 0

76. Answer (2)

Hint: Draw graph of sinx and cosx.

Solution:

y

x

2–1

2

2–

4

2

32

2

∵ 1 22 2

sin1 < sin2

1 2 34

cos1 > cos2 > cos3

∵ 1 sin1 cos14 2

⇒

77. Answer (2)

Hint: x = cos sin 1

sin cos sin cos

and y =

21 sin

coscos cos

Solution: Here,

21 sin

andsin cos cos

x y

3

2 2

3 3

1 sinand

cos cos

x y xy

2 2 2

2 23 3

2 2

1 sin( ) ( ) 1

cos cos

x y xy

78. Answer (4)

Hint: Here, xy and x + y are the roots of the equation.

a2 – 23a + 120 = 0 a = 15, 8


13/16

Solution: ∵ xy(x + y) = 120 and xy + (x + y) = 23.

The equation with roots xy and x + y is

a2 – 23a + 120 = 0.

(a – 15)(a – 8) = 0

xy = 15, x + y = 8

x2 + y2 = (x + y)2 – 2xy = 34

2 2

342

2 15 2

x y

xy

79. Answer (2)

Hint: Put 5x2 – 6x + 8 = t and 5x2 – 6x – 7 = t – 15,

then solve for t and then for x.

Solution: Let 5x2 – 6x + 8 = t


15 1t t

t – 15 = 1 + t – 2 t

t = 8 t = 64

5x2 – 6x + 8 = 64

5x2 – 6x – 56 = 0

5x2 – 20x + 14x – 56 = 0

x = 4, –14

5, but x > 0

x = 4, here x = 4 also satisfies the given equation.

80. Answer (3)


(3x – 9)(3x – 27) = 0 x = 2, 3

Solution: Given equation is 9x – 36 · 3x + 243 = 0

(3x)2 – 36 · 3x + 243 = 0

(3x – 9)(3x – 27) = 0

3x = 32 or 33

x = 2, 3

Sum of roots = 2 + 3 = 5

81. Answer (4)

Hint: 2 2

i

iii e e

Solution:

2

2 2(0 1) cos sin2 2

ii

i ii ii i e e

⎛ ⎞ ⎜ ⎟⎝ ⎠

2e

since i2 = –1

2

2

1ii e

e

2 2 2

2

1

i

iii e e e

e

⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠

82. Answer (1)

Hint:

amp.8 8

sin cos amp(1 ) amp(3 3 )7 7

i i i ⎛ ⎞ ⎜ ⎟

⎝ ⎠

Solution:

amp (z)8 8

amp sin cos amp (1 )7 7

i i ⎛ ⎞ ⎜ ⎟

⎝ ⎠

amp(3 3 )i

amp sin cos amp. (1 )7 7

i i ⎛ ⎞ ⎜ ⎟

⎝ ⎠

amp (3 3 )i

1 1 1 1tan cot ( tan 1 ) tan

7 3

⎛ ⎞ ⎜ ⎟⎝ ⎠

5

14 4 6

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

9

14 4 6

892

84

(For principal value)

79

84

83. Answer (3)

Hint: Critical point is x = –6

5

Now, solve equation for x < –6

5 and for x > –

6

5

Solution: Given inequation is x2 – |5x + 6| > 0

Case I : x < –6

5 x2 + 5x + 6 > 0.

x (– , –3) (–2, )

x (– , –3) 6

2,5

⎛ ⎞ ⎜ ⎟⎝ ⎠

...(i)


14/16

Case II : x > – 6

5 x2 – 5x – 6 > 0.

(x + 1)(x – 6) > 0

x (– , –1) (6, )

x 6, 1

5

⎡ ⎞ ⎟⎢⎣ ⎠

(6, ) ...(ii)

From (i) and (ii),

x (– , –3) (–2, –1) (6, )

[Taking union of (i) and (ii)]

84. Answer (1)

Hint: Put z = a + ib, we get

2a = 22 2( 1)a b 2a = 1 + b2

and arg(z1 – z

2) =

4

|b

1 – b

2| = |a

1 – a

2|

Now, put the values z1 and z

2 in given equation find

arg(z1z

2).

Solution: Here, 2 1z z z

a + ib + a – ib = 2|a + ib – 1|,

where z = a + ib (say)

2a = 22 2( 1)a b

a2 = a2 – 2a + 1 + b2 2a = 1 + b2 ...(i)

Again given, arg(z1 – z

2) =

4

arg[(a1 + ib

1) – (a

2 + ib

2)] =

4

arg[(a1 – a

2) + i(b

1 – b

2)] =

4

tan–11 2 1 2

1 2 1 2

14

b b b b

a a a a

⇒

|b1 – b

2| = |a

1 – a

2| ...(ii)

But z1 = a

1 + ib

1 and z

2 = a

2 + ib

2 satisfy the

equation 2 1z z z

2a1 = 1 + b

1

2 ...(ii)

and 2a2 = 1 + b

2

2 ...(iii)

By [(ii) – (iii)],

2(a1 – a

2) = b

1

2 – b2

2 = (b1 + b

2)(b

1 – b

2)

2|a1 – a

2| = |b

1 – b

2| |b

1 + b

2|

|b1 + b

2| = 2

|Im(z1 + z

2)| = 2

85. Answer (4)

Hint:

Coefficient of x2 = –2 < 0, greatest value = –4

D

a

and least value = 3

2f⎛ ⎞⎜ ⎟⎝ ⎠

in 3, 2

2

⎛ ⎞⎜ ⎟⎝ ⎠

Solution: Coefficient of x2 = –2 and (Discriminant)

= 9 – 4(–2) × 2 = 25

The greatest value = 25 25

4 4( 2) 8

D

a

and least value = 3

2f⎛ ⎞⎜ ⎟⎝ ⎠

= –7

Required sum = 25 31

78 8

O–1

1 2

3

4

3 25,

4 8

⎛ ⎞⎜ ⎟⎝ ⎠

3

2–

X X

Y

86. Answer (2)

Hint: D > 0, 2,2

b

a

4f(2) > 0

Solution: Conditions are D > 0

(–20)2 – 4.4.(252 + 15 – 66) > 0

–16 × 15 + 16 × 66 > 0

< 22

5...(i)

2

b

a

< 2

( 20 )

2 4

< 2

20 < 16

< 4

5...(ii)

4f(2) > 0 f(2) > 0

16 – 40 + 252 + 15 – 66 > 0

252 – 25 – 50 > 0

2 – – 2 > 0


15/16

2 – 2 + – 2 > 0

( + 1)( – 2) > 0

< –1 or > 2

(– , –1) (2, ) ...(iii)

From (i), (ii) and (iii), (– , –1)

87. Answer (1)

Hint: Exp. ax2 + 2hxy + by2 + 2gx + 2fy + c can be

expressed as the product of two linear factors if

0

a h g

h b f

g f c

Solution: The expression ax2 + 2hxy + by2 + 2gx +

2fy + c can be expressed as the product of two linear

factors if

= abc + 2fgh – af2 – bg2 – ch2 = 0

2 × (–6) × (–3) + 2 11 1 121

2 ( 6)2 2 2 4

⎛ ⎞ ⎜ ⎟⎝ ⎠

2 1( 3) 0

4 4

195,

6

= 5 (+ve integral value)

88. Answer (1)

Hint:

3z

z

= 2 2 = 3

z

z

> 3

z

z

> 3

r

r

,

where r = |z| –2 < r – 3

r

< 2

Solution: Here, 2 = 3

z

z

> 3

z

z

3

r

r

< 2, where |z| = r

–2 < r – 3

r

< 2

When, r – 3

r

+ 2 > 0 r2 + 2r – 3 > 0

r2 + 3r – r – 3 > 0

(r + 3)(r – 1) > 0 r < –3 or r > 1

|z| –3 or |z| 1

and when r – 3

r

– 2 < 0 r2 – 2r – 3 < 0

r2 – 3r + r – 3 < 0

(r + 1)(r – 3) < 0

–1 < r < 3

1 < |z| < 3

|z|max

= 3, |z|min

= 1

89. Answer (2)

Hint:

tr

= secr° · sec(r + 1)°

= sin1

sin1 cos cos( 1)r r

sin[( 1) ]

sin1 cos cos( 1)

r r

r r

1[tan( 1) tan1 ]

sin1r

Solution:

Here, tr = secr° · sec(r + 1)°

sin1

sin1 cos cos( 1)r r

1 sin( 1 )

sin1 cos cos( 1)

r r

r r

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

1 sin( 1) cos cos( 1) sin

sin1 cos cos( 1) cos cos( 1)

r r r r

r r r r

⎡ ⎤ ⎢ ⎥ ⎣ ⎦

1(tan( 1) tan )

sin1r r

88

0

sec sec( 1)

r

r r

∑

(tan1 tan0 ) (tan2 tan1 )1

sin1 (tan3 tan2 ) ... tan89 tan88

⎡ ⎤ ⎢ ⎥ ⎣ ⎦

cosec1 tan(90 1 ) cosec1 cot1

90. Answer (4)

Hint: Given exp. = tan ·tan tan3 3

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

cot cot cot3 3

⎡ ⎤ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦


16/16

� � �

Solution: Given exp.

= tan·tan tan3 3

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

cot cot cot3 3

⎡ ⎤ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

tan tan tan3 3

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

3cot cot cot 33 3

⎡ ⎤ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

...(1)

where cot cot cot3 3

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

cos sin sin cos3 3 3 3

sin sin3 3

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

cos

sin

sincos3 3

sinsin sin

3 3

⎛ ⎞ ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

2 2 32sin sin

4cos

sin sin3 3

⎛ ⎞ ⎜ ⎟

⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠

2 23cos cos sin

3

sin sin sin3 3

⎛ ⎞ ⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

3cos cos cos3 3

sin sin sin3 3

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

3cot cot cot3 3

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Test - 2 (Code-D) (Answers) All India Aakash Test Series for JEE (Main)-2020

1/16

1. (4)

2. (2)

3. (1)

4. (3)

5. (1)

6. (2)

7. (2)

8. (4)

9. (4)

10. (2)

11. (1)

12. (3)

13. (2)

14. (1)

15. (3)

16. (2)

17. (3)

18. (4)

19. (4)

20. (3)

21. (4)

22. (3)

23. (1)

24. (3)

25. (4)

26. (3)

27. (3)

28. (4)

29. (1)

30. (2)

PHYSICS CHEMISTRY MATHEMATICS

31. (4)

32. (3)

33. (3)

34. (2)

35. (1)

36. (4)

37. (4)

38. (4)

39. (2)

40. (3)

41. (3)

42. (2)

43. (1)

44. (1)

45. (2)

46. (4)

47. (2)

48. (1)

49. (1)

50. (1)

51. (2)

52. (3)

53. (4)

54. (3)

55. (4)

56. (4)

57. (2)

58. (3)

59. (1)

60. (1)

61. (4)

62. (2)

63. (1)

64. (1)

65. (2)

66. (4)

67. (1)

68. (3)

69. (1)

70. (4)

71. (3)

72. (2)

73. (4)

74. (2)

75. (2)

76. (1)

77. (3)

78. (2)

79. (4)

80. (4)

81. (1)

82. (2)

83. (4)

84. (3)

85. (1)

86. (1)

87. (3)

88. (2)

89. (2)

90. (3)

Test Date : 11/11/2018

ANSWERS

TEST - 2 - Code-D

All India Aakash Test Series for JEE (Main)-2020

All India Aakash Test Series for JEE (Main)-2020 Test - 2 (Code-D) (Hints & Solutions)

2/16

1. Answer (4)

Hint : T1 = 20 + T

2

T2 = 40 + 20

Solution : T1

= 40 + 20 + 20

= 80 N

T2

= 40 + 20

= 60 N

2. Answer (2)

Hint : x x x

v u a t

Solution : 2 (10 2) cos45

Tg

t = T = 2 s

3. Answer (1)

Hint : f2 < (f

2)max

Solution : f1 = 2 N

f2 = 2 N

T = 2 N

4. Answer (3)

Hint : � = constant

Solution : 4a2 + a

3 + a

1 = 0

5. Answer (1)

Hint : f = mg sin

Solution : f = mg sin

= 3

2 105

= 12 N

6. Answer (2)

Hint : u

R

Solution : u

R

2

2

2 cos sin

cos

uR

g

PART - A (PHYSICS)

7. Answer (2)

Hint :

a

ay

Solution : F = 4 + 4a

a

ay

= 4

4 (4)3

g⎛ ⎞ ⎜ ⎟⎝ ⎠

3

4

ya

a

= 12 160

3

4

3

ya

a

⎛ ⎞ ⎜ ⎟⎝ ⎠

= 172

3

⎛ ⎞⎜ ⎟⎝ ⎠

N

8. Answer (4)

Hint : 10 cos37° = v cos53°

Solution : We have,

1 2

42 10

5v v

v1 – 2v

2 = 8

9. Answer (4)

Hint : 10 m/s

v

37°

53°

53°

53°

Solution :

10 m/s

v

37°

53°

53°

53°

3 4( ) (10)

5 5v

40m/s

3v

10. Answer (2)

Hint : f1 – f

2 = 4a

max

Solution : 2

max

8 6 1m/s

4 2a

Test - 2 (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020

3/16

11. Answer (1)

Hint :

22

2 max(1)v

gR

⎛ ⎞ ⎜ ⎟⎜ ⎟

⎝ ⎠

Solution :

22

2 max(1)v

gR

⎛ ⎞ ⎜ ⎟⎜ ⎟

⎝ ⎠

2

2

max4 1

10

v⎛ ⎞ ⎜ ⎟

⎝ ⎠

1/2max

10 3 m/sv

12. Answer (3)

Hint : T2 – T

1 = m2l

Solution : T2 = (m)�2+ T

1

T1 = (2m)(2�)2

T2 = 5m�2

13. Answer (2)

Hint : AB = Range of projectile

Solution : AB = (10) × 2 = 20 m

14. Answer (1)

Hint : Distance = speed × time

Solution :

Take x direction along the flow direction and y along

width

Let swimmer starts with speed u w.r.t

water and makes an angle with flow velocity

ˆ ˆcos sinSR

v u i u j �

0

ˆ ˆcos sinSg

v v u i u j �

sin

at

u

0cos ·

sin

ab v u

u

0

sin cos

avu

b a

f() = bsin – acos will have maximum value of

2 2a b

0

min2 2

3 2012 km/h

5

avu

a b

15. Answer (3)

Hint : 2a BC b � ��

b BC c � ��

Solution :

2a BC b � ��

...(i)

b BC c � ��

...(ii)

From equation (i) and (ii),

2 2a b b c � � � �

1( 2 )

3b a c � � �

AB b a ��

2

3 3

ac a

�

� �

2( )

3c a � �

16. Answer (2)

Hint : 10 sin37° = 4 sin30°

Solution : 3 1

105 2

u

u = 12 m/s

17. Answer (3)

Hint :

53°T T

T

4mg

Solution :

53°T T

T

4mg

2T cos53° = 4mg

3(2 ) 4

5T mg

10

3T mg

cos37 sin37

ˆ ˆT T

a i jm m

�

10 4 10 3ˆ ˆ( )3 5 3 5

g i g j⎛ ⎞ ⎜ ⎟⎝ ⎠

80ˆ ˆ20

3i j


4/16

18. Answer (4)

Hint : 0T mg PF F F � � �

Solution : 0T mg PF F F � � �

T = mg cos

410 8

5

T

m

19. Answer (4)

Hint : dy a

dx g

Solution : dy a

dx g

2dy

kxdx

2 1

2 kg 2 5 10 50

ax

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠

20. Answer (3)

Hint : 2

2

2tan (1 tan )

2

gxy x

u

Solution : 2

2

2tan (1 tan )

2

gxy x

u

tan = 2, 5 m/su

y = 2x – 5x2

21. Answer (4)

Hint : = 21

2t

Solution : = 21

2t

= 21

2

at

R t =

2 R

a

v = at = 2

2R

a Raa

anet

=

2

2 2 Raa

R

⎛ ⎞ ⎜ ⎟⎝ ⎠

= 2 2 24a a

= 21 4a

22. Answer (3)

Hint : (arel

) along incline plane = 0

Solution :

310

4 52 10 12 m

4510

5

PQ

⎛ ⎞ ⎜ ⎟⎝ ⎠

23. Answer (1)

Hint : x = 5 tan

Solution : x = 5 tan

2(5 sec )dx d

vdt dt

2

52

4

5

⎛ ⎞⎜ ⎟⎝ ⎠

1252

16

125m/s

8

24. Answer (3)

Hint : F = mg

Solution : F = mg

25. Answer (4)

Hint : 110 60

6a

Solution : 250

m/s6

ma

26. Answer (3)

Hint : T = 20

2m v

l

Solution : T = 20

2m v

l

27. Answer (3)

Hint : 1 cos cos

24 2

tg g

⎛ ⎞ ⎜ ⎟⎝ ⎠

� �

Sol. : 1 cos cos

24 2

tg g

⎛ ⎞ ⎜ ⎟⎝ ⎠

� �

28. Answer (4)

Hint : Since both collides, then we have

0

0

0

4 1005 2

5

vv

g v

⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠


5/16

PART - B (CHEMISTRY)

31. Answer (4)

Hint: mol of H2 = x

mol of O2 = y

mol of He = z

x + y + z = 5

Due to combustion H2 and O2 consumed so moles

decreased at constant volume and temperature

decreases the pressure.

Solution: x + y + z = 5

2 2 2

1H O H O

2

x y 0

x 2y 0

x – 2y + z

3y = 3, y = 1

Add amount of O2: 3 mol

2 2 2

1H O H O

2

x 2y 3 mol

x 2yx 2y 1.5

2

2x – 4y + x – 2y = 3

3x – 6y = 3

3x – 6 = 3

3x = 9

x = 3

z = 1

32. Answer (3)

Hint: Pressure mole of gas

Solution: Let mol of C4H8 = x

mol of O2 = y

4 1x y 0.1218

0.0821 400

C4H8 + 6O2 4CO2(g) + 4H2O(g)

x y – –

1x y

6 4

y6

4y

6

Total mol = 4 4 1

x y y y6 6 6

Final total moles = x + y + y

6

= 4.5 1

0.0821 400

1

3

x

y

33. Answer (3)

Hint: x = 1

3RT

M

28RT

yM

32RT

zM

Solution : Since both collides, then we have

0

0

0

4 1005 2

5

vv

g v

⎛ ⎞ ⇒ ⎜ ⎟⎝ ⎠

02 4

(5 2) (3) 120 m10

vR

⎡ ⎤ ⎢ ⎥⎣ ⎦

29. Answer (1)

Hint : ˆ( ) ( | | cos )R P Q P Q P � ��

Solution : We have,

ˆ( ) ( | | cos )R P Q P Q P � ��

ˆ | | cosR Q P Q � ��

30. Answer (2)

Hint :

( )

| | | |

A B B A

A B B A

v v r r

v v r r

� ��

� ��

Solution :

( )

| | | |

A B B A

A B B A

v v r r

v v r r

� ��

� ��

2 2

ˆ ˆ ˆ ˆ ˆ3 4 ( 1) ( 2)

5 1 ( 1) ( 2)

i j i y j z k

y z

2

3 12 and

5 1 ( 1) 0z

y

2 251 ( 1)

9y


6/16

Solution: For same value of x = y = z

31 22RT3RT 8RT

M M M

2

1 3

8T3T 2T

3T1 = 2.547T2 = 2T3

So, T1 < T2 < T3

34. Answer (2)

Hint:C C

8aT V 3b

27Rb

C 2

aP

27b

Solution:

2

C

C

T 8a 27b 8b

P 27Rb a R

Higher the value of C

C

T

P

⎛ ⎞⎜ ⎟⎝ ⎠

, higher will be the value of b,

so higher will be size of molecule and VC.

C

C

T 8b

P R

For NO = 177 8b

2.7364.85 R

For CCl4 = 556 8b

12.0645.6 R

So, b of CCl4 is more than NO.

35. Answer (1)

Hint: Equation is virial equation for the van der

Waal's gas. So,

B

aT

Rb

Solution: dZ

0 assume P 0dP

2 3

m m m

a 1 b cZ 1 b

RT V V V

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

as P 0

PVm = RT

m

P 1

RT V

2

2

a P bPZ 1 b

RT RT (RT)

⎛ ⎞ ⎜ ⎟⎝ ⎠

�

B

dZ a 1b 0 (at T T )

dP RT RT

⎛ ⎞ ⎜ ⎟⎝ ⎠

B

aT

Rb

36. Answer (4)

Hint: ClO4– has the greatest number of resonating

structures.

Solution: As resonance increases, stability

increases.

37. Answer (4)

Hint: H-bonding

Solution: M.P. : H2O > NH3 > HF

B.P. : H2O > HF > NH3

38. Answer (4)

Hint:C

1b V

3

a = PC 27b2

Solution: b = 4Vm

V volume of 1 mol of molecules

C

8aT

27Rb

VC = 3b

3 1

C

1b V 49.3 cm mol

3

= 0.0493 dm3 mol

–1

3

A

4b 4 r N

3

⎛ ⎞ ⎜ ⎟⎝ ⎠

r = 1.69 × 10–8

cm = 1.69 Å

a = 27Pcb2

= 27 (48.20) (0.0493)

= 3.16 dm6 atm mol

–2

C

8(3.16)T 231K

27(0.082) 0.0493

39. Answer (2)

Hint: HF > HCl > HBr > HI

Solution: As we move from HF to HI bond length

increases but polarity decrease so order of dipole

moment is

HF > HCl > HBr > HI


7/16

40. Answer (3)

Hint: I2Cl6 has 8 atom in the plane.

Solution: I2Cl6 is a planar molecule.

I

Cl

Cl

Cl

Cl

I

Cl

Cl

41. Answer (3)

Hint: sp3 and sp

3

Solution:

O O

P P

OH HOH OHH H,

(H PO )3 3

(H PO )3 2

42. Answer (2)

Hint:

A

F

F

F

F

Solution: 2 lone pair and 4 bond pair so A belongs

to 18th

group.

From the shape we can say that A has 8 valence

electron.

So AF3+ has T-shape with axial and equatorial bond

but AF5+ has octahedral geometry with one lone pair

so all bonds are identical.

43. Answer (1)

Hint: Law of diffusion.

Solution:

I II

4

4 4

CHHe He

CH CH He

Mr P 2 16 1= = =

r P M 4 4 1

1

1 Becomes the mole ratio in container (II)

II III

4

He

CH

r 1 16 2=

r 1 4 1

III IV

4

He

CH

r 2 16 4=

r 1 4 1

IV Container

4

He

CH

r 4 16 8=

r 1 4 1

44. Answer (1)

Hint: Bond order depends on resonance also.

Solution:

NO2–

1.5

NO3–

1.33

O3 1.5

CO32–

1.33

O2+

2.5

N2+

2.5

45. Answer (2)

Hints: Bent's rule

Solution: H—O—O—H F—O—O—F

Due to high electronegativity of F, p-character in OF

bond increases so s-character increases in O—O

bond.

As percentage of p-character increases, bond length

increases. If percentage of s-character increases,

bond length decreases.

y y

H—O—O—H F—O—O—F

y > x

46. Answer (4)

Hint: Al2Br6 has 3c – 4e– bond

Solution: (AlH3)n 3c – 2e– bond + 2c – 2e

–

B2H6 3c – 2e– bond + 2c – 2e

–

Al2(CH3)6 3c – 2e– bond + 2c – 2e

–

Al2Br6 3c – 4e– bond + 2c – 2e

–

Al

Br

Br

Br

Br

Al

Br

Br

The bridge bond of Al2Br6 are 3c – 4e– bond.


8/16

47. Answer (2)

Hint: Orbital is bonding.

Solution:

+

–

+

+

– +

– +

–+

–

The orbital is bonding.

48. Answer (1)

Hints: 2 > 1

4 > 3

Solution:

3

4

O

FF

O

HH

Due to high electronegativity of F bond angle

reduces

O

SiH3

H3Si

Due to back bonding bond angle increases

49. Answer (1)

Hint: Hybridisation of 'B' in BF3 and BF4– is sp

2 and

sp3 respectively

Solution: In BF3, due to back bonding bond length

of B – F bond reduces.

50. Answer (1)

Hint:3 3

C(CH ) is pyramidal

3C(CN) is planar

Solution: 3

C(CN) is planar due to back bonding.

51. Answer (2)

Hint: Strength of -bond decreases as internuclear

distance increases.

Solution: Strength of -bond depend on internuclear

distance.

2p – 2p > 2p – 3d > 2p – 3p > 3p – 3p

Due to inclined nature of d-orbitals 2p – 3d overlap

is more so 2p – 3d is more stronger than 2p – 3p.

52. Answer (3)

Hint: Bent's rule

Solution:

H

C

Cl

ClCl

x

H

C

H

ClH

y

I II

s-character of hybrid orbital involved in bonding with

H-atom is more in CHCl3 as compared to that in

CH3Cl.

C–H bond length in CHCl3 is less as compared

to C–H bond length in CH3Cl.

x < y

53. Answer (4)

Hint: If temperature remain same, then R.M.S.

speed does not change.

Solution: Due to mixing,

T = constant so individual pressure is

P1V1 = P2V2

V2 = 12.5 L

He f

2 2.5P 0.4 atm

12.5

∵ T = constant.

K.E. remains same and rms speed also

remains same.

Due to mixing pressure of both gases will decrease

due to increase in volume.

54. Answer (3)

Hint: PV = nRT

PVn

RT

Solution: If V, P and T are same then number of

molecules of all gases must be same.

1 1 1 2

2 2 2 1

n P V T

n P V T

P1 = P2, V1 = V2, T1 = T2

55. Answer (4)

Hint: TiC > ScN > MgO > NaF

Solution: Lattice energy Charge

1

Size

Na+, F

– = NaF

Mg2+

, O2–

= MgO

Sc3+

, N3–

= ScN

Ti4 + C

4 = TiC

So order of lattice energy is

NaF < MgO < ScN < TiC


9/16

PART - C (MATHEMATICS)

61. Answer (4)

Hint: Given exp. = tan ·tan tan3 3

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

cot cot cot3 3

⎡ ⎤ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

Solution: Given exp.

= tan·tan tan3 3

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

cot cot cot3 3

⎡ ⎤ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

tan tan tan3 3

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

3cot cot cot 33 3

⎡ ⎤ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

...(1)

where cot cot cot3 3

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

cos sin sin cos3 3 3 3

sin sin3 3

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

cos

sin

sincos3 3

sinsin sin

3 3

⎛ ⎞ ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

2 2 32sin sin

4cos

sin sin3 3

⎛ ⎞ ⎜ ⎟

⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠

56. Answer (4)

Hint: real

ideal

VZ

V

Solution: Vreal < Videal

means force of attraction is dominant

Z = 0.88

m

ZRTV 1.204 L

P

57. Answer (2)

Hint:

C

A

10 cm

Solution: PB = PC = PA + gh

3

5

10 10 9.8 0.1 760gh 74.48 torr

10

PA + gh = 760

PA = 760 – gh

= 760 – 74.48 = 685.52

58. Answer (3)

Hint: Covalent character increases if polarising

power of cation increases.

Solution: Order of covalent character

ZnCl2 < CdCl2 < HgCl2

As we move Zn2+

to Hg2+

size of cation increases but

due poor shielding of inner electrons Zeff increases so

polarizing power increases.

Zn2+

< Cd2+

< Hg2+

59. Answer (1)

Hint: At constant volume mean free path remain

constant.

Solution: T

P

at constant volume

= constant

60. Answer (1)

Hint: 1 1 2 2

1 2

P V P V

T T

Solution:

2

1 6 250V 12.5 L

0.4 300


10/16

2 23cos cos sin

3

sin sin sin3 3

⎛ ⎞ ⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

3cos cos cos3 3

sin sin sin3 3

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

3cot cot cot3 3

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

62. Answer (2)

Hint:

tr

= secr° · sec(r + 1)°

= sin1

sin1 cos cos( 1)r r

sin[( 1) ]

sin1 cos cos( 1)

r r

r r

1[tan( 1) tan1 ]

sin1r

Solution:

Here, tr = secr° · sec(r + 1)°

sin1

sin1 cos cos( 1)r r

1 sin( 1 )

sin1 cos cos( 1)

r r

r r

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

1 sin( 1) cos cos( 1) sin

sin1 cos cos( 1) cos cos( 1)

r r r r

r r r r

⎡ ⎤ ⎢ ⎥ ⎣ ⎦

1(tan( 1) tan )

sin1r r

88

0

sec sec( 1)

r

r r

∑

(tan1 tan0 ) (tan2 tan1 )1

sin1 (tan3 tan2 ) ... tan89 tan88

⎡ ⎤ ⎢ ⎥ ⎣ ⎦

cosec1 tan(90 1 ) cosec1 cot1

63. Answer (1)

Hint:

3z

z

= 2 2 = 3

z

z

> 3

z

z

> 3

r

r

,

where r = |z| –2 < r – 3

r

< 2

Solution: Here, 2 = 3

z

z

> 3

z

z

3r

r

< 2, where |z| = r

–2 < r – 3

r

< 2

When, r – 3

r

+ 2 > 0 r2 + 2r – 3 > 0

r2 + 3r – r – 3 > 0

(r + 3)(r – 1) > 0 r < –3 or r > 1

|z| –3 or |z| 1

and when r – 3

r

– 2 < 0 r2 – 2r – 3 < 0

r2 – 3r + r – 3 < 0

(r + 1)(r – 3) < 0

–1 < r < 3

1 < |z| < 3

|z|max

= 3, |z|min

= 1

64. Answer (1)

Hint: Exp. ax2 + 2hxy + by2 + 2gx + 2fy + c can be

expressed as the product of two linear factors if

0

a h g

h b f

g f c

Solution: The expression ax2 + 2hxy + by2 + 2gx +

2fy + c can be expressed as the product of two linear

factors if

= abc + 2fgh – af2 – bg2 – ch2 = 0

2 × (–6) × (–3) + 2 11 1 121

2 ( 6)2 2 2 4

⎛ ⎞ ⎜ ⎟⎝ ⎠

2 1( 3) 0

4 4

195,

6

= 5 (+ve integral value)


11/16

65. Answer (2)

Hint: D > 0, 2,2

b

a

4f(2) > 0

Solution: Conditions are D > 0

(–20)2 – 4.4.(252 + 15 – 66) > 0

–16 × 15 + 16 × 66 > 0

< 22

5...(i)

2

b

a

< 2

( 20 )

2 4

< 2

20 < 16

< 4

5...(ii)

4f(2) > 0 f(2) > 0

16 – 40 + 252 + 15 – 66 > 0

252 – 25 – 50 > 0

2 – – 2 > 0

2 – 2 + – 2 > 0

( + 1)( – 2) > 0

< –1 or > 2

(– , –1) (2, ) ...(iii)

From (i), (ii) and (iii), (– , –1)

66. Answer (4)

Hint:

Coefficient of x2 = –2 < 0, greatest value = –4

D

a

and least value = 3

2f⎛ ⎞⎜ ⎟⎝ ⎠

in 3, 2

2

⎛ ⎞⎜ ⎟⎝ ⎠

Solution: Coefficient of x2 = –2 and (Discriminant)

= 9 – 4(–2) × 2 = 25

The greatest value = 25 25

4 4( 2) 8

D

a

and least value = 3

2f⎛ ⎞⎜ ⎟⎝ ⎠

= –7

Required sum = 25 31

78 8

O–1

1 2

3

4

3 25,

4 8

⎛ ⎞⎜ ⎟⎝ ⎠

3

2–

X X

Y

67. Answer (1)

Hint: Put z = a + ib, we get

2a = 22 2( 1)a b 2a = 1 + b2

and arg(z1 – z

2) =

4

|b

1 – b

2| = |a

1 – a

2|

Now, put the values z1 and z

2 in given equation find

arg(z1z

2).

Solution: Here, 2 1z z z

a + ib + a – ib = 2|a + ib – 1|,

where z = a + ib (say)

2a = 22 2( 1)a b

a2 = a2 – 2a + 1 + b2 2a = 1 + b2 ...(i)

Again given, arg(z1 – z

2) =

4

arg[(a1 + ib

1) – (a

2 + ib

2)] =

4

arg[(a1 – a

2) + i(b

1 – b

2)] =

4

tan–1 1 2 1 2

1 2 1 2

14

b b b b

a a a a

⇒

|b1 – b

2| = |a

1 – a

2| ...(ii)

But z1 = a

1 + ib

1 and z

2 = a

2 + ib

2 satisfy the

equation 2 1z z z

2a1 = 1 + b

1

2 ...(ii)

and 2a2 = 1 + b

2

2 ...(iii)

By [(ii) – (iii)],

2(a1 – a

2) = b

1

2 – b2

2 = (b1 + b

2)(b

1 – b

2)

2|a1 – a

2| = |b

1 – b

2| |b

1 + b

2|

|b1 + b

2| = 2

|Im(z1 + z

2)| = 2


12/16

68. Answer (3)

Hint: Critical point is x = –6

5

Now, solve equation for x < –6

5 and for x > –

6

5

Solution: Given inequation is x2 – |5x + 6| > 0

Case I : x < –6

5 x2 + 5x + 6 > 0.

x (– , –3) (–2, )

x (– , –3) 6

2,5

⎛ ⎞ ⎜ ⎟⎝ ⎠

...(i)

Case II : x > – 6

5 x2 – 5x – 6 > 0.

(x + 1)(x – 6) > 0

x (– , –1) (6, )

x 6, 1

5

⎡ ⎞ ⎟⎢⎣ ⎠

(6, ) ...(ii)

From (i) and (ii),

x (– , –3) (–2, –1) (6, )

[Taking union of (i) and (ii)]

69. Answer (1)

Hint:

amp.8 8

sin cos amp(1 ) amp(3 3 )7 7

i i i ⎛ ⎞ ⎜ ⎟

⎝ ⎠

Solution:

amp (z)8 8

amp sin cos amp (1 )7 7

i i ⎛ ⎞ ⎜ ⎟

⎝ ⎠

amp(3 3 )i

amp sin cos amp. (1 )7 7

i i ⎛ ⎞ ⎜ ⎟

⎝ ⎠

amp (3 3 )i

1 1 1 1tan cot ( tan 1 ) tan

7 3

⎛ ⎞ ⎜ ⎟⎝ ⎠

5

14 4 6

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

9

14 4 6

892

84

(For principal value)

79

84

70. Answer (4)

Hint: 2 2

i

iii e e

Solution:

2

2 2(0 1) cos sin2 2

ii

i ii ii i e e

⎛ ⎞ ⎜ ⎟⎝ ⎠

2e

since i2 = –1

2

2

1ii e

e

2 2 2

2

1

i

iii e e e

e

⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠

71. Answer (3)


(3x – 9)(3x – 27) = 0 x = 2, 3

Solution: Given equation is 9x – 36 · 3x + 243 = 0

(3x)2 – 36 · 3x + 243 = 0

(3x – 9)(3x – 27) = 0

3x = 32 or 33

x = 2, 3

Sum of roots = 2 + 3 = 5

72. Answer (2)

Hint: Put 5x2 – 6x + 8 = t and 5x2 – 6x – 7 = t – 15,

then solve for t and then for x.

Solution: Let 5x2 – 6x + 8 = t


15 1t t

t – 15 = 1 + t – 2 t

t = 8 t = 64

5x2 – 6x + 8 = 64

5x2 – 6x – 56 = 0

5x2 – 20x + 14x – 56 = 0

x = 4, –14

5, but x > 0

x = 4, here x = 4 also satisfies the given equation.


13/16

73. Answer (4)

Hint: Here, xy and x + y are the roots of the equation.

a2 – 23a + 120 = 0 a = 15, 8

Solution: ∵ xy(x + y) = 120 and xy + (x + y) = 23.

The equation with roots xy and x + y is

a2 – 23a + 120 = 0.

(a – 15)(a – 8) = 0

xy = 15, x + y = 8

x2 + y2 = (x + y)2 – 2xy = 34

2 2

342

2 15 2

x y

xy

74. Answer (2)

Hint: x = cos sin 1

sin cos sin cos

and y =

21 sin

coscos cos

Solution: Here,

21 sin

andsin cos cos

x y

3

2 2

3 3

1 sinand

cos cos

x y xy

2 2 2

2 23 3

2 2

1 sin( ) ( ) 1

cos cos

x y xy

75. Answer (2)

Hint: Draw graph of sinx and cosx.

Solution:

y

x

2–1

2

2–

4

2

32

2

∵ 1 22 2

sin1 < sin2

1 2 34

cos1 > cos2 > cos3

∵ 1 sin1 cos14 2

⇒

76. Answer (1)


sinx = 1 x = (4n + 1)2


sinx = 1 x = (4n + 1)2

cos cos odd multiple of 02

⎛ ⎞ ⎜ ⎟⎝ ⎠

x

tan2x = tan 2 (4 1) 02

n⎛ ⎞ ⎜ ⎟

⎝ ⎠

cot3x = cot 3 (4 1)2

n⎛ ⎞ ⎜ ⎟

⎝ ⎠

cot odd multiple of 02

⎛ ⎞ ⎜ ⎟⎝ ⎠

sin4x = sin 4 (4 1)2

n⎡ ⎤ ⎢ ⎥

⎣ ⎦ = 0

Required sum = 0 + 0 + 0 + 0 = 0

77. Answer (3)

Hint: Use transformation formula.

Solution:

cot70° + 4 cos70°cos70 4cos70 sin70

sin70

cos70 2sin140

sin(90 20 )

sin20 2sin40

cos20

2cos10 sin30 sin40

cos20

cos10 cos50

cos20

2 cos30 cos20

cos20

3 cos203

cos20


14/16

78. Answer (2)

Hint: Find cos2x = a + 3 or –1, by solving equation

and then 0 < a + 3 < 1.


cos2x =

2( 2) 4 4 4 12

2

a a a a

( 2) ( 4)

2

a a

cos2x = a + 3 or –1

But cos2x 0 and 0 < cos2x < 1

0 < a + 3 < 1 a [–3, –2]

79. Answer (4)

Hint: From given inequation,

(x2 + 1) < 3x2 – 7x + 8 < 2(x2 + 1)

Solving both inequations x [1, 6]

Solution: Here, x2 + 1 < 3x2 – 7x + 8 < 2(x2 + 1)

2x2 – 7x + 7 > 0 x R

and x2 – 7x + 6 < 0 x [1, 6]

Required solution set is

x [1, 6] R x [1, 6]

80. Answer (4)

Hint: p N, cot x = ([cot2x] – p) I

p = cot x (cot x – 1)

= product of two consecutive integers.

Solution: Here, [cot2x] = I (an integer) and p N

cot x is also an integer.


cot2x – cot x – p = 0 p = cot x(cot x – 1)

Here, p can be the product of two consecutive

integers.

p = 2 × 1, 3 × 2, 4 × 3, 5 × 4, 6 × 5, 7 × 6,

8 × 7, 9 × 8, 10 × 9

Total number of elements in set S = 9.

81. Answer (1)

Hint: Find critical points and they are 1, 2.

Now solve the equation for x < 1, 1 x < 2 and x 2

Solution: Here, critical points are given by x – 1 = 0

and x – 2 = 0 x = 1 and x = 2

+ve

Put = 3x

+ve +ve–ve

–ve–ve– +1 2

Sign for – 2x

Sign for – 1x

Case I : When – < x < 1

x – 1 < 0 and x – 2 < 0


1 – x + 2 – x > 5

3 – 5 > 2x

x < –1

Common values of x are given by – < x < –1.

Case II : When 1 x < 2

x – 1 0 and x – 2 < 0


x – 1 + 2 – x > 5 1 > 5 (absurd)

Case III : When 2 x <

Then from given inequation,

x – 1 + x – 2 > 5

2x > 8 x > 4

Common values are given by 4 < x <

Required solution set is (–, –1] [4, )

82. Answer (2)

Hint: z is purely imaginary, then use 0z z and

find value of sin.

Solution: Given z = 3 2 sin

1 2 sin

i

i

∵ z is purely imaginary

0z z

3 2 sin 3 2 sin0

1 2 sin 1 2 sin

i i

i i

6 – 8 sin2 = 0

sin2 = 3

4, cos2 =

3 11

4 4 , tan2 = 3

cosec2 + tan2 = 4

3 + 3 =

13

3


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83. Answer (4)

Hint:

Put n = 1, 2 and then check it for n = 3

Solution:

Given, pn + qn = rn + sn ...(i)

Put n = 1, p + q = r + s ...(ii)

Put n = 2, p2 + q2 = r2 + s2 ...(iii)

From (ii),

(p + q)2 = (r + s)2

p2 + q2 + 2pq = r2 + s2 + 2rs

pq = rs ...(iv) [From (iii)]

Also from (ii),

(p + q)3 = (r + s)3 ...(iv)

p3 + q3 + 3pq(p + q) = r3 + s3 + 3rs(r + s)

p3 + q3 = r3 + s3 [From (ii) and (iv)]

(i) is true for n = 3 also.

Again from (iii),

p2 – s2 = r2 – q2

(p – s)(p + s) = (r – q)(r + q)

[∵ From (ii), p – s = r – q]

p + s = r + q

p – q = r – s ...(v)

Multiply (ii) and (v),

(p + q)(p – q) = (r + s)(r – s)

p2 – q2 = r2 – s2

84. Answer (3)

Hint:

f(x, ) = cos2x + cos(x + )[cos(x + ) – 2 cos cosx]

= cos2x – cos(x + ) · cos( – x)

Solution: f(x, ) = cos2x + cos2(x + ) –

2 cos · cosx · cos( + x)

= cos2x – cos( + x) · cos( – x)

= cos2x – (cos2x – sin2)

= sin2

2

23 1 1, sin

4 4 4 22f

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85. Answer (1)

Hint: Use, cosx – cosy = 2 sin 2

x y· sin

2

y x

and sinx + siny = 2 sin 2

x y· cos

2

x y = 0

Solution: cosx – cosy = 2 sin sin2 2

x y y x = 0

sin x + sin y = 2 sin cos2 2

x y x y = 0

Here, sin2

y x and cos

2

x y are not simultaneously

zero.

Hence, sin2

x y = 0 x + y = 2n, n I

sin2020x + sin2020y

= 2 sin1010(x + y) · cos1010(x – y) = 0

86. Answer (1)

Hint: Use sin =

2

2 2

2tan /2 1 tan /2, cos

1 tan /2 1 tan /2

Solution: Given equation is sin + cos = 2

3

2

2 2

2tan 1 tan22 2

31 tan 1 tan

2 2

25 tan 6 tan 1 0

2 2

6 36 4 5 ( 1)tan , 0

2 10

0 tan 02 2 2

⇒

6 56 3 14tan

2 10 5

87. Answer (3)

Hint: Use condition for both common roots.

Solution: 2x2 – 5x + 8 = 0 has both roots non-real,

since discriminant = (–5)2 – 4 × 2 × 8 < 0

Both roots of given two equations are common.


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� � �

, say2 5 8

a b c

a = 2, b = –5, c = 8

(2 8)1

2 10

a c

b

88. Answer (2)

Hint: Solve the expression on L.H.S. of equation and

then find + and .

Solution: From the given equation,

3x2 + 12x – 1 = 0

+ = –4, = –1

3

1 1 1

3( 2)( 2) 3( 3)( 3) 3( 5)( 5)

1 1

3 6( ) 12 3 9( ) 27

1

3 15( ) 75

1 1 1

1 24 12 1 36 27 1 60 75

1 1 1 48

13 10 14 455

89. Answer (2)

Hint: Use sum and product of roots.

Solution: 2p = 1 + q and + = –p, = q

= 2p – 1 = –2( + ) – 1

2 + 2 + = –1 ( + 2)( + 2) = 3

Either + 2 = ±1, + 2 = ±3

or + 2 = ±3, + 2 = ±1

So, the polynomials are x2 – 1 = 0, x2 + 8x + 15 = 0.

90. Answer (3)

Hint: Use cot(A + B) = cot3

4

Solution: Given A + B = 3

4

cot(A + B) = cot3

4

cot cot 11

cot cot

A B

B A

⇒

cot A cot B – 1 + cot B + cot A = 0

cot B(cot A + 1) + (cot A + 1) = 2

(cot A + 1)(cot B + 1) = 2 = n (given)

2 3sin sin

1 3 2

n

n

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test - 2 (code-c) (answers) all india aakash test series ......6. answer (4) hint : 110 60 6 a...

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