Test - 3A (Paper - 1) (Code-A) (Answers) All India Aakash ...€¦ · All India Aakash Test Series for JEE (Advanced)-2019 Test - 3A (Paper - 1) (Code-A) (Hints & Solutions) 6/17

Test - 3A (Paper - 1) (Code-A) (Answers) All India Aakash Test Series for JEE (Advanced)-2019

1/17

1. (C)

2. (D)

3. Delete

4. (C)

5. (C)

6. (B)

7. (A, C)

8. (A, C)

9. (A, D)

10. (A, C)

11. (A, C)

12. (A, C, D)

13. (C, D)

14. (C)

15. (A, B, D)

16. A(P, R, T)

B(Q, R)

C(Q, R)

D(Q, R, T)

17. A(Q)

B(Q, R)

C(T)

D(P, S)

18. (03)

19. (07)

20. (06)

21. (C)

22. (C)

23. (D)

24. (B)

25. (B)

26. (C)

27. (C)

28. (B)

29. (B, C, D)

30. (B)

31. (B, D)

32. (C)

33. (A, B, D)

34. (A, B, C, D)

35. (A, B, C, D)

36. A(P, R, T)

B(S)

C(R, T)

D(Q, S)

37. A(P, T)

B(Q)

C(R)

D(S)

38. (01)

39. (08)

40. (08)

41. (A)

42. (D)

43. (A)

44. (A)

45. (D)

46. (D)

47. (B, D)

48. (A, D)

49. (A, C)

50. (A, C)

51. (A, D)

52. (A, C)

53. (B, C)

54. (B, C)

55. (B, C)

56. A (Q, T)

B (R, S, T)

C (P, T)

D (Q, T)

57. A (P, R)

B (P, R, T)

C (P, R)

D (P, Q, R, S)

58. (02)

59. (07)

60. (08)

ANSWERS

Test Date: 25/11/2018

PHYSICS CHEMISTRY MATHEMATICS

TEST - 3A (Paper-1) - Code-A

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

All India Aakash Test Series for JEE (Advanced)-2019 Test - 3A (Paper - 1) (Code-A) (Hints & Solutions)

2/17

PART - I (PHYSICS)

HINTS & SOLUTIONS

1. Answer (C)

Hint:

Apply Snell’s law and trace the path.

Solution:75º

30º

45º

x

– cot 75º2 2

R Rx

1– tan15º2

R

2. Answer (D)

Hint:

Add angle of deviation at individual surface and

minimize.

Solution:

i

= (i – ) + ( – 2) + (i – )

= (2i + – 4)

sin = sini

cos2 – 4

cos

d i

di

2

–1 4 –sin

3

i

3. (Delete)

4. Answer (C)

Hint:

Apply lens equation for individual lens and trace the

image.

Solution:

After refraction by the first lens, image will be formed

in focal plane.

2f

f

1 4cos ,2

I f f

5. Answer (C)

Hint:

2

2

1

2K V

Solution:

2

2

1

2K V

Vr has to be same

1 1

0 0

2 K m

K m

Test - 3A (Paper - 1) (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019

3/17

6. Answer (B)

Hint:

Based on photoelectric effect.

Solution:

2

2

152.244 13.6 1–

z

n

2

2 2

1 11.224 13.6 –

( – 1)

z

n n

z = 2, n = 5

7. Answer (A, C)

8. Answer (A, C)

9. Answer (A, D)

Hint and solution of Q. Nos. 7 to 9

Hint :

Image will be formed due to combination of lenses +

single lens B.

Solution:

230 6030

d

r

x

1

1 1 1

30 15

V

V1 = 30 cm

2

1 1 1

60 40

V

V2 = 120 cm

120 cm from lens B

Ray diagram is shown in diagram. For image formed

due to lens 1 and 2 image is formed at 120 cm from

lens B. For image due to single lens 2, image is formed

at 60 cm from lens B.

Minimum radius will be at point, where both intersect

60 – 25

60 120

x

x

x = 10 cm

120

603030

10. Answer (A, C)

11. Answer (A, C)

12. Answer (A, C, D)


Hint :

Based on successive disintegration.

Solution:

0– A

A

dNN

dt

0–

0

1–

t

AN e

0 0– 2 B

A B

dNN N

dt

0 0 02 – 20

0

1–

t t t

B

de N e e

dt

0 0 02 2

0

–

t

t t t

Be N e e dt

0 0

0

2

2

0 0

–1 –1–

2

t t

B t

e eN

e

0 0

–2

0

1 – 22

t te e

0–

0 0

0 0

( ) 1–

t

tE t E e dt

= 0

–

0 0

0

|

t

teE t

= 0

–

0

0 0

1–

te

E t

= 0–0

0

0

–1

tE

t e


4/17

13. Answer (C, D)

14. Answer (C)

15. Answer (A, B, D)


Hint :

Image will be alternatively real and virtual. And image

will be always enlarged. And, speed is minimum when

object crosses principal axis. And, lens of image can

be hyperbolic.

Solution:

Lens of image will be like this

F

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

Hint:

Emission will stop, when potential of sphere is equal

to stopping potential.

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

Hint:

Image will be formed on object when ray retraces its

path.

Solution:

Ray should fall normally or on the pole of mirror to

retrace its path.

18. Answer (03)

Hint:

2

2

1 11–

Rzn

Solution:

2

2

1 11–

Rzn

2

2 3

2d Rz

n

2

3 2

2 1

d Rz

n RZ

3 x n

d

19. Answer (07)

Hint:

fabsorption

freflection

IP

c

Solution:

0

reflection

1

2

I Af

c

0absorption

2

I Af

c

02 cos 30º2

I AF

c

03

2 I A

C

20. Answer (06)

Hint: Apply lens formula

Solution:

1 1 1

3x x f...(i)

1 1 1

4 2 4x x f...(ii)

From (i) and (ii),

x = 32

3 3 32

24 cm4 4

xf


5/17

21. Answer (C)

Hint:

Protonation occurs such that conjugate acid has

resonance or highest stability

Solution:

CH — C — NH3 2

O

H+

H+

CH — C — NH3 2

O — H(+)

CH — C — NH3 3

+

O

N NH2

H+

H+

N NH2

+

H

N NH3

+ (No resonance)

22. Answer (C)

Hint:

Rate of P > Q

Rate of R > S

As positive charge on — C —

O

reduces, rate of reaction

decrease.

Solution:

Ph — C — N

O

Ph — C == N

O–

+

I II

Contribution of II is very less because angle strain

increases.

Ph — C — NH2

O

Ph — C == NH2

O

+

III IV

There is no such problem in IV so it donates its lone

pair for resonance and reduce the positive charge of

— C —

O

group.

In experiment II, molecule ‘C’ has bridge head nitrogen,

so it does not donate its lone pair to — C —

O

so in ‘C’

there is highly positive charge on — C —

O

.

23. Answer (D)

Hint:

As stability of enol form increases, equilibrium constant

of keto to enol increases.

Solution:

N C — CH — CO C H2 2 2 5

O

NC == CH — CO C H2 2 5

O

H

Due to hydrogen bonding, K1 is more than K

2 and K

3.

O OH

K4

OH

K5

O

Due to extended conjugation, K5 is larger than K

4.

PART - II (CHEMISTRY)


6/17

24. Answer (B)

Hint:

Ethyl acetoacetate

CH — C — CH — C — OC H3 2 2 5

O O

Methyl Vinyl Ketone CH == CH — C — CH2 3

O

So, P = H C3

C — OC H2 5

O

O

Q =

CH3

CH3

Solution:

O O

C H ONa2 5

O O

O

O == C — OC H2 5

CH

CH2

CH2

C

O

CH3

CH — C3

O

C H ONa2 5

C H — O — C 2 5

OCH

3

O

25. Answer (B)

Hint:

Ph — C — Ph

O18

Peroxy acidPh — C — O — Ph

O18

Solution:

Ph — C — Ph

O18

H+

Ph — C — Ph

18O

H

CH COOO3

Ph — C — O — O — C — CH 3

18OH O

Phenyl migration

Ph — C — O — Ph

18OH

(+)

–H(+)


18O

(+)

Ph

26. Answer (C)

Hint:

cis + syn Meso

cis + anti Racemic

Trans + syn Racemic mixture

Trans + anti Meso

Solution:

(I) OsO4

(ii) H O, NaHSO2 3

Meso(A)

Racemic mixture(B)

(i) MCPBA

(ii) H O2

Baeyer's reagent

Cold KMnO4

Racemic

Meso(i) MCPBA (Cl—mC H COOOH)

6 4

(ii) H O2

(C)

(D)

27. Answer (C)

28. Answer (B)


7/17

29. Answer (B, C, D)


Hint :

O

O

N K– +

OO

Br

EtO OEt +

C

C

C

O

O

N — C — H

C — OEt

C — OEt

O

O

NaOEt

O

O

N — C — CH2

C

C

O

O

N — C

C OEt

C OEt

O

O

CH — S — CH2 3

C — OEt

O

C — OEt

O

X = Cl

SCH3

(i) dil. NaOH

(ii) HCl

COOH

COOH

+ CH — S — CH —CH — C — NH3 2 2 2

H

COOH

By changing the X we can change the -amino acid.

Solution:

Methionine CH — S — CH — CH — C — NH3 2 2 2

H

COOH

To prepare methionine

X = CH3 — S — CH

2 — CH

2 — Cl

If X = Br

then R = ( )CH CH — CH — C — NH3 2 2 2

COOH

H

E =

COOH

COOH

E on reaction with CH — CH2 2

OH OH

form glyptal.

30. Answer (B)

31. Answer (B, D)

32. Answer (C)


Hint :

P = — C — H

O

Q = — C — C —

O OH

R = — C — C —

O O

S = — C — COOH

OH

T = O

OH


8/17

Solution:

C

O

H CN

–

Benzoincondensation

— C — C —

O OH

H

(A) (B)

Oxidation

— C — C —

O O

OH –

Benzilic acidrearrangement

Ph — C — COOH

OH

Ph

O

OH

(i) MeCO O/MeCO( )2 2

–

(ii) H O3

+Ph — C — H

OPerkin reaction


Hint:

CH3

CH3

OH

H

OH

H

HO

H

Meso

Solution:

NH2

H

NH2

H ,

C == C == C

CH3

H

H

CH3

are optically active

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

Hint:

For bridge head only cis is possible so optical isomer

less than 2n

Solution:

H

O

Me Me

Me

Only cis form is possible so only one pair of enantiomer

is possible.

CH3

CH3

+

CH3

CH3

(Meso) (d + l)

So, 3 isomers


Hint:

Cl

Cl

Cl

Cl

Cl

Cl

9 isomers

OH

OH

OH

Total 4 isomers

Racemic mixture does not show optical activity but

has chiral compound.

Solution:

OH

OH

Meso Meso

OH

OH

OH

OH

OH

OH

(d + l)

OH

For

Cl

Cl

Cl

Cl

Cl

Cl

7 Meso + 1 dl pair


9/17

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

Hint:

Hydroboration oxidation gives syn addition; inmercuration - demercuration no carbocationrearrangement takes place, 3º alcohol give immediatewhite turbidity with lucas reagent.

Alkene on reaction with dil H2SO

4 gives rearranged

product.

Solution:

H.B.O.(i) BH - THF

3

(ii) H O ; OH2 2

–

HCH3

HOH

+ enantiomer

Mercuration-demercurationOH

Mercuration-demercuration

OH

dil.H SO2 4

(+)OH

H O2

(+)

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

Hint:

Aldehyde reduce the Tollens reagent, ketone, ether

does not reduce Fehling reagent and — C — CH

O

group

give positive iodoform test.

Solution:

C5H

11 — OH 8 alcohol

1º alcohol C4H

9 — CH

2 — OH

CH3CH

2CH

2CH

2CH

2OH

CH — CH — CH CH OH 3 2 2

CH3

CH — C — CH OH 3 2

CH3

CH3

CH — CH — CH — CH OH3 2 2

CH3

(Optically active)

These alcohol on oxidation form aldehyde that reduce

Tollens as well as Fehling reagent but not give iodoform

test.

CH CH CH CH CH3 2 2 3

OH,

CH — CH — CH CH CH3 2 2 3

OH

(optically active)

CH — C — CH — CH 3 3

H

CH3

OH

(optically active)

CH — C — OH 3

CH3

CH3

Ether:

CH3CH

2CH

2CH

2 — OCH

3

CH — CH — CH — OCH3 2 3

CH3

CH — C — OCH3 3

CH3

CH3

CH — CH — CH — OCH3 2 3

CH3

(optically active ether)

CH3CH

2CH

2 — OC

2H

5

CH — CH — OC H3 2 5

CH3

Total isomer = 14

38. Answer (01)

Hint:

PhN2

+ attack at position (1).

Solution:

OH

PhN2

+

OH

N = N — Ph


10/17

41. Answer (A)

Hint:

Dividing both sides by d

dx

equation becomes linear

differential equation, then use process of linear differentialequation.

Solution:

Given equation is

dy d dy x

dx dx dx


dx

, we get

( )dy

y xd

...(i)

1.I.F.

de e

Solution of equation (i) is given by

y e e d

– 1y e e d e d

1y e e e C e C

1y Ce

1

x

y x Ce

42. Answer (D)

Hint:

Differentiate both sides w.r.t. x and then put f(x) = y,

dyf x

dx our equation becomes 2

2

2

1

dy xy y

dx x

Dividing both sides by y2 and putting 1

zy

, our

equation reduces to linear differential equation.

Solution:

From given differential equation

2

2 2

0

32 2

xf x f t dt

x t

,

where ( ) 2 3 0 6f x

Differentiate both sides w.r.t. x

2 2

2 22

2 – 2 ( )0

22

x f x x f x f x

xx

2

2

2–

2

xf x f x f x

x

2

2

2

2

dy xy y

dx x

2 2

1 2 11

2

dy x

dx yy x

...(i)

Put2

1 1 dy dzz

y dx dxy

Equation (i) becomes

2

21

2

dz xz

dx x

...(ii)

I.F. = 22

2

log (2 ) 222e

xdx

xxe e x

Solution of equation (ii) is

2 22 2z x x dx

PART - III (MATHEMATICS)

39. Answer (08)

Hint:

Anisole, m-cresol and p-cresol is weaker acid than

phenol

Solution:

O

HO OH

HO

HO

O

O

Triangular acid Squaric acid

NO2

OH

O N2

NO2

Picric acid

40. Answer (08)

Hint:

a, b, c, d, e, g, h and i are correct.

Solution:

Glucose 2 2Br /H O gluconic acid


11/17

3

22 2

3

xz x x c

3

212 2

3

xx x c

y

where x = 0, (0)

12 0 0 0 c

y

2 1

6 3c

3

21 12 2

3 3

xx x

y

2 32 6 1

3

x x x

y

23

3 2

6 1

xy f x

x x

3 2 4 182

8 12 1 19f

43. Answer (A)

Hint:

Volume of tetrahedron is obtained by

2 1

36

a a a b a c

V b a b b b c

c a c b c c

��

� � � ��

��

Solution:

Volume of tetrahedron, 1

6V a b c

��

2

2 1

36V a b c

��

1

36

a a a b a c

b a b b b c

c a c b c c

��

� � � ��

��

...(i)

But 2 2 2| | 1, | | 1, | | 1a a a b b b c c c � � �

� � � � � �

1| || | cos60

2b a a b a b � � �

� � �

1| || | cos60

2b c c b b c � � �

� � �

1| | 1| | cos60

2a c c a c a � � � � � �

From (i)

2

1 11

2 2

1 1 11

36 2 2

1 11

2 2

V

1 1 1 1 1 1 1 11 1

36 4 2 2 4 2 4 2

1 3 1 2 1 1 1 2

36 4 2 4 2 4

1 3 1 1 1 3 1

36 4 8 8 36 4 4

1 2 1

36 4 72

1

6 2V cubic units

44. Answer (A)

Hint:

First use the formula for cross product of 4 vectorsand then use the concept for scalar triple product.

Solution:

a b b c p b c p c b p b c � � � � �

� � � � � � � �

–a b c b a b b c

� � � ��

a b c b

� ��

Similarly, –b c c a b c a c b c c a

� � ��

a b c c

��

and –c a a b c a b a c a a b

� � ��

a b c a

��

, ,a b b c b c c a c a a b

� � � ��

, ,a b c b a b c c a b c a

� � � ��

a b c b a b c c a b c a

� � � ��


12/17

2

a b c a b c b c a

� � ��

3

a b c b c a

� ��

4

a b c

��

43 , since 3a b c

��

= 81

45. Answer (D)

Hint:

First find the coordinates of point P using condition

for intersection of line and plane and then point Q.

Then find distance between P and Q using distance

formula.

Solution:

Here equation of line is

2 3 4

1 2 3

x y z say ...(i)

Hence any point on the line (i) will be p( + 2, 2 –

3, –3 + 4)

Given line (i) intersect the given plane

2x + 3y – z = at the point P

2( + 2) +3(2–3) + 3 – 4 = 13

11 – 9 = 13

11 = 22

= 2

P = (4, 1, –2)

Also equation (i) intersects the YZ – plane at Q

i.e. x = 0

+ 2 = 0 = –2

Q (0, –7, 10)

PQ = 2 2 24 0 1 7 2 10

16 64 144

224 16 14 4 14 a b

4 146

3 3

a b

46. Answer (D)

Hint:

Use Bayes' theorem.

Solution:

Let E be the event that visiting lecturer visit the college

late and let E1, E

2, E

3, E

4 be the events that the lecturer

comes by train, bus, scooter and other means of

transport respectively.

1 2 3 4

3 1 1 2, , ,

10 5 10 5P E P E P E P E

1

EP

E

= probability that the lecturer arriving late

by coming train = 1

4

2 3 4

1 1, and P 0

3 12

E E EP P

E E E

, same

he is not late if he comes by other means oftransport.

By Bayes’ theorem

1

11

1 2

1 2

3 4

3 4

EP E P

EEP

E E EP E P P E P

E E

E EP E P P E P

E E

3 1

10 4

3 1 1 1 1 1 20

10 4 5 3 10 12 5

3 3

40 40

3 1 1 9 8 1

40 15 120 120

3 120 1

40 18 2

47. Answer (B, D)

Hint:

First find centroid of G of QRS using formula as

3

a b c �

� �

if vertices are , ,a b c�

� �

and then find PG��

and

its magnitude.


13/17

Solution:

(1, 0, 2)

Q(–2, 3, 4)

R(1, 3, 5)

S(0, –2, 3)

P

h

x PQ PR ��

�

2

G

1

L

M

Here G is the centroid of QRS

Positive vector of G is

ˆ ˆ ˆ2 1 0 3 3 2 4 5 3

3

i j k

i.e. ˆ ˆ ˆ4 12

3

i j k

ˆ ˆ ˆ4 12ˆ ˆ2

3

i j kPG i k

��

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ4 12 3 – 6 4 4 6

3 3

i j k i k i j k

16 16 36 68 2 17| |

9 9 9 9 3PG ��

which is

irrational number.

48. Answer (A, D)

Hint:

Area of PQR = 1| |

2PQ PR��

Solution:

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ2 3 4 – 4 3 3 2PQ i j k i k i j k ��

ˆ ˆ ˆ ˆ ˆ ˆ ˆ3 5 2 3 3PR i j k i k j k ��

ˆ ˆ ˆ

3 3 2

0 3 3

i j k

PQ PR ��

ˆ ˆ ˆ9 6 – 9 0 9 0i j k

ˆ ˆ ˆ3 9 9i j k x �

Area of PQR = 1

2PQ QR��

19 81 81

2

1 1 19 162 171 9 19

2 2 2

319

2 square units which is irrational number.

49. Answer (A, C)

Hint:

Length of perpendicular from vertex S opposite face =

| |

PS n

n

��

�

Solution:

Length of the perpendicular from vertex S to opposite

face = h = | Projection of |PS on x��

ˆ ˆ ˆ ˆ ˆ ˆ ˆ2 3 2 3 9 – 9

| | 9 81 81

j k i k i j kPS x

x

��

�

�

3 18 9

90 81

30 3010 10

171 3 19 19

50. Answer (A, C)

Hint:

– 0b c a �

� �

Solution:

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ– 2 – 2 3b c i j k i j k j k �

�

and ˆ ˆ ˆ ˆ– 3 3 3 3 0b c a j k i j k � � � �

–b c�

�

is orthogonal to a�

51. Answer (A, D)

Hint:

–a b c a c b b c a � � �

� � � � � �

Solution:

a b c �

� �

–c a b �

� �

– –c b a c a b

� ��

Comparing, we get 1 2 2 1b c �

�

1 3 2 6a c � �

0

1 6 0 7 which is a prime number.


14/17

52. Answer (A, C)

Hint:

Let ˆd l b c m c a n a b � �

� � � �

, then find l, m,

n as ˆ ˆ ˆ

, ,a d b d c d

l m n

a b c a b c a b c

��

� � ��

ˆ ˆ ˆ â b c d a d b c b d c a c d a b

� � � ��

Solution:

Here

1 3 1

1 2 1

1 1 2

a b c

��

1 4 2 3 2 1 1 1 2

5 – 9 – 1 –5 0

,, ca b�

��

are non-coplanar

, ,b c c a a b � �

� � � �

also are non-coplanar

ˆd l b c m c a n a b � �

� � � �

ˆ ˆ ˆ

, ,

��

� � ��

a d b d c dl m n

a b c a b c a b c

ˆ ˆ ˆd a b c a d b c b d c a c d a b

� � � � ��

ˆ ˆ â d b c b d c a c d a b � � �

� � � � � �

ˆd a b c

��

a b c

��

= modulus of

1 3 1

1 2 1

1 1 2

1 4 1 3 2 1 1 1 2

5 9 1 5 which is a prime number

53. Answer (B, C)

Hint:

Required prob. = 1

2 3

1 1

6

3

x x x

x

C C C

C

Solution:

Bag

x red balls

2 green ballsx

3 blue ballsx

P(A)| all different colour

2 3

1 1 1

6

3

x x x

x

C C C

C

2 3

6 6 1 6 2

3!

x x x

x x x

3 2

6 6 6

6 6 1 6 2 6 1 2 3 1

x x

x x x x x

23

0.3 given6 1 3 1

x

x x

2

2

2

3 38 9 1 0

1018 9 1

xx x

x x

1 8 1 0x x

1

1,8

x

But x is number of balls

1

8x

x = 1

54. Answer (B, C)

55. Answer (B, C)

Solution for Q. Nos. 54 and 55

Blue

(3 )xGreen

(2 )xRed

( )x

22

01

10

01

22

12

10

01

22


15/17

3 2 2 3 3 2

2 1 1 2 1 1 2 1 1

6

3

( )

x x x x x x x x x

x

C C C C C C C C CP B

C

3 3 1 3 2 2 1 4 1 5

2 2 2

6 6 1 6 2

6

x x x x x x x x x

x x x

2

27 9 16 8 5 52

6 1 6 2

xx x x

x x x

48 2224 112

6 1 2 3 1 2 6 1 3 1

xx

x x

x x x x

when x = 2, P(B) =

2 48 11 37

2 11 5 55

when x = 1, P(B) = 13 13

2 5 2 20

when x = 3, P(B) = 183

272

P(C) =

2

2

5 1

5 12

2 24 1124 11

x x

x

xx x

when x = 1, P(C) = 0

when x = 2, P(C) = 10 5

2 74 74

when x = 3, P(C) = 5 2 5

2 61 61

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

(A) Hint:

Use –a b c a c b a b c � � � � �

� � � �

Solution:

a a b b b c � � �

� � �

– –a b a a a b b c b b b c � � � � � �

� � � � � �

...(i)

Taking dot product of both sides of (i) by b c�

�

, we

have – 0 0 0a b a b c

� ��

But 0 0a b a b c

� ��

(B) Hint:

Use formula for a b c �

� �

Solution:

Given 2a b

22

2a b �

�

2 2| | 2 2a b a b � �

� �

1 1 2 2 0a b a b � �

� �

Now a b c a c a b a b

� � ��

a b a b a b a b � � � �

� � � �

a b a a b b � � �

� � �

a a b b a b � � �

� � �

– –a b a a b b b b a b a b

� � � � � � ��

0 0 0b a a b

� ��

∵

a b �

�

2 22

2 2a b c a b � �

� � �

(C) Hint:

a b c a b c

� ��

Solution:

a a bb a ba b

� � � ��

a a b b a b a b � � � �

� � � �

0 � � � �

� � �

a a b b a b

, ,a b a b a b b a b

� � � � ��

0a b a b � �

� �

22

| || | sin90a b a b � �

� �

21.1.1 1


16/17

(D) Hint:

Use dot product of two vectors and find p + q + r = 0

Solution:

Here 0r a b c �

� � �

0r a r b r c �

� � � � �

0p b c a q c a b r a b c

� � ��

0a b c p q r

��

2 0p q r

p + q + r = 0

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

(A) Hint:

1 1( ) (6)

2, 6P H P

Solution:

Required probability =

2 2 31 5 1 5 1

....2 6 2 6 2

6

7

(B) Hint:

Required prob. 10

3

4

C

Solution:

Here 1, 2, 4; 1, 3, 9; 2,4,8; 4,6,9; are 4 GP’s

Required probability 10

3

4 4

10 9 8 7!

7! 3!

C

4 6

10 9 8

1

30

p

q (given)

q – 2p = 30 – 2 = 28 which is divisible by 2, 4, 7.

(C) Hint:

Required prob. = 10

2

15 14

3

pp q

qC

Solution:

Possible ways of selections of x and y can be

1, 4, 7, 10 1st row

2, 5, 8 2nd row

3, 6, 9 3rd row

Now, if x3 + y3 = (x + y) (x2 – xy + y2) is divisible by 3

x + y must be also divisible by 3. Then x can be

taken from 1st row above and y can be taken from 2nd

row or x can be taken from 2nd row and y can be taken

from 1st row or x and y both can be taken from 3rd row.

4 3 3

1 1 11 4 3 3 15n E C C C

10

2n S C


2

15 15

10!

2! 8!

C

15 15 1

10 9 8! 5 9 3

2 8!

p

q

(given)

1 3 4p q which is divisible by 2 and 4

(D) Hint:

Required prob. = 9

6

4

4

6

C

Solution:

n(E) = Coefficient of x10 in (x + x2 + x3 +..........+ x6)4

= Coefficient of x6 in (1+ x + x2

+ ........+ x5)4

= Coefficient of x6 in

46

4

1

1

x

x

4 6

1Coefficient in 1 ......C x

4 5 2

1 21 ....... C x C x

6 6Coefficient of in 1 4 ......yx x

4 5 2 6 3 7 4 8 5 9 6

1 2 3 4 5 61 . ..... C x C x C x C x C x C x

9

6

9 8 7 6!4 4 84 4 80

6 6!C

n(S) = 6 × 6 × 6 × 6 = 64

Required probability = 4

80

6

n E

n S

80 5(given)

6 6 6 6 81

p

q

813 3 5 27 15 12

3 3

qp which is

divisible by 2, 3, 4, 6


17/17

58. Answer (02)

Hint & Solution:

2 2 22 cot 4 cot 4

2

y x y x ydy

dx

cot cosecy x x

cot cosecdy

x x dxy

log log sin logtan log2

xy x c

tan2

orsin

sin tan2

xc

cy y

xxx

or1 cos 1 cos

c cy y

x x

1 1or

1 cos 1 cosy y

x x

2 2 or 2 24 4

y y

59. Answer (07)

Hint:


� �

Solution:

Given expression

2 – 3a b a b a a b b � � � �

� � � �

2 3a b a a b b a b � � � �

� � � �

2 3 –a b a b a a a b b b a b a b � � � � � � �

� � � � � � �

Here 2| | 1a a a � � �

2| | 1b b b � � �

3 30

10 35a b b a

� �

� �

2 0 3 0a b b a � � � �

� �

222 3 6 5a b a b a a b b

� � � ��

= 6.12 + 0 + 1 = 7

60. Answer (08)

Hint:

Distance between two planes = distance of the point

(2, 3, 4) from the plane 2x – 4y + 2z – k.

Solution:

Let l, m, n be the d.c.’s of normal to the plane of given

lines.

l.3 + m.4 + n.5 = 0

l.4 + m.5 + n.6 = 0

24 25 18 20 15 16

l m n

1 2 1

l m n

But the plane ax – 4y + z = k must be parallel to this

plane.

4 12

1 2 1

aa

distance between the planes = 2 6

= distance of the planes 2x – 4y + 2z – k = 0 from the

point (2, 3, 4)

4 12 8 | |

4 16 4 2 6

k k

4 × 6 = |k|

| | 248

3 3

k

��

Test - 3A (Paper - 1) (Code-B) (Answers) All India Aakash Test Series for JEE (Advanced)-2019

1/17

1. (B)

2. (C)

3. (C)

4. Delete

5. (D)

6. (C)

7. (A, C)

8. (A, C)

9. (A, D)

10. (A, C)

11. (A, C)

12. (A, C, D)

13. (C, D)

14. (C)

15. (A, B, D)

16. A(Q)

B(Q, R)

C(T)

D(P, S)

17. A(P, R, T)

B(Q, R)

C(Q, R)

D(Q, R, T)

18. (06)

19. (07)

20. (03)

21. (C)

22. (B)

23. (B)

24. (D)

25. (C)

26. (C)

27. (C)

28. (B)

29. (B, C, D)

30. (B)

31. (B, D)

32. (C)

33. (A, B, D)

34. (A, B, C, D)

35. (A, B, C, D)

36. A(P, T)

B(Q)

C(R)

D(S)

37. A(P, R, T)

B(S)

C(R, T)

D(Q, S)

38. (08)

39. (08)

40. (01)

41. (D)

42. (D)

43. (A)

44. (A)

45. (D)

46. (A)

47. (B, D)

48. (A, D)

49. (A, C)

50. (A, C)

51. (A, D)

52. (A, C)

53. (B, C)

54. (B, C)

55. (B, C)

56. A (P, R)

B (P, R, T)

C (P, R)

D (P, Q, R, S)

57. A (Q, T)

B (R, S, T)

C (P, T)

D (Q, T)

58. (08)

59. (07)

60. (02)

ANSWERS

Test Date: 25/11/2018

PHYSICS CHEMISTRY MATHEMATICS

TEST - 3A (Paper-1) - Code-B

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

All India Aakash Test Series for JEE (Advanced)-2019 Test - 3A (Paper - 1) (Code-B) (Hints & Solutions)

2/17

PART - I (PHYSICS)

HINTS & SOLUTIONS

1. Answer (B)

Hint:

Based on photoelectric effect.

Solution:

2

2

152.244 13.6 1–

z

n

2

2 2

1 11.224 13.6 –

( – 1)

z

n n

z = 2, n = 5

2. Answer (C)

Hint:

2

2

1

2K V

Solution:

2

2

1

2K V

Vr has to be same

1 1

0 0

2 K m

K m

3. Answer (C)

Hint:

Apply lens equation for individual lens and trace the

image.

Solution:

After refraction by the first lens, image will be formed

in focal plane.

2f

f

1 4cos ,2

I f f

4. (Delete)

5. Answer (D)

Hint:

Add angle of deviation at individual surface and

minimize.

Solution:

i

= (i – ) + ( – 2) + (i – )

= (2i + – 4)

sin = sini

cos2 – 4

cos

d i

di

2

–1 4 –sin

3

i

6. Answer (C)

Hint:

Apply Snell’s law and trace the path.

Test - 3A (Paper - 1) (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2019

3/17

Solution:

75º

30º

45º

x

– cot 75º2 2

R Rx

1– tan15º2

R

7. Answer (A, C)

8. Answer (A, C)

9. Answer (A, D)


Hint :

Image will be formed due to combination of lenses +

single lens B.

Solution:

230 6030

d

r

x

1

1 1 1

30 15

V

V1 = 30 cm

2

1 1 1

60 40

V

V2 = 120 cm

120 cm from lens B

Ray diagram is shown in diagram. For image formed

due to lens 1 and 2 image is formed at 120 cm from

lens B. For image due to single lens 2, image is formed

at 60 cm from lens B.

Minimum radius will be at point, where both intersect

60 – 25

60 120

x

x

x = 10 cm

120

603030

10. Answer (A, C)

11. Answer (A, C)

12. Answer (A, C, D)


Hint :

Based on successive disintegration.

Solution:

0– A

A

dNN

dt

0–

0

1–

t

AN e

0 0– 2 B

A B

dNN N

dt

0 0 02 – 20

0

1–

t t t

B

de N e e

dt

0 0 02 2

0

–

t

t t t

Be N e e dt

0 0

0

2

2

0 0

–1 –1–

2

t t

B t

e eN

e

0 0

–2

0

1 – 22

t te e

0–

0 0

0 0

( ) 1–

t

tE t E e dt

= 0

–

0 0

0

|

t

teE t


4/17

= 0

–

0

0 0

1–

te

E t

= 0–0

0

0

–1

tE

t e

13. Answer (C, D)

14. Answer (C)



Hint :

Image will be alternatively real and virtual. And image

will be always enlarged. And, speed is minimum when

object crosses principal axis. And, lens of image can

be hyperbolic.

Solution:

Lens of image will be like this

F

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

Hint:

Image will be formed on object when ray retraces its

path.

Solution:

Ray should fall normally or on the pole of mirror to

retrace its path.

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

Hint:

Emission will stop, when potential of sphere is equal

to stopping potential.

18. Answer (06)

Hint: Apply lens formula

Solution:

1 1 1

3x x f...(i)

1 1 1

4 2 4x x f...(ii)

From (i) and (ii),

x = 32

3 3 32

24 cm4 4

xf

19. Answer (07)

Hint:

fabsorption

freflection

IP

c

Solution:

0

reflection

1

2

I Af

c

0absorption

2

I Af

c

02 cos 30º2

I AF

c

03

2 I A

C

20. Answer (03)

Hint:

2

2

1 11–

Rzn

Solution:

2

2

1 11–

Rzn

2

2 3

2d Rz

n

2

3 2

2 1

d Rz

n RZ

3 x n

d


5/17

21. Answer (C)

Hint:

cis + syn Meso

cis + anti Racemic

Trans + syn Racemic mixture

Trans + anti Meso

Solution:

(I) OsO4

(ii) H O, NaHSO2 3

Meso(A)

Racemic mixture(B)

(i) MCPBA

(ii) H O2

Baeyer's reagent

Cold KMnO4

Racemic

Meso(i) MCPBA (Cl—mC H COOOH)

6 4

(ii) H O2

(C)

(D)

22. Answer (B)

Hint:

Ph — C — Ph

O18

Peroxy acidPh — C — O — Ph

O18

Solution:

Ph — C — Ph

O18

H+

Ph — C — Ph

18O

H

CH COOO3

Ph — C — O — O — C — CH 3

18OH O

Phenyl migration


18OH

(+)

–H(+)


18O

(+)

Ph

23. Answer (B)

Hint:

Ethyl acetoacetate

CH — C — CH — C — OC H3 2 2 5

O O

Methyl Vinyl Ketone CH == CH — C — CH2 3

O

So, P = H C3

C — OC H2 5

O

O

Q =

CH3

CH3

Solution:

O O

C H ONa2 5

O O

O

O == C — OC H2 5

CH

CH2

CH2

C

O

CH3

CH — C3

O

C H ONa2 5

C H — O — C 2 5

OCH

3

O

PART - II (CHEMISTRY)


6/17

24. Answer (D)

Hint:

As stability of enol form increases, equilibrium constant

of keto to enol increases.

Solution:

N C — CH — CO C H2 2 2 5

O

NC == CH — CO C H2 2 5

O

H

Due to hydrogen bonding, K1 is more than K

2 and K

3.

O OH

K4

OH

K5

O

Due to extended conjugation, K5 is larger than K

4.

25. Answer (C)

Hint:

Rate of P > Q

Rate of R > S

As positive charge on — C —

O

reduces, rate of reaction

decrease.

Solution:

Ph — C — N

O

Ph — C == N

O–

+

I II

Contribution of II is very less because angle strain

increases.

Ph — C — NH2

O

Ph — C == NH2

O

+

III IV

There is no such problem in IV so it donates its lone

pair for resonance and reduce the positive charge of

— C —

O

group.

In experiment II, molecule ‘C’ has bridge head nitrogen,

so it does not donate its lone pair to — C —

O

so in ‘C’

there is highly positive charge on — C —

O

.

26. Answer (C)

Hint:

Protonation occurs such that conjugate acid has

resonance or highest stability

Solution:

CH — C — NH3 2

O

H+

H+

CH — C — NH3 2

O — H(+)

CH — C — NH3 3

+

O

N NH2

H+

H+

N NH2

+

H

N NH3

+ (No resonance)

27. Answer (C)

28. Answer (B)


7/17

29. Answer (B, C, D)


Hint :

O

O

N K– +

OO

Br

EtO OEt +

C

C

C

O

O

N — C — H

C — OEt

C — OEt

O

O

NaOEt

O

O

N — C — CH2

C

C

O

O

N — C

C OEt

C OEt

O

O

CH — S — CH2 3

C — OEt

O

C — OEt

O

X = Cl

SCH3

(i) dil. NaOH

(ii) HCl

COOH

COOH

+ CH — S — CH —CH — C — NH3 2 2 2

H

COOH

By changing the X we can change the -amino acid.

Solution:

Methionine CH — S — CH — CH — C — NH3 2 2 2

H

COOH

To prepare methionine

X = CH3 — S — CH

2 — CH

2 — Cl

If X = Br

then R = ( )CH CH — CH — C — NH3 2 2 2

COOH

H

E =

COOH

COOH

E on reaction with CH — CH2 2

OH OH

form glyptal.

30. Answer (B)

31. Answer (B, D)

32. Answer (C)


Hint :

P = — C — H

O

Q = — C — C —

O OH

R = — C — C —

O O

S = — C — COOH

OH

T = O

OH


8/17

Solution:

C

O

H CN

–

Benzoincondensation

— C — C —

O OH

H

(A) (B)

Oxidation

— C — C —

O O

OH –

Benzilic acidrearrangement

Ph — C — COOH

OH

Ph

O

OH

(i) MeCO O/MeCO( )2 2

–

(ii) H O3

+Ph — C — H

OPerkin reaction


Hint:

CH3

CH3

OH

H

OH

H

HO

H

Meso

Solution:

NH2

H

NH2

H ,

C == C == C

CH3

H

H

CH3

are optically active


Hint:

For bridge head only cis is possible so optical isomer

less than 2n

Solution:

H

O

Me Me

Me

Only cis form is possible so only one pair of enantiomer

is possible.

CH3

CH3

+

CH3

CH3

(Meso) (d + l)

So, 3 isomers


Hint:

Cl

Cl

Cl

Cl

Cl

Cl

9 isomers

OH

OH

OH

Total 4 isomers

Racemic mixture does not show optical activity but

has chiral compound.

Solution:

OH

OH

Meso Meso

OH

OH

OH

OH

OH

OH

(d + l)

OH

For

Cl

Cl

Cl

Cl

Cl

Cl

7 Meso + 1 dl pair


9/17

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

Hint:

Aldehyde reduce the Tollens reagent, ketone, ether

does not reduce Fehling reagent and — C — CH

O

group

give positive iodoform test.

Solution:

C5H

11 — OH 8 alcohol

1º alcohol C4H

9 — CH

2 — OH

CH3CH

2CH

2CH

2CH

2OH

CH — CH — CH CH OH 3 2 2

CH3

CH — C — CH OH 3 2

CH3

CH3

CH — CH — CH — CH OH3 2 2

CH3

(Optically active)

These alcohol on oxidation form aldehyde that reduce

Tollens as well as Fehling reagent but not give iodoform

test.

CH CH CH CH CH3 2 2 3

OH,

CH — CH — CH CH CH3 2 2 3

OH

(optically active)

CH — C — CH — CH 3 3

H

CH3

OH

(optically active)

CH — C — OH 3

CH3

CH3

Ether:

CH3CH

2CH

2CH

2 — OCH

3

CH — CH — CH — OCH3 2 3

CH3

CH — C — OCH3 3

CH3

CH3

CH — CH — CH — OCH3 2 3

CH3

(optically active ether)

CH3CH

2CH

2 — OC

2H

5

CH — CH — OC H3 2 5

CH3

Total isomer = 14

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

Hint:

Hydroboration oxidation gives syn addition; inmercuration - demercuration no carbocationrearrangement takes place, 3º alcohol give immediatewhite turbidity with lucas reagent.

Alkene on reaction with dil H2SO

4 gives rearranged

product.

Solution:

H.B.O.(i) BH - THF

3

(ii) H O ; OH2 2

–

HCH3

HOH

+ enantiomer

Mercuration-demercurationOH

Mercuration-demercuration

OH

dil.H SO2 4

(+)OH

H O2

(+)

38. Answer (08)

Hint:

a, b, c, d, e, g, h and i are correct.

Solution:

Glucose 2 2Br /H O gluconic acid

39. Answer (08)

Hint:

Anisole, m-cresol and p-cresol is weaker acid than

phenol


10/17

41. Answer (D)

Hint:

Use Bayes' theorem.

Solution:

Let E be the event that visiting lecturer visit the college

late and let E1, E

2, E

3, E

4 be the events that the lecturer

comes by train, bus, scooter and other means of

transport respectively.

1 2 3 4

3 1 1 2, , ,

10 5 10 5P E P E P E P E

1

EP

E

= probability that the lecturer arriving late

by coming train = 1

4

2 3 4

1 1, and P 0

3 12

E E EP P

E E E

, same

he is not late if he comes by other means oftransport.

By Bayes’ theorem

1

11

1 2

1 2

3 4

3 4

EP E P

EEP

E E EP E P P E P

E E

E EP E P P E P

E E

3 1

10 4

3 1 1 1 1 1 20

10 4 5 3 10 12 5

3 3

40 40

3 1 1 9 8 1

40 15 120 120

3 120 1

40 18 2

42. Answer (D)

Hint:

First find the coordinates of point P using condition

for intersection of line and plane and then point Q.

Then find distance between P and Q using distance

formula.

Solution:

Here equation of line is

2 3 4

1 2 3

x y z say ...(i)

Hence any point on the line (i) will be p( + 2, 2 –

3, –3 + 4)

Given line (i) intersect the given plane

2x + 3y – z = at the point P

2( + 2) +3(2–3) + 3 – 4 = 13

11 – 9 = 13

11 = 22

= 2

P = (4, 1, –2)

Also equation (i) intersects the YZ – plane at Q

i.e. x = 0

PART - III (MATHEMATICS)

Solution:

O

HO OH

HO

HO

O

O

Triangular acid Squaric acid

NO2

OH

O N2

NO2

Picric acid

40. Answer (01)

Hint:

PhN2

+ attack at position (1).

Solution:

OH

PhN2

+

OH

N = N — Ph


11/17

+ 2 = 0 = –2

Q (0, –7, 10)

PQ = 2 2 24 0 1 7 2 10

16 64 144

224 16 14 4 14 a b

4 146

3 3

a b

43. Answer (A)

Hint:

First use the formula for cross product of 4 vectorsand then use the concept for scalar triple product.

Solution:

a b b c p b c p c b p b c � � � � �

� � � � � � � �

–a b c b a b b c

� � � ��

a b c b

� ��

Similarly, –b c c a b c a c b c c a

� � ��

a b c c

��

and –c a a b c a b a c a a b

� � ��

a b c a

��

, ,a b b c b c c a c a a b

� � � ��

, ,a b c b a b c c a b c a

� � � ��

a b c b a b c c a b c a

� � � ��

2

a b c a b c b c a

� � ��

3

a b c b c a

� ��

4

a b c

��

43 , since 3a b c

��

= 81

44. Answer (A)

Hint:

Volume of tetrahedron is obtained by

2 1

36

a a a b a c

V b a b b b c

c a c b c c

��

� � � ��

��

Solution:

Volume of tetrahedron, 1

6V a b c

��

2

2 1

36V a b c

��

1

36

a a a b a c

b a b b b c

c a c b c c

��

� � � ��

��

...(i)

But 2 2 2| | 1, | | 1, | | 1a a a b b b c c c � � �

� � � � � �

1| || | cos60

2b a a b a b � � �

� � �

1| || | cos60

2b c c b b c � � �

� � �

1| | 1| | cos60

2a c c a c a � � � � � �

From (i)

2

1 11

2 2

1 1 11

36 2 2

1 11

2 2

V

1 1 1 1 1 1 1 11 1

36 4 2 2 4 2 4 2

1 3 1 2 1 1 1 2

36 4 2 4 2 4

1 3 1 1 1 3 1

36 4 8 8 36 4 4

1 2 1

36 4 72

1

6 2V cubic units


12/17

45. Answer (D)

Hint:

Differentiate both sides w.r.t. x and then put f(x) = y,

dyf x

dx our equation becomes 2

2

2

1

dy xy y

dx x

Dividing both sides by y2 and putting 1

zy

, our

equation reduces to linear differential equation.

Solution:

From given differential equation

2

2 2

0

32 2

xf x f t dt

x t

,

where ( ) 2 3 0 6f x

Differentiate both sides w.r.t. x

2 2

2 22

2 – 2 ( )0

22

x f x x f x f x

xx

2

2

2–

2

xf x f x f x

x

2

2

2

2

dy xy y

dx x

2 2

1 2 11

2

dy x

dx yy x

...(i)

Put2

1 1 dy dzz

y dx dxy

Equation (i) becomes

2

21

2

dz xz

dx x

...(ii)

I.F. = 22

2

log (2 ) 222e

xdx

xxe e x

Solution of equation (ii) is

2 22 2z x x dx

3

22 2

3

xz x x c

3

212 2

3

xx x c

y

where x = 0, (0)

12 0 0 0 c

y

2 1

6 3c

3

21 12 2

3 3

xx x

y

2 32 6 1

3

x x x

y

23

3 2

6 1

xy f x

x x

3 2 4 182

8 12 1 19f

46. Answer (A)

Hint:


dx

equation becomes linear

differential equation, then use process of linear differentialequation.

Solution:

Given equation is

dy d dy x

dx dx dx


dx

, we get

( )dy

y xd

...(i)

1.I.F.

de e

Solution of equation (i) is given by

y e e d

– 1y e e d e d

1y e e e C e C

1y Ce

1

x

y x Ce

47. Answer (B, D)

Hint:

First find centroid of G of QRS using formula as

3

a b c �

� �

if vertices are , ,a b c�

� �

and then find PG��

and

its magnitude.


13/17

Solution:

(1, 0, 2)

Q(–2, 3, 4)

R(1, 3, 5)

S(0, –2, 3)

P

h

x PQ PR ��

�

2

G

1

L

M

Here G is the centroid of QRS

Positive vector of G is

ˆ ˆ ˆ2 1 0 3 3 2 4 5 3

3

i j k

i.e. ˆ ˆ ˆ4 12

3

i j k

ˆ ˆ ˆ4 12ˆ ˆ2

3

i j kPG i k

��

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ4 12 3 – 6 4 4 6

3 3

i j k i k i j k

16 16 36 68 2 17| |

9 9 9 9 3PG ��

which is

irrational number.

48. Answer (A, D)

Hint:

Area of PQR = 1| |

2PQ PR��

Solution:

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ2 3 4 – 4 3 3 2PQ i j k i k i j k ��

ˆ ˆ ˆ ˆ ˆ ˆ ˆ3 5 2 3 3PR i j k i k j k ��

ˆ ˆ ˆ

3 3 2

0 3 3

i j k

PQ PR ��

ˆ ˆ ˆ9 6 – 9 0 9 0i j k

ˆ ˆ ˆ3 9 9i j k x �

Area of PQR = 1

2PQ QR��

19 81 81

2

1 1 19 162 171 9 19

2 2 2

319

2 square units which is irrational number.

49. Answer (A, C)

Hint:

Length of perpendicular from vertex S opposite face =

| |

PS n

n

��

�

Solution:

Length of the perpendicular from vertex S to opposite

face = h = | Projection of |PS on x��

ˆ ˆ ˆ ˆ ˆ ˆ ˆ2 3 2 3 9 – 9

| | 9 81 81

j k i k i j kPS x

x

��

�

�

3 18 9

90 81

30 3010 10

171 3 19 19

50. Answer (A, C)

Hint:

– 0b c a �

� �

Solution:

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ– 2 – 2 3b c i j k i j k j k �

�

and ˆ ˆ ˆ ˆ– 3 3 3 3 0b c a j k i j k � � � �

–b c�

�

is orthogonal to a�

51. Answer (A, D)

Hint:

–a b c a c b b c a � � �

� � � � � �

Solution:

a b c �

� �

–c a b �

� �

– –c b a c a b

� ��

Comparing, we get 1 2 2 1b c �

�

1 3 2 6a c � �

0

1 6 0 7 which is a prime number.


14/17

52. Answer (A, C)

Hint:

Let ˆd l b c m c a n a b � �

� � � �

, then find l, m,

n as ˆ ˆ ˆ

, ,a d b d c d

l m n

a b c a b c a b c

��

� � ��

ˆ ˆ ˆ â b c d a d b c b d c a c d a b

� � � ��

Solution:

Here

1 3 1

1 2 1

1 1 2

a b c

��

1 4 2 3 2 1 1 1 2

5 – 9 – 1 –5 0

,, ca b�

��

are non-coplanar

, ,b c c a a b � �

� � � �

also are non-coplanar

ˆd l b c m c a n a b � �

� � � �

ˆ ˆ ˆ

, ,

��

� � ��

a d b d c dl m n

a b c a b c a b c

ˆ ˆ ˆd a b c a d b c b d c a c d a b

� � � � ��

ˆ ˆ â d b c b d c a c d a b � � �

� � � � � �

ˆd a b c

��

a b c

��

= modulus of

1 3 1

1 2 1

1 1 2

1 4 1 3 2 1 1 1 2

5 9 1 5 which is a prime number

53. Answer (B, C)

Hint:

Required prob. = 1

2 3

1 1

6

3

x x x

x

C C C

C

Solution:

Bag

x red balls

2 green ballsx

3 blue ballsx

P(A)| all different colour

2 3

1 1 1

6

3

x x x

x

C C C

C

2 3

6 6 1 6 2

3!

x x x

x x x

3 2

6 6 6

6 6 1 6 2 6 1 2 3 1

x x

x x x x x

23

0.3 given6 1 3 1

x

x x

2

2

2

3 38 9 1 0

1018 9 1

xx x

x x

1 8 1 0x x

1

1,8

x

But x is number of balls

1

8x

x = 1

54. Answer (B, C)

55. Answer (B, C)

Solution for Q. Nos. 54 and 55

Blue

(3 )xGreen

(2 )xRed

( )x

22

01

10

01

22

12

10

01

22


15/17

3 2 2 3 3 2

2 1 1 2 1 1 2 1 1

6

3

( )

x x x x x x x x x

x

C C C C C C C C CP B

C

3 3 1 3 2 2 1 4 1 5

2 2 2

6 6 1 6 2

6

x x x x x x x x x

x x x

2

27 9 16 8 5 52

6 1 6 2

xx x x

x x x

48 2224 112

6 1 2 3 1 2 6 1 3 1

xx

x x

x x x x

when x = 2, P(B) =

2 48 11 37

2 11 5 55

when x = 1, P(B) = 13 13

2 5 2 20

when x = 3, P(B) = 183

272

P(C) =

2

2

5 1

5 12

2 24 1124 11

x x

x

xx x

when x = 1, P(C) = 0

when x = 2, P(C) = 10 5

2 74 74

when x = 3, P(C) = 5 2 5

2 61 61

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

(A) Hint:

1 1( ) (6)

2, 6P H P

Solution:

Required probability =

2 2 31 5 1 5 1

....2 6 2 6 2

6

7

(B) Hint:

Required prob. 10

3

4

C

Solution:

Here 1, 2, 4; 1, 3, 9; 2,4,8; 4,6,9; are 4 GP’s


3

4 4

10 9 8 7!

7! 3!

C

4 6

10 9 8

1

30

p

q (given)

q – 2p = 30 – 2 = 28 which is divisible by 2, 4, 7.

(C) Hint:

Required prob. = 10

2

15 14

3

pp q

qC

Solution:

Possible ways of selections of x and y can be

1, 4, 7, 10 1st row

2, 5, 8 2nd row

3, 6, 9 3rd row

Now, if x3 + y3 = (x + y) (x2 – xy + y2) is divisible by 3

x + y must be also divisible by 3. Then x can be

taken from 1st row above and y can be taken from 2nd

row or x can be taken from 2nd row and y can be taken

from 1st row or x and y both can be taken from 3rd row.

4 3 3

1 1 11 4 3 3 15n E C C C

10

2n S C


2

15 15

10!

2! 8!

C

15 15 1

10 9 8! 5 9 3

2 8!

p

q

(given)

1 3 4p q which is divisible by 2 and 4

(D) Hint:

Required prob. = 9

6

4

4

6

C

Solution:

n(E) = Coefficient of x10 in (x + x2 + x3 +..........+ x6)4

= Coefficient of x6 in (1+ x + x2

+ ........+ x5)4

= Coefficient of x6 in

46

4

1

1

x

x


16/17

4 6

1Coefficient in 1 ......C x

4 5 2

1 21 ....... C x C x

6 6Coefficient of in 1 4 ......yx x

4 5 2 6 3 7 4 8 5 9 6

1 2 3 4 5 61 . ..... C x C x C x C x C x C x

9

6

9 8 7 6!4 4 84 4 80

6 6!C

n(S) = 6 × 6 × 6 × 6 = 64

Required probability = 4

80

6

n E

n S

80 5(given)

6 6 6 6 81

p

q

813 3 5 27 15 12

3 3

qp which is

divisible by 2, 3, 4, 6

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

(A) Hint:

Use –a b c a c b a b c � � � � �

� � � �

Solution:

a a b b b c � � �

� � �

– –a b a a a b b c b b b c � � � � � �

� � � � � �

...(i)

Taking dot product of both sides of (i) by b c�

�

, we

have – 0 0 0a b a b c

� ��

But 0 0a b a b c

� ��

(B) Hint:


� �

Solution:

Given 2a b

22

2a b �

�

2 2| | 2 2a b a b � �

� �

1 1 2 2 0a b a b � �

� �

Now a b c a c a b a b

� � ��

a b a b a b a b � � � �

� � � �

a b a a b b � � �

� � �

a a b b a b � � �

� � �

– –a b a a b b b b a b a b

� � � � � � ��

0 0 0b a a b

� ��

∵

a b �

�

2 22

2 2a b c a b � �

� � �

(C) Hint:

a b c a b c

� ��

Solution:

a a bb a ba b

� � � ��

a a b b a b a b � � � �

� � � �

0 � � � �

� � �

a a b b a b

, ,a b a b a b b a b

� � � � ��

0a b a b � �

� �

22

| || | sin90a b a b � �

� �

21.1.1 1

(D) Hint:

Use dot product of two vectors and find p + q + r = 0

Solution:

Here 0r a b c �

� � �

0r a r b r c �

� � � � �

0p b c a q c a b r a b c

� � ��

0a b c p q r

��

2 0p q r

p + q + r = 0


17/17

��

58. Answer (08)

Hint:

Distance between two planes = distance of the point

(2, 3, 4) from the plane 2x – 4y + 2z – k.

Solution:

Let l, m, n be the d.c.’s of normal to the plane of given

lines.

l.3 + m.4 + n.5 = 0

l.4 + m.5 + n.6 = 0

24 25 18 20 15 16

l m n

1 2 1

l m n

But the plane ax – 4y + z = k must be parallel to this

plane.

4 12

1 2 1

aa

distance between the planes = 2 6

= distance of the planes 2x – 4y + 2z – k = 0 from the

point (2, 3, 4)

4 12 8 | |

4 16 4 2 6

k k

4 × 6 = |k|

| | 248

3 3

k

59. Answer (07)

Hint:


� �

Solution:

Given expression

2 – 3a b a b a a b b � � � �

� � � �

2 3a b a a b b a b � � � �

� � � �

2 3 –a b a b a a a b b b a b a b � � � � � � �

� � � � � � �

Here 2| | 1a a a � � �

2| | 1b b b � � �

3 30

10 35a b b a

� �

� �

2 0 3 0a b b a � � � �

� �

222 3 6 5a b a b a a b b

� � � ��

= 6.12 + 0 + 1 = 7

60. Answer (02)

Hint & Solutions

2 2 22 cot 4 cot 4

2

y x y x ydy

dx

cot cosecy x x

cot cosecdy

x x dxy

log log sin logtan log2

xy x c

tan2

orsin

sin tan2

xc

cy y

xxx

or1 cos 1 cos

c cy y

x x

1 1or

1 cos 1 cosy y

x x

2 2 or 2 24 4

y y

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