All India Aakash Test Series for JEE (Advanced)-2020 MOCK ... › Aakash › s3fs...Mar 05, 2020...

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

Test Date : 03/05/2020

ANSWERS

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

MOCK TEST - 5 (Paper-2) - Code-C

Mock Test - 5 (Paper-2) (Code-C)_(Answers) All India Aakash Test Series for JEE (Advanced)-2020

1/13

PHYSICS CHEMISTRY MATHEMATICS

1. (A, D)

2. (A, C, D)

3. (A, C, D)

4. (A, B, C)

5. (A, B)

6. (A, B, C, D)

7. (75)

8. (10)

9. (24)

10. (12)

11. (51)

12. (16)

13. (40)

14. (20)

15. (C)

16. (D)

17. (B)

18. (A)

19. (B, D)

20. (B, C, D)

21. (A, B, C, D)

22. (A, C, D)

23. (B, C, D)

24. (A, B, D)

25. (17)

26. (22)

27. (15)

28. (40)

29. (30)

30. (12)

31. (21)

32. (90)

33. (A)

34. (D)

35. (B)

36. (A)

37. (A, B)

38. (A, C, D)

39. (A, B)

40. (B, D)

41. (B, C)

42. (A, B, D)

43. (20)

44. (48)

45. (02)

46. (27)

47. (04)

48. (84)

49. (65)

50. (25)

51. (C)

52. (B)

53. (D)

54. (A)


All India Aakash Test Series for JEE (Advanced)-2020 Mock Test-5 (Paper-2) (Code-C)_(Hints & Solutions)

2/13

PART - I (PHYSICS)

1. Answer (A, D)

Hint : ,

2m C

Pv

m

=

Solution : ,

2m C

Pv

m

=

22

, 2

/ 2 2

m CV PT m m

m

= =

( )

2

2

PT

m

=

and acceleration of m1 and m2 are same as

( )

2

22

Pa

m

=

2. Answer (A, C, D)

Hint : Both capacitors will behave like parallelly connected capacitor. Solution : Both capacitors will behave like parallelly connected capacitor.

So potential difference finally is ( )04

Qv

r C =

+

So charge on capacitor ‘C’ is ( )04

QCq

r C=

+

Total energy dissipated

2 2

0 02 4 2 4

Q QE

r r C = −

+

2

0 02 4 4

Q CE

r r C =

+

Energy dissipated across (R1)

( )( )( )

2

11

0 1 2 08 4

Q CRR

r R R r C =

+ +

3. Answer (A, C, D)

Hint : 0A

Cd

= ; charge remains constant.

Solution : Strength of electric field remains same

but ‘d’ increases. So, v Edr = increases and capacitance decreases. Change in electric field energy is contributed by external agent.

4. Answer (A, B, C)

Hint : ( )1 2 11 1 2

v u R

− = −

Solution : First image will be formed of 60 cm so, if another lens is placed at 120 cm then for 2nd lens the object will be at its focus and hence the

final emerging rays will be parallel to x-axis, or the image will be formed at infinity. In case if whole of the system is placed inside a medium of refractive

index of = 9/8 then effective focal length comes out to be 30. So, this time again the image will be formed at infinity. And if a concave lens of focal length 60 cm is placed in contact with the first lens then also image will be formed at infinity.

5. Answer (A, B)

Hint : mT = constant, 0.01 TA4 = 0.81 TB4.

Solution : eA = 0.01 and eB = 0.81

0.01 TA4 = 0.81 TB4

3

AB

TT =

58021934 K

3BT = =

Now by Wien's displacement law

3constant 2.73 10mT mk− = =

0.5 mmA =

1 m 1.5 mmB mA = + =

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

Hint : Standing waves.

2 2

,Tk w

= =

Solution : There would be 7 pressure node (including both ends) and 6 displacement node.

2 2 24m

K

= = =

Particles initial phase of oscillation with respect to

time is /6 so after covering a phase of 5/6 all the particles will be at their mean value of

oscillation 5

6

=

t

Since adjacent loops vibrates with a phase

difference of , so the phase difference of

oscillation at point x = 2.5 m and at x = 4.5 m is .

7. Answer (75)

Hint : (m1 + m3) vc = mv

Solution : As there is no friction so m2 will remain intact.

( ) 21

2=KE i mv

vc (m + 3m) = mv vc = v/4

( )2 21 1

.4 . .2 16 2 4

v mvKE f m= =

2 21 1 3 11 .2 4 4 2

KE mv mv

= − =

% decrease 2

2

3 1. 100 75%

14 2

2

mv

mv

= =


Mock Test-5 (Paper-2) (Code-C)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020

3/13

8. Answer (10)

Hint : Find the phase of interference

Solution : 2

2 2

3 4 21

as p s p D a D D

b

− = + − = +

2 21

2 2 = − =

a aD D

D D

Phase difference between s3p – s4p = /2

2 22

2 4

a a

D D

= =

Now if we take the reference phase for s4p to be zero then phase of

4 3

5 6

2 1

0 ,2

, 22

,2

s p s p

s p s p

s p s p

→ →

→ →

→ →

For 2

6 2

4. 1

as p D D

D + −

2 2

2

2 21

+ − =

a aD D

DD

For 2

1 2

91

as p D D

D + −

2 2

2

9 9 91

2 42

a aD D

DD

+ − =

So phase or sum of

0 2 102 2 2

A A A A A A A

+ + + + + =

210 10QI A I = =

9. Answer (24) Hint : Conservation of angular momentum about point of contact.

Solution : Angular momentum about point ‘P’

2

0

2.

2 3 =

l Mmv w

03

4 =

mvw

M

So, .cos4522

= =oyw

v w

0 03 3

.4 2 8

= =y

mv mvv

M M

03

8y y

mvN dt Mv = =

3 8 8

248

YN dt

= =

10. Answer (12)

Hint : NIA =

Solution : Radius of coil ‘R’

2R.N =

2

RN

=

( )2

2

2 2

.. .

4 .

NIN I R

N

= =

2

44

Ik

N = =

So 3k = 12

11. Answer (51)

Hint : ( ) 022

EPE n

−= =

Solution : ( ) 022

EPE n

−= =

And ( ) 048

EPE n

−= =

If PE (n = 2) is taken as zero reference

Then 0 00 0

3

8 2 8

E EPE E

= − + =

0

3 13.65.1e V

8PE

= =

10(PE0) = 51 eV

12. Answer (16)

Hint : 3 30 0

4 46

3 3 = + R g R g Rv

Solution : Initial acceleration ( )0

0

− =

a g

And terminal velocity ( )2 0

0

2

9

R gv

− =

( )

( )2 0

0 0

2

9Avg

R gva

t t

− = =

( )0

8Avg

aa =

( ) ( )2 0 0

0

2

9 8

R g g

t

− − =

2

0

16

9

=

Rt

N = 16

13. Answer (40)



4/13

Hint : T RKE KE KE= +

Solution : 2 2 20

1 1 2. .

2 2 3+ =mv mR mgh

201 2

12 3

mv mgh

+ =

20

1 3

2 5 = =Tmv KE mgh

21 2

2 3 = =RI KE mgh

100 40%RKE

KE =

14. Answer (20)

Hint : H = R + R cos + ( )

2sin

2

v

g

Solution : Vertical component of velocity of ‘P’

vy = (v sin )

2 2sin

cos2

vH R R

g

= + +

Solving it for maxima

2

sin .2sin cos 02

dH vR

d g= − + =

100 sin

2sin .cos 02 10 2

− =

1

sin 10cos 02

− =

1

cos20

=

sec = 20

15. Answer (C)

Hint : Nuclear reaction.

Solution : In -decay:- mass decrease by 4 unit

and atomic number decreases by 2 unit. 2He4

nucleus emits with some energy.

In -decay:- atomic number increase by 1 unit

and mass number remain same. One electron

emits from nucleus accompanying with

antineutrino and some energy. In nuclear fusion

two light nucleus fuse to form a heavy nucleus.

In nuclear fission: A heavy nucleus (unstable)

breaks into two lighter nuclei.

16. Answer (D)

Hint : ( )

2

2 ;2

red

eq eq

mgmT PE

k K=

Solution : For (P); ( )

1 2

1 24 2= =

+eq eq

k k kK k

k k

( )

2

12 ;12

= =mgm

T PEk k

For (Q): 1 2

1 2

32

4 3eq

k k k mK T

k k k= = =

+

( )

2

2

mgPE

k=

For (R): 1 2 3

2 3 3 1 1 2

3

4 4 13eq

k k k kK

k k k k k k= =

+ +

( )

2

132 ;

3 2

mgmT PE

k k = =

For (S):

2eqm

K K Tk

= =

( )

2

12

mgPE

k=

17. Answer (B)

Hint : ; = = vBl F idl B .

Solution :

For (P): After some time the speed of both the rod becomes same and they will be moving in same direction.

For (Q): They may collide or even before collision the speed of I of the rod may change its direction and finally both of the rod will move with same speed in same direction.

For (R): There final separation becomes infinite.

For (S): There will collide and stick together and stop.

18. Answer (A)

Hint : 1

v

RC C

n= −

−for nPv = constant

Solution : For 3/2Pv = constant

Monoatomic ideal gas 3

2vC R= and C = –R/2

3

; ; 22 2

NR TQ U NR T W NR T

= − = = −



5/13

For P

v= constant, monoatomic ideal gas

3

2vC R= and C = 2R

32 ; ;

2 2

NQ NR T U NR T W R T = = =

For TV1/3 = constant; diatomic ideal gas

5;

2 2v

RC R C

−= =

5; ; 3

2 2

NR TQ U NR T W NR T

− = = = −

For 3

2

T

P= constant, diatomic ideal gas

5; 2

2vC R C R= =

5

2 ; ;2 2

NR TQ NR T U NR T W

− = = =

PART - II (CHEMISTRY)

19. Answer (B, D)

Hint :

Nucleophilic addition reaction of carbonyl

compounds.

Solution :

The attacking nucleophile is 2–

3SO ion and the

attack takes place through S-atom rather than

O-atom because S-atom is a better nucleophile

than O-atom.

– 2–

3 3HSO H SO+ +

The bisulphite adduct regenerates carbonyl

compound on reaction with (i) acid (ii) base

(i)

(ii)

20. Answer (B, C, D)

Hint :

AgNO3 reacts with excess of I2 to form AgI, HIO3

and HNO3.

Solution :

When excess of I2 reacts with AgNO3 in aqueous

solution, the following reaction takes place

5AgNO3 + 3I2 + 3H2O → HIO3 + 5AgI + 5HNO3

However, when excess of AgNO3 is treated with

I2 in aqueous solution then AgIO3 is formed

instead of HIO3 as per the following reaction.

6AgNO3 + 3I2 + 3H2O → AgIO3 + 5AgI + 6HNO3

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

Hint :

2

cell cell 2

0.06 ZnE E – log

2 Cu

+

+= .

Solution :

The cell reaction is

2 2Zn Cu Zn Cu+ ++ ⎯⎯→ +

0.8 M 0.2 M

0.5 M 0.5 M

0.9 M 0.1 M [Q = 0.8 F]

EMF of the cell at the beginning of discharge

Ecell = E°cell

2

2

0.6 Zn– log

2 Cu

+

+

( ) 0.06 0.2

0.34 – –0.76 – log2 0.8

=

0.06 1

1.10 – log 1.118 V2 4

= =

Ecell = E°cell when [Zn2+] = [Cu2+] = 0.5 M

EMF of the cell just after charging

Ecell = 0.06 1

1.10 – log 1.129 V2 9

=

Concentration of Cu2+ and Zn2+ just after charging

[Cu2+] = 0.5 + 0.4 = 0.9 M

[Zn2+] = 0.5 – 0.4 = 0.1 M

Ecell becomes independent of temperature when

Q = 0 i.e., [Zn2+] = [Cu2+]



6/13

22. Answer (A, C, D)

Hint :

Use the relation of 1

2

t and K for zero order,

1st order and 2nd order reaction.

Solution :

For zero order reaction (I)

1

02

at

2K= …(i)

For first order reaction (II)

1

12

0.693t

K= …(ii)

For second order reaction (III)

1

22

1t

aK= …(iii)

Initial concentration, a = 1 mol/litre

From (i), (ii) and (iii)

K1 = 1.386 K0; K1 = 0.693 K2 and K2 = 2K0

23. Answer (B, C, D)

Hint :

Preparation of N2 gas.

Solution :

(A) NaNO2 + 3Zn + 5NaOH + 5H2O → NH3 +

3Na2 [Zn(OH)4]

(B) 3 2 2

2NH 3CuO N 3H O 3Cu

+ ⎯⎯→ + +

(C) ( ) ( )6 24 2 33K Fe CN N Fe C 2 CN 12KCN C

⎯⎯→ + + + +

(D) 3 2

2NaN 2Na 3N⎯⎯→ +

24. Answer (A, B, D)

Hint :

Br2(H2O) oxidises aldoses and hemiacetals but

not acetals.

Solution :

Br2 water oxidises aldoses and hemiacetals but

does not oxidise acetals. Aldose are oxidised to

aldonic acids. The hemiacetals are first

hydrolysed to hydroxy aldehydes and then the

aldehyde part is oxidised to carboxylic acid.

Option (A) and (B) are hemiacetals and option (D)

is aldose. Option (C) is acetal.

25. Answer (17)

Hint :

a m

m

K;

C

= =

Solution :

–

3 3CH COOH H CH COO+ +

C(1 – ) C C

C22 a m

a

KCK C ;

1– C

= = =

C –5

–2m 1.764 10 4.2 10405 0.01

= =

Cm

17 = ohm–1 cm2 mol–1

26. Answer (22)

Hint :

For bcc unit cell, Density = ( )3

0

2 M

N a

.

Solution :

For bcc unit cell, Z = 2

Edge length, a = 300 pm = 3 × 10–8 cm

Density of the element, = 7.0 gm/cc

( )3

0

ZM

N a =

( )

323 8

2 M7.0 ; M 56.7

6 10 3 10−

= =

No. of atoms present in 208 gm of the element

23

236 10 208 22 1056.7

= =

x = 22

27. Answer (15)

Hint :

Reaction of CuSO4 with KCN.

Solution :

CuSO4 reacts with KCN as given below

( )24 2 4CuSO 2KCN Cu CN K SO 2+ ⎯⎯→ +

( ) ( ) ( )2 2 222Cu CN Cu CN CN⎯⎯→ +

( ) ( )

( ) ( )

2 42 3

4 24 3 2 4

Cu CN 6 KCN 2K Cu CN

________________________________________________

2CuSO 10 KCN 2K Cu CN 2K SO CN

+ ⎯⎯→

+ ⎯⎯→ + +

Total number of moles of distinct species

= 2(3 + 1) + 2 (2 + 1) + 1 = 15



7/13

28. Answer (40)

Hint :

Acetaldehyde reduces Fehling’s solution to give

red ppt. of Cu2O.

Solution :

CH3CHO + 2Cu2+ + 5OH– → CH3COO– +Cu2O +

3H2O

Red

No. of moles of CH3CHO = No. of moles of Cu2O

30.718

5 10143.6

−= =

Mass of CH3CHO = 5 × 10–3 × 44 = 0.22 gm

Percentage of CH3CHO in the mixture

0.22 100

40%0.55

= =

29. Answer (30)

Hint :

Cyanides do not react with HNO2 or acetyl

chloride.

Solution :

Since (A) and (B) do not react with HNO2 and

acetyl chloride. They cannot be amines. So they

must be cyanides.

Degree of unsaturation of (A) or (B)

= ( )2 4 2 – 7 – 1

2

+

= 2

So, compounds (A) and (B) are CH3CH2CH2CN and respectively. They undergo

hydrolysis on reaction with dil HCl to give corresponding acids.

( ) ( )

dil HCl

3 2 2 3 2 2A C

CH CH CH CN CH CH CH COOH

⎯⎯⎯⎯→

No. of moles of 0.66 gm of (C) or (D)

= –30.66

7.5 1088

=

Volume of 0.25 N — NaOH required to react with 0.66 gm of (C) or (D) completely,

–3V 0.25

7.5 10 ; V 30 ml1000

= =

30. Answer (12)

Hint :

Group-2 cations get precipitated by passing H2S in presence of dil. HCl.

Solution :

The cation which get precipitated by passing H2S gas in presence of dil HCl are Cu2+, Ag+, Pb2+, Hg2+, Cd2+, As3+, Sn2+, Sb3+, Bi3+

Rest of the given cations do not get precipitated with H2S in presence of dil. HCl.

No. of cations which will not get precipitated

= 21 – 9 = 12

31. Answer (21)

Hint :

At constant pressure 1 1 2 2

1 2

n T n T

V V= .

Solution :

Let the no. of moles of air initially present in the

vessel of volume (V) at 300 K be n.

No. of moles of air at 459 K = n – 7

Volume of the vessel at 459 K = 1.02 V

Since pressure is constant,

1 1 2 2

1 2

n T n T

V V=

( )n 300 n – 7 459

V 1.02 V

=

n = 21

32. Answer (90)

Hint :

Specific rotation is related to angle of rotation :

tDC I

=

Solution :

Specific rotation, tD is related to observed

angle of rotation as tD

C l

=

Where, C is the concentration in g/ml and l is the

length of polarimeter tube in dm

10

C 0.02 g / ml500

= =

l = 2 dm. and = –5.40°

D–5.40

–1350.02 2

= =



8/13

The optical purity of the final solution is 10/15 or

66.7% because 10 gm of optically pure fructose

has been mixed with 5 gm of racemic fructose.

The specific rotation of the mixture will be (10/15)

of the specific rotation of optically pure fructose.

t tD D2

– 135 –90 ; 903

= = =

33. Answer (A)

Hint :

CS2, C3O2, HgCl2 : Planar with sp hybridised

central atom.

–

4 3 3ICl , BCl , BrCl : Planar

Solution :

2–

2 2 3SnCl ,SO ,SO : Polar

2 2

ClO , NO , NO : Paramagnetic, planar and

polar.

34. Answer (D)

Hint :

Ag : Cyanide Process

Zr : van Arkel Process

Solution :

Ni : Mond’s Process

Al : Bayer’s Process

35. Answer (B)

Hint :

Cottrell precipitator : Electrical precipitation of

smoke

Micelles : Associated Colloid

Solution :

Multimolecular Colloid : Gold Sol

Butter : Emulsion

36. Answer (A)

Hint :

1– n

1 0

2

t A

Solution :

P : Zero order

Q : Third order

R : First order

S : Second order

PART - III (MATHEMATICS)

37. Answer (A, B)

Hint : Orthocentre (ABC) Incentre (DEF)

Solution: Here DEF is the pedal triangle of

ABC

Orthocentre of ABC is the incentre of DEF

Position vector of orthocentre

( ) ( ) ( )x EF y DF z DE

EF DF DE

+ +=

+ +

Also, OD = 2RcosB.cosC

i.e.,

( ) ( ) ( )2 cos cos

x EF y DF z DEx R B C

EF DF DE

+ +− =

+ +

( ) ( ) 2 cos cosDF y x DE z x PR B C − + − =

38. Answer (A, C, D)

Hint : Draw graph

Solution : The graph of ( )1cos cosy x−= is

The graph of ( )1sin siny x−= is

The graph of ( )1cos cosy x−= is



9/13

The graph of ( )1sin siny x−= is

39. Answer (A, B)

Hint : Assume parabolas ( )2 4y a x k= − and

( )2 4y b x k= − +

Solution : The equation of parabolas can be

taken as ( )2 4y a x k= − and ( )2 4y b x k= − +

A line parallel to the common axis is y = h

Then 2

,4

hA k h

a

+

and 2

,4

hB k h

b

− −

If ( ),P , then

2 21

,2 4 4

h hk k h

a b

= + − − =

2 1 1

24

h

a b

= −

i.e., locus of P is 2

24

y b ax

ab

− =

40. Answer (B, D)

Hint : Area of (ABCD)max = (Area of ACD)max

Solution : Area of quadrilateral ABCD is

maximum when area of ACD is maximum

distance of D from AC is maximum

i.e., cos – sin is maximum

2 cos4

= +

is maximum

7

4

=

and area 6

2 2 122

= = sq. units

(Since ABCD is a rectangle)

41. Answer (B, C)

Hint : G(x) = f(x) – f(x + 1)

Solution : Let G(x) = f(x) – f(x + 1)

So, G(0) = f(0) – f(1)

and G(1) = f(1) – f(2)

G(0) + G(1) = 0

G(0) and G(1) are of opposite sign

f(x) = f(x + 1) at least one in [0,1]

42. Answer (A, B, D)

Hint : Solve differential equation

Solution : 1/

2 2 1/

2 2

xxdy dy ex y e dx

dx y x= =

1/1 xe C

y − = − +

As ( )0

lim 1 1x

f x C→ −

= = −

So, 1/

1/

1 11

1

x

xe y

y e− = − − =

+

( )

1/

22 1/1

x

x

dy e

dx x e

=

+

0 0dy

x Rdx

−

i.e., 1/

1 1lim

21 xx e→=

+

and 1/

0

1lim 0

1 xx e→ +=

+



10/13

Graph of the function is

43. Answer (20)

Hint : Find range of g(x)

Solution : We have

( ) 1 12 2

2 22 sin 2 sin

1 1

x xg x

x x

− − = + = + + +

As, 1

2

2sin ,

2 21

x

x

− − +

So, 12

2sin 2, 1,0,1

1

x

x

− = − − +

Range of g(x) = {0, 1, 2, 3} for ( )( ) 0f g x x R

f(0) < 0 and f(3) < 0

Now, f(0) < 0 a – 2 < 0 a < 2

and ( )7

3 0 9 6 2 05

f a a a − + −

7

,25

a

Hence 1 27

, 25

k k= =

(10k1 + 3k2) = 14+ 6 = 20

44. Answer (48)

Hint : ( )3 3

3 32f x dx dx

− − −

Solution : Given ( ) '' 2 3,3f x x − −

( )3 3

3 3'' 2f x dx dx

− − −

f(3) – f(–3) ≥ –12

( ) '' 2 3,3f x x = − −

i.e., f(x) = –2x + C

As f(3) = 0 C = 6

f(x) = 6 – 2x

f(x) = 6x – x2 + C

As, f(0) = C = – 4

Hence f(x) = 6x – x2 – 4

Now, ( ) ( ) ( )20 0

6 4x x

g x f t dt t t dt= = − + −

( )3

23 43

xg x x x

− = + − in [–3, 3]

i.e., ( ) 2' 6 4 0g x x x= − + − = (put)

3 – 5, 3 5x = +

Now, g(–3) = 48, which is the maximum values in

[–3, 3]

45. Answer (02)

Hint : ( ) ( ) ( )2

0 0 02

a a af x dx f x dx f a x dx= + −

Solution : Let

( ) 20 3

3 cos 4sin sin 41 sin

xI x x x dx

x

= + +

+

( )( )

( ) ( )( )

2

20 3

2

3

3 cos 4sin sin 4

1 sin

3 cos 4sin sin 4

1 sin

x x x x

x

x x x xdx

x

+ += +

+

− − + + +

+

( ) 2 32

0 3

2 3 cos sin4 1 sin

1 sin

x x xI x dx

x

− = + +

+

( )

( )

232

0 3

'

6 sin cos4 1 cos

1 cos f xxf x

x xI x dx

x

− = + +

+

23 20

4 1 cos 2I x x

= + =

k = 2

46. Answer (27)

Hint : Find BD

Solution : In OED



11/13

1/2 1

cos2 4

= =

Also, 2 = −

2cot 2cot2 2

BD

= = −

= tan

2 15= (from BDC)

So, area of ABC = 1

2AC BD

1

3 2 15 3 152

= =

47. Answer (04)

Hint : Point of intersection (at1t2, a(t1 + t2)

Solution : ( ) ( )1 20, ; 0,Q at R at

( )( ) ( )1 2 1 2, ,P at t a t t h k +

Perpendicular through P to QR bisects QR

Area of PQR = ( )1 2 1 21

2a t t at t−

( )2

21 2 1 2

2

aC t t t t = −

( )( )4

24 2 21 2 1 2 1 24

4

aC t t t t t t = + −

4 2 2

2 2

4

4

a h k h

aa a

= −

( )4 2 24 4C h k ah = −

Locus is x2(y2 – 4ax) = 4C4

x = 4

48. Answer (84)

Hint : Use partial fraction

Solution :

( )( ) ( )( )

4 41 16 1 1

2 1 2 1 16 2 1 2 1r

r rT

r r r r

− + = =

− + − +

( )( )( )

21 14 116 2 1 2 1

rr r

= + +

− +

( )21 1 1 14 116 2 2 1 2 1

rr r

= + + −

− +

( ) ( )21 1 14 16

1 1 1

32 2 1 2 1

nS r

r r

= + +

−

− +

( )( )1 2 1 1 1

124 16 32 2 1

n n n n

n

+ + = + + −

+

3 2 5 1

12 8 48 16 2 1

n n n n

n

= + + +

+

So, A = 12, B = 8, C = 48, D = 16

i.e., A + B + C + D = 84

49. Answer (65)

Hint : ( ) ( )1f x x A B= + + , where ( )1

0A tf t dt=

and ( )1 2

0B t f t dt=

Solution : ( ) ( ) ( )1 1 2

0 0f x x x t f t dt t f t dt= + +

( ) ( )1f x x A B = + +

Where ( )1

0A tf t dt= and ( )

1 2

0B t f t dt=

Now,

( )( ) ( )1 13

1 2

000

1 13 2

t BA t t A B dt A t

= + + = + +

4A – 3B = 2 ……….. (i)

Again, ( ) 1 2

0

11

4 3

A BB t t A B dt

+= + + = +

8B – 3A = 3…………(ii)

Solving 25 18

,23 23

A B= =



12/13

i.e., ( )1 1

0 0

25 181

23 23f x dx x dx

= + +

p + q = 65

50. Answer (25)

Hint : Find '2

y

Solution : Let 2

11a C− = and 2

21a a C+ − =

i.e. C2 – C1 = a

1

1 1 2

2 sintan

cos

x xy

C C C x

− = − +

( )

( )

22

2 21 1 2

2

cos cos sin1 2 1

cossin1

cos

dy

dx

C x x x

C C C xx

C x

=

+ + −

+ +

2

1 1/4 2

1 2 1

1x

dy

dx C C C=

= − +

222

1 2

11

1

C

C C

− = +

Putting a = 2,

2x

dy

dx =

( )

( )

2

2

2 3 11

3 2 3 1

+ −=

+ +

1

0.502

= =

51. Answer (C)

Hint : Make perfect square

Solution :

(P) Let a, ar, ar2, …….. are in G.P.

Now, ar4 = 7! and ar7 = 8!

on dividing, we get r = 2 a = 315

Now, ( )315 2 1

2205 32 1

n

n−

= =−

(Q) 2 29cos sec 6cos 4sec 5 0x y x y+ − − + =

( ) ( )2 2

3cos 1 sec 2 0x y − + − =

1 1

cos and cos3 2

x y = =

No. of distinct ordered pairs (x, y) = 2 × 2 = 4

(R) Domain of f(x) is ( ) ( )0 ,1 1,

As ( ) ( )log 1 constantf x f x e= = =

f’(x) = 0 for all ( ) ( )0 ,1 1,x

(S) Given

( ) ( )( )2 2 0 0a c b a b c a b c+ − − + + +

f(–1) . f(1) < 0

So there must exist at least one solution in (–1, 1)

where f(x) = 0

As f(x) is a quadratic equation so has exactly one

solution.

52. Answer (B)

Hint : Put 1

x tx

+ =

Solution :

(P) ( ) ( )

( )

( )

22 2

2 1

22 1 2

22 1

22 1 2

1 1

1

11

1

x xI

x

xdx

x

+

−

+

−

+ − −= =

+

−

− +

( )

22 1

22 1 2

1

12

1

I

xdx

x

+

−

−= −

+

where ( )

( )

22

1 2 22

1/ 1/

111

11

a a

a a

dxx dxx

I

x xx

− −

= =

+ +

=

1

10

1

a

a

xx

− = +

(Putting 1

x tx

+ = )

So, I = 2.



13/13

(Q) Clearly (1,0) is the points of intersection of given

curves

Now, ( ) ( )( )2

' 2 log 2 logx

xf x xx

= +

Slope of tangent to two curve f(x) at (1,0) is m1 = 2

Similarly

( ) ( ) ( )2 log 2' 1 2 2logx x xdg x e x xdx

= − = +

Slope of tangent to the curve g(x) at (1,0) is 2

Two curves touch each other, so angle

between them is 0.

Hence cos = 1

(R) 22

3 ' 3 3 '0 'y

y y y xy yy x

− − =−

for y = 0 y = 0, no real x

for 2 3' 1 1y y x y y= = = =

The point is (1, 1)

(S) Put ( )3 5m = − + in ( )6 5 2 0n m n+ − =

We get, ( )( )6 5 3 5 2 0n kl n n+ + + =

2 230 15 45 0n n + + =

2 22 3 0n n + + =

1

1,2

n − = −

If 2n m= − = using 2 2 2 1m n+ + =

1

6 =

If 2 2

n m= − = − using 2 2 2 1m n+ + =

2

3 =

So 1

cos6

=

53. Answer (D)

Hint : 0 90oAB BC B = =

Solution :

(P) AB BC AC+ =

u v

BCu v

= +

Now, 0u v u v

AB BCu v u v

= − + =

o90B =

So, cos2A + cos2B + cos2C = –1 – 4cosA. cosB. cosC = –1

(Q) , ,OA a OB b OC c= = = and OD d=

So, AB CD BC AD CA BD + +

4a b b c c a= + + = area of ABC

k = 4

(R) As , ,a b c are coplanar

So, ( ) ( ) ( ) 0a b b c c a =

(S) Required probability 2

2

32

n

n

n C

C

+= where n = 668

54. Answer (A)

Hint : A3 = 0 index is 3

Solution:

(P) Each number in the product should have the last

digit as one of the digits 1,3,7,9

Required probability 4

4

4 16

62510= =

(Q) Let probability of showing even and odd number

be p and q respectively

i.e.,1

2p q= =

Probability that faces with even number show odd

number of times

2 1 2 2 1 2 2 31 3

2 1 22 1.......

n n n n

n nn

C p q C p q

C q

+ + −

++

= + +

+

1

2=

(R) Let (,, ) be the point of intersections of lines

i.e., (sinA + sinB + sinC) = 2d2

and (sin2A + sin2B + sin2C) =d2

sin2 sin2 sin2 1

sin sin sin 2

A B C

A B C

+ + =

+ +

1

sin sin sin2 2 2 16

A B C =

(S) A3 = 0, so index is 3

❑ ❑ ❑

All India Aakash Test Series for JEE (Advanced)-2020 MOCK ... › Aakash › s3fs...Mar 05, 2020...

Documents

Transcript of All India Aakash Test Series for JEE (Advanced)-2020 MOCK ... › Aakash › s3fs...Mar 05, 2020...