Test - 1A (Paper - 2) (Code-G) (Answers) All India Aakash ... · Hint : Velocity of cylinder along...

Test - 1A (Paper - 2) (Code-G) (Answers) All India Aakash Test Series for JEE (Advanced)-2020

1/12

PHYSICS

1. (A, B, C)

2. (A, B, D)

3. (A, B)

4. (A, B, D)

5. (A, D)

6. (A, B)

7. (18)

8. (16)

9. (12)

10. (15)

11. (02)

12. (12)

13. (15)

14. (03)

15. (A)

16. (A)

17. (B)

18. (B)

CHEMISTRY

19. (A, C, D)

20. (A, B, C, D)

21. (A, B)

22. (A, D)

23. (A)

24. (D)

25. (03)

26. (15)

27. (06)

28. (20)

29. (40)

30. (43)

31. (65)

32. (98)

33. (C)

34. (D)

35. (A)

36. (B)

MATHEMATICS

37. (C, D)

38. (B, C)

39. (A, B)

40. (A, D)

41. (B, C)

42. (B, C)

43. (81)

44. (02)

45. (01)

46. (00)

47. (11)

48. (11)

49. (31)

50. (01)

51. (C)

52. (B)

53. (D)

54. (C)

Test Date : 18/11/2018

ANSWERS

TEST - 1A (Paper-2) - Code-G

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

All India Aakash Test Series for JEE (Advanced)-2020 Test - 1A (Paper - 2) (Code-G) (Hints & Solutions)

2/12

PART - I (PHYSICS)

HINTS & SOLUTIONS

1. Answer (A, B, C)

Hint : Write down the x coordinate of A3 in terms of

� and . dx

vdt

Solution :

x = 2(3l cos + 2l cos + l cos )

x = 12l cos

v = 1

12 sinsin

d dl

dt dt

⎛ ⎞ ⇒ ⎜ ⎟ ⎝ ⎠ �

2 3B

dv l

dt

⎛ ⎞ ⎜ ⎟⎝ ⎠

6 m/s4sin

v

2. Answer (A, B, D)

Hint : Write down the expression for tangential

acceleration and acceleration in y-direction. Use the

fact that aj = a

t

Solution :

( sin )tdv

g gdt

( sin )y

dvg g

dt

v

d

ytdvdv

dt dt

vT – v

0 = –v

0sin v

T – v

0 =

03

5

v

vT =

02

5

v

3. Answer (A, B)

Hint : Velocity of cylinder along normal relative to

connected bodies is zero.

Solution :

3

ˆ ˆ

v xi yj �

x = +10

2 2

ˆ

v v i�

v2

v2

sin 37° = x sin 37° + y cos 37°

23 4

65 5

v y

3v2

= 30 + 4y

4. Answer (A, B, D)

Hint : Draw F.B.D. and equation of motion. Write

down the constraint motion.

Solution :

T cos 37° = ma1

3mg – T = 3ma2

1 2

4

5a a⎛ ⎞ ⎜ ⎟⎝ ⎠

1

5

4T ma

3mg – T = 1

12

5ma

3mg = 1

73

20ma

1

60

73

ga

5. Answer (A, D)

Hint : Draw FBD and write equation of motion with

constrain relation

Solution :

2mg – T1 = 2ma

1

T1 – T

2 = 2ma

1

2

02

Tma

2a1 = a

0

1

4

ga

6. Answer (A, B)

Hint : Acceleration of block B with respect to A is

2

0v

�

Solution :

T – 0.05 (2 mg) = 2 ma0

T + ma0

=

2mv

�

22

T mgT mg

⎛ ⎞ ⎜ ⎟⎝ ⎠

T

ma0

5

3

mgT

03

ga

Test - 1A (Paper - 2) (Code-G) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020

3/12

7. Answer (18)

Hint : Along an incline maximum range is

2

(1 sin )

u

g

Solution :

2

10(1 sin )

u

g

2 4100 1

5u

⎛ ⎞ ⎜ ⎟⎝ ⎠

u

2 = 20 × 9

8. Answer (16)

Hint : Use the concept that

2

c

va

R

Solution :

216 m/s

dy

dt

2

2

216 m/s

x

d xa

dt

2

16

x

vR

a

9. Answer (12)

Hint : Use addition of vector, net force will be along

PO.

Solution :

2

202 1

10k OP

��

= 40

k = 2 N/m

10. Answer (15)

Hint : Direction of friction will change, hence

magnitude of acceleration

Solution :

2

16 m/s

2

Mg Mga

M

0

1

1

vt

a

2

0

12

vs

a

2

2 0

2 2

1

1

2 2

va t

a

0

2

1 2

vt

a a

0

1 1 2

1 1vT

a a a

⎡ ⎤ ⎢ ⎥

⎢ ⎥⎣ ⎦

0

1 2

1 2

va a

a a

11. Answer (02)

Hint : Apply NSL to get acceleration of each and find

relative acceleration along y direction.

Solution :

2 mg sin37° – mg sin 53° = 3 ma

23

5

ga

2

15

ga

21 3 4

2 5 5a t h⎛ ⎞ ⎜ ⎟⎝ ⎠

210 7

15 5t h

⎛ ⎞ ⎜ ⎟⎝ ⎠

214

15

th

12. Answer (12)

Hint : Apply concept of similar triangle

Solution :

h Y

x

�

hY

x �

2

dY hv

dt x �

4hv

⎛ ⎞ ⎜ ⎟⎝ ⎠�


4/12

13. Answer (15)

Hint : Write down the dimension of each and

compare.

Solution :

= M1L–3

� = M0L1T0

V = M0L1T–1

d = M0L1T0

= M1L–1T–1

P = ML–1T–2

14. Answer (03)

Hint : Length of string remains constant.

Solution :

2 22( ) ( )L h y h x

0dL

dt

2 2

1

2

A

B

xvV

h x

0

3

1 4

52

4

v

h

0

3

10v

15. Answer (A)

Hint : Write down the dimension and compare.

Solution :

dvF A

dx

1F A T

1

2

vA

A

1

2f A v

b c1 3 1 1 2 2 1 1

ML MLT L LT ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

a = 1

c = –2

b = –1

16. Answer (A)

Hint : Initial static friction will act and then kinetic

friction will act.

Solution :

fs = applied force

fk = N

N will decrease linearly with time.

17. Answer (B)

Hint : Write down vAB

which is constant

Solution :

At the time of minimum separation vAB

is

perpendicular to relative separation.

(4, 3)

37°53°

vA

8°

(3, –4)B

10 2 sin(8)l

0

10 2 cos(8)

5t

2 2 cos8

18. Answer (B)

Hint : Relative motion will start when s

f f

Solution :

(f1)max

= × 1 × 10

= 10

Write down the equation of motion of each.

A Ba a also

maxsf f


5/12

PART - II (CHEMISTRY)

19. Answer (A, C, D)

Hint : In T.B.P. geometry, more electronegative atom

attached at axial position.

Solution :

P

B

B

A

A

A

On replacementof A by C

0 = 0

P

C

C

C

B

B

On replacementof A by C P

C

C

B

B

B

= 0

P

B

B

B

A

A 0

20. Answer (A, B, C, D)

Hint : N – N x

H

H

H

H

N – N y

F

F

F

F

y < x

as E.N atom has 'p' character so 's' character in

N–N bond and bond length decrease.

Solution : N2H

4 > N

2F

4 i.e.,

NF N

F

F

F

H

N

H

N H

H

<

1. It can be explained on the basis of Bent's rule. In

N2F

4, N – N bond has more s-character hence

bond strength increases and bond length

decreases.

While in N2H

4, N – N bond has less s-character

hence bond length increases.

2. As the positive charge increases (Zeff

) nuclear

charge increases.

3. As the negative charge increases, ionic radius

increases.

21. Answer (A, B)

Hint : In NO, HOMO is 2 = 2p px y

* * while in N2 it is

2pz.

Solution : In NO, HOMO is 2 = 2p px y

* * while in N2

it is 2pz.

22. Answer (A, D)

Hint : 2H2 + O

2

At 100ºC 2H2O

at 100ºC H2O exist as water.

Solution :

At 100ºC

2 2 2(g) (g) (g)

2H O 2H O

Before reaction 2 2 0

volume in litre

After reaction 0 1 2

volume in litre

(A) Mole of H2O formed

2PV 1 2n 8.9 10 mole

RT 0.0821 273

Mass of H2O formed = 8.9 × 10–2 × 18

= 160.2 × 10–2 g = 1.602 g

(B) O2 is left as unreacted = 1 litre

At STP mole of O2 left

= 21

4.4 10 mole22.4

Mass of O2 left = 4.4 × 10–2 × 32 = 1.408 g

(C) At 100ºC H2O also exists as vapours

Total mole present at 100ºC

2 2

H O On n left

2 2 28.9 10 4.4 10 13.3 10 moles

Volume of the vessel = 2 L

2nRT 13.3 10 0.0821 373

P 2.03 atmV 2


6/12

(D) Volume of O2 used for formation of H

2O = 1 L

2

2

O

1n 4.4 10 mole

22.4

23. Answer (A)

Hint : 2

aP V b RT

V

⎛ ⎞ ⎜ ⎟⎝ ⎠

Solution : If a = 0 if b = 0

Z = Pb

1RT

Z = aP

1RT

24. Answer (D)

Hint : If labelling of oleum is (100 + x)% the free SO3 (in g)

is x 80

18

Solution : Initial moles of free SO3 present in oleum

= No. of moles H2O used

12 2moles

18 3

Moles of free SO3 combined with H

2O after 4.5 g of

H2O added

3SO

4.5 1n mole

18 4

Moles of free SO3 remains

2 1 5

3 4 12 moles.

Volume of free SO3 at STP

522.4 9.33 L

12

25. Answer (03)

Hint : If labelling of H2SO

4 is (100 + x)% then mole

of SO3

x

18

mol of H2SO

4 =

x100 80

18

98

⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

Solution : 109% labelled oleum means it has 9 g

H2O which reacts with free SO

3 to give H

2SO

4 i.e.

SO3 + H

2O H

2SO

4.

Mass of SO3 which will react with 8 g

H2O

80 940 g

18

Let 100 g sample of 109% labelled oleum contains

40 g SO3 and 60 + 38 = 98 g of H

2SO

4.

Moles of H2SO

4 = x = 1.0 mole

Moles of SO3 = y = 0.5 mole

x + y = 1.5 moles

x – y = 0.5 mole

x y

x y

= 03

26. Answer (15)

Hint : For photoelectric effect.

Ein = E

Thr + K.E.

Solution :

According to photoelectric effect.

E = E0 + K.E.

h = h0 + K.E.

h1

= K.E1

+ h

0.....(i)

h2

= K.E2

+ h

0.....(ii)

2h1

– 2h2 = h

0(∵ 2 K.E

1 = K.E

2)

(21

– 2) =

0

0

= 1 2 1 2

2c c 2 1c

⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦

0

=

8

10

3 10 2 1

2000 180010

⎡ ⎤⎢ ⎥⎣ ⎦

(∵ 1 Å = 10–10 m).

0

= 1.33 × 1015 s–1.

27. Answer (06)

Hint : Number of spectral line = n (n 1)

2

Solution :

2

n 2

13.6 ZE eV

n

Also E = En – E

0

(E = Ionisation potential)

Energy absorb by 1 atom 51

12.754

2

13.612.75 ( 13.6)

n

2

13.612.75 13.6

n

n2 = 16 n = 4

Thus, the number of spectral lines emitted during the

de-excitation will be n(n 1) 4(4 1)

62 2


7/12

28. Answer (20)

Hint & Solution :

10.50 0.39Al 0.39 1

27 0.39 ⇒

15.1 0.39K 0.39 1

39 0.39 ⇒

24.96 0.78S 0.78 2

32 0.39 ⇒

49.92 3.12O 3.12 8

16 0.39 ⇒

The empirical formula of the salt is KAlS2O

8

Molecular formula of hydrated salt is KAlS2O

8 yH

2O

When

18 y100 45.6

39 27 128 64 18y

y = 12

So the value of x + y = 8 + 12 = 20

29. Answer (40)

Hint : PM = dRT

Molar mass of mixture = M

Solution :

dRT 600 1 0.08 300P

M 760 M

⇒

M = 30.4 gm/mole

mol of N2 = x, mole of O

2 = 1 – x

30.4 = 32 (1 – x) + 28 (x)

x = 0.4

30. Answer (43)

Hint :

Rate of diffusion

1

Molar mass

Solution : Now for diffusion of gaseous mixture and

pure O2.

2

2

O mix

mix O

r M

r M

(or) 2

2 2

O M Mix

M O O

V T M

V T M

mixM1 232

1 200 32

Mmix

= 43

31. Answer (65)

Hint : For van der Waal's gas

2

2

n aP (V nb) nRT

V

⎛ ⎞ ⎜ ⎟

⎝ ⎠

Solution :

Given n = 4 moles, V = 1 L, T = 37 + 273 = 310 K

a = 3.592 : b = 0.0427

using van der Waal's equation for 4 moles

2

2

n aP (V nb) nRT

V

⎛ ⎞ ⎜ ⎟

⎝ ⎠

24 3.592P (1 4 0.0427) 4 0.0821 310

1

⎛ ⎞ ⎜ ⎟⎝ ⎠

P = 65.3 65 atm

32. Answer (98)

Hint : Mass of solution = (v × d)solution

= 1000 × 1.26

Mass of H2SO

4 = 10.15 × 98

% H2SO

4 =

10.15 98 100

1000 1.26

Solution :

% by mass = 79

Molarity = 10.15

d = 1.26 gm/ml

M =

%by mass 10 d 79 10 1.2610.15

Molar mass Molar mass

⇒

Molar Mass = 98

33. Answer (C)

Hint : Covalent character decrease as size of cation

increase.

Solution : Number of d – p = P4O

10 > SO

3 > SO

2

4 2 1

Number of unpaired electron : 1 2

2 2 2O O O

Covalent character : 2 2 2

MgCl CaCl SrCl


8/12

34. Answer (D)

Hint & solution :

N

SiH3

SiH3

H Si3

sp2

sp3

2NO sp

, sp2

3NO

sp sp3 3

3 3N(CH ) ,

sp d sp3 3

6 4PCl , PCl

35. Answer (A)

Hint & solution :

BF3 Non-polar, planar

N3H Polar, planar

NH3 Polar, non-planar

36. Answer (B)

Hint : Radius =

2

0

na

Z

⎛ ⎞⎜ ⎟⎝ ⎠

Separation energy =

2

0 2

ZE

n

Solution : Radius of 4th orbit of Be+3

=

2

0 0

4a 4a

4

Separation energy of electron in 2nd orbit of Li2+ ion

=

2

0 02

3 9E E

42

PART - III (MATHEMATICS)

37. Answer (C, D)

Hint : 8pq = P

Solution :

2 4 68 sin sin sin

7 7 7pq

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 3sin sin sin

7 7 7p

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

1

8q

Let 2 3

cos , cos , cos7 7 7

a b c ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠,

1

8abc

2 2 4 68 1 cos 1 cos 1 cos

7 7 7p

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

= 2 3

1 cos 1 cos 1 cos7 7 7

⎛ ⎞⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

= (1 – b) (1 + c) (1 + a)

= 1 + a – b + c + ac – ab – bc – abc

ac – ab – bc = 1 3 2

2cos cos 2cos cos2 7 7 7 7

⎡ ⎢⎣

2 32cos cos

7 7

⎤ ⎥⎦

= 1 4 2 3

cos cos cos cos2 7 7 7 7

⎡ ⎢⎣

5cos cos

7 7

⎤ ⎥⎦

=

2 3cos cos cos

7 7 7b a c

a – b + c + ac – ab – bc = 0

8p2

=

1 7 71

8 8 8p ⇒

38. Answer (B, C)

Hint : Write cos7 in terms of cos

Solution :

∵ cos7 = 64cos7 – 112cos5 + 56 cos3 – 7cos

cos (7) = cos (2)

7 = 2n ± 2

Taking + sign

= 2

5

n

= 2 4

0, , ,...5 5

for +ve values

Taking – sign

= 2

9

n

= 2 4

0 , , ,...9 9

for +ve values


9/12

39. Answer (A, B)

Hint : 2

sinlog (cos ) 0, 1

xx

Solution :

y

5 –1

2y

2x

O

y x= sin

2

For intersection points of f(x) and g(x),

logsinx

(cos2x) = sgn(tan x)

If tan x = 0, cos2x = 1

sinx = 0, no solution

if tanx > 0, cos2x = sinx sin2x + sinx – 1 = 0

sinx = 1 1 4

2

sinx =

1 5

2

sinx = 5 1

2

Only one solution

If tanx < 0, cos2 x = cosec x, (no solution)

40. Answer (A, D)

Hint : Put z = x + iy

Solution :

Put z = x + iy; x, y R in 2z z

z

z z

2 2

2 2

2 2

22 2( )

x yz zz x iy

zz x y

⇒

z1 = 1

Put z = x + iy in 1

Re( ) 2z z z i

z z

1( ) ( ) 2

( ) ( )x x iy x iy i

x iy x iy

x = 0, 2 22

iiy i

y 4y

2 – 4y + 1 = 0

1

2y

41. Answer (B, C)

Hint : If coefficients are real, then complex roots

exists in pair which are conjugate to each other.

Solution :

x2 + bx + c = 0

roots are 1 + i and 1 – i

b = –2, c = 2

Six quadratic equations are possible

2– 2 2 0x x ...... (1)

22 – 2 0x x ...... (2)

22 – 2 0x x .......(3)

22 – 2 1 0x x .......(4)

2–2 2 0x x .......(5)

2–2 2 1 0x x .......(6)

(1) & (4) have imaginary roots. (2), (3), (5), (6) have

irrational roots.

42. Answer (B, C)

Hint : Draw graph

Solution :

–2 –1 O 1 2 3 4

y = 2

y

f x = |x – 2x |( ) – 12

2g x( ) 2 sin x

⎛ ⎞ ⎜ ⎟

⎝ ⎠

x

2 2 + 1– 2 + 1


10/12

43. Answer (81)

Hint : Make perfect square

Solution :

2 3sin 2cos

2a a

= 2 23 1

1 cos 2cos (cos 1)2 2

a a a

cos2x + 4sinx + 12 = 1 – sin2

x + 4 sinx + 12

= 17 – (sinx – 2)2

f(x) is maximum when cos a = –1 and sin x = –1

f(x)max

=

1ln

12–

ln8 3

= 1

–3

–4 = 81

44. Answer (02)

Hint : Case 1 : Both roots are negative

Case 2 : Exactly one root is negative

Solution :

D = (– 3)2 – 4= 2 – 9 – 6– 4= (– 5)2 – 16

= (– 9) (– 1)

D 0, (–, 1] [9, )

Case-1: If both roots are negative

D 0, –3 < > 0 (0, 1]

Case-2: If exactly one root is negative and other root

+ve or zero.

D 0, 0 (–, 0]

(–, 0] (0, 1]

(–, 1]

45. Answer (01)

Hint : Draw graphs

Solution :

y

(0, 1)

(1, 0)

(0, –1)

(–1, 0) Ox

A B = P B (A ) = { } ( (A )) = 1n P B

(1, 1)(–1, 1)

x

(–1, –1)

y

(1, –1)

46. Answer (00)

Hint : cos 0, 12x

⎛ ⎞ ⎜ ⎟⎝ ⎠

Solution :

When 1, 3, 5, ..... cos 02

x x⎛ ⎞ ⎜ ⎟

⎝ ⎠

When 0, 2, 4, ..... cos 12

x x⎛ ⎞ ⎜ ⎟

⎝ ⎠

there is no integral value of x belonging to the

domain of f(x)

47. Answer (11)

Hint : ( ) 4sin 3cos6 6

f x x x ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Solution :

cos sin3 6

x x ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠∵ and

sin cos3 6

x x ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

f(x) = 4sin 3cos6 6

x x ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

–5 f(x) 5

48. Answer (11)

Hint : Solve for sin x

Solution :

4(1– sin2x)sinx – 2sin2

x = 3sinx

4sinx – 4sin3x – 2sin2

x = 3sinx

sinx = 0 OR 4sin2x + 2sinx – 1 = 0

sinx = 2 4 16 1 5

8 4

y

5 –1

4y

2x

O–2

5 –1

4y –

49. Answer (31)

Hint : If z = x + iy has negative principal argument

then x > 0, y < 0 or x < 0, y < 0


11/12

Solution :

z = (3 – 42 + 2) + i (2 – 18 – 175)

z has negative principal argument

3 – 42 + 2 < 0 and 2 – 18 – 175 < 0

OR 3 – 42 + 2 > 0 and 2 – 18 – 175 < 0

2 – 18 – 175 < 0

2 – 25 – 7 – 175 < 0

( – 25) + 7( – 25) < 0 (–7, 25)

50. Answer (01)

Hint : Solve for z

Solution :

z4 + z3 + 2z

2 + z + 1 = 0

z4 + z2 + z3 + z + z2 + 1 = 0

z2(z2 + 1) + z(z2 + 1) + 1(z2 + 1) = 0

(z2 + 1)(z2 + z + 1) = 0

z2 + 1 = 0 OR z2 + z + 1 = 0

z = ± i OR z = 1 1 4 1 3

2 2

i

z1

= i, z2

= –i, z3

= 1 3

2

i , z

4

= 1 3

2

i

4

4 4 4 4

1 2 3 4

1 31 1 2Re

2

iz z z z

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠

1 31 1 2Re

2 2

i⎛ ⎞ ⎜ ⎟⎜ ⎟

⎝ ⎠

= 1 + 1 – 1 = 1

51. Answer (C)

Hint : Separate coefficients of a, b and c

Solution :

(P)

2

01 1

x xa b c

x x

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

1 1

xx

x

⇒ ⇒

(Q)

2

1 10

x xa b c

x x

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

1 1

1

xx

x

⇒ ⇒

(R)

2

01 1

x xa b c

x x

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

1 1

xx

x

⇒ ⇒

(S)

2

1 10

x xa b c

x x

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

1 1

1

xx

x

⇒ ⇒

52. Answer (B)

Hint : Use properties of functions to find domain

Solution :

(P) 1 – x2 0 and 10 – x2 0 1 x2 10

Integral values of x are ± 1, ± 2, ± 3

(Q) 4[x] – [x]2 – 1 0 [x]2 – 4[x] + 1 0

([x] – 2)2 3

3 [ ] 2 3 2 3 [ ] 3 2x x ⇒

Integral values of x are 1, 2, 3

(R) x3 + 1 > 0,

(x + 1)(x2 – x + 1) > 0

x > –1

x3 + 1 1,

x 0

2x – x2 + 9 > 1

x2 – 2x

– 8 < 0

(x – 1)2 < 9

–3 < x – 1 < 3 –2 < x < 4 and x > –1

–1 < x < 4 and x 0


(S) 34{f(x)} = 36 – 2x 1 36 – 2x < 34

∵ {f(x)} [0, 1)

2x 35 .....(i)

and

2x > 2 .....(ii)

Integral values of x satisfying both (i) & (ii) are

2, 3, 4, 5


12/12

53. Answer (D)

Hint : Find general solution

Solution :

(P) sin3x – sinx = 0

sinx (sin2x – 1) = –1

cos2x = cosec x

(Not possible)

(Q) cos3x – cosx +1 = 0

cosx (cos2x – 1) = –1

sin2x = sec x

(Not possible)

(R) tanx tan2x – 1 = 0

tan2x = cot x = tan2

x⎛ ⎞⎜ ⎟

⎝ ⎠

2x = 2

n x

(2 1)

,6

nx n I

5 7 3 11, , , , ,

6 2 6 6 2 6x

But 3

,2 2

x

(S) tan2x + sin2

x – sec2x – sinx – 1 = 0

sin2x – sin x – 2 = 0

sin x = – 1, sin x = 2

When sin x = –1, sec x is not defined.

54. Answer (C)

Hint : If cos(+) = 1 then sin(+) = 0

Solution :

z1z

2 = cos( + ) + i sin( + )

Re(z1z

2) = cos( + ) = 1

+ = 2 = 2 –

z1 =

cos – i sin , z

2 = cos + i sin

� � �

Test - 1A (Paper - 2) (Code-H) (Answers) All India Aakash Test Series for JEE (Advanced)-2020

1/12

PHYSICS

1. (A, B)

2. (A, D)

3. (A, B, D)

4. (A, B)

5. (A, B, D)

6. (A, B, C)

7. (03)

8. (15)

9. (12)

10. (02)

11. (15)

12. (12)

13. (16)

14. (18)

15. (B)

16. (B)

17. (A)

18. (A)

CHEMISTRY

19. (D)

20. (A)

21. (A, D)

22. (A, B)

23. (A, B, C, D)

24. (A, C, D)

25. (98)

26. (65)

27. (43)

28. (40)

29. (20)

30. (06)

31. (15)

32. (03)

33. (B)

34. (A)

35. (D)

36. (C)

MATHEMATICS

37. (B, C)

38. (B, C)

39. (A, D)

40. (A, B)

41. (B, C)

42. (C, D)

43. (01)

44. (31)

45. (11)

46. (11)

47. (00)

48. (01)

49. (02)

50. (81)

51. (C)

52. (D)

53. (B)

54. (C)

Test Date : 18/11/2018

ANSWERS

TEST - 1A (Paper-2) - Code-H

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

All India Aakash Test Series for JEE (Advanced)-2020 Test - 1A (Paper - 2) (Code-H) (Hints & Solutions)

2/12

PART - I (PHYSICS)

HINTS & SOLUTIONS

1. Answer (A, B)

Hint : Acceleration of block B with respect to A is

2

0v

�

Solution :

T – 0.05 (2 mg) = 2 ma0

T + ma0

=

2mv

�

22

T mgT mg

⎛ ⎞ ⎜ ⎟⎝ ⎠

T

ma0

5

3

mgT

03

ga

2. Answer (A, D)

Hint : Draw FBD and write equation of motion with

constrain relation

Solution :

2mg – T1 = 2ma

1

T1 – T

2 = 2ma

1

2

02

Tma

2a1 = a

0

1

4

ga

3. Answer (A, B, D)

Hint : Draw F.B.D. and equation of motion. Write

down the constraint motion.

Solution :

T cos 37° = ma1

3mg – T = 3ma2

1 2

4

5a a⎛ ⎞ ⎜ ⎟⎝ ⎠

1

5

4T ma

3mg – T = 1

12

5ma

3mg = 1

73

20ma

1

60

73

ga

4. Answer (A, B)

Hint : Velocity of cylinder along normal relative to

connected bodies is zero.

Solution :

3

ˆ ˆ

v xi yj �

x = +10

2 2

ˆ

v v i�

v2

v2

sin 37° = x sin 37° + y cos 37°

23 4

65 5

v y

3v2

= 30 + 4y

5. Answer (A, B, D)

Hint : Write down the expression for tangential

acceleration and acceleration in y-direction. Use the

fact that aj = a

t

Solution :

( sin )tdv

g gdt

( sin )y

dvg g

dt

v

d

ytdvdv

dt dt

vT – v

0 = –v

0sin v

T – v

0 =

03

5

v

vT =

02

5

v

6. Answer (A, B, C)

Hint : Write down the x coordinate of A3 in terms of

� and . dx

vdt

Solution :

x = 2(3l cos + 2l cos + l cos )

x = 12l cos

Test - 1A (Paper - 2) (Code-H) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020

3/12

v = 1

12 sinsin

d dl

dt dt

⎛ ⎞ ⇒ ⎜ ⎟ ⎝ ⎠ �

2 3B

dv l

dt

⎛ ⎞ ⎜ ⎟⎝ ⎠

6 m/s4sin

v

7. Answer (03)

Hint : Length of string remains constant.

Solution :

2 22( ) ( )L h y h x

0dL

dt

2 2

1

2

A

B

xvV

h x

0

3

1 4

52

4

v

h

0

3

10v

8. Answer (15)

Hint : Write down the dimension of each and

compare.

Solution :

= M1L–3

� = M0L1T0

V = M0L1T–1

d = M0L1T0

= M1L–1T–1

P = ML–1T–2

9. Answer (12)

Hint : Apply concept of similar triangle

Solution :

h Y

x

�

hY

x �

2

dY hv

dt x �

4hv

⎛ ⎞ ⎜ ⎟⎝ ⎠�

10. Answer (02)

Hint : Apply NSL to get acceleration of each and find

relative acceleration along y direction.

Solution :

2 mg sin37° – mg sin 53° = 3 ma

23

5

ga

2

15

ga

21 3 4

2 5 5a t h⎛ ⎞ ⎜ ⎟⎝ ⎠

210 7

15 5t h

⎛ ⎞ ⎜ ⎟⎝ ⎠

214

15

th

11. Answer (15)

Hint : Direction of friction will change, hence

magnitude of acceleration

Solution :

2

16 m/s

2

Mg Mga

M

0

1

1

vt

a

2

0

12

vs

a

2

2 0

2 2

1

1

2 2

va t

a

0

2

1 2

vt

a a

0

1 1 2

1 1vT

a a a

⎡ ⎤ ⎢ ⎥

⎢ ⎥⎣ ⎦

0

1 2

1 2

v

a a

a a


4/12

12. Answer (12)

Hint : Use addition of vector, net force will be along

PO.

Solution :

2

202 1

10k OP

��

= 40

k = 2 N/m

13. Answer (16)

Hint : Use the concept that

2

c

va

R

Solution :

216 m/s

dy

dt

2

2

216 m/s

x

d xa

dt

2

16

x

vR

a

14. Answer (18)

Hint : Along an incline maximum range is

2

(1 sin )

u

g

Solution :

2

10(1 sin )

u

g

2 4100 1

5u

⎛ ⎞ ⎜ ⎟⎝ ⎠

u2 = 20 × 9

15. Answer (B)

Hint : Relative motion will start when s

f f

Solution :

(f1)max

= × 1 × 10

= 10

Write down the equation of motion of each.

A Ba a also

maxsf f

16. Answer (B)

Hint : Write down vAB

which is constant

Solution :

At the time of minimum separation vAB

is

perpendicular to relative separation.

(4, 3)

37°53°

vA

8°

(3, –4)B

10 2 sin(8)l

0

10 2 cos(8)

5t

2 2 cos8

17. Answer (A)

Hint : Initial static friction will act and then kinetic

friction will act.

Solution :

fs = applied force

fk = N

N will decrease linearly with time.

18. Answer (A)

Hint : Write down the dimension and compare.

Solution :

dvF A

dx

1F AT

1

2

vA

A

1

2f A v

b c1 3 1 1 2 2 1 1

ML MLT L LT ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

a = 1

c = –2

b = –1


5/12

PART - II (CHEMISTRY)

19. Answer (D)

Hint : If labelling of oleum is (100 + x)% the free SO3 (in g)

is x 80

18

Solution : Initial moles of free SO3 present in oleum

= No. of moles H2O used

12 2moles

18 3

Moles of free SO3 combined with H

2O after 4.5 g of

H2O added

3SO

4.5 1n mole

18 4

Moles of free SO3 remains

2 1 5

3 4 12 moles.

Volume of free SO3 at STP

522.4 9.33 L

12

20. Answer (A)

Hint : 2

aP V b RT

V

⎛ ⎞ ⎜ ⎟⎝ ⎠

Solution : If a = 0 if b = 0

Z = Pb

1RT

Z = aP

1RT

21. Answer (A, D)

Hint : 2H2 + O

2

At 100ºC 2H2O

at 100ºC H2O exist as water.

Solution :

At 100ºC

2 2 2(g) (g) (g)

2H O 2H O

Before reaction 2 2 0

volume in litre

After reaction 0 1 2

volume in litre

(A) Mole of H2O formed

2PV 1 2n 8.9 10 mole

RT 0.0821 273

Mass of H2O formed = 8.9 × 10–2 × 18

= 160.2 × 10–2 g = 1.602 g

(B) O2 is left as unreacted = 1 litre

At STP mole of O2 left

= 21

4.4 10 mole22.4

Mass of O2 left = 4.4 × 10–2 × 32 = 1.408 g

(C) At 100ºC H2O also exists as vapours

Total mole present at 100ºC

2 2

H O On n left

2 2 28.9 10 4.4 10 13.3 10 moles

Volume of the vessel = 2 L

2nRT 13.3 10 0.0821 373

P 2.03 atmV 2

(D) Volume of O2 used for formation of H

2O = 1 L

2

2

O

1n 4.4 10 mole

22.4

22. Answer (A, B)

Hint : In NO, HOMO is 2 = 2p px y

* * while in N2 it is

2pz.

Solution : In NO, HOMO is 2 = 2p px y

* * while in N2

it is 2pz.

23. Answer (A, B, C, D)

Hint : N – N x

H

H

H

H

N – N y

F

F

F

F

y < x

as E.N atom has 'p' character so 's' character in

N–N bond and bond length decrease.

Solution : N2H

4 > N

2F

4 i.e.,

NF N

F

F

F

H

N

H

N H

H

<

1. It can be explained on the basis of Bent's rule. In

N2F

4, N – N bond has more s-character hence

bond strength increases and bond length

decreases.

While in N2H

4, N – N bond has less s-character

hence bond length increases.

2. As the positive charge increases (Zeff

) nuclear

charge increases.

3. As the negative charge increases, ionic radius

increases.


6/12

24. Answer (A, C, D)

Hint : In T.B.P. geometry, more electronegative atom

attached at axial position.

Solution :

P

B

B

A

A

A

On replacementof A by C

0 = 0

P

C

C

C

B

B

On replacementof A by C P

C

C

B

B

B

= 0

P

B

B

B

A

A 0

25. Answer (98)

Hint : Mass of solution = (v × d)solution

= 1000 × 1.26

Mass of H2SO

4 = 10.15 × 98

% H2SO

4 =

10.15 98 100

1000 1.26

Solution :

% by mass = 79

Molarity = 10.15

d = 1.26 gm/ml

M =

%by mass 10 d 79 10 1.2610.15

Molar mass Molar mass

⇒

Molar Mass = 98

26. Answer (65)

Hint : For van der Waal's gas

2

2

n aP (V nb) nRT

V

⎛ ⎞ ⎜ ⎟

⎝ ⎠

Solution :

Given n = 4 moles, V = 1 L, T = 37 + 273 = 310 K

a = 3.592 : b = 0.0427

using van der Waal's equation for 4 moles

2

2

n aP (V nb) nRT

V

⎛ ⎞ ⎜ ⎟

⎝ ⎠

24 3.592P (1 4 0.0427) 4 0.0821 310

1

⎛ ⎞ ⎜ ⎟⎝ ⎠

P = 65.3 65 atm

27. Answer (43)

Hint :

Rate of diffusion

1

Molar mass

Solution : Now for diffusion of gaseous mixture and

pure O2.

2

2

O mix

mix O

r M

r M

(or) 2

2 2

O M Mix

M O O

V T M

V T M

mixM1 232

1 200 32

Mmix

= 43

28. Answer (40)

Hint : PM = dRT

Molar mass of mixture = M

Solution :

dRT 600 1 0.08 300P

M 760 M

⇒

M = 30.4 gm/mole

mol of N2 = x, mole of O

2 = 1 – x

30.4 = 32 (1 – x) + 28 (x)

x = 0.4

29. Answer (20)

Hint & Solution :

10.50 0.39Al 0.39 1

27 0.39 ⇒


7/12

15.1 0.39K 0.39 1

39 0.39 ⇒

24.96 0.78S 0.78 2

32 0.39 ⇒

49.92 3.12O 3.12 8

16 0.39 ⇒

The empirical formula of the salt is KAlS2O

8

Molecular formula of hydrated salt is KAlS2O

8 yH

2O

When

18 y100 45.6

39 27 128 64 18y

y = 12

So the value of x + y = 8 + 12 = 20

30. Answer (06)

Hint : Number of spectral line = n (n 1)

2

Solution :

2

n 2

13.6 ZE eV

n

Also E = En – E

0

(E = Ionisation potential)

Energy absorb by 1 atom 51

12.754

2

13.612.75 ( 13.6)

n

2

13.612.75 13.6

n

n2 = 16 n = 4

Thus, the number of spectral lines emitted during the

de-excitation will be n(n 1) 4(4 1)

62 2

31. Answer (15)

Hint : For photoelectric effect.

Ein = E

Thr + K.E.

Solution :

According to photoelectric effect.

E = E0 + K.E.

h = h0 + K.E.

h1

= K.E1

+ h

0.....(i)

h2

= K.E2

+ h

0.....(ii)

2h1

– 2h2 = h

0(∵ 2 K.E

1 = K.E

2)

(21

– 2) =

0

0

= 1 2 1 2

2c c 2 1c

⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦

0

=

8

10

3 10 2 1

2000 180010

⎡ ⎤⎢ ⎥⎣ ⎦

(∵ 1 Å = 10–10 m).

0

= 1.33 × 1015 s–1.

32. Answer (03)

Hint : If labelling of H2SO

4 is (100 + x)% then mole

of SO3

x

18

mol of H2SO

4 =

x100 80

18

98

⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

Solution : 109% labelled oleum means it has 9 g

H2O which reacts with free SO

3 to give H

2SO

4 i.e.

SO3 + H

2O H

2SO

4.

Mass of SO3 which will react with 8 g

H2O

80 940 g

18

Let 100 g sample of 109% labelled oleum contains

40 g SO3 and 60 + 38 = 98 g of H

2SO

4.

Moles of H2SO

4 = x = 1.0 mole

Moles of SO3 = y = 0.5 mole

x + y = 1.5 moles

x – y = 0.5 mole

x y

x y

= 03

33. Answer (B)

Hint : Radius =

2

0

na

Z

⎛ ⎞⎜ ⎟⎝ ⎠

Separation energy =

2

0 2

ZE

n

Solution : Radius of 4th orbit of Be+3

=

2

0 0

4a 4a

4


8/12

PART - III (MATHEMATICS)

37. Answer (B, C)

Hint : Draw graph

Solution :

–2 –1 O 1 2 3 4

y = 2

y

f x = |x – 2x |( ) – 12

2g x( ) 2 sin x

⎛ ⎞ ⎜ ⎟

⎝ ⎠

x

2 2 + 1– 2 + 1

38. Answer (B, C)

Hint : If coefficients are real, then complex roots

exists in pair which are conjugate to each other.

Solution :

x2 + bx + c = 0

roots are 1 + i and 1 – i

b = –2, c = 2

Separation energy of electron in 2nd orbit of Li2+ ion

=

2

0 02

3 9E E

42

34. Answer (A)

Hint & solution :

BF3 Non-polar, planar

N3H Polar, planar

NH3 Polar, non-planar

35. Answer (D)

Hint & solution :

N

SiH3

SiH3

H Si3

sp2

sp3

2NO sp

, sp2

3NO

sp sp3 3

3 3N(CH ) ,

sp d sp3 3

6 4PCl , PCl

36. Answer (C)

Hint : Covalent character decrease as size of cation

increase.

Solution : Number of d – p = P4O

10 > SO

3 > SO

2

4 2 1

Number of unpaired electron : 1 2

2 2 2O O O

Covalent character : 2 2 2

MgCl CaCl SrCl

Six quadratic equations are possible

2– 2 2 0x x ...... (1)

22 – 2 0x x ...... (2)

22 – 2 0x x .......(3)

22 – 2 1 0x x .......(4)

2–2 2 0x x .......(5)

2–2 2 1 0x x .......(6)

(1) & (4) have imaginary roots. (2), (3), (5), (6) have

irrational roots.

39. Answer (A, D)

Hint : Put z = x + iy

Solution :

Put z = x + iy; x, y R in 2z z

z

z z

2 2

2 2

2 2

22 2( )

x yz zz x iy

zz x y

⇒

z1 = 1

Put z = x + iy in 1

Re( ) 2z z z i

z z

1( ) ( ) 2

( ) ( )x x iy x iy i

x iy x iy


9/12

x = 0, 2 22

iiy i

y 4y2 – 4y + 1 = 0

1

2y

40. Answer (A, B)

Hint : 2

sinlog (cos ) 0, 1

xx

Solution :

y

5 –1

2y

2x

O

y x= sin

2

For intersection points of f(x) and g(x),

logsinx

(cos2x) = sgn(tan x)

If tan x = 0, cos2x = 1

sinx = 0, no solution

if tanx > 0, cos2x = sinx sin2x + sinx – 1 = 0

sinx = 1 1 4

2

sinx =

1 5

2

sinx = 5 1

2

Only one solution

If tanx < 0, cos2 x = cosec x, (no solution)

41. Answer (B, C)

Hint : Write cos7 in terms of cos

Solution :

∵ cos7 = 64cos7 – 112cos5 + 56 cos3 – 7cos

cos (7) = cos (2)

7 = 2n ± 2

Taking + sign

= 2

5

n

= 2 4

0, , ,...5 5

for +ve values

Taking – sign

= 2

9

n

= 2 4

0 , , ,...9 9

for +ve values

42. Answer (C, D)

Hint : 8pq = P

Solution :

2 4 68 sin sin sin

7 7 7pq

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 3sin sin sin

7 7 7p

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

1

8q

Let 2 3

cos , cos , cos7 7 7

a b c ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠,

1

8abc

2 2 4 68 1 cos 1 cos 1 cos

7 7 7p

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

= 2 3

1 cos 1 cos 1 cos7 7 7

⎛ ⎞⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

= (1 – b) (1 + c) (1 + a)

= 1 + a – b + c + ac – ab – bc – abc

ac – ab – bc = 1 3 2

2cos cos 2cos cos2 7 7 7 7

⎡ ⎢⎣

2 32cos cos

7 7

⎤ ⎥⎦

= 1 4 2 3

cos cos cos cos2 7 7 7 7

⎡ ⎢⎣

5cos cos

7 7

⎤ ⎥⎦

=

2 3cos cos cos

7 7 7b a c

a – b + c + ac – ab – bc = 0

8p2

=

1 7 71

8 8 8p ⇒

43. Answer (01)

Hint : Solve for z

Solution :

z4 + z3 + 2z2 + z + 1 = 0


10/12

z4 + z2 + z3 + z + z2 + 1 = 0

z2(z2 + 1) + z(z2 + 1) + 1(z2 + 1) = 0

(z2 + 1)(z2 + z + 1) = 0

z2 + 1 = 0 OR z2 + z + 1 = 0

z = ± i OR z = 1 1 4 1 3

2 2

i

z1

= i, z2

= –i, z3

= 1 3

2

i , z

4

= 1 3

2

i

4

4 4 4 4

1 2 3 4

1 31 1 2Re

2

iz z z z

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠

1 31 1 2Re

2 2

i⎛ ⎞ ⎜ ⎟⎜ ⎟

⎝ ⎠

= 1 + 1 – 1 = 1

44. Answer (31)

Hint : If z = x + iy has negative principal argument

then x > 0, y < 0 or x < 0, y < 0

Solution :

z = (3 – 42 + 2) + i (2 – 18 – 175)

z has negative principal argument

3 – 42 + 2 < 0 and 2 – 18 – 175 < 0

OR 3 – 42 + 2 > 0 and 2 – 18 – 175 < 0

2 – 18 – 175 < 0

2 – 25 – 7 – 175 < 0

( – 25) + 7( – 25) < 0 (–7, 25)

45. Answer (11)

Hint : Solve for sin x

Solution :

4(1– sin2x)sinx – 2sin2x = 3sinx

4sinx – 4sin3x – 2sin2x = 3sinx

sinx = 0 OR 4sin2x + 2sinx – 1 = 0

sinx = 2 4 16 1 5

8 4

y

5 –1

4y

2x

O–2

5 –1

4y –

46. Answer (11)

Hint : ( ) 4sin 3cos6 6

f x x x ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Solution :

cos sin3 6

x x ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠∵ and

sin cos3 6

x x ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

f(x) = 4sin 3cos6 6

x x ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

–5 f(x) 5

47. Answer (00)

Hint : cos 0, 12x

⎛ ⎞ ⎜ ⎟⎝ ⎠

Solution :

When 1, 3, 5, ..... cos 02

x x⎛ ⎞ ⎜ ⎟

⎝ ⎠

When 0, 2, 4, ..... cos 12

x x⎛ ⎞ ⎜ ⎟

⎝ ⎠

there is no integral value of x belonging to the

domain of f(x)

48. Answer (01)

Hint : Draw graphs

Solution :

y

(0, 1)

(1, 0)

(0, –1)

(–1, 0) Ox

A B = P B (A ) = { } ( (A )) = 1n P B

(1, 1)(–1, 1)

x

(–1, –1)

y

(1, –1)

49. Answer (02)

Hint : Case 1 : Both roots are negative

Case 2 : Exactly one root is negative

Solution :

D = (– 3)2 – 4= 2 – 9 – 6– 4= (– 5)2 – 16

= (– 9) (– 1)

D 0, (–, 1] [9, )


11/12

Case-1: If both roots are negative

D 0, –3 < > 0 (0, 1]

Case-2: If exactly one root is negative and other root

+ve or zero.

D 0, 0 (–, 0]

(–, 0] (0, 1]

(–, 1]

50. Answer (81)

Hint : Make perfect square

Solution :

2 3sin 2cos

2a a

= 2 23 1

1 cos 2cos (cos 1)2 2

a a a

cos2x + 4sinx + 12 = 1 – sin2x + 4 sinx + 12

= 17 – (sinx – 2)2

f(x) is maximum when cos a = –1 and sin x = –1

f(x)max

=

1ln

12–

ln8 3

= 1

–3

–4 = 81

51. Answer (C)

Hint : If cos(+) = 1 then sin(+) = 0

Solution :

z1z

2 = cos( + ) + i sin( + )

Re(z1z

2) = cos( + ) = 1

+ = 2 = 2 – z

1 =

cos – i sin , z

2 = cos + i sin

52. Answer (D)

Hint : Find general solution

Solution :

(P) sin3x – sinx = 0

sinx (sin2x – 1) = –1

cos2x = cosec x

(Not possible)

(Q) cos3x – cosx +1 = 0

cosx (cos2x – 1) = –1

sin2x = sec x

(Not possible)

(R) tanx tan2x – 1 = 0

tan2x = cot x = tan2

x⎛ ⎞⎜ ⎟

⎝ ⎠

2x = 2

n x

(2 1)

,6

nx n I

5 7 3 11, , , , ,

6 2 6 6 2 6x

But 3

,2 2

x

(S) tan2x + sin2x – sec2x – sinx – 1 = 0

sin2x – sin x – 2 = 0

sin x = – 1, sin x = 2

When sin x = –1, sec x is not defined.

53. Answer (B)

Hint : Use properties of functions to find domain

Solution :

(P) 1 – x2 0 and 10 – x2 0 1 x2 10

Integral values of x are ± 1, ± 2, ± 3

(Q) 4[x] – [x]2 – 1 0 [x]2 – 4[x] + 1 0

([x] – 2)2 3

3 [ ] 2 3 2 3 [ ] 3 2x x ⇒


(R) x3 + 1 > 0,

(x + 1)(x2 – x + 1) > 0

x > –1

x3 + 1 1,

x 0

2x – x2 + 9 > 1

x2 – 2x – 8 < 0

(x – 1)2 < 9

–3 < x – 1 < 3 –2 < x < 4 and x > –1

–1 < x < 4 and x 0



12/12

� � �

(S) 34{f(x)} = 36 – 2x 1 36 – 2x < 34

∵ {f(x)} [0, 1)

2x 35 .....(i)

and

2x > 2 .....(ii)

Integral values of x satisfying both (i) & (ii) are

2, 3, 4, 5

54. Answer (C)

Hint : Separate coefficients of a, b and c

Solution :

(P)

2

01 1

x xa b c

x x

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

1 1

xx

x

⇒ ⇒

(Q)

2

1 10

x xa b c

x x

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

1 1

1

xx

x

⇒ ⇒

(R)

2

01 1

x xa b c

x x

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

1 1

xx

x

⇒ ⇒

(S)

2

1 10

x xa b c

x x

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

1 1

1

xx

x

⇒ ⇒

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