Test - 4A (Paper - 2) (Code-C) (Hints & Solutions) All ...€¦ · Test - 4A (Paper - 2) (Code-C)...

Test - 4A (Paper - 2) (Code-C) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019

1/13

1. (B, C)

2. (B, C)

3. (A, C)

4. (A, C)

5. (A, C, D)

6. (A, D)

7. (A, B)

8. (A, C, D)

9. (A, D)

10. (A, B, D)

11. (B, D)

12. (A, C)

13. (A, B)

14. (B, D)

15. (C, D)

16. A (P, S)

B (R, S)

C (P, S)

D (R, T)

17. A (Q, R)

B (S)

C (P)

D (T)

18. (30)

19. (16)

20. (45)

21. (A, D)

22. (A, B, C, D)

23. (A, B, C, D)

24. (A, C)

25. (A, C)

26. (A, C)

27. (A, B, D)

28. (A, B, C, D)

29. (A, C)

30. (A, D)

31. (C)

32. (A, B)

33. (A, B, D)

34. (A, B, D)

35. (A, B, C)

36. A (R, S)

B (R, S)

C (P, Q, T)

D (P, Q, T)

37. A (P, R, T)

B (Q, R)

C (Q, R)

D (R, T)

38. (08)

39. (28)

40. (25)

41. (A, B, C)

42. (B, C, D)

43. (C, D)

44. (A, C)

45. (A, B, D)

46. (A, B, C)

47. (A, B, C)

48. (A, B, C)

49. (A, C, D)

50. (A, B, D)

51. (A)

52. (B, C)

53. (A, D)

54. (A, B, C)

55. (A, C)

56. A (T)

B (Q, S)

C (R, S, T)

D (P, T)

57. A (P, Q)

B (P, Q, R, S)

C (R, S, T)

D (T)

58. (64)

59. (71)

60. (32)

PHYSICS CHEMISTRY MATHEMATICS

Test Date : 27/01/2019

ANSWERS

TEST - 4A (Paper-2) - Code-C

All India Aakash Test Series for JEE (Advanced)-2019

All India Aakash Test Series for JEE (Advanced)-2019 Test - 4A (Paper - 2) (Code-C) (Hints & Solutions)

2/13

PART - I (PHYSICS)

HINTS & SOLUTIONS

1. Answer (B, C)

Hint:

2dvv kvdx

Solution:

2dvv kvdx

ln2 = kD

ln2D

k

2dvkv

dt

0 2

0

2

v

v

dvkt

v

0 0

2 1kt

v v

0

1t

kv

2. Answer (B, C)

Hint:

Friction force will oppose the relative motion

Solution:

|vrel

| = 10 m/s

2(10)20 m

12 10

4

d

b = d × 0.8

3. Answer (A, C)

Hint:

E = 2

fnRT

Solution:

U = 3 5

22 2RT RT

U = 13

2RT U

avg =

13

6

RT

rms He

3( )

4

RTV

m

2rms N

3( )

28

RTV

m

2

rms (He)

rms (N )

28

4

V m

V m

= 7

4. Answer (A, C)

Hint:

Momentum will get transferred from lighter to heavier

every time.

Solution:

2 2

1 2

1 1

2 2MV mV mgh

mV2 = MV

1

V2 =

2Mgh

M m

For next time

mV2 = (M + m) V

3

2 2

2 3

1 1( )

2 2mV mgh M m V

2

2

1 2 1 2

2 ( ) 2 ( )

Mgh Mghm m mgh

M m M m

h =

2

2( )

hM

M m

5. Answer (A, C, D)

Hint:

Velocity of A will first increase and finally it is zero.

Solution:

At = 90°, velocity of end A is zero, it becomes I AOR.


3/13

221

2 3 2

MMg � �

3g �

cm

3 3

2 4

g gv � �

�

6. Answer (A, D)

Hint and Solution:

F = Weight of liquid above cylinder.

R

2

23

2

Rg R g F

� �

22

32

R gF R g

��

2

32

F R g

�

7. Answer (A, B)

Hint:

eff

2Tg

�

Solution:

eff

2Tg

�

8. Answer (A, C, D)

Hint:

( )

Tv

x

Solution:

0

Tv

x

�

0

dx T

dt x

�

00

Tx dx dt

�

�

3/2

0

23

Tt

� �

9. Answer (A, D)

Hint:

PV = constant

Solution:

= 5

3

0

2

V

P T0 0, 32 , 32P T

0 0

V1

V2

0

2

V

For left part : 0

0 1( )

2

VP P V

For right part : 0

0 232 ( )

2

VP P V

2

1

32V

V

5 3

2

1

32V

V

3/52

1

(32)V

V

V2 = 8V

1Also V

1 + V

2 = V

0

V1 + 8V

1 = V

0 V

1 =

0

9

V and V

2 =

08

9

V

Final pressure P =

5/3 5/3

0 0

5/3 5/3

0

( ) (9)

(2) ( )

P V

V

P =

5/3

0

9

2P

Now

1

10

0 1 1( )

2

VT T V

T

1 =

1/3

0

81

4T

And

1 1

0 0

0 2

832

2 9

V VT T

1/3

2 0

818

4T T

2

1

8T

T


4/13

10. Answer (A, B, D)

Hint:

Equivalent thermal resistance is 8

7

R where R is the

thermal resistance of each rod.

Solution:


7

R where R is the


Power =

eq

Temp. difference 80 720 W

8R R

TP = 50°C and T

q = 30°C

11. Answer (B, D)

12. Answer (A, C)

13. Answer (A, B)

Hint & Solution Q.Nos. 11 to 13

Hint:

5kx5 = 4kx

4 + 3kx

3

Solution:

5kx5 = 4kx

4 + 3kx

3

5x5 = 4x

4 + 3x

3...(i)

4kx4

= 3kx3 = 2kx

2 = kx

1

4x4

= 3x3 = 2x

2 = x

1

x4

= 3

4x3

5x5

= 43

4x3 + 3x

3 = 6x

3

14. Answer (B, D)

15. Answer (C, D)

Hint & Solution for Q.No. 14 and 15

Hint:

Apply the corresponding condition

Solution:

0

0sin

yvdxv

dt d

0cos

dyv

dt

y = 0cosv t

0

0

0

cossin

x

v tdx v dt

d

2 2

0

0

cos0 sin

2

v tv t

d

0cos

dt

v

2 2

0 0

2 200

cos sin0

cos2 cos

v d v d

vd v

= 30°

2

3 3

y yx

d

2

max4 3 2 3

d dx

d

4 3

d

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

Hint:

COM will move if there is external force.

Solution:


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

Hint:

Apply Pascal law to know pressure variation

Solution:


18. Answer (30)

Hint:

Draw vector diagram of acceleration

Solution:

A Ba a� �

= 10 m/s2

g s

in37°

19. Answer (16)

Hint:

Based on Archimedes principle

Solution:

mg = w A(h

2 – h

1) g

mg = wVg +

wA(h

3 – h

1) g

wVg +

wA(h

3 – h

1)g =

wA(h

2 – h

1) g

V = A(h2 – h

3)

A(h2 – h

3) =

wA(h

2 – h

1)

12 1 2

12 3 4w

h h

h h

2

w


5/13

20. Answer (45)

Hint:

Based on vector addition

Solution:

It form a sum of 6 vector

O

O

A0

A0

A0

OO =

0sin

2

sin2

nA

r

A = OO =

0

30sin 6

2

sin15

A

=

0

sin15

A

I 2

2 0

2( )

sin 15r

AA

=

0 014.9 15I I

As I = nI0 = 15I

0 n = 15

3n = 45

PART - II (CHEMISTRY)

21. Answer (A, D)

Hint:

Shape of 2z

d is different than other four orbital.

Solution:

All five d-orbital are degenerate.

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

Hint:

Cu

3 3 2 2 3573

4

CH Cl Si Me SiCl Me SiCl MeSiCl

Me Si

Solution:

Me2SiCl

2 + H

2O Me

2Si(OH)

2

HO OHSi

Me

Me

+ HO OHSi

Me

Me

HO OSi

CH3

CH3

Si

CH3

O

CH3

CH3 OSi

CH3

CH3

Si

CH3

CH3

O Si

CH3

CH3

Si

CH3

CH3

O CH3

Me SiCl + H O3 2

Silicones are surrounded by non-polar group so they

are water repelling.


Hint:

fH for 1st group element E

EF > ECl > EBr > ESolution:

fH for fluoride become less negative from Li to Cs

while reverse is true for Br, Cl and .

24. Answer (A, C)

Hint:

Superoxide are coloured while oxide and peroxide are

colourless. Element form coloured compound so it can

be Rb, Cs, K.

Solution:

MO2 + H

2O MOH + H

2O

2 + O

2

25. Answer (A, C)

Hint:

H H

OH

O

H

HO

H

O

H

H

O

H

4-hydrogen bond

Solution:

In ice each oxygen atom is surrounded tetrahedrally

by four other oxygen atoms at same distance.


6/13

26. Answer (A, C)

Hint:

internuclear axis

pyQ

pyT

TQy y

andp p form -bond

Solution:

+ bond

pz

pz


Hint:

P is SNF3.

Solution:

S : N : F

0.971875 0.969 2.912

1 1 3

Empirical formula mass = 103 g

So, formula is SNF3.

Structure:

N S

F

F

F


Hint:

Mendeleev law: properties of elements are function of

atomic weight.

Solution:

Mendeleev left the gap for Ga and Ge.

29. Answer (A, C)

Hint:

Na2O is only basic oxide.

Solution:

Amphoteric Neutral Acidic Basic

Al2O

3, As

2O

3, CO, NO, N

2O Cl

2O

3, N

2O

5Na

2O

SnO2, PbO

2, GeO

2

SnO, PbO CO2

30. Answer (A, D)

Hint:

In molten state C.N. of Al is four.

Solution:

State Co-ordination number

Crystalline AlCl3

6

molten AlCl3

4

31. Answer (C)

Hint:

Suniverse

= Ssurr

+ SSystem

surr

qS –

T

2 1System p

1 2

T PS nC ln n R ln

T P

Solution:

Pi = 10 atm

Pf = 1 atm

i

i

i

nRT 1 R 300V 30 R

P 10

f

f

f

nRT 1 R 300V 300 R

P 1

external pressure = 1 atm

W = – Pext

(V2 – V

i) = – 1 atm (300 R – 30 R)

= –270 × 0.0821 × 1 × 101.3

= –2245.52 J

Temperature is constant so T = 0

q = – W = 2245.52 J

surr

q –2245.52S –

T 300 = –7.49 J/K

2 1System p

1 2

T PS nC ln n R ln

T P

= 1 × 8.31 × 2.303 log 10

1

= 19.137 J/K

Suniverse

= +11.65 J/K

32. Answer (A, B)

Hint:

Wir = –P

ext (V

2 – V

1)

Solution:

Wir = – 1atm (300 R – 30 R)

= –2245.52 J


7/13

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

Hint:

Cis + Anti Racemic mxiture

Cis + Syn Meso

Trans + Anti Meso

Trans + Syn Racemic mixture

Solution:

dil. KMnO4/OH–/H

2O Syn

OsO4/NaHSO

3/H

2O Syn

Br2/CCl

4 Anti

MCPBA/H3O+ Anti

D2/Pt Syn

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

Hint:

3 3 2

Salt

CH COOH NaOH CH COONa H O

1mol1mol

Salt

CH3COOH + CH

3COONa : Buffer

CH3COOH + NH

4OH CH

3COONH

4 : Buffer

Solution:

(A) Acidic buffer:

a

SaltpH pK log

Acid

if H+ added pH

(B) Basic Salt

a

1 1pH 7 pK logC

2 2

as C pH

pH > 7

(C) CH3COOH + NaOH CH

3COONa + H

2O

1 2 mol 0 -

0 1 1 -

act as strong base, pH as H+ is added.

38. Answer (08)

Hint:

X = 4

Y = 3

Z = 1


Hint:

Ssys

> 0 Ssurr

< 0

Solution:

Temperature remain constant so value of H is zero.

2 1sys p

1 2

T PS nC ln nR ln

T P

2

1

T1

T So,

2

1

Tln 0

T

P1 > P

2

So 1

2

Pn R ln 0

P

at constant temperature, gas is expanding so it must

absorb the heat so q is positive for system.


35. Answer (A, B, C)

Hint and solution for Q. Nos. 34 and 35

Hint:

Let Fe3O

4 contain x mol of FeO and y mol of Fe

2O

3

2

a a

FeO CO Fe CO

2 3 2

3bb

Fe O 3CO 2Fe 3CO

Solution:

2

15.68mol of CO 0.7

22.4

72a + 160b = 39.2 ...(i)

a + 3b = 0.7 ...(ii)

b = 0.2

a = 0.1

mol of FeO = 0.1 mol

mol of Fe2O

3 = 0.2 mol

KMnO4 or K

2Cr

2O

7 oxidise only FeO.

meq of FeO = meq of K2Cr

2O

7 = meq of KMnO

4

0.1 × 1000 × 1 = 0.1 × 6 × y = 0.1 × 5 × x

x = 200

y = 166.67


8/13

Solution:

1° Amine CH3 – CH

2 – CH

2 – CH

2 – NH

2

CH – CH – CH – NH 3 2 2

CH – CH – CH – NH 3 2 2

CH – C – NH 3 2

CH3

CH3

CH3

CH3

x = 4

2° Amine CH3 – NH – CH

2 – CH

2 – CH

3

CH – NH – CH – CH 3 3

CH3

CH – CH – NH – CH 3 2 2

– CH3

y = 3

3° Amine CH – N – CH – CH 3 2 3

CH3

z = 1

39. Answer (28)

Hint:

Molar mass of A = 30

Molar mass of B = 58

Solution:

CH CH COONa3 2

Kolbe’s electrolysis

C H4 10

(B)

C H2 6

Sodalime

40. Answer (25)

Hint:

CH3

P =

COOH

Q =

Solution:

CH3 COOH

(P) (Q)

4KMnO /H2 3Cr On-heptane

PART - III (MATHEMATICS)


Hint:

Common roots will be get by subtracting the given two

equations.

Solution:

The equations are 2x4 – 2x3 + x2 + 3x – 6 = 0 ...(i)

and 4x4 – 2x3 + 3x – 9 = 0 ...(ii)

eq. (ii) – eq. (i): 2x4 – x2 – 3 = 0

(x2 + 1) (2x2 – 3) = 0

2x4 – 2x3 + x2 + 3x – 6 = (2x2 – 3) (x2 – x + 2)

and 4x4 – 2x3 + 3x – 9 = (2x2 – 3) (2x2 – x + 3)

Product of common roots = 3

2

Sum of all non-common roots = 1 3

12 2

Product of all non-common roots = 3

2 32

42. Answer (B, C, D)

Hint:

4 4 4 4 4 4 4 4

1 1 1 1 1 1 1 11 ... 1 ...

2 3 2 3 4 5 6 7

4 4 4

2 4 81 ...

2 4 8


9/13

Solution:

4 4 4 4 4 4 4 4

1 1 1 1 1 1 1 11 ... 1 ...

2 3 2 3 4 5 6 7

4 4 4

2 4 81 ...

2 4 8

= 3 3 2 3 3

1 1 11 ...

2 (2 ) (2 )

= 1 8

1 71

8

and 4 4

1 11 ...

2 3

= 4 4 4 4 4 4 4

1 1 1 1 1 1 11 ...

2 3 4 5 6 7 8

3 4 4

1 1 2 41 ...

2 2 4 8

= 3 3 2 3 3

1 1 1 1 1 11 ...

2 2 22 (2 ) (2 )

= 3

3

1

1 21

121

2

= 15

14

∵ 4 4 4

1 1 11 ...

2 3 4K

4 4 4 4 4

1 1 1 1 11 ... ...

3 5 2 4 6K

4 4 4 4 4

1 1 1 1 1 11 ... ...

163 5 1 2 3K

4 4

1 1 11 ...

163 5K K

= 15

16K

43. Answer (C, D)

Hint:

PS PS K represents hyperbola eccentricity

= SS

K

.

Solution:

Centre and radius of S1 = 0 is (–3, 0) and 1

and centre of radius of S2 = 0 is (2, 0) and 2

Required locus of centre of circles is:

2 22 22 3 1x y x y

2A = 1, SS = 2Ae = 5

which is a hyperbola with eccentricity = 5

44. Answer (A, C)

Hint:

Use of expansion formula

21

1 1 ...2!

n n n

x nx x

Solution:

1 3 1 2

2

2 3 2

3 4 2 2 4lim 1 1 2n

n

n nn n n

=

1 3 1 22 3 2

20

1 (3 4 2 ) 1 1 ( 2 4 ) 1

limh

h h h h h

h

=

2 3 2

20

1 11

1 1 3 3lim (3 4 2 ) (3 4

3 2!h

h h h h hh

23 2 2 2

1 11

1 2 22 ) ... 2 4 2 4 ...

2 2!h h h h h

= 11

6


Hint:

Use the sum of infinite series.


10/13

Solution:

Let y = sec tanx x y

y2 = (sec tan )x x y

2

2

1sec tan and sec tanx x y y x x

y y

sec (sec tan )

2 1

dy x x x

dx y

=

4 3 22 1

2(2 1)

y y y

y


Hint:

Rearrange the coefficient then use the law of expansion.

Solution:

We have,

4

2 4 2

1 2

0

(1 2 ) (1 ... )n

K n n n

K

K

a x x x C x C x x

On equating we get:

1 1 2 2 3 1 3., 2 , 2

n n n na C a C a C C

∵ a1 + a

3 = 2a

2

n3 – 9n2 + 26n – 24 = 0

n = 2, 3, 4


Hint:

Equation of circle which touching the line lx + my

+ n = 0 at (x1, y

1) is

(x – x1)2 + (y – y

1)2 + (lx + my + x) = 0

Solution:

Equation of tangent to parabola at 6 8,

5 5

is

4x – 3y = 0

Equation of circle is

2 26 8

(4 3 ) 05 5

x y x y

∵ Circle touches the x-axis:

= 2 8

or5 5

S1 x2 + y2 – 4x – 2y + 4 = 0

and S2 x2 + y2 + 4x – 8y + 4 = 0


Hint:

Equation of circle passing through the point of intersection

of two curves S = 0 and S = 0 is S + S = 0

Solution:

The equation of possible circle is:

x2 – y2 – a2 + (x2 – y) = 0 where is parameter

1 + = –1 = –2

Equation of circle is: x2 + y2 – 2y + a2 = 0

For possible point of intersection:

x2 – x4 = a2

x2 =

21 1 4

2

a

1 1,

2 2a

49. Answer (A, C, D)

Hint:

Use of modulus and general solution.

Solution:

f(x) = 2 2sin sin 1 cos2 sin 2

4x x x

= sin sin 2 2 cos24

x x

= sin sin |sin |2

x x

If f(x) = 0 then x = 0, , 3

,2

2, 3, 3 ,

2

4, 5,

52

If f(x) = 2 then x = , 2 , 4

2 2 2


Hint:

Napier analogy tan cot2 2

A B a b C

a b


11/13

Solution:

∵ tan cot2 2

A B a b C

a b

9 3

cot2 63 7

C

2

2

1 tan12

cos8

1 tan2

C

CC

and C2 = a2 + b2 – 2ab cosC

C = 6

r = s

, where

1 63 15 75 4

2 8 4

and semiperimeter (s) = 5 4 6

2

= 15

2

51. Answer (A)

Hint:

General equation of tangent is y = a

mx

m

and locus

is elimination of (x1, y

1)

Solution:

∵ Line y = a

mx

m

is always a tangent to parabola

∵ It pass through P(x1, y

1)

m2x1 – my

1 + a = 0

∵ m1, m

2 are root of the equation

m1 + m

2 =

1

1

y

xand m

1, m

2 =

1

a

x

2

1 1

1

4tan

6

y ax

a x

Equation of locus of P is: y2 – 4ax = 21

( )3

a x

(x + 7a)2 – 3y2 = 48a2

2 2

2 2

( 7 )1

48 16

x a y

a a

Eccentricity of hyperbola =

2

2

16 21

48 3

a

a

Directrices are: x = –13a, x = –a

and foci are (a, 0), (–15a, 0)

52. Answer (B, C)

Hint:

22

1

2 211

2 21 2

2

1

4

4 16

y a

xxa a l C

m m a a a

x

Solution:

y = a

mx

m

be a tangent passing through point

P(x1, y

1)

m2x1 – my

1 + a = 0

Let two tangents for m1 and m

2 intersect the y-axis at

(0, ) and (0, )

( – ) = 4C

1 2

4a a

Cm m

2

1

2 211

2 2

2

1

4

16

y a

xx C

a a

x

Required locus is:

22 4

4C

y a xa

53. Answer (A, D)

Hint:

cos1 cos

2 = 2 2

1 21 1m m

Solution:

y = a

mx

m

is tangent to parabola and it passes

through (x1, y

1) then

m2x1 – my

1 + a = 0

∵ m1 = tan

1 and m

2 = tan

2

then tan1 + tan

2 =

1

1

y

x and tan

1 tan

2 =

1

a

x


12/13

∵ cos1 cos

2 =

3

5

sec21 sec2

2 =

5

3

(1 + tan21) (1 + tan2

2) =

5

3

2 2 2 2

1 2 1 2

2

3m m m m

22

1

2 2

11 1

2 2

3

ya a

xx x

Required locus is: 2 2 22

23x ax y a

2

2

2 2

3

21

15 5

4 2

ax

y

a a


Hint:

Rearrange as (1 + x2 – x3)9 = {(1 + x2) – x3}9

Solution:

(1 + x2 – x3)9 = {(1 + x2) – x3}9

= 9C0 (1 + x2)9 – 9C

1 (1 + x2)8 x3 + 9C

2 (1 + x2)7 x6...

Total number of terms = 27 (since the term containing

x is not exist).

Coefficient of x8 = Coefficient of x8 in (1 + x2)9 +9C

2 {Coefficient gx2 in (1 + x2)8}

= 9C4 + 9C

2 7C

1 = 378

and a0 + a

2 + a

4 + ... = 9 91

1 32

=

93 1

2

55. Answer (A, C)

Hint:

Number of terms = n + r – 1Cr – 1

and general terms = 1 2

!

! ! ... !n

n

n n n

Solution:

Number of distinct terms = 15+4–1C4–1

= 18C3 = 816

Coefficient of 12

1 2 40a a a

The greatest coefficient = 1 3 3

15! 15!

(3!) (4!) (3!) (4!)

The coefficient of 15

2

15!1

0! 15! 0! 0!a

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

Hint:

Distribution of alike and different things among different

groups.

Solution:

(A) Number of selection of 4 different flavours out of 6

different flavours ice cream = 6C4 = 15

(B) 4 combination of 6 distinct objects in which

repetitions are allowed = 6–1+4C4 = 9C

4 = 126

(C) Selection of 4 cones of exactly three different

flavours = (The no. of ways to select 3 different

flavours) × (the no. of ways to select 4 cones of 3

different flavours)

= 6C3 × 3 = 60

(D) Similarly as (C) number of ways of selecting 2 or

flavours = 60 + 45 = 105

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

Hint:

Firstly check for domain then use the properties to solve

each of them.

Solution:

(A) –4x2 + 12x – 8 > 0, –4x2 + 12x – 8 1 and 4x – 5 0

31, 2 and

2x x

and 4x – 5 < 1 6

1,4

x

3 5

1,2 4

x

(B) |x – 1| + |x – 4| 3 x [1, 4]

(C) The possible values of x are 2, 3, 4, 11

(D) The values of x are 1 and 11

58. Answer (64)

Hint:

P(AB) = P(A) + P(B) – P(AB)


13/13

Solution:

Let A be the event that first die shows a number greater

than 3 and B the event that the second die shows a

number greater than 3.

P(A) = 3 1

6 2 and P(B) =

3 1

6 2

P(AB) = P(A) + P(B) – P(AB)

= 1 1 1

2 2 4

= 3

4

qp = 43 = 64

59. Answer (71)

Hint:

Let x = 2 1n and 2 1y n

then f(x) =

2 2x y xy

x y

Solution:

Let x = 2 1 and 2 1 thenn y n

x2 + y2 = 4n and x2 – y2 = 2

xy = 24 1n

f(n) =

2 2x y xy

x y

=

3 3

2 2

x y

x y

= 3 2 3 212 1 2 1

2n n

24

3 2

1

1( ) 49 1

2K

f K

= 31

(7 1)2

= 171

N – 100 = 71

60. Answer (32)

Hint:

x = 9K, y = 12K, x + y = 16K

2

4 41 0

3 3

K K

Solution:

∵ log9 x = log

12y = log

16 (x + y) = K (Say),

x = 9K, y = 12K, x + y = 16K

We have to find 4

3

yK

x

9K + 12K = 16K

12 16

19 9

K K

24 4

1 03 3

K K

5 1

2

y

x

1 5

is rejected2

∵

p = 5 and q = 2

qp = 25 = 32

��

Test - 4A (Paper - 2) (Code-D) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019

1/13

1. (A, B, D)

2. (A, D)

3. (A, C, D)

4. (A, B)

5. (A, D)

6. (A, C, D)

7. (A, C)

8. (A, C)

9. (B, C)

10. (B, C)

11. (B, D)

12. (A, C)

13. (A, B)

14. (B, D)

15. (C, D)

16. A (Q, R)

B (S)

C (P)

D (T)

17. A (P, S)

B (R, S)

C (P, S)

D (R, T)

18. (45)

19. (16)

20. (30)

21. (A, D)

22. (A, C)

23. (A, B, C, D)

24. (A, B, D)

25. (A, C)

26. (A, C)

27. (A, C)

28. (A, B, C, D)

29. (A, B, C, D)

30. (A, D)

31. (C)

32. (A, B)

33. (A, B, D)

34. (A, B, D)

35. (A, B, C)

36. A (P, R, T)

B (Q, R)

C (Q, R)

D (R, T)

37. A (R, S)

B (R, S)

C (P, Q, T)

D (P, Q, T)

38. (25)

39. (28)

40. (08)

41. (A, B, D)

42. (A, C, D)

43. (A, B, C)

44. (A, B, C)

45. (A, B, C)

46. (A, B, D)

47. (A, C)

48. (C, D)

49. (B, C, D)

50. (A, B, C)

51. (A)

52. (B, C)

53. (A, D)

54. (A, B, C)

55. (A, C)

56. A (P, Q)

B (P, Q, R, S)

C (R, S, T)

D (T)

57. A (T)

B (Q, S)

C (R, S, T)

D (P, T)

58. (32)

59. (71)

60. (64)

PHYSICS CHEMISTRY MATHEMATICS

Test Date : 27/01/2019

ANSWERS

TEST - 4A (Paper-2) - Code-D

All India Aakash Test Series for JEE (Advanced)-2019

All India Aakash Test Series for JEE (Advanced)-2019 Test - 4A (Paper - 2) (Code-D) (Hints & Solutions)

2/13

PART - I (PHYSICS)

HINTS & SOLUTIONS

1. Answer (A, B, D)

Hint:


7

R where R is the


Solution:


7

R where R is the


Power =

eq

Temp. difference 80 720 W

8R R

TP = 50°C and T

q = 30°C

2. Answer (A, D)

Hint:

PV = constant

Solution:

= 5

3

0

2

V

P T0 0, 32 , 32P T

0 0

V1

V2

0

2

V

For left part : 0

0 1( )

2

VP P V

For right part : 0

0 232 ( )

2

VP P V

2

1

32V

V

5 3

2

1

32V

V

3/52

1

(32)V

V

V2 = 8V

1Also V

1 + V

2 = V

0

V1 + 8V

1 = V

0 V

1 =

0

9

V and V

2 =

08

9

V

Final pressure P =

5/3 5/3

0 0

5/3 5/3

0

( ) (9)

(2) ( )

P V

V

P =

5/3

0

9

2P

Now

1

10

0 1 1( )

2

VT T V

T

1 =

1/3

0

81

4T

And

1 1

0 0

0 2

832

2 9

V VT T

1/3

2 0

818

4T T

2

1

8T

T

3. Answer (A, C, D)

Hint:

( )

Tv

x

Solution:

0

Tv

x

�

0

dx T

dt x

�

00

Tx dx dt

�

�

3/2

0

23

Tt

� �

4. Answer (A, B)

Hint:

eff

2Tg

�

Solution:

eff

2Tg

�


3/13

5. Answer (A, D)

Hint and Solution:

F = Weight of liquid above cylinder.

R

22

32

Rg R g F

� �

22

32

R gF R g

��

2

32

F R g

�

6. Answer (A, C, D)

Hint:

Velocity of A will first increase and finally it is zero.

Solution:

At = 90°, velocity of end A is zero, it becomes I AOR.

221

2 3 2

MMg � �

3g �

cm

3 3

2 4

g gv � �

�

7. Answer (A, C)

Hint:

Momentum will get transferred from lighter to heavier

every time.

Solution:

2 2

1 2

1 1

2 2MV mV mgh

mV2 = MV

1

V2 =

2Mgh

M m

For next time

mV2 = (M + m) V

3

2 2

2 3

1 1( )

2 2mV mgh M m V

2

2

1 2 1 2

2 ( ) 2 ( )

Mgh Mghm m mgh

M m M m

h =

2

2( )

hM

M m

8. Answer (A, C)

Hint:

E = 2

fnRT

Solution:

U = 3 5

22 2RT RT

U = 13

2RT U

avg =

13

6

RT

rms He

3( )

4

RTV

m

2rms N

3( )

28

RTV

m

2

rms (He)

rms (N )

28

4

V m

V m

= 7

9. Answer (B, C)

Hint:

Friction force will oppose the relative motion

Solution:

|vrel

| = 10 m/s

2(10)20 m

12 10

4

d

b = d × 0.8

10. Answer (B, C)

Hint:

2dvv kvdx

Solution:

2dvv kvdx


4/13

ln2 = kD

ln2D

k

2dvkv

dt

0 2

0

2

v

v

dvkt

v

0 0

2 1kt

v v

0

1t

kv

11. Answer (B, D)

12. Answer (A, C)

13. Answer (A, B)

Hint & Solution Q.Nos. 11 to 13

Hint:

5kx5 = 4kx

4 + 3kx

3

Solution:

5kx5 = 4kx

4 + 3kx

3

5x5 = 4x

4 + 3x

3...(i)

4kx4

= 3kx3 = 2kx

2 = kx

1

4x4

= 3x3 = 2x

2 = x

1

x4

= 3

4x3

5x5

= 43

4x3 + 3x

3 = 6x

3

14. Answer (B, D)

15. Answer (C, D)

Hint & Solution for Q.No. 14 and 15

Hint:

Apply the corresponding condition

Solution:

0

0sin

yvdxv

dt d

0cos

dyv

dt

y = 0cosv t

0

0

0

cossin

x

v tdx v dt

d

2 2

0

0

cos0 sin

2

v tv t

d

0cos

dt

v

2 2

0 0

2 200

cos sin0

cos2 cos

v d v d

vd v

= 30°

2

3 3

y yx

d

2

max4 3 2 3

d dx

d

4 3

d

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

Hint:


Solution:


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

Hint:


Solution:


18. Answer (45)

Hint:

Based on vector addition

Solution:

It form a sum of 6 vector

O

O

A0

A0

A0

OO =

0sin

2

sin2

nA

r

A = OO =

0

30sin 6

2

sin15

A

=

0

sin15

A

I 2

2 0

2( )

sin 15r

AA

=

0 014.9 15I I

As I = nI0 = 15I

0 n = 15

3n = 45


5/13

PART - II (CHEMISTRY)

21. Answer (A, D)

Hint:

In molten state C.N. of Al is four.

Solution:

State Co-ordination number

Crystalline AlCl3

6

molten AlCl3

4

22. Answer (A, C)

Hint:

Na2O is only basic oxide.

Solution:

Amphoteric Neutral Acidic Basic

Al2O

3, As

2O

3, CO, NO, N

2O Cl

2O

3, N

2O

5Na

2O

SnO2, PbO

2, GeO

2

SnO, PbO CO2


Hint:

Mendeleev law: properties of elements are function of

atomic weight.

Solution:

Mendeleev left the gap for Ga and Ge.


Hint:

P is SNF3.

Solution:

S : N : F

0.971875 0.969 2.912

1 1 3

Empirical formula mass = 103 g

So, formula is SNF3.

Structure:

N S

F

F

F

25. Answer (A, C)

Hint:

internuclear axis

pyQ

pyT

TQy y

andp p form -bond

Solution:

+ bond

pz

pz

26. Answer (A, C)

Hint:

H H

OH

O

H

HO

H

O

H

H

O

H

4-hydrogen bond

Solution:

In ice each oxygen atom is surrounded tetrahedrally

by four other oxygen atoms at same distance.

19. Answer (16)

Hint:

Based on Archimedes principle

Solution:

mg = w A(h

2 – h

1) g

mg = wVg +

wA(h

3 – h

1) g

wVg +

wA(h

3 – h

1)g =

wA(h

2 – h

1) g

V = A(h2 – h

3)

A(h2 – h

3) =

wA(h

2 – h

1)

12 1 2

12 3 4w

h h

h h

2

w

20. Answer (30)

Hint:

Draw vector diagram of acceleration

Solution:

A Ba a� �

= 10 m/s2

g s

in37°


6/13

27. Answer (A, C)

Hint:

Superoxide are coloured while oxide and peroxide are

colourless. Element form coloured compound so it can

be Rb, Cs, K.

Solution:

MO2 + H

2O MOH + H

2O

2 + O

2


Hint:

fH for 1st group element E

EF > ECl > EBr > ESolution:

fH for fluoride become less negative from Li to Cs

while reverse is true for Br, Cl and .


Hint:

Cu

3 3 2 2 3573

4

CH Cl Si Me SiCl Me SiCl MeSiCl

Me Si

Solution:

Me2SiCl

2 + H

2O Me

2Si(OH)

2

HO OHSi

Me

Me

+ HO OHSi

Me

Me

HO OSi

CH3

CH3

Si

CH3

O

CH3

CH3 OSi

CH3

CH3

Si

CH3

CH3

O Si

CH3

CH3

Si

CH3

CH3

O CH3

Me SiCl + H O3 2

Silicones are surrounded by non-polar group so they

are water repelling.

30. Answer (A, D)

Hint:

Shape of 2z

d is different than other four orbital.

Solution:

All five d-orbital are degenerate.

31. Answer (C)

Hint:

Suniverse

= Ssurr

+ SSystem

surr

qS –

T

2 1System p

1 2

T PS nC ln n R ln

T P

Solution:

Pi = 10 atm

Pf = 1 atm

i

i

i

nRT 1 R 300V 30 R

P 10

f

f

f

nRT 1 R 300V 300 R

P 1

external pressure = 1 atm

W = – Pext

(V2 – V

i) = – 1 atm (300 R – 30 R)

= –270 × 0.0821 × 1 × 101.3

= –2245.52 J

Temperature is constant so T = 0

q = – W = 2245.52 J

surr

q –2245.52S –

T 300 = –7.49 J/K

2 1System p

1 2

T PS nC ln n R ln

T P

= 1 × 8.31 × 2.303 log 10

1

= 19.137 J/K

Suniverse

= +11.65 J/K

32. Answer (A, B)

Hint:

Wir = –P

ext (V

2 – V

1)

Solution:

Wir = – 1atm (300 R – 30 R)

= –2245.52 J


7/13

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

Hint:

3 3 2

Salt

CH COOH NaOH CH COONa H O

1mol1mol

Salt

CH3COOH + CH

3COONa : Buffer

CH3COOH + NH

4OH CH

3COONH

4 : Buffer

Solution:

(A) Acidic buffer:

a

SaltpH pK log

Acid

if H+ added pH

(B) Basic Salt

a

1 1pH 7 pK logC

2 2

as C pH

pH > 7

(C) CH3COOH + NaOH CH

3COONa + H

2O

1 2 mol 0 -

0 1 1 -

act as strong base, pH as H+ is added.

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

Hint:

Cis + Anti Racemic mxiture

Cis + Syn Meso

Trans + Anti Meso

Trans + Syn Racemic mixture

Solution:

dil. KMnO4/OH–/H

2O Syn

OsO4/NaHSO

3/H

2O Syn

Br2/CCl

4 Anti

MCPBA/H3O+ Anti

D2/Pt Syn

38. Answer (25)

Hint:

CH3

P =


Hint:

Ssys

> 0 Ssurr

< 0

Solution:

Temperature remain constant so value of H is zero.

2 1sys p

1 2

T PS nC ln nR ln

T P

2

1

T1

T So,

2

1

Tln 0

T

P1 > P

2

So 1

2

Pn R ln 0

P

at constant temperature, gas is expanding so it must

absorb the heat so q is positive for system.



Hint and solution for Q. Nos. 34 and 35

Hint:

Let Fe3O

4 contain x mol of FeO and y mol of Fe

2O

3

2

a a

FeO CO Fe CO

2 3 2

3bb

Fe O 3CO 2Fe 3CO

Solution:

2

15.68mol of CO 0.7

22.4

72a + 160b = 39.2 ...(i)

a + 3b = 0.7 ...(ii)

b = 0.2

a = 0.1

mol of FeO = 0.1 mol

mol of Fe2O

3 = 0.2 mol

KMnO4 or K

2Cr

2O

7 oxidise only FeO.

meq of FeO = meq of K2Cr

2O

7 = meq of KMnO

4

0.1 × 1000 × 1 = 0.1 × 6 × y = 0.1 × 5 × x

x = 200

y = 166.67


8/13

COOH

Q =

Solution:

CH3 COOH

(P) (Q)

4KMnO /H2 3Cr On-heptane

39. Answer (28)

Hint:

Molar mass of A = 30

Molar mass of B = 58

Solution:

CH CH COONa3 2

Kolbe’s electrolysis

C H4 10

(B)

C H2 6

Sodalime

40. Answer (08)

Hint:

X = 4

Y = 3

Z = 1

PART - III (MATHEMATICS)


Hint:

Napier analogy tan cot2 2

A B a b C

a b

Solution:

∵ tan cot2 2

A B a b C

a b

9 3

cot2 63 7

C

2

2

1 tan12

cos8

1 tan2

C

CC

and C2 = a2 + b2 – 2ab cosC

C = 6

r = s

, where

1 63 15 75 4

2 8 4

and semiperimeter (s) = 5 4 6

2

= 15

2

42. Answer (A, C, D)

Hint:

Use of modulus and general solution.

Solution:

f(x) = 2 2sin sin 1 cos2 sin 2

4x x x

Solution:

1° Amine CH3 – CH

2 – CH

2 – CH

2 – NH

2

CH – CH – CH – NH 3 2 2

CH – CH – CH – NH 3 2 2

CH – C – NH 3 2

CH3

CH3

CH3

CH3

x = 4

2° Amine CH3 – NH – CH

2 – CH

2 – CH

3

CH – NH – CH – CH 3 3

CH3

CH – CH – NH – CH 3 2 2

– CH3

y = 3

3° Amine CH – N – CH – CH 3 2 3

CH3

z = 1


9/13

= sin sin 2 2 cos24

x x

= sin sin |sin |2

x x

If f(x) = 0 then x = 0, , 3

,2

2, 3, 3 ,

2

4, 5,

52

If f(x) = 2 then x = , 2 , 4

2 2 2


Hint:

Equation of circle passing through the point of intersection

of two curves S = 0 and S = 0 is S + S = 0

Solution:

The equation of possible circle is:

x2 – y2 – a2 + (x2 – y) = 0 where is parameter

1 + = –1 = –2

Equation of circle is: x2 + y2 – 2y + a2 = 0

For possible point of intersection:

x2 – x4 = a2

x2 =

21 1 4

2

a

1 1,

2 2a


Hint:

Equation of circle which touching the line lx + my

+ n = 0 at (x1, y

1) is

(x – x1)2 + (y – y

1)2 + (lx + my + x) = 0

Solution:

Equation of tangent to parabola at 6 8,

5 5

is

4x – 3y = 0

Equation of circle is

2 26 8

(4 3 ) 05 5

x y x y

∵ Circle touches the x-axis:

= 2 8

or5 5

S1 x2 + y2 – 4x – 2y + 4 = 0

and S2 x2 + y2 + 4x – 8y + 4 = 0


Hint:

Rearrange the coefficient then use the law of expansion.

Solution:

We have,

4

2 4 2

1 2

0

(1 2 ) (1 ... )n

K n n n

K

K

a x x x C x C x x

On equating we get:

1 1 2 2 3 1 3., 2 , 2

n n n na C a C a C C

∵ a1 + a

3 = 2a

2

n3 – 9n2 + 26n – 24 = 0

n = 2, 3, 4


Hint:

Use the sum of infinite series.

Solution:

Let y = sec tanx x y

y2 = (sec tan )x x y

2

2

1sec tan and sec tanx x y y x x

y y

sec (sec tan )

2 1

dy x x x

dx y

=

4 3 22 1

2(2 1)

y y y

y

47. Answer (A, C)

Hint:

Use of expansion formula

21

1 1 ...2!

n n n

x nx x


10/13

Solution:

1 3 1 2

2

2 3 2

3 4 2 2 4lim 1 1 2n

n

n nn n n

=

1 3 1 22 3 2

20

1 (3 4 2 ) 1 1 ( 2 4 ) 1

limh

h h h h h

h

=

2 3 2

20

1 11

1 1 3 3lim (3 4 2 ) (3 4

3 2!h

h h h h h

h

23 2 2 2

1 11

1 2 22 ) ... 2 4 2 4 ...

2 2!h h h h h

= 11

6

48. Answer (C, D)

Hint:

PS PS K represents hyperbola eccentricity

= SS

K

.

Solution:

Centre and radius of S1 = 0 is (–3, 0) and 1

and centre of radius of S2 = 0 is (2, 0) and 2

Required locus of centre of circles is:

2 22 22 3 1x y x y

2A = 1, SS = 2Ae = 5

which is a hyperbola with eccentricity = 5

49. Answer (B, C, D)

Hint:

4 4 4 4 4 4 4 4

1 1 1 1 1 1 1 11 ... 1 ...

2 3 2 3 4 5 6 7

4 4 4

2 4 81 ...

2 4 8

Solution:

4 4 4 4 4 4 4 4

1 1 1 1 1 1 1 11 ... 1 ...

2 3 2 3 4 5 6 7

4 4 4

2 4 81 ...

2 4 8

= 3 3 2 3 3

1 1 11 ...

2 (2 ) (2 )

= 1 8

1 71

8

and 4 4

1 11 ...

2 3

= 4 4 4 4 4 4 4

1 1 1 1 1 1 11 ...

2 3 4 5 6 7 8

3 4 4

1 1 2 41 ...

2 2 4 8

= 3 3 2 3 3

1 1 1 1 1 11 ...

2 2 22 (2 ) (2 )

= 3

3

1

1 21

121

2

= 15

14

∵ 4 4 4

1 1 11 ...

2 3 4K

4 4 4 4 4

1 1 1 1 11 ... ...

3 5 2 4 6K

4 4 4 4 4

1 1 1 1 1 11 ... ...

163 5 1 2 3K

4 4

1 1 11 ...

163 5K K

= 15

16K


Hint:

Common roots will be get by subtracting the given two

equations.


11/13

Solution:

The equations are 2x4 – 2x3 + x2 + 3x – 6 = 0 ...(i)

and 4x4 – 2x3 + 3x – 9 = 0 ...(ii)

eq. (ii) – eq. (i): 2x4 – x2 – 3 = 0

(x2 + 1) (2x2 – 3) = 0

2x4 – 2x3 + x2 + 3x – 6 = (2x2 – 3) (x2 – x + 2)

and 4x4 – 2x3 + 3x – 9 = (2x2 – 3) (2x2 – x + 3)

Product of common roots = 3

2

Sum of all non-common roots = 1 3

12 2

Product of all non-common roots = 3

2 32

51. Answer (A)

Hint:

General equation of tangent is y = a

mx

m

and locus

is elimination of (x1, y

1)

Solution:

∵ Line y = a

mx

m

is always a tangent to parabola

∵ It pass through P(x1, y

1)

m2x1 – my

1 + a = 0

∵ m1, m

2 are root of the equation

m1 + m

2 =

1

1

y

xand m

1, m

2 =

1

a

x

2

1 1

1

4tan

6

y ax

a x

Equation of locus of P is: y2 – 4ax = 21

( )3

a x

(x + 7a)2 – 3y2 = 48a2

2 2

2 2

( 7 )1

48 16

x a y

a a

Eccentricity of hyperbola =

2

2

16 21

48 3

a

a

Directrices are: x = –13a, x = –a

and foci are (a, 0), (–15a, 0)

52. Answer (B, C)

Hint:

22

1

2 211

2 21 2

2

1

4

4 16

y a

xxa a l C

m m a a a

x

Solution:

y = a

mx

m

be a tangent passing through point

P(x1, y

1)

m2x1 – my

1 + a = 0

Let two tangents for m1 and m

2 intersect the y-axis at

(0, ) and (0, )

( – ) = 4C

1 2

4a a

Cm m

2

1

2 211

2 2

2

1

4

16

y a

xx C

a a

x

Required locus is:

22 4

4C

y a xa

53. Answer (A, D)

Hint:

cos1 cos

2 = 2 2

1 21 1m m

Solution:

y = a

mx

m

is tangent to parabola and it passes

through (x1, y

1) then

m2x1 – my

1 + a = 0

∵ m1 = tan

1 and m

2 = tan

2

then tan1 + tan

2 =

1

1

y

x and tan

1 tan

2 =

1

a

x

∵ cos1 cos

2 =

3

5

sec21 sec2

2 =

5

3


12/13

(1 + tan21) (1 + tan2

2) =

5

3

2 2 2 2

1 2 1 2

2

3m m m m

22

1

2 211 1

2 2

3

ya a

xx x

Required locus is: 2 2 22

23x ax y a

2

2

2 2

3

21

15 5

4 2

ax

y

a a


Hint:

Rearrange as (1 + x2 – x3)9 = {(1 + x2) – x3}9

Solution:

(1 + x2 – x3)9 = {(1 + x2) – x3}9

= 9C0 (1 + x2)9 – 9C

1 (1 + x2)8 x3 + 9C

2 (1 + x2)7 x6...

Total number of terms = 27 (since the term containing

x is not exist).

Coefficient of x8 = Coefficient of x8 in (1 + x2)9 +9C

2 {Coefficient gx2 in (1 + x2)8}

= 9C4 + 9C

2 7C

1 = 378

and a0 + a

2 + a

4 + ... = 9 91

1 32

=

93 1

2

55. Answer (A, C)

Hint:

Number of terms = n + r – 1Cr – 1

and general terms = 1 2

!

! ! ... !n

n

n n n

Solution:

Number of distinct terms = 15+4–1C4–1

= 18C3 = 816

Coefficient of 12

1 2 40a a a

The greatest coefficient = 1 3 3

15! 15!

(3!) (4!) (3!) (4!)

The coefficient of 15

2

15!1

0! 15! 0! 0!a

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

Hint:

Firstly check for domain then use the properties to solve

each of them.

Solution:

(A) –4x2 + 12x – 8 > 0, –4x2 + 12x – 8 1 and 4x – 5 0

31, 2 and

2x x

and 4x – 5 < 1

61,4

x

3 5

1,2 4

x

(B) |x – 1| + |x – 4| 3 x [1, 4]

(C) The possible values of x are 2, 3, 4, 11

(D) The values of x are 1 and 11

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

Hint:

Distribution of alike and different things among different

groups.

Solution:

(A) Number of selection of 4 different flavours out of 6

different flavours ice cream = 6C4 = 15

(B) 4 combination of 6 distinct objects in which

repetitions are allowed = 6–1+4C4 = 9C

4 = 126

(C) Selection of 4 cones of exactly three different

flavours = (The no. of ways to select 3 different

flavours) × (the no. of ways to select 4 cones of 3

different flavours)

= 6C3 × 3 = 60

(D) Similarly as (C) number of ways of selecting 2 or

flavours = 60 + 45 = 105

58. Answer (32)

Hint:

x = 9K, y = 12K, x + y = 16K

2

4 41 0

3 3

K K

Solution:

∵ log9 x = log

12y = log

16 (x + y) = K (Say),

x = 9K, y = 12K, x + y = 16K

We have to find 4

3

yK

x


13/13

9K + 12K = 16K

12 16

19 9

K K

24 4

1 03 3

K K

5 1

2

y

x

1 5

is rejected2

∵

p = 5 and q = 2

qp = 25 = 32

59. Answer (71)

Hint:

Let x = 2 1n and 2 1y n

then f(x) =

2 2x y xy

x y

Solution:

Let x = 2 1 and 2 1 thenn y n

x2 + y2 = 4n and x2 – y2 = 2

xy = 24 1n

f(n) =

2 2x y xy

x y

=

3 3

2 2

x y

x y

= 3 2 3 212 1 2 1

2n n

24

3 2

1

1( ) 49 1

2K

f K

= 31

(7 1)2

= 171

N – 100 = 71

60. Answer (64)

Hint:

P(AB) = P(A) + P(B) – P(AB)

Solution:

Let A be the event that first die shows a number greater

than 3 and B the event that the second die shows a

number greater than 3.

P(A) = 3 1

6 2 and P(B) =

3 1

6 2

P(AB) = P(A) + P(B) – P(AB)

= 1 1 1

2 2 4

= 3

4

qp = 43 = 64

��

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