All India Aakash Test Series for JEE (Advanced)-2020 TEST - 3A …€¦ · 10/06/2019 · B → (P, R, S, T) C → (R, T) D → (R, S, T) 58. (19) 59. (30) 60. (32) Aakash Educational

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

Test Date : 06/10/2019

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TEST - 3A (Paper-2) - Code-C

Test - 3A (Paper-2) (Code-C)_(Answers) All India Aakash Test Series for JEE (Advanced)-2020

1/12

PHYSICS CHEMISTRY MATHEMATICS

1. (C)

2. (A)

3. (A)

4. (C)

5. (A)

6. (A, D)

7. (A, C)

8. (A, B)

9. (A, C)

10. (C)

11. (A, D)

12. (A, C)

13. (A, C, D)

14. (A, C)

15. (A, C)

16. A → (Q)

B → (P, T)

C → (P, S)

D → (P, R, T)

17. A → (Q, T)

B → (R, S, T)

C → (P, R, S)

D → (P, Q)

18. (25)

19. (20)

20. (22)

21. (D)

22. (A)

23. (C)

24. (D)

25. (B)

26. (A, C, D)

27. (A, C, D)

28. (B, C, D)

29. (A, B, D)

30. (A, B, C)

31. (D)

32. (D)

33. (A)

34. (A, B, D)

35. (A, B, C)

36. A → (Q, R, S, T)

B → (P, Q, R)

C → (P, Q, R)

D → (P, R, T)

37. A → (P, R, T)

B → (Q, T)

C → (P, S, T)

D → (P, Q, T)

38. (07)

39. (03)

40. (13)

41. (B)

42. (A)

43. (B)

44. (A)

45. (A)

46. (C, D)

47. (A, B, C)

48. (A, D)

49. (C, D)

50. (A, B, D)

51. (A, C)

52. (C, D)

53. (A, B, C)

54. (A, B, C)

55. (C, D)

56. A → (P, S)

B → (S)

C → (Q, S, T)

D → (Q, S, T)

57. A → (P, Q)

B → (P, R, S, T)

C → (R, T)

D → (R, S, T)

58. (19)

59. (30)

60. (32)


All India Aakash Test Series for JEE (Advanced)-2020 Test - 3A (Paper-2) (Code-C)_(Hints & Solutions)

2/13

PART - I (PHYSICS)

1. Answer (C)

Hint : Displacement method.

Solution :

1

150 –=

xm

x

2 150 –

=x

mx

( )2

1

22

16 150 –

1= =

m x

m x

150 –

4=x

x

x = 30 cm

1 1 1

120 30+ =

f

1 5

120=

f

24 cm=f

2. Answer (A)

Hint : At the instant of sharp change, the flux

would remain same.

Solution : Just after changing flux would remain

same

3

LL i

R

=

0

33i l

R

= =

Now, – – 03

=L di

Ridt

( )–3

=L di

Ridt

00 3

3

–=

t i

i

dt di

L Ri

–3 –

ln–2

=

Rt Ri

L

3–

– 2 =

Rt

LRi e

2–

1 2

Rt

Li eR

= +

3. Answer (A)

Hint : eq 1 2 3

1 1 1 1

L L L L= + +

Solution :

eff 1 2 3

1 1 1 1= + +

L L L L

eff

2

3=L H

itotal = 2 + 3 + 5 = 10 A

2

21

2 2=

qLi

C

2 –6210 10 2 10

3= q

–320 20

10 C mC3 3

= =q

4. Answer (C)

Hint : AC circuit Analysis.

Solution : 2

=A

VV

and ( )

–B 22

VRV

R L

=

+

tan

=L

R

For R → 0 VB = V

2

=V

v

For R → VB = V

2

=V

v

At any time V

( )( )22– – 2 cos = = +

A B A B A BV V V V V V V

( )

2 2 22

22

2–

4 2

V V R VV

R L

= +

+

( ) ( )2 22 2

V R R

R L R L

+ +


Test - 3A (Paper-2) (Code-C)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020

3/13

2

2

4 =

VV

2

=V

V

5. Answer (A)

Hint : Induced electric field is non-conservative

in nature.

Solution : Induced electric field are produced by

changing magnetic field and they form a closed

loop,

So 0E dl and potential can’t be defined.

6. Answer (A, D)

Hint : Image by reflection.

Solution : As the object is not on the bisector,

the polygon will be irregular.

7. Answer (A, C)

Hint : 1 1 1

;v

mv u f u

− = =

Solution : Let x be the object distance from lens

and d be the distance between lens and mirror.

Then for image by lens 1

1 1 1+ =

v x f

1

1 –x f

v xf=

11 –

= =v f

mx x f

1 –

=

x fv

x f

For 2nd time object distance is 2 ––

xfd

x f

2

1 1 1

2 ––

+ =xfv f

dx f

2

2 – ––

=f

mxf

d fx f

1 2. .

–2 – –

–

=

f fm m

xfx fd f

x f

( )

( )( )

2

2

–

– 2 – 2 – –

f x f

x f xd df xf xf f=

+

( )

2

2 – 2 2 –=

+

fm

f df x d f

For m to be independent of x

df = 0 d = f

So, 2

2 21

– 2= =

fm

f f

8. Answer (A, B)

Hint : It could be real image or virtual image.

Solution : –

2–

=+

f

f u and

––2

–=

+

f

f u

2

=f

u and 3

2=

fu

9. Answer (A, C)

Hint : rms rms

; cosV

Z P i VI

= =

Solution : = 100

400sin6

V t

= +

volt

I = 600 sin(t) mA

3400 10 2000

600 3

VZ

I

= = =

Power factor = 3

2

Average power dissipation:

–3

rms rms

400 600 10 3cos

22 2i V

=

avg 60 3=P watt.

10. Answer (C)

Hint : L-R circuit.

Solution : 5

=L

R

iL(t = 0) = 0

( )

4=

=

L ti

R

54

1

− = −

tR

LLi e

R

11. Answer (A, D)

12. Answer (A, C)

13. Answer (A, C, D)

Hint for Q.Nos. 11 to 13 :

1 1 1

v u f+ =



4/13

Solution for Q.Nos. 11 to 13 :

At any time t, cos2

fx t=

object distance from mirror u = (2f – x )

(towards left)

Let v is the position of image w.r.t. mirror then

1 1 2

v u R+ = −

1 1 1

2v f x f

−− =

−

( )

1 2

2

− +=

−

f f x

v f f x

( )2 −

=−

f f xv

x f

Distance of image from mirror

2 cos2

cos2

ff f t

vf

t f

−

=

−

( )

( )

( )2 4 cos 2 4 cos

2 cos 2 2 cosIM

f t f tv

f t t

− − = =

− −

Now position of image w.r.t. origin is

( )

0 0

22

I IM M

f f xr r r f

x f

−= + = +

−

0

2 22 2 2− + − = =

− −I

f fx fx f f xr

x f x f

( )0

2coscos2

cos 2cos

2

= = −

−I

ff t

f tr

f f tt f

0

cos

2 cos

= −

− I

f tr

t (oscillating but not SHM)

Now velocity of image is

( ) ( )

( )

0

2

2 cos sin cos sin

2 cos

I

I

rt f t f t t

d vdt t

− + = =

−

( )

( ) ( )2 2

sin 2 cos cos 2 sin

2 cos 2 cos

− + = =

− − I

f t t t f tv

t t

At 3

t

=

2

2 3 3 4

912 2

2

= =

−

I

f fv

Velocity of object 0 sin 32 4

= − = −

dx f ft

dt

( )rel 2 1

4 3 3 4 13

9 4 9 4

= − = + = +

f fV v v f

rel

253

36v f=

Magnification is always negative so always real image

will be formed. For the time when object lies between

2f and f from mirror it will produce magnified image.

For rest of the time it will be diminished image.

14. Answer (A, C)

15. Answer (A, C)

Hint for Q.Nos. 14 and 15 :

2= eE m x

Solution for Q.Nos. 14 and 15 :

Because of pseudo force, free charge would try

to shift outward since free charges are electrons.

So Fout = m2x. Because of that electric field

would induce in a way that 2eE m x=

2m

E xe

= (from A to B as electron has

shifted towards B)

2+

= =

l L

AB

l

mV Edx xdx

e

( )2 2

2 22 2

l L

lAB

m mV x L L l

e e

+ = = +

As we see 2m

E xe

= implies that the field

region (in rod and outside of it) would exist in x

direction as per the equation then there must be

uniform charge distribution.

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

Hint : Req(t = 0) = 40

Req(t = ) =80

3

transient LR circuit.

Solution :

Req (t = 0) = 40



5/13

( )eq

80

3R t = =

Clearly at t = 0 current 1

1002.5 A

40i = =

and at t = 1

51.25 A

4l = =

Current i2 at t = 0 is zero at inductor and it will

oppose the sudden change of current.

And i2 at t = is 10

2.5 A4

=

Power delivered by battery p = i

As i increases from 2.5 A to 3.75 A

So power delivered by battery increases from

250 watt to 375 watt.

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

Hint : For lens 1 1 1

v u f− =

For mirror 1 1 1

v u f+ =

Solution : For lens 1 1 1

v u f− =

For mirror 1 1 1

v u f+ =

18. Answer (25)

Hint : 1 1 1

5 5 10Z j j= +

−

Solution :

Let Z be the impedance across MN.

Then 1 1 1

5 5 10z j j= +

−

1 10 5 5 5 5

50 50 50 50

j j j

Z j j

+ − += =

+ +

10Z =

1

20C

=

1

25 Hz2 20

= =

fC

19. Answer (20)

Hint : After first refraction from the plane surface

the image seems to be at 3

2L from plane surface.

Solution : After first refraction from the plane

surface p seems to be at 3


Now for 2nd surface ( )11

3

2

v RR L

− + =

− +

v =

1

3

2

RR L

−=

+

L = 20 cm

20. Answer (22)

Hint : 2 1 2 1

v u R

− − =

Solution :

1

2 1

71

7 1 4

4 24 6from water surface

4 7

4 7 3 4

3 4

−

+ =

−

− =

v

v v

2

4 1 3

3 24 24v+ =

2

4 1

3 12v=

v2 = 16 cm

Distance from bottom of tank is 38 – 16 = 22 cm.

PART - II (CHEMISTRY)

21. Answer (D)

Hint : Spectator ligand will affect the C — O bond

length. Order of ligand field strength of the given

ligand is

CO PF3 > PCl3 > PAr3 > PMe3



6/13

Solution : The ligand PPh3 is a weaker

acceptor than CO. As a result d electrons of

metal in such mixed carbonyl will be drawn more

towards CO than in pure metal carbonyl. If the Ph

groups of PPh3 are replaced by more electron

attracting Cl or F groups, the tendency of d

electrons of metal to move towards P increases.

And if Ph groups of PPh3 are replaced by electron

releasing Me groups, the tendency of d

electrons of metal to move towards P further

decreases. As d electrons of metal move

towards P more and less towards CO, the CO

bond Order will be more or CO bond length will

be less and vice versa.

22. Answer (A)

Hint : For the fast rate, back side of LG Should

be less hindered.

Solution : In an SN2 reaction, the leaving group

must be in an axial position in order to allow

backside attack to occur without steric

hinderance from the cyclohexane ring. When the

Br-atom is in axial position in the all cis-isomer,

both the methyl groups are in equatorial

positions, the structure (II) is most reactive as the

approaching nucleophile experiences least

crowding. In the all trans isomer (IV) both the

ethyl groups are in axial positions providing the

maximum crowding to the approaching

nucleophile. Thus structure (IV) is least reactive.

Structure (I) and (III) are cis-trans type. Their

reactivity lies between those of (II) and (IV),

structure (I) is less reactive than (III)as it has

bulky ethyl group at the axial position

23. Answer (C)

Hint : Fe(OH)3 and Na2CrO4 will be formed.

Solution : Na2O2 + 2H2O → 2NaOH + H2O2

( )4 2 2 2 43(Brown)

2FeSO 4NaOH H O 2Fe OH 2Na SO→ + + +

( )2 4 2 2 4 23Al SO NaOH 2NaAlO 3Na SO 4H O+ → + +

( )2 4 2 23

2 4 2 4 2

Cr SO 10NaOH 3H O

2Na CrO 3Na SO 8H O

+ + →

+ +

Fe(OH)3 is a brown residue, NaAlO2 is a colourless solution and Na2CrO4 is a yellow solution.

24. Answer (D)

Hint : Cation present is Na+.

Solution : Fe(CH3COO)3 is red in colour.

25. Answer (B)

Hint : Dispersion of charge decreases the

energy.

Solution :

The general energy diagram of the above reaction

is

On increasing polarity of the solvent, all charged

species will get solvated. Thus their energies will be

lowered. In the above reaction the reactants do not

carry any charge and hence their energy remains

unaffected by the increase in polarity of the solvent.

But the energy of transition state is lowered due to

its solvation. This results in decrease of energy

barrier and hence increase in the rate of reaction.


Hint : CFSE stability.

Solution :

1. In a transition group, stability increases down

the group due to increase in effective nuclear

charge.

2. NO2– is stronger ligand than NH3.

3. Chelate complexes are more stable and as

the number of cyclic rings increases, the

stability of the complex increases.

4. For the same metal ion, stability of the

complex increases with increase in oxidation

state of the metal ion.


Hint : Fact based.



7/13

Solution :

( ) ( ) 2

6 64 2White ppt.

2Zn K Fe CN Zn Fe CN 4K+

++ → +

( ) ( ) 2

6 64 2White ppt.

2Cd K Fe CN Cd Fe CN 4K+

++ → +

( ) ( ) 2

6 64 2Brown ppt.

2Cu K Fe CN Cu Fe CN 4K+ ++ → +

Al3+ ion does not form any precipitate with

K4[Fe(CN)6].

28. Answer (B, C, D)

Hint : With Fe, NO exist in +1 O.S.

Solution :

(I) ( )2

4 2 2 5(X)

FeSO 5H O NO Fe H O NO+

+ + →

Oxidation state of Fe changes from +2 to +1

due to transfer of an electron from NO to Fe+2.

Electronic configuration of Fe+ is 3d7. It has

three unpaired electrons and hence magnetic

moment of (X) is 15 BM .

(II) ( ) ( )52 2 4 5(Y)

Na Fe CN NO Na S Na Fe CN NOS + →

Oxidation state and hybridisation of Fe in the

reactant and product (Y) of reaction

(II) remains unchanged i.e. +2 and d2sp3

respectively.

29. Answer (A, B, D)

Hint : Less hindered more rate.

Solution :

3 3 3 2 2 2 3 2 3

N

(CH ) C — Br CH CH CH CH — Br CH — CH — Br CH — Br

Rate of S 2 reaction

30. Answer (A, B, C)

Hint :

Less hindered double bond will be more reactive.

Solution :

(A) Loss of Br (a) atom in dehydrobromination

reaction results in the formation of least

stable alkene and hence most reactive

towards hydrogenation.

(B) Dehydrohalogenation occurs.

(C) It has 3 stereocentres which means 8

optically active isomers

(D) The given compound has 3 chiral centres and

disubstituted cyclic ring that show

geometrical isomerism.

31. Answer (D)

Hint :

Solution :

Compound (P) has two dissimilar chiral C-atoms. It has two pair of enantiomers or four pair of diastereomers.

32. Answer (D)

Hint : Compound (Q) has two chiral centres.

Solution :

No. of optical Isomers = 2n

(for unsymmetrical molecule) = 2n = 4

33. Answer (A)

Hint : Saytzeff elimination.

Solution : R is Ph — CD = CH — CH3



Hint : Sodium salt gives violet colour solution with

sodium nitroprusside.

Solution : P, Q and R are Na2SO3, Na2S2O3 and

Na2S Respectively

( ) ( )

( )2 3 2 2

XP

Na SO 2HCl 2NaCl H O SO g+ → + +

( )

( )22 2 3 2(X)

Q

Na S O 2HCl 2NaCl SO g S H O+ → + + +



8/13

( )

( )( )

2 2R Y

Na S 2HCl 2NaCl H S g+ → +

SO2 + 2H2S → 3S + 2H2O

2H2S + O2 → 2H2O + 2S

H2S + 2FeCl3 → 2FeCl2 + 2HCl + S

( )2 3 32 BlackH S CH COO Pb PbS 2CH COOH+ → +

( ) ( )2 2 45 5Violet colour

Na S Na Fe CN NO Na Fe CN NOS → +

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

D(P, R, T)

Hint : Fact based

Solution :

( ) 2–2

4Blue colour

Co 4KSCN Co SCN 4K+ ++ → +

( )( )3 2 3 3Blue

Co(NO ) NaOH Co OH NO NaNO+ → +

( )( ) ( )Warm

3 32Excess PinkCo OH NO NaOH Co OH NaNO+ ⎯⎯⎯⎯→ +

( )2 –

2 3 2 26Yellow

Co 7NO 2H 3K K Co NO NO H O+ + + + + + → + +

( ) ( ) 26 64 2

Chocolate brown


( )22

Bluish white

Cu 2NaOH Cu OH 2Na+ ++ → +

( ) ( )22 2

WhiteBlack2Cu 4KSCN 2Cu SCN 2CuSCN SCN++ → → +

( ) ( ) 36 6 34 4

Pr ussian Blue

4Fe 3K Fe CN Fe Fe CN 12K+ ++ → +

( )3 –3

Deep red colour

Fe 3SCN Fe SCN++ ⎯⎯→

( )33

Brown

Fe 3NaOH Fe OH 3Na+ ++ ⎯⎯→ +

( ) ( ) 26 6 24 2 3

White

3Zn 2K Fe CN K Zn Fe CN 6K+ ++ → +

( )22

WhiteZn 2NaOH Zn OH 2Na+ ++ → +

( )( )

( ) 2 4Excess Water soluble

Zn OH 2NaOH Na Zn OH+ →

( ) ( ) ( ) 24 44 42

Zn NH Hg SCN Zn Hg SCN 2NH+ ++ → +

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

Hint :

The possible stereoisomer of the given complexes are

(A)

(B)

Solution :

(C)

(D)



9/13

38. Answer (07)

Hint : D. B. E = 1.

Solution :

39. Answer (03)

Hint : PbBr2 → Colourless

MnS → Buff

Solution : As2S5, PbI2 and AgI are yellow

coloured compound.

40. Answer (13)

Hint : Pt form square planar complex.

Solution :

x = 7

y = 6

x + y = 13

PART - III (MATHEMATICS)

41. Answer (B)

Hint : Put secx + tanx = t

Solution :

secdt

x dxt

= and 1

sec tanx xt

− =

( )( ) ( )1 24

1212 1

0 1

22sec 2secI x x dx t t dt

t

+

−= = +

42. Answer (A)

Hint : + = –t2 and = –2t

Solution :

2 2 2

2

2 2 2 2

1

1 1( )f t x x dx

−

+ = + + +

( )

243 2

2 2

1

2 2 1 1

3 4 2 4 2

t tx xx

t t t−

− − = + + −

2

2

8 1 1 3 1 13

3 3 4 2 4 2

t

t t t

= + + + + −

2

2

3 33

8 4

t

t= + +

( )14

3

6 60 2

8 4

tf t t

t = − = =

( )min

3 2 3 3 23 3

8 44 2f t = + + = +

43. Answer (B)

Hint : ( )3

232 32

2

99 16 16x x

x

− = −

Solution :

3

23

2

916

dx

xx

−

Put 2

2

916 t

x− =

3

182dx t dt

x− =

3

1 2

18

t dt

t −

( )

12 2

1

9 9 9 16

xc c

t x

= + = +

−



10/13

44. Answer (A)

Hint : Degree is power of highest differential

coefficient when expressed as polynomial in

differential coefficients

Solution :

1

2 2

2sin

1 1

x Ay

x x

−

= −− −

2 11 2siny x x A− − = −

( )2

2 2

2 21

1 1

y xy x

x x

− − + =

− −

( )21 2 2y x xy − − =

Hence degree = 1

45. Answer (A)

Hint : Area between f(x) and g(x) is

( ) ( )( )f x g x dx

− , where and are point of

intersection of the curves Solution :

( )( )1 0

2

2 1 2 1

2 3 1A x dy y y y dy− + − +

= = − + + − −

( )( )1

2

0

2 3 1y y y dy+ − + + − +

( ) ( ) ( )1 0 1

2

01 2 1 2

4 1 1 1y dy y dy y dy− −

− − + − − +

( )1

2 1

1 2

1 4 14 1 sin

2 2 2

y yy −

−

− − − − +

0 12 2

1 2 02 2

y yy y

−

+ − − +

( ) 11 10 0 0 2sin 0

2 2

− + − − + − + +

1 2 2 2 2 2 2 11 0

2 2

+ − − + − + −

1 3

2 1 sq. units4 2 2 2

− − + − −

46. Answer (C, D)

Hint : Substitute x = sint

Solution :

( ) ( )1 2

0 0

sin cosf x dx f t tdt I

= =

( )2

0

cos sinf t tdt I

= =

( ) ( )( )2

0

2 sin cos cos sinI f t t f t t dt

= +

2

0

2 1I dx

4

I


Hint : Form linear differential equation

Solution :

( )1tan 21y dy

e x ydx

−

− = +

1tan

21

ydx e x

dy y

−

−=

+

1tan

2 21 1

ydx x e

dy y y

−

+ =+ +

I.F 12 tan1

dy

yye e−

+= =

( )1

12tan

tan

21

yy e

d x e dyy

−

−

=+

1 1tan 2 tan1

2

y yx e e c− −

= +

1

, 02

1tan2 yx e

−

=

3

02

ex

=

48. Answer (A, D)

Hint : Multiply and divide by sec2x

Solution :

( )4

11 13 3

4

secsin cos

sec

xI x x dx

x

− −=

( )2 2

113

1 tan sec

tan

x x dxI

x

+=

Put tanx = t sec2x dx = dt

2

11 113 3

1 tI dt

t t

= +

( ) ( )8 2

3 33 3

tan tan8 2

x x c− −

= − − +



11/13

49. Answer (C, D)


( ) ( )( )f x g x dx


intersection of f(x) and g(x)

Solution :

( )

22

1

4 24

sin cos sin cos sinA x x dx x dx x x

= − + = − −

( )2

1 1cos 1 1 0 2

2 2x

+ − = − + + + − =

( )

55 4

4

1 2

44

And sin cos cos sinA A x x dx x x

+ = − = − −

1 1 1 12 2

2 2 2 2= + + + =

2 1 1 22 and 2 2A A A A= = =


Hint : If x (0, 1) x2 > x3 and if x (1, 2) x3 > x2

Solution :

( ) ( ) ( )2 3

If 0, 1 2020 2020x x

x

( ) ( ) ( )3 2

and if 1, 2 2020 2020x x

x

I4 > I3 > I1 > I2

51. Answer (A, C)

52. Answer (C, D)



xdy + ydx = d(xy) and 2

xdy ydx yd

x x

− =


( )32

2 2

y x dy y dxxy dx x dy

x x

−+=

( ) 2 2 y yd xy x y d

x x

=

( )

( )2

d xy y yd

x xxy

=

21 1

2

4, 2

yc

xy x

− = +

−

c = 0

( ) ( )132f x y x = −

Also g(x) = sin–1sinx·2sinxcosx

+ cos–1cosx 2cosx(–sinx) = 0

g(x) = constant

Put x = 4

we get g(x) y =

30,

16 2

x

( ) ( )( )2

0

= −A g x f x dx

4 2

3 331

8 4

= +

A


55. Answer (C, D)


Substitute expression in tan.

Solution for Q.Nos. 54 and Q.55 :

( )

( )

3 32 2

3 3

sin cos

sin cos sin

d +

+

( )

32

332

sin

sin cos sin cos cos sin

d

+

( )

32

332

cos

cos sin sin cos cos sin

d +

+

( )

1

2cos tan cos sin

d

I

+

( )2

2sin cos cos sin

d

I

+

+



12/13

For I1 put tan cos + sin = t2

cos sec2 d = 2t dt

For I2 put cos + cot sin = 2

–cosec2·sin d = 2 d

2 2

cos sin

t dt d

t

−

2sec sin tan cos 2cosec + −

cos cot sin c + +

f() = sin + cos tan

and g() = cos + cot sin

f() = 2sin (0, 2)

g() = 2cos

( ) ( ) (2, 2 2 +

f g

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

(A) Hint : Convert all terms into sinx and cosx

( )2 2

4 2

sin cos

sin cos

x x dxI

x x

+=

2

2 2

4

cossec 2cosec

sin

xx x dx

x

= + +

3cot

tan 2cot3

xx x k= − − +

1

1, 2 and3

A B c= = − = −

4A + B + 3C = 1

(B) Hint : Distribute x27

in both brackets

( ) ( ) ( )6

4 5 6 5 4 36 5 4g x x x x x x x dx= + + + +

Put x6 + x5 + x4 = t

(6x5 + 5x4 + 4x3)dx = dt

( )7

6 4 51

7x x x c= + + +

As g(0) = 0 ( )( )

76 4 5

7

x x xg x

+ +=

( )73

17

g =

(C) 2

3

0

112sin 12 3

4x G xd x

= =

(D) ( ) ( )

2 2

35 2

5 3 2

143

x xI e dx k e dx−

+ −

−

= +

I1 + I2

In I1 put x + 5 = y and in I2 put 3x – 2 = –t

( )2 2

0 0

1 1

03

y ykI e dy e dt= + − =

k = 3

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

Hint : Applying properties of integrals

Solution :

(A) ( )1

1 1

0 0 6

5 sin 5 sin 5 cosx dx x dx x= = −

( )5 1 cos1= −

(B) ( )5

5

1 10dx−

− =

(C) 14 5

4

51 0

1lim

5 5

n

nr

r xx dx

n→=

= = =

(D) ( )2

0

1lim 2

x

nx

ex

x→

−

2 21 4lim

2

x

nn

e x

x x→

−=

n 3 for limit to be finite

58. Answer (19)

Hint : Integration by parts

Solution :

( )

1

20

11

1n n

I dxx

= +

( ) ( )

1

1 2

12 2

00

2

1 1n n

x x dxn

x x+

= ++ +

( ) ( )

1 1

12 2

0 0

12

2 1 1n nn

dx dxn

x x+

= + − + +

1

12 2

2n nn

n I n I+

= + −

2n In + 1 = 2–n + (2n – 1)In

Put n = 10, we get

20 I11 = 2–10 + 19 I10



13/13

59. Answer (30)

Hint : Distance from origin distance from line

x = 3

Solution :

Given 2 2 3x y x+ −

x2 + y2 x2 – 6x + 9

y2 – 6x + 9

2 3

62

y x

− −

32

0

9 6A x dx= −

( )

( )

32

32

0

9 6

36

2

x−=

−

( )

329 27

0 3 sq. units9 9

= + = =

10A = 30

60. Answer (32)

Hint : ( ) 0a

a

f x dx−

= if f(x) is odd.

Solution :

4

2

4

secI x dx

−

=

4

2

0

2 sec x dx

=

4

02tan 2x

= =

I5 = 25 = 32

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

Test Date : 06/10/2019

ANSWERS


TEST - 3A (Paper-2) - Code-D

Test - 3A (Paper-2) (Code-D)_(Answers) All India Aakash Test Series for JEE (Advanced)-2020

1/12

PHYSICS CHEMISTRY MATHEMATICS

1. (A)

2. (C)

3. (A)

4. (A)

5. (C)

6. (C)

7. (A, C)

8. (A, B)

9. (A, C)

10. (A, D)

11. (A, D)

12. (A, C)

13. (A, C, D)

14. (A, C)

15. (A, C)

16. A → (Q, T)

B → (R, S, T)

C → (P, R, S)

D → (P, Q)

17. A → (Q)

B → (P, T)

C → (P, S)

D → (P, R, T)

18. (22)

19. (20)

20. (25)

21. (B)

22. (D)

23. (C)

24. (A)

25. (D)

26. (A, B, C)

27. (A, B, D)

28. (B, C, D)

29. (A, C, D)

30. (A, C, D)

31. (D)

32. (D)

33. (A)

34. (A, B, D)

35. (A, B, C)

36. A → (P, R, T)

B → (Q, T)

C → (P, S, T)

D → (P, Q, T)

37. A → (Q, R, S, T)

B → (P, Q, R)

C → (P, Q, R)

D → (P, R, T)

38. (13)

39. (03)

40. (07)

41. (A)

42. (A)

43. (B)

44. (A)

45. (B)

46. (A, B, D)

47. (C, D)

48. (A, D)

49. (A, B, C)

50. (C, D)

51. (A, C)

52. (C, D)

53. (A, B, C)

54. (A, B, C)

55. (C, D)

56. A → (P, Q)

B → (P, R, S, T)

C → (R, T)

D → (R, S, T)

57. A → (P, S)

B → (S)

C → (Q, S, T)

D → (Q, S, T)

58. (32)

59. (30)

60. (19)


All India Aakash Test Series for JEE (Advanced)-2020 Test - 3A (Paper-2) (Code-D)_(Hints & Solutions)

2/13

PART - I (PHYSICS)

1. Answer (A)

Hint : Induced electric field is non-conservative

in nature.

Solution : Induced electric field are produced by

changing magnetic field and they form a closed

loop,

So 0E dl and potential can’t be defined.

2. Answer (C)

Hint : AC circuit Analysis.

Solution : 2

=A

VV

and ( )

–B 22

VRV

R L

=

+

tan

=L

R

For R → 0 VB = V

2

=V

v

For R → VB = V

2

=V

v

At any time V

( )( )22– – 2 cos = = +

A B A B A BV V V V V V V

( )

2 2 22

22

2–

4 2

V V R VV

R L

= +

+

( ) ( )2 22 2

V R R

R L R L

+ +

2

2

4 =

VV

2

=V

V

3. Answer (A)

Hint : eq 1 2 3

1 1 1 1

L L L L= + +

Solution :

eff 1 2 3

1 1 1 1= + +

L L L L

eff

2H

3L =

itotal = 2 + 3 + 5 = 10 amp

2

21

2 2=

qLi

C

2 –6210 10 2 10

3= q

–320 20

10 C mC3 3

= =q

4. Answer (A)

Hint : At the instant of sharp change, the flux

would remain same.

Solution : Just after changing flux would remain

same

3

LL i

R

=

0

33i l

R

= =

Now, – – 03

=L di

Ridt

( )–3

=L di

Ridt

00 3

3

–=

t i

i

dt di

L Ri

–3 –

ln–2

=

Rt Ri

L

3–

– 2 =

Rt

LRi e

2–

1 2

Rt

Li eR

= +

5. Answer (C)

Hint : Displacement method.

Solution :

1

150 –=

xm

x

2 150 –

=x

mx

( )2

1

22

16 150 –

1= =

m x

m x


Test - 3A (Paper-2) (Code-D)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020

3/13

150 –

4=x

x

x = 30 cm

1 1 1

120 30+ =

f

1 5

120=

f

24 cm=f

6. Answer (C)

Hint : L-R circuit.

Solution : 5

=L

R

iL(t = 0) = 0

( )

4=

=

L ti

R

54

1

− = −

tR

LLi e

R

7. Answer (A, C)

Hint : rms rms

; cosV

Z P i VI

= =

Solution : = 100

400sin6

V t

= +

volt

I = 600 sin(t) mA

3400 10 2000

600 3

VZ

I

= = =

Power factor = 3

2

Average power dissipation:

–3

rms rms

400 600 10 3cos

22 2i V

=

avg 60 3=P watt.

8. Answer (A, B)

Hint : It could be real image or virtual image.

Solution : –

2–

=+

f

f u and

––2

–=

+

f

f u

2

=f

u and 3

2=

fu

9. Answer (A, C)

Hint : 1 1 1

;v

mv u f u

− = =

Solution : Let x be the object distance from lens

and d be the distance between lens and mirror.

Then for image by lens 1

1 1 1+ =

v x f

1

1 –x f

v xf=

11 –

= =v f

mx x f

1 –

=

x fv

x f

For 2nd time object distance is 2 ––

xfd

x f

2

1 1 1

2 ––

+ =xfv f

dx f

2

2 – ––

=f

mxf

d fx f

1 2. .

–2 – –

–

=

f fm m

xfx fd f

x f

( )

( )( )

2

2

–

– 2 – 2 – –

f x f

x f xd df xf xf f=

+

( )

2

2 – 2 2 –=

+

fm

f df x d f

For m to be independent of x

df = 0 d = f

So, 2

2 21

– 2= =

fm

f f

10. Answer (A, D)

Hint : Image by reflection.

Solution : As the object is not on the bisector,

the polygon will be irregular.

11. Answer (A, D)

12. Answer (A, C)



1 1 1

v u f+ =



4/13


At any time t, cos2

fx t=

object distance from mirror u = (2f – x )

(towards left)

Let v is the position of image w.r.t. mirror then

1 1 2

v u R+ = −

1 1 1

2v f x f

−− =

−

( )

1 2

2

− +=

−

f f x

v f f x

( )2 −

=−

f f xv

x f

Distance of image from mirror

2 cos2

cos2

ff f t

vf

t f

−

=

−

( )

( )

( )2 4 cos 2 4 cos

2 cos 2 2 cosIM

f t f tv

f t t

− − = =

− −

Now position of image w.r.t. origin is

( )

0 0

22

I IM M

f f xr r r f

x f

−= + = +

−

0

2 22 2 2− + − = =

− −I

f fx fx f f xr

x f x f

( )0

2coscos2

cos 2cos

2

= = −

−I

ff t

f tr

f f tt f

0

cos

2 cos

= −

− I

f tr

t (oscillating but not SHM)

Now velocity of image is

( ) ( )

( )

0

2

2 cos sin cos sin

2 cos

I

I

rt f t f t t

d vdt t

− + = =

−

( )

( ) ( )2 2

sin 2 cos cos 2 sin

2 cos 2 cos

− + = =

− − I

f t t t f tv

t t

At 3

t

=

2

2 3 3 4

912 2

2

= =

−

I

f fv

Velocity of object 0 sin 32 4

= − = −

dx f ft

dt

( )rel 2 1

4 3 3 4 13

9 4 9 4

= − = + = +

f fV v v f

rel

253

36v f=

Magnification is always negative so always real image

will be formed. For the time when object lies between

2f and f from mirror it will produce magnified image.

For rest of the time it will be diminished image.

14. Answer (A, C)

15. Answer (A, C)


2= eE m x

Solution for Q.Nos. 14 and 15 :

Because of pseudo force, free charge would try

to shift outward since free charges are electrons.

So Fout = m2x. Because of that electric field

would induce in a way that 2eE m x=

2m

E xe

= (from A to B as electron has

shifted towards B)

2+

= =

l L

AB

l

mV Edx xdx

e

( )2 2

2 22 2

l L

lAB

m mV x L L l

e e

+ = = +

As we see 2m

E xe

= implies that the field

region (in rod and outside of it) would exist in x

direction as per the equation then there must be

uniform charge distribution.

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

Hint : For lens 1 1 1

v u f− =

For mirror 1 1 1

v u f+ =

Solution : For lens 1 1 1

v u f− =

For mirror 1 1 1

v u f+ =



5/13

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

Hint : Req(t = 0) = 40

Req(t = ) =80

3

transient LR circuit.

Solution :

Req (t = 0) = 40

( )eq

80

3R t = =

Clearly at t = 0 current 1

1002.5 A

40i = =

and at t = 1

51.25 A

4l = =

Current i2 at t = 0 is zero at inductor and it will

oppose the sudden change of current.

And i2 at t = is 10

2.5 A4

=

Power delivered by battery p = i

As i increases from 2.5 A to 3.75 A

So power delivered by battery increases from

250 watt to 375 watt.

18. Answer (22)

Hint : 2 1 2 1

v u R

− − =

Solution :

1

2 1

71

7 1 4

4 24 6from water surface

4 7

4 7 3 4

3 4

−

+ =

−

− =

v

v v

2

4 1 3

3 24 24v+ =

2

4 1

3 12v=

v2 = 16 cm

Distance from bottom of tank is 38 – 16 = 22 cm.

19. Answer (20)

Hint : After first refraction from the plane surface

the image seems to be at 3


Solution : After first refraction from the plane

surface p seems to be at 3


Now for 2nd surface ( )11

3

2

v RR L

− + =

− +

v =

1

3

2

RR L

−=

+

L = 20 cm

20. Answer (25)

Hint : 1 1 1

5 5 10z j j= +

−

Solution :

Let Z be the impedance across MN.

Then 1 1 1

5 5 10Z j j= +

−

1 10 5 5 5 5

50 50 50 50

j j j

Z j j

+ − += =

+ +

10Z =

1

20C

=

1

25 Hz2 20

= =

fC

PART - II (CHEMISTRY)

21. Answer (B)

Hint : Dispersion of charge decreases the

energy.

Solution :



6/13

The general energy diagram of the above reaction

is

On increasing polarity of the solvent, all charged

species will get solvated. Thus their energies will be

lowered. In the above reaction the reactants do not

carry any charge and hence their energy remains

unaffected by the increase in polarity of the solvent.

But the energy of transition state is lowered due to

its solvation. This results in decrease of energy

barrier and hence increase in the rate of reaction.

22. Answer (D)

Hint : Cation present is Na+.

Solution : Fe(CH3COO)3 is red in colour.

23. Answer (C)

Hint : Fe(OH)3 and Na2CrO4 will be formed.

Solution : Na2O2 + 2H2O → 2NaOH + H2O2

( )4 2 2 2 43(Brown)

2FeSO 4NaOH H O 2Fe OH 2Na SO→ + + +

( )2 4 2 2 4 23Al SO NaOH 2NaAlO 3Na SO 4H O+ → + +

( )2 4 2 23

2 4 2 4 2

Cr SO 10NaOH 3H O

2Na CrO 3Na SO 8H O

+ + →

+ +

Fe(OH)3 is a brown residue, NaAlO2 is a

colourless solution and Na2CrO4 is a yellow

solution.

24. Answer (A)

Hint : For the fast rate, back side of LG Should

be less hindered.

Solution : In an SN2 reaction, the leaving group

must be in an axial position in order to allow

backside attack to occur without steric

hinderance from the cyclohexane ring. When the

Br-atom is in axial position in the all cis-isomer,

both the methyl groups are in equatorial

positions, the structure (II) is most reactive as the

approaching nucleophile experiences least

crowding. In the all trans isomer (IV) both the

ethyl groups are in axial positions providing the

maximum crowding to the approaching

nucleophile. Thus structure (IV) is least reactive.

Structure (I) and (III) are cis-trans type. Their

reactivity lies between those of (II) and (IV),

structure (I) is less reactive than (III)as it has

bulky ethyl group at the axial position

25. Answer (D)

Hint : Spectator ligand will affect the C — O bond

length. Order of ligand field strength of the given

ligand is

CO PF3 > PCl3 > PAr3 > PMe3

Solution : The ligand PPh3 is a weaker

acceptor than CO. As a result d electrons of

metal in such mixed carbonyl will be drawn more

towards CO than in pure metal carbonyl. If the Ph

groups of PPh3 are replaced by more electron

attracting Cl or F groups, the tendency of d

electrons of metal to move towards P increases.

And if Ph groups of PPh3 are replaced by electron

releasing Me groups, the tendency of d

electrons of metal to move towards P further

decreases. As d electrons of metal move

towards P more and less towards CO, the CO

bond Order will be more or CO bond length will

be less and vice versa.


Hint :

Less hindered double bond will be more reactive.

Solution :

(A) Loss of Br (a) atom in dehydrobromination

reaction results in the formation of least

stable alkene and hence most reactive

towards hydrogenation.

(B) Dehydrohalogenation occurs.

(C) It has 3 stereocentres which means 8

optically active isomers

(D) The given compound has 3 chiral centres and

disubstituted cyclic ring that show

geometrical isomerism.



7/13


Hint : Less hindered more rate.

Solution :

3 3 3 2 2 2 3 2 3

N

(CH ) C — Br CH CH CH CH — Br CH — CH — Br CH — Br

Rate of S 2 reaction

28. Answer (B, C, D)

Hint : With Fe, NO exist in +1 O.S.

Solution :

(I) ( )2

4 2 2 5(X)

FeSO 5H O NO Fe H O NO+

+ + →

Oxidation state of Fe changes from +2 to +1

due to transfer of an electron from NO to Fe+2.

Electronic configuration of Fe+ is 3d7. It has

three unpaired electrons and hence magnetic

moment of (X) is 15 BM .

(II) ( ) ( )52 2 4 5(Y)

Na Fe CN NO Na S Na Fe CN NOS + →

Oxidation state and hybridisation of Fe in the

reactant and product (Y) of reaction

(II) remains unchanged i.e. +2 and d2sp3

respectively.


Hint : Fact based.

Solution :

( ) ( ) 2

6 64 2White ppt.

2Zn K Fe CN Zn Fe CN 4K+

++ → +

( ) ( ) 2

6 64 2White ppt.

2Cd K Fe CN Cd Fe CN 4K+

++ → +

( ) ( ) 2

6 64 2Brown ppt.


Al3+ ion does not form any precipitate with

K4[Fe(CN)6].


Hint : CFSE stability.

Solution :

1. In a transition group, stability increases down

the group due to increase in effective nuclear

charge.

2. NO2– is stronger ligand than NH3.

3. Chelate complexes are more stable and as

the number of cyclic rings increases, the

stability of the complex increases.

4. For the same metal ion, stability of the

complex increases with increase in oxidation

state of the metal ion.

31. Answer (D)

Hint :

Solution :

Compound (P) has two dissimilar chiral

C-atoms. It has two pair of enantiomers or four

pair of diastereomers.

32. Answer (D)

Hint : Compound (Q) has two chiral centres.

Solution :

No. of optical Isomers = 2n

(for unsymmetrical molecule) = 2n = 4

33. Answer (A)

Hint : Saytzeff elimination.

Solution : R is Ph — CD = CH — CH3



Hint : Sodium salt gives violet colour solution with

sodium nitroprusside.

Solution : P, Q and R are Na2SO3, Na2S2O3 and

Na2S Respectively

( ) ( )

( )2 3 2 2

XP

Na SO 2HCl 2NaCl H O SO g+ → + +

( )

( )22 2 3 2(X)

Q

Na S O 2HCl 2NaCl SO g S H O+ → + + +

( )

( )( )

2 2R Y

Na S 2HCl 2NaCl H S g+ → +

SO2 + 2H2S → 3S + 2H2O



8/13

2H2S + O2 → 2H2O + 2S

H2S + 2FeCl3 → 2FeCl2 + 2HCl + S

( )2 3 32 BlackH S CH COO Pb PbS 2CH COOH+ → +

( ) ( )2 2 45 5Violet colour

Na S Na Fe CN NO Na Fe CN NOS → +

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

Hint :

The possible stereoisomer of the given complexes are

(A)

(B)

Solution :

(C)

(D)

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

D(P, R, T)

Hint : Fact based

Solution :

( ) 2–2

4Blue colour

Co 4KSCN Co SCN 4K+ ++ → +

( )( )3 2 3 3Blue

Co(NO ) NaOH Co OH NO NaNO+ → +

( )( ) ( )Warm

3 32Excess PinkCo OH NO NaOH Co OH NaNO+ ⎯⎯⎯⎯→ +

( )2 –

2 3 2 26Yellow

Co 7NO 2H 3K K Co NO NO H O+ + + + + + → + +

( ) ( ) 26 64 2

Chocolate brown


( )22

Bluish white

Cu 2NaOH Cu OH 2Na+ ++ → +

( ) ( )22 2

WhiteBlack2Cu 4KSCN 2Cu SCN 2CuSCN SCN++ → → +

( ) ( ) 36 6 34 4

Pr ussian Blue

4Fe 3K Fe CN Fe Fe CN 12K+ ++ → +

( )3 –3

Deep red colour

Fe 3SCN Fe SCN++ ⎯⎯→

( )33

Brown

Fe 3NaOH Fe OH 3Na+ ++ ⎯⎯→ +

( ) ( ) 26 6 24 2 3

White

3Zn 2K Fe CN K Zn Fe CN 6K+ ++ → +

( )22

WhiteZn 2NaOH Zn OH 2Na+ ++ → +

( )( )

( ) 2 4Excess Water soluble

Zn OH 2NaOH Na Zn OH+ →

( ) ( ) ( ) 24 44 42

Zn NH Hg SCN Zn Hg SCN 2NH+ ++ → +



9/13

38. Answer (13)

Hint : Pt form square planar complex.

Solution :

x = 7

y = 6

x + y = 13

39. Answer (03)

Hint : PbBr2 → Colourless

MnS → Buff

Solution : As2S5, PbI2 and AgI are yellow

coloured compound.

40. Answer (07)

Hint : D. B. E = 1.

Solution :

PART - III (MATHEMATICS)

41. Answer (A)


( ) ( )( )f x g x dx


intersection of the curves Solution :

( )( )1 0

2

2 1 2 1

2 3 1A x dy y y y dy− + − +

= = − + + − −

( )( )1

2

0

2 3 1y y y dy+ − + + − +

( ) ( ) ( )1 0 1

2

01 2 1 2

4 1 1 1y dy y dy y dy− −

− − + − − +

( )1

2 1

1 2

1 4 14 1 sin

2 2 2

y yy −

−

− − − − +

0 12 2

1 2 02 2

y yy y

−

+ − − +

( ) 11 10 0 0 2sin 0

2 2

− + − − + − + +

1 2 2 2 2 2 2 11 0

2 2

+ − − + − + −

1 3

2 1 sq. units4 2 2 2

− − + − −

42. Answer (A)

Hint : Degree is power of highest differential

coefficient when expressed as polynomial in

differential coefficients

Solution :

1

2 2

2sin

1 1

x Ay

x x

−

= −− −

2 11 2siny x x A− − = −

( )2

2 2

2 21

1 1

y xy x

x x

− − + =

− −

( )21 2 2y x xy − − =

Hence degree = 1



10/13

43. Answer (B)

Hint : ( )3

232 32

2

99 16 16x x

x

− = −

Solution :

3

23

2

916

dx

xx

−

Put 2

2

916 t

x− =

3

182dx t dt

x− =

3

1 2

18

t dt

t −

( )

12 2

1

9 9 9 16

xc c

t x

= + = +

−

44. Answer (A)

Hint : + = –t2 and = –2t

Solution :

2 2 2

2

2 2 2 2

1

1 1( )f t x x dx

−

+ = + + +

( )

243 2

2 2

1

2 2 1 1

3 4 2 4 2

t tx xx

t t t−

− − = + + −

2

2

8 1 1 3 1 13

3 3 4 2 4 2

t

t t t

= + + + + −

2

2

3 33

8 4

t

t= + +

( )14

3

6 60 2

8 4

tf t t

t = − = =

( )min

3 2 3 3 23 3

8 44 2f t = + + = +

45. Answer (B)

Hint : Put secx + tanx = t

Solution :

secdt

x dxt

= and 1

sec tanx xt

− =

( )( ) ( )1 24

1212 1

0 1

22sec 2secI x x dx t t dt

t

+

−= = +


Hint : If x (0, 1) x2 > x3 and if x (1, 2) x3 > x2

Solution :

( ) ( ) ( )2 3

If 0, 1 2020 2020x x

x

( ) ( ) ( )3 2

and if 1, 2 2020 2020x x

x

I4 > I3 > I1 > I2

47. Answer (C, D)


( ) ( )( )f x g x dx


intersection of f(x) and g(x)

Solution :

( )

22

1

4 24

sin cos sin cos sinA x x dx x dx x x

= − + = − −

( )2

1 1cos 1 1 0 2

2 2x

+ − = − + + + − =

( )

55 4

4

1 2

44

And sin cos cos sinA A x x dx x x

+ = − = − −

1 1 1 12 2

2 2 2 2= + + + =

2 1 1 22 and 2 2A A A A= = =

48. Answer (A, D)

Hint : Multiply and divide by sec2x

Solution :

( )4

11 13 3

4

secsin cos

sec

xI x x dx

x

− −=

( )2 2

113

1 tan sec

tan

x x dxI

x

+=

Put tanx = t sec2x dx = dt

2

11 113 3

1 tI dt

t t

= +

( ) ( )8 2

3 33 3

tan tan8 2

x x c− −

= − − +



11/13


Hint : Form linear differential equation

Solution :

( )1tan 21y dy

e x ydx

−

− = +

1tan

21

ydx e x

dy y

−

−=

+

1tan

2 21 1

ydx x e

dy y y

−

+ =+ +

I.F 12 tan1

dy

yye e−

+= =

( )1

12tan

tan

21

yy e

d x e dyy

−

−

=+

1 1tan 2 tan1

2

y yx e e c− −

= +

1

, 02

1tan2 yx e

−

=

3

02

ex

=

50. Answer (C, D)

Hint : Substitute x = sint

Solution :

( ) ( )1 2

0 0

sin cosf x dx f t tdt I

= =

( )2

0

cos sinf t tdt I

= =

( ) ( )( )2

0

2 sin cos cos sinI f t t f t t dt

= +

2

0

2 1I dx

4

I

51. Answer (A, C)

52. Answer (C, D)



xdy + ydx = d(xy) and 2

xdy ydx yd

x x

− =


( )32

2 2

y x dy y dxxy dx x dy

x x

−+=

( ) 2 2 y yd xy x y d

x x

=

( )

( )2

d xy y yd

x xxy

=

21 1

2

4, 2

yc

xy x

− = +

−

c = 0

( ) ( )132f x y x = −

Also g(x) = sin–1sinx·2sinxcosx

+ cos–1cosx 2cosx(–sinx) = 0

g(x) = constant

Put x = 4

we get g(x) y =

30,

16 2

x

( ) ( )( )2

0

= −A g x f x dx

4 2

3 331

8 4

= +

A


55. Answer (C, D)


Substitute expression in tan.

Solution for Q.Nos. 54 and Q.55 :

( )

( )

3 32 2

3 3

sin cos

sin cos sin

d +

+



12/13

( )

32

332

sin

sin cos sin cos cos sin

d

+

( )

32

332

cos

cos sin sin cos cos sin

d +

+

( )

1

2cos tan cos sin

d

I

+

( )2

2sin cos cos sin

d

I

+

+

For I1 put tan cos + sin = t2

cos sec2 d = 2t dt

For I2 put cos + cot sin = 2

–cosec2·sin d = 2 d

2 2

cos sin

t dt d

t

−

2sec sin tan cos 2cosec + −

cos cot sin c + +

f() = sin + cos tan

and g() = cos + cot sin

f() = 2sin (0, 2)

g() = 2cos

( ) ( ) (2, 2 2 +

f g

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

Hint : Applying properties of integrals

Solution :

(A) ( )1

1 1

0 0 6

5 sin 5 sin 5 cosx dx x dx x= = −

( )5 1 cos1= −

(B) ( )5

5

1 10dx−

− =

(C) 14 5

4

51 0

1lim

5 5

n

nr

r xx dx

n→=

= = =

(D) ( )2

0

1lim 2

x

nx

ex

x→

−

2 21 4lim

2

x

nn

e x

x x→

−=

n 3 for limit to be finite

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

(A) Hint : Convert all terms into sinx and cosx

( )2 2

4 2

sin cos

sin cos

x x dxI

x x

+=

2

2 2

4

cossec 2cosec

sin

xx x dx

x

= + +

3cot

tan 2cot3

xx x k= − − +

1

1, 2 and3

A B c= = − = −

4A + B + 3C = 1

(B) Hint : Distribute x27

in both brackets

( ) ( ) ( )6

4 5 6 5 4 36 5 4g x x x x x x x dx= + + + +

Put x6 + x5 + x4 = t

(6x5 + 5x4 + 4x3)dx = dt

( )7

6 4 51

7x x x c= + + +

As g(0) = 0 ( )( )

76 4 5

7

x x xg x

+ +=

( )73

17

g =

(C) 2

3

0

112sin 12 3

4x G xd x

= =

(D) ( ) ( )

2 2

35 2

5 3 2

143

x xI e dx k e dx−

+ −

−

= +

I1 + I2

In I1 put x + 5 = y and in I2 put 3x – 2 = –t

( )2 2

0 0

1 1

03

y ykI e dy e dt= + − =

k = 3



13/13

58. Answer (32)

Hint : ( ) 0a

a

f x dx−

= if f(x) is odd.

Solution :

4

2

4

secI x dx

−

=

4

2

0

2 sec x dx

=

4

02tan 2x

= =

I5 = 25 = 32

59. Answer (30)

Hint : Distance from origin distance from line

x = 3

Solution :

Given 2 2 3x y x+ −

x2 + y2 x2 – 6x + 9

y2 – 6x + 9

2 3

62

y x

− −

32

0

9 6A x dx= −

( )

( )

32

32

0

9 6

36

2

x−=

−

( )

329 27

0 3 sq. units9 9

= + = =

10A = 30

60. Answer (19)

Hint : Integration by parts

Solution :

( )

1

20

11

1n n

I dxx

= +

( ) ( )

1

1 2

12 2

00

2

1 1n n

x x dxn

x x+

= ++ +

( ) ( )

1 1

12 2

0 0

12

2 1 1n nn

dx dxn

x x+

= + − + +

1

12 2

2n nn

n I n I+

= + −

2n In + 1 = 2–n + (2n – 1)In

Put n = 10, we get

20 I11 = 2–10 + 19 I10

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All India Aakash Test Series for JEE (Advanced)-2020 TEST - 3A …€¦ · 10/06/2019 · B → (P, R, S, T) C → (R, T) D → (R, S, T) 58. (19) 59. (30) 60. (32) Aakash Educational