all india aakash test series for jee (advanced)-2021 test ......2020/06/21 · defects (b) unit...

All India Aakash Test Series for JEE (Advanced)-2021

Test Date : 21/06/2020

ANSWERS

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

Test -1A (Paper-2) (Code-C)_(Answers) All India Aakash Test Series for JEE (Advanced)-2021

1/13

PHYSICS CHEMISTRY MATHEMATICS

1. (A, B, C)

2. (A, C)

3. (A, C)

4. (A, C)

5. (C, D)

6. (B)

7. (A)

8. (C)

9. (C)

10. (B)

11. (A)

12. (B)

13. (D)

14. (C)

15. (D)

16. A → (Q, R, T)

B → (P, S)

C → (Q, S)

D → (Q, R)

17. A → (R)

B → (T)

C → (T)

D → (S)

18. (03)

19. (02)

20. (03)

21. (A, C, D)

22. (A, C, D)

23. (A, B, D)

24. (B)

25. (A, C)

26. (A)

27. (A)

28. (C)

29. (B)

30. (B)

31. (A)

32. (A)

33. (B)

34. (B)

35. (C)

36. A → (P, S)

B → (P, T)

C → (Q, R)

D → (P, S, T)

37. A → (P, R, T)

B → (Q, R, S)

C → (P)

D → (Q)

38. (08)

39. (04)

40. (03)

41. (B, C, D)

42. (A, B, D)

43. (A, B, C)

44. (B, C)

45. (A, B, D)

46. (A)

47. (D)

48. (C)

49. (B)

50. (D)

51. (C)

52. (D)

53. (A)

54. (C)

55. (B)

56. A → (P, S)

B → (Q, R, T)

C → (R, T)

D → (T)

57. A → (P, R, S)

B → (R, S)

C → (T)

D → (T)

58. (02)

59. (04)

60. (06)


All India Aakash Test Series for JEE (Advanced)-2021 Test -1A (Paper-2) (Code-C)_(Hints & Solutions)

2/13

HINTS & SOLUTIONS PART - I (PHYSICS)

1. Answer (A, B, C)

Hint :

Let angle subtended by equal to Q, at Point P

q1 = r1Qλ

q2 = r2Qλ

1 1

2 2

=q r

q r

Solution :

V due to the elements are 1 2

2 2

andkq kq

r r

Since 1 2

1 2

=q q

r r, hence potentials are equal

⇒ Electric feed at point P has magnitude

1 21 22 2

1 2

and= =kq kq

E Er r

Hence,

1

1 11 2

22 1

2 2

1

1

= ⇒

q

r rE r

qE r

r r

2. Answer (A, C)

Hint :

Now electric field at any distance x from centre

on axis of hole be

20 0 2

– 1–2 2σ σ = ε ε +

xE

x R

2 2

02σ

=ε +

x

R x

Solution :

2 20

–

2

σ=

ε +

e xF

x R

2 20

–

2

σ=

ε +

dv e xmv

dx x R

0

2 20 3 0

–

2

σ=

ε +∫ ∫v

R

e xdxmvdv

x R

2 02 2

30

–2 2

σ = + ε R

mv ex R

[ ]0 0

–– 2

2 2σ σ

= =ε εe eR

R R

⇒ 0

σ=

εe R

vm

3. Answer (A, C)

Hint :

On closing the switches and earthing, charges on left-most and right-most surfaces will become = 0

Solution :

E = 0 inside the conductor

Since A and B connected, hence

VA = VC ( connected)

⇒ VA – VB = VC – VB

E1d = E2d

⇒ E1 = E2

⇒ 0 0

=ε εx y

A A

⇒ x = y …(i)

Also VB = VD (both earthed)

VC – VD = VC – VB

E3d = E2d

⇒ y = z …(ii)

By charge conservation,

x + y + z = Q0

⇒ 3x = Q0


Test -1A (Paper-2) (Code-C)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2021

3/13

0

3= = =

Qx y z

Hence charge flown through

( ) 0 01 0

2 5

2 3= + + = + =

Q QS x y Q

Magnitude of charge flown through

0 02 0

72

3 3= + =

Q QS Q

4. Answer (A, C)

Hint :

Let speeds of A and B just before collision

By conservation of linear momentum

mVA = 2mVB

⇒ VA = 2VB …(i)

Solution :

Hence

2

2

122

1 122

= =AA

BB

mVK

K mV

By conservation of total mechanical energy

Ui + Ki = Uf + Kf

⇒ 2 2– –

010 2

+ = + +A BKQ KQ

K KR R

⇒ 22

5+ =A B

KQK K

R

5. Answer (C, D)

Hint :

Conductors are equipotential, hence VA = VB…

Solution :

in

0 0

lim→∞

⋅ = =ε ε∫

x

q qE ds

1

∝ σ ∝ER

Hence E will not be equal

6. Answer (B)

Hint :

( )3

2 2 204

=

πε +

QxE

R x

Solution :

The net force on the dipole isrE

Prx

( ) ( )

( )

3 12 2 2 2 22 2

022 2

32 – 2

4 2

+ +

πε =+

Qqa R x R x x

R x

( ) [ ]

[ ]

12 2 2 22

32 20

– 22

4+

=πε +

Q R x R xqa

R x

[ ]

[ ]

2 2

50 2 2 2

– 22

4=

πε+

Q R xF qa

R x

7. Answer (A)

Hint :

W = ∆u

Ui = – PEcos0

Uf = – PEcosπ

Solution :

( )

( ) ( )3 1

2 2 2 22 20 0

2. 22

4

= = =

πε + πε +

qa Qx aqQxW PE

R x R x

8. Answer (C)

Hint :

2= πI

TPE

Solution :

( )

2

32 2 2

0

22

2

4

= π πε +

mT

Qxqa

R x

( )

32 2 2

042

πε += π

R x maT

qQx

9. Answer (C)

Hint :

Find arrangement of charge before and after closing switch

Solution :

Before



4/13

Before

Considering isolated plates

Hence charge flown =5

CE

10. Answer (B)

Hint :

WDcell = qsupply× emf of cell

Solution :

WDcell = qsupply × emf of cell

12 2 12

–5 3 7

= + CE CE CE

2105

=CE

2

cell2105

=CE

WD

11. Answer (A)

Hint :

H = WDcell – ∆u

Solution :

H = WDcell – ∆u

2 2

22 1 12 1 50– –

105 2 5 2 21

= × ×

CE CE

CE

2 2 22 6 25

–105 5 21

= +CE CE CE

2

22 – 126 125105 105

+= =

CECE

12. Answer (B)

Hint :

Apply superposition principle.

Solution :

By superposition principle the given system is

equivalent to

13. Answer (D)

Hint :

( ) ( )( )

( )

0 0

,

0 0,

, – , –= ⋅∫

x y

x y

V x y V x y E dr

Solution :

( ) ( )( )

( )

0 0

,

0 0,

, – , –= ⋅∫

x y

x y

V x y V x y E dr

Let (x0, y0) be (Q, 0) and V(x0, y0) = 0

⇒ ( ) ( ),

0,0

, –= +∫x y

V x y ydx xdy

( ) ( ),

0,0

, – –= =∫x y

V x y d xy yx

Hence equation of equipotential surface is like

xy = constant which represent a hyperbola.

14. Answer (C)

Hint :

Apply Gauss’s Law.

Solution :

As position of charge q is changed, electric field

and charge distribution on inner walls will

change. Electric field outside depends upon the

rotate distance from centre of sphere for any

point that lies outside the sphere.

15. Answer (D)

Hint :

Electrostatic flied are conservative, hence WD

in closed path = 0 and also WD is path

independent



5/13

Solution :

Electrostatic flied are conservative, hence WD in

closed path = 0 and also WD is path

independent

∴ 0⋅ =∫

E dl or2 2 2

1 1 1

, ,

, ,

0⋅ =∫

x y z

x y z

E dr

for (x1, y1, z1) = (x2, y2, z2)

( )1 2 3 0⋅ = + + ≠∫ ∫

E dl xydx yzdy xzdz

( )2 22

ˆ2 2⋅ = + + +∫ ∫

E dl y dx xy z dy yzk

( ) ( )2 2= +∫ ∫d xy d z y

2 2 2

1 1 1

, ,2 2, , 0 = + =

x y z

x y zxy z y

Hence only 2

E is conservative hence

electrostatic field

16. Answer A(Q, R, T); B(P, S); C(Q, S); D(Q, R)

Hint :

Use Field formula.

Solution :

(A)

(B)

(C)

(D)

17. Answer A(R); B(T); C(T); D(S)

Hint :

Apply KVL.

Solution :

By KVL in loop-1

( )– 2 – –

– 01 3 2

+ =x x Q Q x

⇒ –6x – 4x + 2Q + 3Q – 3x = 0

⇒ –5Q = 13x …(i)

By KVL in loop-2

( )–

10 – – 02 1

=Q x x

20 – Q + x – 2x = 0

20 – Q = x …(ii)

From equation (i) and (ii)

50

20 –13

=Q

18

2013

=Q

260 13018 9

= =Q

⇒ 5 130 50

13 9 9= × =x

Now –2

=B CQx

V V

130 50– 409 9– 02 9

= =BV

409

=BV

50

–1 9

= =D Cx

V V

509

=DV

Charge on

100 130 30 103 2 – – – –

9 9 9 3µ = = = =F x Q



6/13

18. Answer (03)

Hint :

Use COE

Solution :

By energy conservation

( )

22 2

30 0

13 – 0

2 4 48

+ = +

πε + πε

Qq Qq Rkx R

R x R

( )

2

0 0

1 112 4 32

+ =πε + πε

Qq Qqkx

R x R

( )

2

0 0

1 11– , since

2 32 4= >>

πε πε +Qq Qq

kx R xR R x

2

0 0 0

1 11 3–

2 32 4 32= =

πε πε πεQq Qq Qq

kxR R R

0

316

=πεQq

xR

Comparing R = 3

19. Answer (02)

Hint :

Now since electric field between the plates is

uniform and equal to 0

σε

, hence motion of

charged particle is SHM.

Solution :

22

= πm

Tk

( springs are in parallel)

The point of released of charged particle will be extreme position, since its velocity = 0 at this position mean position is where Fnet = 0

⇒ 0

2σ

=εq

kA

02

σ=

εq

Ak

, hence b = 2

20. Answer (03)

Hint :

2 2

,= =+

o Pkq kq

V Vr a r

Solution :

=okq

Vr

2 2

=+

Pkq

Va r

2 2

10+

= =o

P

V a rV r

⇒ 2 2

210

+=

a r

r

a2 = 90

2

3=ar

PART - II (CHEMISTRY)

21. Answer (A, C, D)

Hint :

Unit cell characteristics

(A) AgBr shows both schottky and frenkel

defects

Solution :

(B) Unit cell having crystal parameter a ≠ b ≠ c,

α=γ = 90° and β≠ 90° is monoclinic

(C) In Al2O3, O2–

ion form hexagonal unit cell

and Al3+

ion occupy 2/3rd of octahedral voids.

So each Al3+

ion is surrounded by 6 O2–

ion,

each O2–

ions is surrounded by 4 Al3+

ions

(D) In ZnS structure the Zn2+

ions occupy

alternate tetrahedral voids. The shortest

distance between two Zn2+

ions is given by

2 2a a ad

4 4 2= + =

22. Answer (A, C, D)

Hint :

The nearest and next nearest neighbours of a

lattice point in rock salt structure.

Solution :

This distance of 2nd co-ordination B–

is a

2other

statements are fact based.

23. Answer (A, B, D)

Hint :

BCC unit cell with the lattice point at the centre

of cube missing.



7/13

Solution :

If body centre atom is removed, effective no. of

atom remain one (i.e. half of original)

So density become half

PE become half

24. Answer (B)

Hint :

Factors on which V. P of a liquid does not

depend.

Solution :

Surface area doesn’t effect V.P

25. Answer (A, C)

Hint :

Clausius clapeyron equation.

Solution :

2

1 1 2

P H 1 1log –

P 2.303R T T∆ =

3

i

760 4.606 10 1 1log –

7.6 2.303 2 T 300× = ×

3

1

1 1log100 10 –

T 300 =

1

2 1 11000 300 T

+ =

13000

T 187.5 K16

= =

= –85.5°C

26. Answer (A)

Hint :

Simple cubic unit cell.

Solution :

It is simple cubic packing

PE = 52.4%

27. Answer (A)

Hint :

Density = ( )3

A

1 Mgm/cc

N a

×

Solution :

d = 2.08 gm/cc

28. Answer (C)

Hint :

Nearest neighbours in simple cubic unit cell.

Solution :

Co-ordination

No. = 6

29. Answer (B)

Hint :

Lowering in V.P. = P°– P

Solution :

( )A

A

0.01 0.2 1i

0.01 0.2

× + α=

×= 1 + 0.08

iA = 1.08

BBi 1–

2

α=

A A

A A A

B BB B

B

P – P

P i n

i nP – P

P

°

°

°

°

=

B A A

B BA

P i n

i nP

°

°=

B

30 1.08 0.2100

1– 0.82

×=

α ×

αB = 0.2 ⇒ 20%



8/13

30. Answer (B)

Hint :

Liquid B has more relative lowering of V.P than

liquid A.

Solution :

B has more relative lowering in V.P

31. Answer (A)

Hint :

Liquid A is denser than liquid B.

Solution :

Liquid A comes 1st as it has high density

32. Answer (A)

Hint :

Two dimensional solid having a lattice point at

the centre of a square unit cell.

Solution :

2a 4r=

4

a r2

=

2 216a r

2=

= 8r2

2 2

2 2

2 r 2 rPE

4a 8r

π π π= = =

% PE = 78.5%

33. Answer (B)

Hint :

NaCl lattice with certain lattice points missing

which will change its formula.

Solution :

Removal of all ions present at one C3-axis

1

Na 124

+ = ×

34. Answer (B)

Hint :

Micro-organism do not grow in a pickle due to dehydration.

Solution :

Due to dehydration of microorganism exosmosis.

35. Answer (C)

Hint :

Osmotic pressure of a solution at a certain

temperature ∝ ic.

Solution :

1 2

1 1 2 2i C i C

π π=

2

2.5 4.52 0.1 i 0.2

=× ×

24.5

i 1.82.5

= =

–3 3CH COOH CH COO H++

i = 1 – α + 2α

i = 1 + α

1.8 = 1 + α

α= 0.8

[H+] = Cα

= 0.2 × 0.8

= 0.16

[H+] = 16 × 10

–2

pH = – log16 + 2

= –4 × 0.3 + 2

= –1.2 + 2

= 0.8

36. Answer A(P, S); B(P, T); C(Q, R); D(P, S, T)

Hint :

In ionic compounds, generally anions form

space lattice and cation occupy voids except

CaF2



9/13

Solution :

(A) P, S (B) P, T (C) Q, R (D) P, S, T

37. Answer A(P, R, T); B(Q, R, S); C(P); D(Q)

Hint :

Concentration terms not involving volume are independent of temperature.

Solution :

(A) P, R, T (B) Q, R, S (C) P (D) Q

38. Answer (08)

Hint :

Boron exists as B12 icosahedron molecular form.

Solution :

B12 has 12 corners and 20 triangular faces

∴ X = 20 and Y = 12

X – Y = 08

39. Answer (04)

Hint :

vant Hoff factor takes into account any dissociation or association of solute particles.

Solution :

60400

3 2–2 3

0 020

A B 2A 3B+ +

( )200 54

2 3 2 3 420 0

4 A B A B

=

40 60 5

a20 20

+ +=

+

= 2.625

a + 1.375 = 2.625 + 1.375 = 4

40. Answer (03)

Hint :

∆Tf = ikfm

Solution :

CoCl3.yNH3

∆Tf = ikfm

0.0558 = i × 1.86 × 0.01

i = 3

PART - III (MATHEMATICS) 41. Answer (B, C, D)

Hint :

Draw the graphs of different functions.

Solution :

(A) Take –1 –11 1sin tan , – ,

2 2x xπ π = = θ θ ∈

⇒ 1

sin tan 0x

θ = = θ ≠

⇒ cos 1 – ,2 2 2π π π θ = ⇒ θ = ∈

∴ no solution

(B) Given ( )–1 –11 1 1sin cos ; 0, 1

x x x= ∈

–1 –11 1sin sin

2x xπ

+ =

⇒ –1 –11 12sin sin

2 4x xπ π

= ⇒ =

⇒ 2x = only solution

(C) –1 –11 1 1sin cosec –1or 1

x x x= ⇒ =

x = –1

or 1 ⇒ 2 solutions

(D) Take –1 –11 1sin cot ; 0,

2x xπ = = θ θ ∈

⇒ [ )1sin cot 0; 1,x

xθ = θ = > ∈ ∞

⇒ cos2θ + cos θ –1 = 0

⇒ 5 – 1 –1– 5

cos cos –12 2

θ = θ ≠ <

⇒ 2

5 – 1 5 – 1sin 1–

2 2

θ = =

⇒ 2

5 – 1x = is only solution


Hint :

Domain of two functions in addition form is

intersection of domains of given two functions.



10/13

Solution :

( ) –12sin –2

f x xπ

=

–1 –1 1: 2sin – 0 sin 1

2 4 2fD x x xπ π

≥ ⇒ ≥ ⇒ ≤ ≤

( ) –12 tan –2

g x xπ

=

–1: tan 14gD x xπ

≥ ⇒ ≤ ≤ ∞

( ) –12sec –2

h x xπ

=

–1: sec4hD xπ

≥ ( ] )– , – 1 2,x ⇒ ∈ ∞ ∪ ∞

A. {1}f gD + =

B. )2,g hD += ∞

C. n gD + = φ

D. f g hD + + = φ

43. Answer (A, B, C)

Hint :

Use replacement property

Solution :

Putting x = 25 and y = 2

(f(50))2 = 25.(f(2)

2

= 25.6

2

∴ f(50) = 30 or –30

and by putting 1

xy

= we get

( ) ( )2 211f f y

y=

also f(2) = 6 so by putting y = 2

we have ( ) ( )2 21 11 2 36 18

2 2f f= = × =

( ) 31 18 2f = =

∴ ( ) 3 2f x x=

44. Answer (B, C)

Hint :

2

–1 –1 –13 – 3 1cos cos – cos

2 2 2x x

x + =

.

When12

x<

Solution :

Put x = cosθ

( ) ( )–1 –1cos cos cos cos –3

f x π = θ + θ

, 0

3 3

2 – ,3 3

π π < θ <= π π θ < θ ≤ π


Hint :

Odd power of odd function is odd and even

power of odd function is even.

Solution :

( ) log cot –4 2

n

f x π π =

46. Answer (A)

Hint :

Use replacement property two time to get

function

47. Answer (D)

Hint :

( ) 1 1 – 1–

2 1–x

f x xx x

= +

48. Answer (C)

Hint :

Put 12

x = in used functions

Solution for Q. No. 46, 47 & 48

Given that ( ) – 1xf x f x

x + =

…(i)

For all x≠ 0, 1

Replacing x with ( )– 1x

x both sides, we get that

( )

– 1– 1

– 1 – 1– 1

+ =

xx xxf f

xx xx

That is – 1 1 – 1

1–x x

f fx x x

+ =

…(ii)

Again replacing x with ( )– 1x

xin this, we get



11/13

( )1 11– 1–

f f xx x

+ =

…(iii)

Then by taking equation (i) + equation (iii) – equation (ii), we get that

( ) 1 – 12 –

1–x

f x xx x

= +

or ( ) 1 1 – 1–

2 1–x

f x xx x

= +

Substituting the values x = –1 and 12

in above

equation we get the solution for (ii) and (iii).

49. Answer (B)

Hint :

a3 + b

3 = (a + b)

3 – 3ab(a + b)

Solution :

( ) ( )3–1 –1 –1 –1 –1 –1tan cot – 3 tan cot tan cot= + +A x x x x x x

3

–1 –13– tan – tan

3 2 2x x

π π π =

23

–13tan –

3 2 4x

π π π = +

3 3

3 8A

π π≤ <

50. Answer (D)

Hint :

a2 + b

2 = (a + b)

2 – 2ab

Solution :

( )2–1 –1 –1 –1sin cos – 2sin – sin2

B t t t tπ = +

22

–12 sin –8 4

tπ π = +

∴ 2 2 2

max 28 16 4

Bπ π π

= + =

51. Answer (C)

Hint :

cot–1

(cotx) = x if x∈ (0, π)

Solution :

Here 4

λ π=

µ

∴ –1 –1– 3cot cot cot cot –

4 4 λ µπ π π = = π

52. Answer (D)

Hint :

log log logb c ca b a⋅ =

Solution :

Given ( )( )

2 2 32

log log log 11 11–3

f xf x

fx

= ⋅ +

⇒ ( )( )3

2 2log log log 11

f xf x

fx

= ⋅ +

⇒ ( )( )

( ) ( )1 11

1f x

f x f x f f f xx xf

x

= + ⇒ ⋅ = +

⇒ f(x) = 1 ± x4

∴ f(5) =1 ± 54 = 126

∴ f(x) = 1 + x3

∴ f(10) = 1001

53. Answer (A)

Hint :

–1 –1

2sin tan

1–

xx

x=

Solution :

( )

–1 –1– – 1 – – 1sin tan

1 – 11

r r r r

r rr r

= ++

–1 –1tan – tan – 1r r=

∴

( )

( )–1 –1 –1

1 1

– – 1sin tan – tan – 1

1

n n

r r

r rr r

r r= =

= +

∑ ∑

–1tan n=

54. Answer (C)

Hint :

Cubic function may have their maximum and

minimum values

Solution :

Let ( ) ( )3 – 3 6

, 0 12

x xg x g x x

+ ′= = ⇒ = ±



12/13

and g(1) = 2 and g(–1) = 4

∴ y = 2, y = 4 touch y = g(x) at x = 1 and

x = –1 respectively

and g(x) = 2 ⇒ x3 – 3x + 2 = 0

⇒ (x – 1)2 (x + 2) = 0

and g(x) = 4 ⇒ x3 – 3x – 2 = 0

⇒ (x + 1)2 (x – 2) = 0

and g(x) = 2 ⇒ x = 1, 1, –2

and g(x) = 4 ⇒ x = –1, –1, 2

Now due to (x – α) a void, will be

created in y = g(x), for the given

Condition to be true that void can be created

( ) ( )– , – 2 2,x∀ ∈ ∞ ∪ ∞

∴ α∈ [–2, 2]

55. Answer (B)

Hint :

Let 2

– 1

– 1

xy

p x=

+ then find range of f(x)

Solution :

( )22

– 1– 1 – 1 0

– 1

xy x y x y p

p x= ⇒ + + =

+

A3x ∈R, ∴ D≥ 0 ⇒ 1 + 4y(y(p + 1)+1)≥0

⇒ 4y2(p + 1) + 4y + 1 > 0

∴ 1

–1, –3

y ∉

So, ( )2 14 1 4 1 0 –1, –

3y p y y + + + < ∀ ∈

⇒ (2y + 1)2 + 4y2p< 0

⇒ 22 1 1

– –1, –2 3

+ < ∀ ∈

yp y

y

Hence, 2

min

2 1–

2y

py

+ <

Now, maximum value of 22 1

2y

y+

occurs at

y = –1 and is equal to 1

–4

∴ 1

–4

p <

56. Answer A(P, S); B(Q, R, T); C(R, T); D(T)

Hint :

Draw graph of the functions

Solution :

(A) ( ) [ ]–13 sin ; –1, 1 ; odd function= + =ff x x x D

Also, f(x) is an increasing so, f(x) is one-one

function

(B) ( ) 1– | |sgn ;

1 | | fx

f x D Rx

= = +

Rf = {–1, 0, 1} : even function

(C) ( ) 8 – 2 – 2f x x x=

For domain of f(x), 8 – 2x – x2≥ 0

⇒ x2 + 2x – 8 ≤ 0 ⇒ (x + 4) (x – 2) ≤ 0

⇒ x ∈ [–4, 2]

∴ Rf = [0, 3]

(D) ( )[ ]–

| | – | |{ }

2– 2 2 – 2 0 0

2

xx x x

xf x x= = = ∀ ≤

57. Answer A(P, R, S); B(R, S); C(T); D(T)

Hint :

Type of functions property will be used.



13/13

Solution :

(A) ( )–

cos or2

x xb bf x ax

+= where b> 0

∴ f(x) is even, into and many one function

(B) f(x) is many one, into function

(C) f(x) = x3 + 1, which is bijective function

(D) f(x) = 2x + sinx is a bijective function

58. Answer (02)

Hint :

Here

1993

( ) –2

f x x =

Solution :

( )194 2 –2 91

4cos – 2cos – 4 2sin –2

f x x x x x x = +

( )192 91

1 cos2 – 2cos2 – cos4 –2

x x x x = +

192 2 2 2 91

1 cos 2 2cos – 2cos – cos 22

x x x x x = + + +

1993

–2

x =

∴ f(f(x)) = x

Then f(f(2)) = 2

59. Answer (04)

Hint :

–1 –1 –1tan tan tan1–x y

x yxy

+ + =

if xy < 1

Solution :

–1 –13sin2 tantan tan

5 3cos2 4α α + + α

–1 –12

6 tan tantan tan

48 2tan

α α = + + α

2

2

2 tan1

16 4 tan

α<

+ α

= tan–1(tanα)

= α

60. Answer (06)

Hint :

( ) ( ) ( ) ( )2 – 4

2 – 4 2 – 4x x

x x y y

f yf xf x k= ⇒ =

Solution :

Given, f(x + y) = 2xf(y) + 4

yf(x) …(i)

Also, f(y + x) = 2yf(x) + 4

xf(y) …(ii)

⇒ ( ) ( )

4 – 2 4 – 2x x y y

f yf x= = λ

⇒ f(x) = λ(4x – 2x)

f(1) = 2 = k .2 ⇒k = 1

∴ f(x) = 4x – 2x

f′(n) = 4xlog4 – 2xlog2

f′(2) = 16log4 – 4log2 = 28 log2

∴ k = 28

So, no. of factors of k are 6

all india aakash test series for jee (advanced)-2021 test ......2020/06/21 · defects (b) unit...

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