mock test - 1 (paper-1) - code-a...mock test - 1 (paper - 1) (code-a) (hints & solutions) all...

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

1/16

1. (7)

2. (4)

3. (4)

4. (5)

5. (6)

6. (8)

7. (3)

8. (4)

9. (A, B, D)

10. (A, B, C)

11. (A, C)

12. (A, C)

13. (B, C)

14. (A, B, D)

15. (A, B, C)

16. (B, D)

17. (B, D)

18. (A, D)

19. A (Q)

B (P, R)

C (T)

D (R, S)

20. A (T)

B (P, R)

C (Q)

D (S)

21. (8)

22. (9)

23. (6)

24. (5)

25. (0)

26. (1)

27. (3)

28. (6)

29. (A, B)

30. (B, D)

31. (A, C)

32. (A, C)

33. (A, B, C)

34. (A, B, C)

35. (A, D)

36. (A, B, C, D)

37. (A, B, C, D)

38. (A, B, C, D)

39. A (P, R, S, T)

B (P, R, S, T)

C (P, R, S, T)

D (Q, S, T)

40. A (P, Q, T)

B (P, Q, R, S, T)

C (R, S)

D (P, Q, T)

41. (4)

42. (6)

43. (8)

44. (7)

45. (9)

46. (6)

47. (4)

48. (4)

49. (A, B)

50. (A, B, C, D)

51. (A, C, D)

52. (A, B, C)

53. (A, C)

54. (A, B, C)

55. (B)

56. (A, D)

57. (B, D)

58. (A, B, D)

59. A (Q, S)

B (Q, R, S, T)

C (P, Q)

D (R)

60. A (P, Q, R, S)

B (P, Q, R, S)

C (P, Q, R, S, T)

D (R, S)

ANSWERS

PHYSICS CHEMISTRY MATHEMATICS

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

Test Date: 17/02/2019

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

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

2/16

PART - I (PHYSICS)

HINTS & SOLUTIONS

1. Answer (7)

Hint: Apply concept of solid angle

Solution :

h

2

d

2

2

22 (1 cos60 ) 2 14 2

2

d

dh

7

6

dh

x = 7

2. Answer (4)

Hint: PV = constant

Solution :

1

4 TP

R

2

2

0

2

2

TP

R

PV = constant

015 2

8

T

R

n = 4

3. Answer (4)

Hint: ( ) � � �

F q E V B

Solution :

y

x

0ˆE j

0 0ˆ ˆ ˆ ˆ( ) ( )

x y

dvF m qE j q v i v j B k

dt

�

�

0 x

y

dvm qv B

dt

0 0 y

x

dvm qE qv B

dt

0 0

0

y

y

dv qB qyBmv qE

dy m

2 2 2

0

00 0

2y

q B yv qE y

m

0

2

0

2mEy

qB

0

0 0 2

0

2mEK qE y qE

qB

2

0

2

0

2E

K mB

4. Answer (5)

Hint: Potential across capacitor is same as potential

across 4 resistor.

Solution :

I2

I1I

C

3I1 = I

2

IC + I

1 = I

2

IC = 2I

1

3

0 42

13

t

CC

V Cq e

3 3

0 04 42 2 3 1

1 43 3 4 2

t t

C CV V C t

e eC

3 3

04 40

31

2

t t

C CV

V e e

3

42 5

t

Ce

4 5In

3 2

Ct

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

3/16

5. Answer (6)

Hint: 2

mp

RTV

M

Solution : 2 2

3

RT RT

M M

3

2

n = 6

6. Answer (8)

Hint: vr = constant

Solution :

00 0 1

2

rv r v

v1 = (2v

0)

22

01

0 0

8

2

mvmvT

r r

7. Answer (3)

Hint: Work done by force is, 0

x

dw Fdx

Solution :

Let block be displaced by x, liquid rises by y

2

( )4

RF x y g

2

( )4

R g

dw x y dx

3 x

y

3 /82

0

4

4 3

h

R gw xdx

2 23

128w R gh

8. Answer (4)

Hint: Based on beat formation

Solution :

1 2sin( )Y A t

2 2sin( )Y A t

1 2 1 2

1 2

( ) ( )2 sin cos

2 2

t tY Y A

(2 ) 42 cos sin 2 (302)

2

tA t

2

20

44cos

13

44

I

I A

9. Answer (A, B, D)

Hint: 0 (1 cos )2

CVQ t

Solution : max 0di

idt

At that moment, potential across capacitor is same

10. Answer (A, B, C)

Hint: sin 0.6 sr

r

V

V

Solution :

sin sr

r

v

v

601000

100020

sr

dv

t

[vr = v

sr sin ]

11. Answer (A, C)

Hint: Image formed by lens should fall either normally

or on the pole of mirror.

Solution :

Image formed by lens should be either on pole or fall

normally on mirror.

12. Answer (A, C)

Hint: At steady state current through circuit is zero

Solution :

2

0

1 3

2 4

CH V

13. Answer (B, C)

Hint: Draw phasor diagram to derive effective current


4/16

Solution :

0 0CV

V0

0

2 2 2

0

V

R L

0

0 02 2 2

0

sinVCV

R L

0 0

0 02 2 2

0

V L

CVR L

2 2 2

0

LR L

C

2

0 2 2

L R

L C L

2

2

1 R

LC L

0 0

2 2 22 20

V R V Ri

LR LR R

C

0CV R

L

14. Answer (A, B, D)

Hint:

dNN

dt

N = N0e–t

Solution :

dNN

dt

N = N0e–t

R0 = log |N

0e–t|

R0 = log |N

0| – t

0

0

Re

N

0

02

R

t


Hint: For equilibrium, mass of liquid displaced is equal

to total mass

Solution :

fb = (m

1 + m

2)g

1 2

3 3 1 2

1 2

( )m m

m m

31 3 1

2

21

( )1

mm

2 31 1

2 2 3 1

( )

( )

m

m

16. Answer (B, D)

Hint: , is critical angle of incidence, 1

1

1sin

Solution :

= – 2

2

d

d

1 1sin 30

2

17. Answer (B, D)

Hint: Velocity of separation

Velocity of approach e

Solution :

v0

1sin

2

0

1 2

3

2

v

v v

0

2 1

3

4

v

v v

0

2

3 3

8

v

v


5/16

18. Answer (A, D)

Hint: Here COG is different from COM

Solution :

2

0( )

eR

e

e e

GM MF dx

R R x

22

e

e

GM MF

R

2

2

2

e

( )

(COG) ( )

2

e

e

R

R

e

y dm GM

y MY dm dy

GM M R

R

Y = 2R ln2 = 1.4R

d = 1.4R – R = 0.4R

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

Hint: To escape, total energy 0

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

Hint: VIM

= –m2VOM

Solution :

VIM

= –m2VOM

| |V

mu

PART - II (CHEMISTRY)

21. Answer (8)

Hint:

n - factor of Ba(MnO4)

2 = 10

Solution :

Normality of H2O

2 =

11.2

5.6= 2 N

meq (Ba(MnO4)

2) = meq of H

2O

2 reacted

= 2 × 100 = 200 meq

= 0.2 eq

Moles of Ba(MnO4)

2 = 0.02

0.02 375% purity 100 92%

8.15

% impurity = 8%

22. Answer (9)

Hint:

aE

–RTK Ae Arrhenius equation

Solution :

aE

–RT

0K A e

ln K = In a

0

EA –

RT

11488log K log 1000 –

8.314 300 2.30

= 3 – 2 = 1

K = 10 hr–1

0

t

AKt In

A

0

t

A2.3 2.3 log

A

0 0

t 0

A A10

A 1 A – x

t

t

E.C9

B.C.C

23. Answer (6)

Hint:

Ni(CO)4 is tetrahedral in shape.

Solution :

Ni(CO)4 is tetrahedral, which has 6 planes of symmetry.

24. Answer (5)

Hint:

CH – C 3

OH O

Show G.I

Solution :

CH – C 3

OOH

CH – C 3

OOH

,

CH = CH2

OH O

CH – C3

OH O

,


6/16

25. Answer (0)

Hint:

All the given ions have different magnetic moment with

Strong field ligand and Weak field ligand

Solution :

Pairing can takes place in all ion with S.F.L

26. Answer (1)

Hint:

At Equilibrium Ecell

= 0

Solution :

Balanced equation is

2 – 2

2 3 2 2Hg NO 3H 2Hg HNO H O

22

cell 32 –

2 3

HgRTE 0 E – log

nFHg NO H

⎡ ⎤⎣ ⎦

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ ⎣ ⎦

2

32 2

10.060.09 log

21 1 H

pH = 1

27. Answer (3)

Hint & Solution :

106 ml of water contains 100 g or 1 mole of CaCO3

= 2 mole of H+, [H+] = 2 × 10–3 M

28. Answer (6)

Hint:

In claisen rearrangement, carbon attached at ortho.

Solution :

O

1

2

34

5

O1

2

3

H

OH 2

3

1

Tautomeric

x = 3

y and z = 2 and 1

29. Answer (A, B)

Hint:

A : FeCr2O

4 , B : Na

2CrO

4 , C : Fe

2O

3 , D : CrO

5 ,

E : CrO8

–3

, F : SO

2 , G : SO

3

Solution :

FeCr O + 2 4

Na CO + O2 3 2

Fe O + 2 3

Na CrO + CO2 4 2

(A) (C) (B)

Na CrO2 4

H O /H2 2

+

CrO5

CrO8

–3

(D)

(E)H O

2 2

Alkaline

30. Answer (B, D)

Hint:

Process is reversible so G = 0 only at constant T

and P.

Solution :

Process A to B is isothermal

so H = U = 0

U = q + w

w = –ve and q = +ve

31. Answer (A, C)

Hint:

B and P has vacant orbital.

Solution :

Charge can be delocalised only in 3 2O –P CH and

3 2O –B CH due to presence of vacant orbital.

32. Answer (A, C)

Hint:

In elecrophilic addition, carbocation is intermediate.

Solution :

Rate of electrophilic addition is proportional to stability

of carbocation.


7/16


Hint:

M : HO – C – C – CH – OH 2

O H

CH3

N : CI – C – C – CH – CI 2

O H

CH3

O : C H – N – C – C – CH – CI 2 5 2

O H

H CH3

Solution :

Structure of M

HO – C – CH – CH – OH 2 2

– CH2

O

Not possible because M does not contain 3° hydrogen

M : HO – C – C – CH – OH 2

O H

CH3

HO – C – C – CH – OH 2

O H

CH3

CI – C – C – CH – CI 2

O H

CH3

SOCI2

C H NH2 5 2

C H N – C – C – CH – CI 2 5 2

O H

CH3

H


Hint :

Aldo

Solution :

(A)

O

(i) CH MgBr3

(ii) H O2OH

CH3

H+

+H

+

O

O

(i) O3

(ii) Zn, H O2

+

+

(B)

CHO(i) CH MgBr

3

(ii) H O2

CH – CH3

OH

H+

+

O

O

(i) O3

(ii) Zn, H O2

+

(C)

OCH MgBr3

H O2

OH

CH3

H+

O3

Zn, H O2

O

O

35. Answer (A, D)

Solution :

(A) Acidic buffer will form (pH < 7)

(B) Both are salt of SASB (pH = 7)

(C) Weak base will form (pH > 7)

(D) Strong acid solution (pH < 7)

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

Hint :

Sulphur show disproportionate reaction with alkali.

Solution :

S8 + 6NaOH Na

2S

2O

3 + 2Na

2S + 3H

2O

Zn + NaOH Na2ZnO

2 + H

2 (g)

AI + NaOH Na2AIO

2 + H

2 (g)


Hint :

P Na2S

2O

3

Solution :

Na2SO

3 + S Na

2S

2O

3

AgX + 2Na2S

2O

3 [Ag(S

2O

3)]3 – + X + 4Na+

X = CI, Br, I

Na2S

2O

3 + CI

2 + H

2O 2HCI + S + Na

2SO

4

2Na2S

2O

3 + I

2 Na

2S

4O

6 + 2NaI


8/16


Hint :

B2H

6 + 6CI

2 2BCI

3 + 6HCI

Solution :

B2H

6 + 6H

2O B(OH)

3 + 6H

2 (g)

B2H

6 + 2NH

3 [H

2B(NH

3)

2]+ + [BH

4]–

B2H

6 + 2(CH

3)

3N 2H

3B N(CH

3)

3

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

S, T)

Hint:

In B2H

6, hybridisation of ‘B’ is sp3.

Solution :

(A) B

H

B BB

H

H

H

H

H

sp3

Empty orbital participate in hybridization

Two types of bonds covalent and 3c – 2e

(B)AI

CI

Cl CI

CI

AI

CI

CI

sp3

Empty orbital of AI participate in hybridization

Two types of bonds covalent and co-ordinate

(C) BeCI2 is polymeric in solid state

BBe x Bey

CI

CI

CI

CI

BBe

sp3 Vacant orbital of Be participate

Two types of bonds covalent of Co-ordinate

(D)

N

H

sp2

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

Hint:

1st group cation form white ppt with dil.HCI

Chromate of Ag+ and 2

2Hg are red while chromate of

Pb2+ is yellow.

Solution :

Ag2S, NiS, CuS, PbS black

22

3 3 4Excess

Cu NH Cu NH

22

3 3 6Excess

Ni NH Ni NH

PART - III (MATHEMATICS)

41. Answer (4)

Hint: Draw graph

Solution :

x

y

(3, )3

(1, 1)

(–1, 3)

(–4, 0) (6, 0)O

x = 0, 1, 3, 6 are points of extremum.

42. Answer (6)

Hint: Here sin 12

a bC

ab

Solution : Given, ( )R a b C ab

( ) (2 sin )R a b R C ab

sin2

a bC

ab

...(1)

But 1

2

a b

ab

and sin C 1

(1) will be true if

2 and sin 1 a b ab C

2( ) 0a b and 2

C

and2

a b c

1sin

2

2

ar C

ra b cS


9/16

A

CB a

b = a

2c a

2

1sin

12 2

2 (2 2)

2

a aa

a a a a

2 2

a

2 + 2 = 22 + 2 = 6

43. Answer (8)

Hint: Find

1 1

1

2 1

sin

tan

lim lim1

K

K

kx x

tdt

t

k C

k

Solution : Here,

1

1

1

1

sin ( )

tan ( )

k

k

k

kxC dx

kx

,

put kx = t

1

dx dtk

when 1

1x

k

, then

1

kt

k

when 1

xk

, t = 1

1 1

1

1

sin ( )

(tan )k

k

t dt

kt

1 1

1

1

1 sin

tank

k

xdx

k x

1 1

2 2

1

1

1 sin4 lim 4 lim

tank

k kk

k

xk C k dx

k x

1 1

1

1

sin4 lim

tankk

k

xk dx

x

1 1

1

1

sin

tan

4 lim1

k

k

k

xdx

x

k

1

21

2

sin( 1) 1 11

0( 1)

tan1

4 lim1k

k

k kk

k k

k

k

12

1

1sin

11

14 lim

11

1tan

11

k

k

k

k

21

1

sin 1 14

1 0tan (1)

424 1 4 1 8

2

4

44. Answer (7)

Hint: ∵ CC1 = r + r

1 and CC

2 = r

2 + r.

Solution : Equation of circle C1 is |z + 1| = 3

|x + iy + 1| = 3

2 2( 1) 3x y

Circle C1 : (x + 1)2 + y2 = 9 having centre C

1(–1, 0)

and radius = 3 = r1

Similarly circle C2 : (x – 2)2 + y2 = 49 having centre

C2 (2, 0) and radius r

2 = 7.

Variable circle is

C : |z – z0| = r

i.e., C: |(x + iy) – (h + ik)| = r2

i.e., C: (x – h)2 + (y – k)2 = r2


10/16

C2(2, 0)

C1(–1,0)

r1

r c

( , )h k

Which touches circle C1 externally and circle C

2

internally.

Now, CC1 = r + r

1

CC2 = r

2 – r

CC1 + CC

2 = r

1 + r

2

Locus of C is a conic E as an ellipse with focus at C1

and C2 respectively

Now r1 + r

2 = 3 + 7 = 10 = sum of focal distances = 2a

Distance between foci = 2ac = 2 2(2 1) 10 0) 3

10e = 3 [∵ 2a = 10]

3

10

pe

q q – p = 10 – 3 = 7

45. Answer (9)

Hint: Let P, Q are end points of chord and from point

(h, x) it is chord of contact.

Solution :

o

P

R h k( , )

Y

Y

Q

XX

Let the point of intersection of tangents be R(h, k)

Equation of chord of contact PQ is

ky = 2(x + h)

Join equation of OP and OQ having ‘Q’ O as the

origin is given by

2 24 0

2

yk xy x

h

2hy 2 – 4kxy + 8x2 = 0 ...(1)

Which is identical with the equation

5x2 + 3y2 + xy = 0 ...(2)

Comparing (1) and (2)

8 2 4

5 3

h k

3 82

5h

Locus of R(h, k) is12

(given)5

px

q

p – q + 2 = 12 – 5 + 2 = 9

46. Answer (6)

Hint: Cut equation of two lines AB, CD then their point

of intersection P.

Solution :

Equations of lines AB and CD are:

ˆ ˆ ˆ ˆ ˆ ˆ: (9 7 ) (4 3 )AB r i j k t i j k �

...(1)

ˆ ˆ ˆ ˆ ˆ ˆ: (7 2 7 ) (2 2 )CD r i j k s i j k �

...(2)

Co-ordinates of point of intersection p of lines

(1) and (2)

9 1 7

4 1 3

x y zt

D

P

A(9, –1, 7)

(7, –2, 7)

B

d.r.'s 4, –1, 3

d.r.'s 2, –1, 2

C

Point on line AB is (9 + 4t, –1–t, 7 + 3t) and point

on line CD are (7 + 2s, –2 –s, 7 + 2s)

9 + 4t = 7 + 2s

2s – 4t = 2

s – 2t = 1 ...(1)

–1 – t = –2 – s

s – t = –1 ...(2)

and 7 + 3t = 7 + 2s

2s – 3t = 0 ...(3)


11/16

From (3), 2

3

st

From (2), 2

1 33

ss s

2 1 3 1 4t s

t = –2

P (9 – 8), –1 + 2, 7 – 6) P (1, 1, 1)

Let Q (x, y, z)

ˆ ˆ ˆ( 1) ( 1) (2 1)PQ x i y j k ��

Given, 15PQ ��

2 2 2( 1) ( 1) (2 1) 15x y

(x – 1)2 + (y – 1)2 + (z – 1)2 = 225 ...(4)

given, andPQ AB PQ CD ��

PQ AB CD��

Where

ˆ ˆ ˆ

4 1 3

2 1 2

i j k

AB CD

��

ˆ ˆ ˆ( 2 3) (8 6) ( 4 2)i j k

ˆ ˆ ˆ2 2i j k

and so 1 1 1

(say)1 2 2

x y z

...(5)

x – 1 = , y – 1= –2, z – 1 = –2Putting these values in (4)

2 + (–2)2 + (–2)2 = 225

92 = 225 = ± 5

when = 5, then from (5)

x = 6, y = –9, z = –9

when = –5, then from (5),

x = –4, y = 11, z = 11

Coordinates of Q can be (6, –9, –9) and (–4, 11, 11)

So that x1 = 6, x

2 = –6, x

3 = –9 and y

1 = –4, y

2 = 11,

y3 = 11

3

1 1 2 2 3 3

1

( ) ( ) ( )i i

i

x y x y x y x y

= (x1 + x

2 + x

3) + (y

1 + y

2 + y

3)

= (6 – 9 – 9) + (–4 + 11 + 11)

= –12 + 18 = 6

47. Answer (4)

Hint: Find a general point on line and its image w.r.t.

given plane

Solution : Any point on given line is (1 + 3r, 3 + 5r, 4 + 2r)

Any two points or given line can be taken as P(1, 3, 4)

and Q(4, 8, 6).

Let image of point P(1, 3, 4) due to plane 2x – y + z + 3

= 0 be R and image of the point Q(4, 8, 6,) due to

same plane be S.

For image of Q.

4 8 6 2(8 8 6 3)

2 1 1 4 1 1

x y z

4 8 6 18

2 1 1 6

x y z

x = 4 – 6 = –2, y = 3 + 8 = 11, z = –3 + 6 = 3

S (–2, 11, 3)

For image of P(1, 3, 4)

1 3 2 4 2(2 3 4 3) 122

2 1 1 6 6

x y

x = –4 + 1 = –3, y = 2 + 3 = 5, z = –2 + 4 = 2

R (–3, 5, 2)

Equation of line RS i.e., equation of line L is

3 5 2

2 3 11 5 3 2

x y z

3 5 2

1 6 1

x y z ...(1)

Now equation of the plane which contains the line L is

a(x + 3) + b(y – 5) + c(z –2) = 0 ...(2)

and a + 6b + c = 0 ...(3)

Also plane (2) is perpendicular to the plane

2x – y + z + 3 = 0

2a – b + c = 0 ...(4)

Solving (3) and (4) for a, b, c we get

,say6 1 1 2 1 12

a b c

a = 7, b = , c = –13

Putting these values of a, b, c in equation (2)


12/16

We have the equation of plane as

7(x + 3) – (y – 5) –13(z – 2) = 0

7(x + 3) + (y – 5) –13(z – 2) = 0

7x + y – 13z + 42 = 0

Comparing it with given equation

p = 1, q = –13, r = 42

p + r + 3q

= 1 + 42 – 39 = 43 – 39 = 4

48. Answer (4)

Hint: Equate two pairs together.

Solution : Let x(x – 1) + 2yz = ...(1)

y (y – 1) + 2zx = ...(2)

and z(z –1) + 2xy = ...(3)

By [(2) – (1)]

y2 – y + 2zx – x2 + x – 2zy = 0

(x – y) + 2z(x – y) – (x2 – y2) = 0

(x – y)[1 + 2z – (x + y)] = 0

(x – y)(x + y – 1 – 2z) = 0 ...(4)

Similarly, (y – z)(y + z – 1 – 2x) = 0 ...(5)

and (z – x)(z + x – 1 – 2y) = 0 ...(6)

Case-I

When x y z

From (4), (5) and (6),

x + y – 1 – 2z = 0

y + z – 1 – 2x = 0

z + x – 1 – 2y = 0

Adding all these, –3 = 0 (a contradiction)

no solution is possible in this case

Case-II

When any two are equal.

Let x = y z, then from (5)

(y – z)(x + z – 1 – 2x) = 0

z – 1 – x = 0 z – x = 1, and z – y = 1 y – z = –1

and x – y = 0

(x – y, y – z, z – x) (0, –1, 1)

Similarly when x y = z, then

(x – y, y – z, z – x) (1, 0, –1)

and x = z y, then (x – y, y – z, z – x) (–1, 1, 0)

Three solutions are possible.

Case-III

When x = y = z, then (x – y, y – z, z – x) (0, 0, 0)

one solution is possible.

Total number of ordered triplets is 4.

49. Answer (A, B)

Hint: Use tree diagram to get result.

Solution :

C1

CoinsCoin C2

C3

,Fair

biased

H-Blue

T-Blue

T-White

H-Red

Tree diagram of the experiment is shown below:

C1

C2

C2C

3

C1C

2

C1

C3

C2

H B( )

H B( )

H R( )H R( )

H B( )

H R( )

T B( )

T W( )

T B( )

T W( )

T B( )

T B( )C

2

C1

C3

C3C

1

C3

Required probability (When both coins show up the

same colour)

1 1 1 1 1 1(1 ) ( ) (1 )(1 ) (1 )

3 2 3 3 3 2k k k k k k

2 22(1 ) 1[ (1 ) ]

3 2 3

kk k

2

2 21 2 3 2 29[1 2 1] (given)

3 3 96

k kk k k k

2 292 3 2

32k k

64k2 – 96k + 35 = 0

64k2 – 56k – 40k + 35 = 0

8k (8k – 7) – 5(8k – 7) = 0

5 7or

8 8k


13/16


Hint: px2 + qy2 = –r represents different conics on

conditions of p, q and r.

Solution : Given equation is

(px2 +qy2 + r)(4x2 + 4y2 – 8x – 5) = 0

Either, 4x2 + 4y2 – 8x – 5 = 0 ...(1)

2 2 52 0

4x y x

2 2 9( 2 1)

4x x y

2 2 9( 1)

4x y

Which is a circle with centre (1, 0) and radius = 3

2

or, px2 + qy2 + r = 0 ...(2)

Here equation (2) will represent a pair of straight lines

when r = 0 and p and q will be opposite signs.

Also equation (2) will represent a circle if p = q and r

will be of sign opposite to that of p.

Again equation (3) will represent a hyperbola when p

and q are of opposite sign and r 0.

Also, equation (3) will represent an ellipse if p and q

are unequal and of same and r has sign opposite to

that of p.

51. Answer (A, C, D)

Hint: Every constant function is periodic.

Solution :

(A) 0x

e x R

f(x) sgn(e–x) = 1

and every constant function is periodic.

f(x) = sgn(e–x) is a periodic function.

(B) Let T > 0 be a rational number.

Then , if

( ) ( ),if

x x Qf T x f x

x x Q

f(x) is a non-periodic function

(C)2 2

4(1 sin 1 sin ) 8( )

1 sin cos

x xf x

x x

2 2 sec x which is a periodic function.

(D) f(x) = {x} + 1

Which is also a periodic function, since {x} is a

periodic function.


Hint: Use the graph.

Solution :

(A) Here [ ] 1

( )[ ] 1

xf x

x x

0 1

0 1x

, if 0 x < 1

1 1

1 1x

, 1 x < 2

2 1 5, 2

2 1 2

x

x

1, 0 1

1

2, 1 2

3 5, 2

1 2

x

x

x

x

x

x

10

y

x2 5/2

1/2

1

2

3

From graph it is discontinuous at the points x = 1, 2

f(x) is discontinuous as well as non-diff.

Also, 1, 3

2

fR = co-domain of f(x)

f(x) is surjective


14/16

Also, 01 1

1 1 1lim ( ) lim lim

( 1) 1 1 2hx x

f xx h

and 01 1

2 2 2lim ( ) lim lim 2

1 1 0hx x

f xx h

1 1

1min. lim ( ), lim ( ) max. ,2

2x x

f x f x

option (C) is correct 1

(1)2

f

53. Answer (A, C)

Hint: Reciprocal inequality.


x3 – x2 + (a4 + 4a2 + 1)x – a2 = 0

= 1 ...(1)

+ + = a4 + 4a2 + 1 ...(2)

= a2 ...(3)

Now,

1

1 1 1

, from (1)

1 1 13

3

4 2

2

4 13

a a

a

2

2

11a

a

3.

Minimum value of

1

3

Now, 3 2

2 2log 8 log 2 3log2 3

4

2 2 2log 16 log 2 4log 2 4

3

3 3 3log 27 log 3 3log 3 3

and 3

2 5 5log 125 log 5 3log 5 3


Hint: Differentiate both sides w.r.t. x.


0 0

( ) ( ) 1

x x

xf t dt t f x t dt e

0 0

( ) ( 0 ) ( 0 ) 1

x x

xf t dt x t f x x t dt e

0 0 0

( ) ( ) ( ) 1

x x x

xf t dt x f t dt t f t dt e

Differentiating both sides w.r.t. x, we get

0

( ) 1 1 ( ) [ ( ) 1 (0) 0]

x

f x f t dt x f x f

( ) ( ) 0 (0) (0) 0 xd dxf x x f e

dx dx

0

( ) ( ) ( ) ( )

x

xf x f t dt xf x xf x e

...(1)

Put x = 0 in (1), we get

0(0) 0 1 (0) 1f e f

Again diff. both sides w.r.t. x, we get

( ) ( ) 1 (0) 0 xf x f x f e

[ ( ) ( )]xe f x f x

[ ( )]xd

e f xdx

Integrating, we get exf(x) = x + c ...(2)

Put x = 0 is (2) e0f (0) = 0 + c c = 1× (–1) = –1

From (2),

f(x) = (x – 1) e–x

f (2) = (2 – 1 )e–2 = e–2 option (A) is correct

( ) ( 1) x xf x x e e

0 0(0) (0 1) 2f e e

f(0) + f (0) = –1 + 2 = 1

Option (B) and (C) are correct


15/16

55. Answer (B)

Hint: Put t = x3

Solution : Put t = x3

dt = 3x2dx

I = 2 2 1 2(1 ) 3x x x dx

= 2

31

dx

x

= 3tan–1x + c

=

1

1 33 tan t c

56. Answer (A, D)

Hint: Use formulae of increase trignometric function.

Solution : 1 2 1( ) tan

1 2

xf x

x x

1 1tan ( 2 1) tan ( ) x

1tan ( )8

x

1( ) 0

2 (1 )

∵f x x

x x

f(x) < 0

f(x) is decreasing function and 1 4

4 5

f

57. Answer (B, D)

Hint: Use general term in the expanssion.

Also nCr +

nC

r + 1 =

n + 1Cr + 1

Solution : Here 100

100100

0

3 4r r

r

r

E C x

100 100( 3) 4 (1 )x x

Coefficient of x2 in E = 100C2 = 4950

1

1

1

n

k

k n k

k

C kA

nC

1 1

0 0

1( 1)

( 1)

n n

k

k k

A kn

1 ( 1)

( 1) 2 2

n n n

n

...(1)

But given

1

3

0

4n

k

k

A

1

0

64

n

k

k

A

...(2)

From (1) and (2),

64 1282

n


Hint: A2 = 4A + 5I3

Solution :

1 2 2 0 0

2 1 2 0 0

2 2 1 0 0

A I

1 2 2

2 1 2

2 2 1

Now

1 2 2

2 1 2 0

2 2 1

A I

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

2 4 2(1 ) 0

(1 – )[1 + 2 – 2 – 4] – 2(2 – 2 – 4) + 2(4 – 2 + 2) = 0

2(1 ) 2 3 2 2 2 2 2 2 0

2 – 2 – 3 – 3 + 22 + 3 + 8 + 8 = 0

–3 + 32 + 9 + 5 = 0

3 – 32 – 9 – 5 = 0

A3 – 3A2 – 9A – 5I3 = 0

(A + I) (A2 – 4A – 5I3) = 0

A2 – 4A – 5I3 = 0 Option (A) is correct

Also, A2 = 4A + 5I3

(A–1A)A = 4(A–1A) + 5(A–1I3)

IA = 4I3 + 5A–1

–1 34

5

A IA

option (B) is correct.

Now 2

1 2 2 1 2 2

2 1 2 2 1 2

2 2 1 2 2 1

A

9 8 8

8 9 8

8 8 9


16/16

and |A2| = 9(81 – 64) – 8(72 – 64) + 8(64 – 72)

= 9 × 17 – 8 × 8 + 8 × (–8)

= 153 – 64 – 64 = 153 – 128 25

A2 invertible and

A3 = A A2 = A (4A + 5I3) = 4A2 + 5A

36 32 32 15 10 10 41 42 42

32 36 32 5 10 5 10 42 41 42

32 32 36 10 10 5 42 42 41

A3 = 41(412 – 422) – 42(42 × 41 – 422) +

42(422 – 42 × 41)

= 41(41 – 42)(41 + 42) – 422(41 – 42)

– 422(41 – 42)

= –[41 × 83 + 422 + 422] 0

A3 is also invertible.

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

Hint: Use graph of y = sin–1(sinx), y = cos–1(cosx),

y = tan–1(tanx)

Solution : (A) f2(x) = f

1(f

1(x)) = x

f3(x) = f

1(f

2(x)) = f

1(x) = x

From given equation

xxx – 25xx + 175x = 375

x3 – 25x2 + 175x – 375 = 0

x3 – 5x2 – 20x2 + 100x + 75x – 375 = 0

x2(x – 5) – 20x(x – 5) + 75(x – 5) = 0

(x – 5) (x2 – 20x + 75) = 0

(x – 5)2 (x – 15) = 0

x = 5, 15

(B) 2 2( 2) 2( 1)2 2 2 52

x x x 2

2 64 3x

x (C) Here domain of f(x) is [–1, 1]

f(x) = 5x + 4x – x = 8x

f(x) [–8, 8]

(D) Here domain of f(x) is [–1, 1]

For positive roots, x [0, 1].

Then from given equation

21

22

xx x

1 + x2 = 2x

x = 1

|10a| = 10

[|10a|] = 10

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

D(R, S)

Hint:

lim

lim

1lim ( )

n

n

b

nb

nr a a

n

rf f x dx

n n

Solution : (A) Given f(x) = 2

1

2 (ln 2ln )

x

x t t dt

Differentiating both sides w.r.t . x,

(ln + 1)2 = 0

1 1

x e

e

= 1

0, 1a be a b

(B) Given 2 2

1 1

2

0 0

2 t tt e dt e dt

2 21 1

0 0

( )( 2 )t tt t e dt e dt

2 2 21 1

1

0

0 0

[( ) ] ( 1)t t tt e e dt e dt

2 21 1

1

0 0

[( 1) 0] t te e dt e dt

= e–1 = + e–1 (given)

= 0, = 1

–= 1 – 0 = 1

(C)

2

20

(cos 1)lim

2

0lim 2 cos 1

2

x

xx

x

x

xe

1 1

2 2

e p qe

p = 0, q = 1 and 1

2 2

p q

(D) tan

2

lim (cosec cos ) Form 1

x

x

l x x

2

cosec cos 1lim

cot

x

x x

x

e

2

2

cosec cot sinlim

cosec

x

x x x

x

e

1 5 q

e p e

p = 0, q = 5

��

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

1/16

1. (4)

2. (3)

3. (8)

4. (6)

5. (5)

6. (4)

7. (4)

8. (7)

9. (A, D)

10. (B, D)

11. (B, D)

12. (A, B, C)

13. (A, B, D)

14. (B, C)

15. (A, C)

16. (A, C)

17. (A, B, C)

18. (A, B, D)

19. A (T)

B (P, R)

C (Q)

D (S)

20. A (Q)

B (P, R)

C (T)

D (R, S)

21. (6)

22. (3)

23. (1)

24. (0)

25. (5)

26. (6)

27. (9)

28. (8)

29. (A, B, C, D)

30. (A, B, C, D)

31. (A, B, C, D)

32. (A, D)

33. (A, B, C)

34. (A, B, C)

35. (A, C)

36. (A, C)

37. (B, D)

38. (A, B)

39. A (P, Q, T)

B (P, Q, R, S, T)

C (R, S)

D (P, Q, T)

40. A (P, R, S, T)

B (P, R, S, T)

C (P, R, S, T)

D (Q, S, T)

41. (4)

42. (4)

43. (6)

44. (9)

45. (7)

46. (8)

47. (6)

48. (4)

49. (A, B, D)

50. (B, D)

51. (A, D)

52. (B)

53. (A, B, C)

54. (A, C)

55. (A, B, C)

56. (A, C, D)

57. (A, B, C, D)

58. (A, B)

59. A (P, Q, R, S)

B (P, Q, R, S)

C (P, Q, R, S, T)

D (R, S)

60. A (Q, S)

B (Q, R, S, T)

C (P, Q)

D (R)

ANSWERS

PHYSICS CHEMISTRY MATHEMATICS

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

Test Date: 17/02/2019

MOCK TEST - 1 (Paper-1) - Code-B

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

2/16

PART - I (PHYSICS)

HINTS & SOLUTIONS

1. Answer (4)

Hint: Based on beat formation

Solution :

1 2sin( )Y A t

2 2sin( )Y A t

1 2 1 2

1 2

( ) ( )2 sin cos

2 2

t tY Y A

(2 ) 42 cos sin 2 (302)

2

tA t

2

20

44cos

13

44

I

I A

2. Answer (3)

Hint: Work done by force is, 0

x

dw Fdx

Solution :

Let block be displaced by x, liquid rises by y

2

( )4

RF x y g

2

( )4

R g

dw x y dx

3 x

y

3 /82

0

4

4 3

h

R gw xdx

2 23

128w R gh

3. Answer (8)

Hint: vr = constant

Solution :

00 0 1

2

rv r v

v1 = (2v

0)

22

01

0 0

8

2

mvmvT

r r

4. Answer (6)

Hint: 2

mp

RTV

M

Solution : 2 2

3

RT RT

M M

3

2

n = 6

5. Answer (5)

Hint: Potential across capacitor is same as potential

across 4 resistor.

Solution :

I2

I1I

C

3I1 = I

2

IC + I

1 = I

2

IC = 2I

1

3

0 42

13

t

CC

V Cq e

3 3

0 04 42 2 3 1

1 43 3 4 2

t t

C CV V C t

e eC

3 3

04 40

31

2

t t

C CV

V e e

3

42 5

t

Ce

4 5In

3 2

Ct

6. Answer (4)

Hint: ( ) � � �

F q E V B

Solution :

y

x

0ˆE j

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

3/16

0 0ˆ ˆ ˆ ˆ( ) ( )

x y

dvF m qE j q v i v j B k

dt

�

�

0 x

y

dvm qv B

dt

0 0 y

x

dvm qE qv B

dt

0 0

0

y

y

dv qB qyBmv qE

dy m

2 2 2

0

00 0

2y

q B yv qE y

m

0

2

0

2mEy

qB

0

0 0 2

0

2mEK qE y qE

qB

2

0

2

0

2E

K mB

7. Answer (4)

Hint: PV = constant

Solution :

1

4 TP

R

2

2

0

2

2

TP

R

PV = constant

015 2

8

T

R

n = 4

8. Answer (7)

Hint: Apply concept of solid angle

Solution :

h

2

d

2

2

22 (1 cos60 ) 2 14 2

2

d

dh

7

6

dh

x = 7

9. Answer (A, D)

Hint: Here COG is different from COM

Solution :

2

0( )

eR

e

e e

GM MF dx

R R x

22

e

e

GM MF

R

2

2

2

e

( )

(COG) ( )

2

e

e

R

R

e

y dm GM

y MY dm dy

GM M R

R

Y = 2R ln2 = 1.4R

d = 1.4R – R = 0.4R

10. Answer (B, D)

Hint: Velocity of separation

Velocity of approach e

Solution :

v0

1sin

2

0

1 2

3

2

v

v v

0

2 1

3

4

v

v v

0

2

3 3

8

v

v


4/16

11. Answer (B, D)

Hint: , is critical angle of incidence, 1

1

1sin

Solution :

= – 2

2

d

d

1 1sin 30

2


Hint: For equilibrium, mass of liquid displaced is equal

to total mass

Solution :

fb = (m

1 + m

2)g

1 2

3 3 1 2

1 2

( )m m

m m

31 3 1

2

21

( )1

mm

2 31 1

2 2 3 1

( )

( )

m

m


Hint:

dNN

dt

N = N0e–t

Solution :

dNN

dt

N = N0e–t

R0 = log |N

0e–t|

R0 = log |N

0| – t

0

0

Re

N

0

02

R

t

14. Answer (B, C)

Hint: Draw phasor diagram to derive effective current

Solution :

0 0CV

V0

0

2 2 2

0

V

R L

0

0 02 2 2

0

sinVCV

R L

0 0

0 02 2 2

0

V L

CVR L

2 2 2

0

LR L

C

2

0 2 2

L R

L C L

2

2

1 R

LC L

0 0

2 2 22 20

V R V Ri

LR LR R

C

0CV R

L

15. Answer (A, C)

Hint: At steady state current through circuit is zero

Solution :

2

0

1 3

2 4

CH V

16. Answer (A, C)

Hint: Image formed by lens should fall either normally

or on the pole of mirror.

Solution :

Image formed by lens should be either on pole or fall

normally on mirror.


5/16

PART - II (CHEMISTRY)

21. Answer (6)

Hint:

In Claisen rearrangement, carbon attached at ortho.

Solution :

O

1

2

34

5

O1

2

3

H

OH 2

3

1

Tautomeric

x = 3

y and z = 2 and 1

22. Answer (3)

Hint & Solution :

106 ml of water contains 100 g or 1 mole of CaCO3

= 2 mole of H+, [H+] = 2 × 10–3 M

23. Answer (1)

Hint:

At equilibrium Ecell

= 0

Solution :

Balanced equation is

2 – 2

2 3 2 2Hg NO 3H 2Hg HNO H O

22

cell 32 –

2 3

HgRTE 0 E – log

nF Hg NO H

2

32 2

10.060.09 log

21 1 H

pH = 1

24. Answer (0)

Hint:

All the given ions have different magnetic moment with

Strong field ligand and Weak field ligand

Solution :

Pairing can takes place in all ion with S.F.L

25. Answer (5)

Hint:

CH – C 3

OH O

Show G.I

Solution :

CH – C 3

OOH

CH – C 3

OOH

,

CH = CH2

OH O

CH – C3

OH O

,


Hint: sin 0.6 sr

r

V

V

Solution :

sin sr

r

v

v

601000

100020

sr

dv

t

[vr = v

sr sin ]


Hint: 0 (1 cos )2

CVQ t

Solution : max 0di

idt

At that moment, potential across capacitor is same

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

Hint: VIM

= –m2VOM

Solution :

VIM

= –m2VOM

| |V

mu

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

Hint: To escape, total energy 0


6/16

26. Answer (6)

Hint:

Ni(CO)4 is tetrahedral in shape.

Solution :

Ni(CO)4 is tetrahedral, which has 6 planes of symmetry.

27. Answer (9)

Hint:

aE

–RTK Ae Arrhenius equation

Solution :

aE

–RT

0K A e

ln K = In a

0

EA –

RT

11488log K log 1000 –

8.314 300 2.30

= 3 – 2 = 1

K = 10 hr–1

0

t

AKt In

A

0

t

A2.3 2.3 log

A

0 0

t 0

A A10

A 1 A – x

t

t

E.C9

B.C.C

28. Answer (8)

Hint:

n - factor of Ba(MnO4)

2 = 10

Solution :

Normality of H2O

2 =

11.2

5.6 = 2 N

meq (Ba(MnO4)

2) = meq of H

2O

2 reacted

= 2 × 100 = 200 meq

= 0.2 eq

Moles of Ba(MnO4)

2 = 0.02

0.02 375% purity 100 92%

8.15

% impurity = 8%


Hint :

B2H

6 + 6CI

2 2BCI

3 + 6HCI

Solution :

B2H

6 + 6H

2O B(OH)

3 + 6H

2 (g)

B2H

6 + 2NH

3 [H

2B(NH

3)

2]+ + [BH

4]–

B2H

6 + 2(CH

3)

3N 2H

3B N(CH

3)

3


Hint :

P Na2S

2O

3

Solution :

Na2SO

3 + S Na

2S

2O

3

AgX + 2Na2S

2O

3 [Ag(S

2O

3)]3 – + X + 4Na+

X = CI, Br, I

Na2S

2O

3 + CI

2 + H

2O 2HCI + S + Na

2SO

4

2Na2S

2O

3 + I

2 Na

2S

4O

6 + 2NaI


Hint :

Sulphur show disproportionate reaction with alkali.

Solution :

S8 + 6NaOH Na

2S

2O

3 + 2Na

2S + 3H

2O

Zn + NaOH Na2ZnO

2 + H

2 (g)

AI + NaOH Na2AIO

2 + H

2 (g)

32. Answer (A, D)

Solution :

(A) Acidic buffer will form (pH < 7)

(B) Both are salt of SASB (pH = 7)

(C) Weak base will form (pH > 7)

(D) Strong acid solution (pH < 7)


Hint :

Aldo

Solution :

(A)

O

(i) CH MgBr3

(ii) H O2OH

CH3

H+

+H

+

O

O

(i) O3

(ii) Zn, H O2

+

+

(B)

CHO(i) CH MgBr

3

(ii) H O2

CH – CH3

OH

H+

+

O

O

(i) O3

(ii) Zn, H O2

+


7/16

(C)

OCH MgBr3

H O2

OH

CH3

H+

O3

Zn, H O2

O

O


Hint:

M : HO – C – C – CH – OH 2

O H

CH3

N : CI – C – C – CH – CI 2

O H

CH3

O : C H – N – C – C – CH – CI 2 5 2

O H

H CH3

Solution :

Structure of M

HO – C – CH – CH – OH 2 2

– CH2

O

Not possible because M does not contain 3° hydrogen

M : HO – C – C – CH – OH 2

O H

CH3

HO – C – C – CH – OH 2

O H

CH3

CI – C – C – CH – CI 2

O H

CH3

SOCI2

C H NH2 5 2

C H N – C – C – CH – CI 2 5 2

O H

CH3

H

35. Answer (A, C)

Hint:

In elecrophilic addition, carbocation is intermediate.

Solution :

Rate of electrophilic addition is proportional to stability

of carbocation.

36. Answer (A, C)

Hint:

B and P has vacant orbital.

Solution :

Charge can be delocalised only in 3 2O –P CH and

3 2O –B CH due to presence of vacant orbital.

37. Answer (B, D)

Hint:

Process is reversible so G = 0 only at constant T

and P.

Solution :

Process A to B is isothermal

so H = U = 0

U = q + w

w = –ve and q = +ve


Hint:

A : FeCr2O

4 , B : Na

2CrO

4 , C : Fe

2O

3 , D : CrO

5 ,

E : CrO8

–3

, F : SO

2 , G : SO

3

Solution :

FeCr O + 2 4

Na CO + O2 3 2

Fe O + 2 3

Na CrO + CO2 4 2

(A) (C) (B)

Na CrO2 4

H O /H2 2

+

CrO5

CrO8

–3

(D)

(E)H O

2 2

Alkaline


8/16

PART - III (MATHEMATICS)

41. Answer (4)

Hint: Equate two pairs together.

Solution : Let x(x – 1) + 2yz = ...(1)

y (y – 1) + 2zx = ...(2)

and z(z –1) + 2xy = ...(3)

By [(2) – (1)]

y2 – y + 2zx – x2 + x – 2zy = 0

(x – y) + 2z(x – y) – (x2 – y2) = 0

(x – y)[1 + 2z – (x + y)] = 0

(x – y)(x + y – 1 – 2z) = 0 ...(4)

Similarly, (y – z)(y + z – 1 – 2x) = 0 ...(5)

and (z – x)(z + x – 1 – 2y) = 0 ...(6)

Case-I

When x y z

From (4), (5) and (6),

x + y – 1 – 2z = 0

y + z – 1 – 2x = 0

z + x – 1 – 2y = 0

Adding all these, –3 = 0 (a contradiction)

no solution is possible in this case

Case-II

When any two are equal.

Let x = y z, then from (5)

(y – z)(x + z – 1 – 2x) = 0

z – 1 – x = 0 z – x = 1, and z – y = 1 y – z = –1

and x – y = 0

(x – y, y – z, z – x) (0, –1, 1)

Similarly when x y = z, then

(x – y, y – z, z – x) (1, 0, –1)

and x = z y, then (x – y, y – z, z – x) (–1, 1, 0)

Three solutions are possible.

Case-III

When x = y = z, then (x – y, y – z, z – x) (0, 0, 0)

one solution is possible.

Total number of ordered triplets is 4.

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

Hint:

1st group cation form white ppt with dil.HCI

Chromate of Ag+ and 2

2Hg are red while chromate of

Pb2+ is yellow.

Solution :

Ag2S, NiS, CuS, PbS black

22

3 3 4Excess

Cu NH Cu NH

22

3 3 6Excess

Ni NH Ni NH

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

S, T)

Hint:

In B2H

6, hybridisation of ‘B’ is sp3.

Solution :

(A) B

H

B BB

H

H

H

H

H

sp3

Empty orbital participate in hybridization

Two types of bonds covalent and 3c – 2e

(B)AI

CI

Cl CI

CI

AI

CI

CI

sp3

Empty orbital of AI participate in hybridization

Two types of bonds covalent and co-ordinate

(C) BeCI2 is polymeric in solid state

BBe x Bey

CI

CI

CI

CI

BBe

sp3 Vacant orbital of Be participate

Two types of bonds covalent of Co-ordinate

(D)

N

H

sp2


9/16

42. Answer (4)

Hint: Find a general point on line and its image w.r.t.

given plane

Solution : Any point on given line is (1 + 3r, 3 + 5r, 4 + 2r)

Any two points or given line can be taken as P(1, 3, 4)

and Q(4, 8, 6).

Let image of point P(1, 3, 4) due to plane 2x – y + z + 3

= 0 be R and image of the point Q(4, 8, 6,) due to

same plane be S.

For image of Q.

4 8 6 2(8 8 6 3)

2 1 1 4 1 1

x y z

4 8 6 18

2 1 1 6

x y z

x = 4 – 6 = –2, y = 3 + 8 = 11, z = –3 + 6 = 3

S (–2, 11, 3)

For image of P(1, 3, 4)

1 3 2 4 2(2 3 4 3) 122

2 1 1 6 6

x y

x = –4 + 1 = –3, y = 2 + 3 = 5, z = –2 + 4 = 2

R (–3, 5, 2)

Equation of line RS i.e., equation of line L is

3 5 2

2 3 11 5 3 2

x y z

3 5 2

1 6 1

x y z ...(1)

Now equation of the plane which contains the line L is

a(x + 3) + b(y – 5) + c(z –2) = 0 ...(2)

and a + 6b + c = 0 ...(3)

Also plane (2) is perpendicular to the plane

2x – y + z + 3 = 0

2a – b + c = 0 ...(4)

Solving (3) and (4) for a, b, c we get

,say6 1 1 2 1 12

a b c

a = 7, b = , c = –13

Putting these values of a, b, c in equation (2)

We have the equation of plane as

7(x + 3) – (y – 5) –13(z – 2) = 0

7(x + 3) + (y – 5) –13(z – 2) = 0

7x + y – 13z + 42 = 0

Comparing it with given equation

p = 1, q = –13, r = 42

p + r + 3q

= 1 + 42 – 39 = 43 – 39 = 4

43. Answer (6)

Hint: Cut equation of two lines AB, CD then their point

of intersection P.

Solution :

Equations of lines AB and CD are:

ˆ ˆ ˆ ˆ ˆ ˆ: (9 7 ) (4 3 )AB r i j k t i j k �

...(1)

ˆ ˆ ˆ ˆ ˆ ˆ: (7 2 7 ) (2 2 )CD r i j k s i j k �

...(2)

Co-ordinates of point of intersection p of lines

(1) and (2)

9 1 7

4 1 3

x y zt

D

P

A(9, –1, 7)

(7, –2, 7)

B

d.r.'s 4, –1, 3

d.r.'s 2, –1, 2

C

Point on line AB is (9 + 4t, –1–t, 7 + 3t) and point

on line CD are (7 + 2s, –2 –s, 7 + 2s)

9 + 4t = 7 + 2s

2s – 4t = 2

s – 2t = 1 ...(1)

–1 – t = –2 – s

s – t = –1 ...(2)

and 7 + 3t = 7 + 2s

2s – 3t = 0 ...(3)

From (3), 2

3

st

From (2), 2

1 33

ss s

2 1 3 1 4t s

t = –2

P (9 – 8), –1 + 2, 7 – 6) P (1, 1, 1)


10/16

Let Q (x, y, z)

ˆ ˆ ˆ( 1) ( 1) (2 1)PQ x i y j k ��

Given, 15PQ ��

2 2 2( 1) ( 1) (2 1) 15x y

(x – 1)2 + (y – 1)2 + (z – 1)2 = 225 ...(4)

given, andPQ AB PQ CD ��

PQ AB CD��

Where

ˆ ˆ ˆ

4 1 3

2 1 2

i j k

AB CD

��

ˆ ˆ ˆ( 2 3) (8 6) ( 4 2)i j k

ˆ ˆ ˆ2 2i j k

and so 1 1 1

(say)1 2 2

x y z

...(5)

x – 1 = , y – 1= –2, z – 1 = –2

Putting these values in (4)

2 + (–2)2 + (–2)2 = 225

92 = 225 = ± 5

when = 5, then from (5)

x = 6, y = –9, z = –9

when = –5, then from (5),

x = –4, y = 11, z = 11

Coordinates of Q can be (6, –9, –9) and (–4, 11, 11)

So that x1 = 6, x

2 = –6, x

3 = –9 and y

1 = –4, y

2 = 11,

y3 = 11

3

1 1 2 2 3 3

1

( ) ( ) ( )i i

i

x y x y x y x y

= (x1 + x

2 + x

3) + (y

1 + y

2 + y

3)

= (6 – 9 – 9) + (–4 + 11 + 11)

= –12 + 18 = 6

44. Answer (9)

Hint: Let P, Q are end points of chord and from point

(h, x) it is chord of contact.

Solution :

o

P

R h k( , )

Y

Y

Q

XX

Let the point of intersection of tangents be R(h, k)

Equation of chord of contact PQ is

ky = 2(x + h)

Join equation of OP and OQ having ‘Q’ O as the

origin is given by

2 24 0

2

yk xy x

h

2hy 2 – 4kxy + 8x2 = 0 ...(1)

Which is identical with the equation

5x2 + 3y2 + xy = 0 ...(2)

Comparing (1) and (2)

8 2 4

5 3

h k

3 82

5h

Locus of R(h, k) is12

(given)5

px

q

p – q + 2 = 12 – 5 + 2 = 9

45. Answer (7)

Hint: ∵ CC1 = r + r

1 and CC

2 = r

2 + r.

Solution : Equation of circle C1 is |z + 1| = 3

|x + iy + 1| = 3

2 2( 1) 3x y

Circle C1 : (x + 1)2 + y2 = 9 having centre C

1(–1, 0)

and radius = 3 = r1

Similarly circle C2 : (x – 2)2 + y 2 = 49 having centre

C2 (2, 0) and radius r

2 = 7.

Variable circle is

C : |z – z0| = r

i.e., C: |(x + iy) – (h + ik)| = r2

i.e., C: (x – h)2 + (y – k)2 = r2


11/16

C2(2, 0)

C1(–1,0)

r1

r c

( , )h k

Which touches circle C1 externally and circle C

2

internally.

Now, CC1 = r + r

1

CC2 = r

2 – r

CC1 + CC

2 = r

1 + r

2

Locus of C is a conic E as an ellipse with focus at C1

and C2 respectively

Now r1 + r

2 = 3 + 7 = 10 = sum of focal distances = 2a

Distance between foci = 2ac = 2 2(2 1) 10 0) 3

10e = 3 [∵ 2a = 10]

3

10

pe

q q – p = 10 – 3 = 7

46. Answer (8)

Hint: Find

1 1

1

2 1

sin

tan

lim lim1

K

K

kx x

tdt

t

k C

k

Solution : Here,

1

1

1

1

sin ( )

tan ( )

k

k

k

kxC dx

kx

,

put kx = t

1

dx dtk

when 1

1x

k

, then

1

kt

k

when 1

xk

, t = 1

1 1

1

1

sin ( )

(tan )k

k

t dt

kt

1 1

1

1

1 sin

tank

k

xdx

k x

1 1

2 2

1

1

1 sin4 lim 4 lim

tank

k kk

k

xk C k dx

k x

1 1

1

1

sin4 lim

tankk

k

xk dx

x

1 1

1

1

sin

tan

4 lim1

k

k

k

xdx

x

k

1

21

2

sin( 1) 1 11

0( 1)

tan1

4 lim1k

k

k kk

k k

k

k

12

1

1sin

11

14 lim

11

1tan

11

k

k

k

k

21

1

sin 1 14

1 0tan (1)

424 1 4 1 8

2

4

47. Answer (6)

Hint: Here sin 12

a bC

ab

Solution : Given, ( )R a b C ab

( ) (2 sin )R a b R C ab

sin2

a bC

ab

...(1)

But 1

2

a b

ab

and sin C 1

(1) will be true if

2 and sin 1 a b ab C

2( ) 0a b and 2

C

and2

a b c


12/16

1sin

2

2

ar C

ra b cS

A

CB a

b = a

2c a

2

1sin

12 2

2 (2 2)

2

a aa

a a a a

2 2

a

2 + 2 = 22 + 2 = 6

48. Answer (4)

Hint: Draw graph

Solution :

x

y

(3, )3

(1, 1)

(–1, 3)

(–4, 0) (6, 0)O

x = 0, 1, 3, 6 are points of extremum.


Hint: A2 = 4A + 5I3

Solution :

1 2 2 0 0

2 1 2 0 0

2 2 1 0 0

A I

1 2 2

2 1 2

2 2 1

Now

1 2 2

2 1 2 0

2 2 1

A I

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

2 4 2(1 ) 0

(1 – )[1 + 2 – 2 – 4] – 2(2 – 2 – 4) + 2(4 – 2 + 2) = 0

2(1 ) 2 3 2 2 2 2 2 2 0

2 – 2 – 3 – 3 + 22 + 3 + 8 + 8 = 0

–3 + 32 + 9 + 5 = 0

3 – 32 – 9 – 5 = 0

A3 – 3A2 – 9A – 5I3 = 0

(A + I) (A2 – 4A – 5I3) = 0

A2 – 4A – 5I3 = 0 Option (A) is correct

Also, A2 = 4A + 5I3

(A–1A)A = 4(A–1A) + 5(A–1I3)

IA = 4I3 + 5A–1

–1 34

5

A IA

option (B) is correct.

Now 2

1 2 2 1 2 2

2 1 2 2 1 2

2 2 1 2 2 1

A

9 8 8

8 9 8

8 8 9

and |A2| = 9(81 – 64) – 8(72 – 64) + 8(64 – 72)

= 9 × 17 – 8 × 8 + 8 × (–8)

= 153 – 64 – 64 = 153 – 128 25

A2 invertible and

A3 = A A2 = A (4A + 5I3) = 4A2 + 5A

36 32 32 15 10 10 41 42 42

32 36 32 5 10 5 10 42 41 42

32 32 36 10 10 5 42 42 41

A3 = 41(412 – 422) – 42(42 × 41 – 422) +

42(422 – 42 × 41)

= 41(41 – 42)(41 + 42) – 422(41 – 42)

– 422(41 – 42)

= –[41 × 83 + 422 + 422] 0

A3 is also invertible.

50. Answer (B, D)

Hint: Use general term in the expanssion.

Also nCr +

nC

r + 1 =

n + 1Cr + 1


13/16

Solution : Here 100

100100

0

3 4r r

r

r

E C x

100 100( 3) 4 (1 )x x

Coefficient of x2 in E = 100C2 = 4950

1

1

1

n

k

k n k

k

C kA

nC

1 1

0 0

1( 1)

( 1)

n n

k

k k

A kn

1 ( 1)

( 1) 2 2

n n n

n

...(1)

But given

1

3

0

4

n

k

k

A

1

0

64

n

k

k

A

...(2)

From (1) and (2),

64 1282

n

51. Answer (A, D)

Hint: Use formulae of increase trignometric function.

Solution : 1 2 1( ) tan

1 2

xf x

x x

1 1tan ( 2 1) tan ( ) x

1tan ( )8

x

1( ) 0

2 (1 )

∵f x x

x x

f(x) < 0

f(x) is decreasing function and 1 4

4 5

f

52. Answer (B)

Hint: Put t = x3

Solution : Put t = x3

dt = 3x2dx

I = 2 2 1 2(1 ) 3x x x dx

= 2

31

dx

x

= 3tan–1x + c

=

1

1 33 tan t c


Hint: Differentiate both sides w.r.t. x.


0 0

( ) ( ) 1

x x

xf t dt t f x t dt e

0 0

( ) ( 0 ) ( 0 ) 1

x x

xf t dt x t f x x t dt e

0 0 0

( ) ( ) ( ) 1

x x x

xf t dt x f t dt t f t dt e

Differentiating both sides w.r.t. x, we get

0

( ) 1 1 ( ) [ ( ) 1 (0) 0]

x

f x f t dt x f x f

( ) ( ) 0 (0) (0) 0 xd dxf x x f e

dx dx

0

( ) ( ) ( ) ( )

x

xf x f t dt xf x xf x e

...(1)

Put x = 0 in (1), we get

0(0) 0 1 (0) 1f e f

Again diff. both sides w.r.t. x, we get

( ) ( ) 1 (0) 0 xf x f x f e

[ ( ) ( )]xe f x f x

[ ( )]xd

e f xdx

Integrating, we get exf(x) = x + c ...(2)

Put x = 0 is (2) e0f (0) = 0 + c c = 1× (–1) = –1


14/16

From (2),

f(x) = (x – 1) e–x

f (2) = (2 – 1 )e–2 = e–2 option (A) is correct

( ) ( 1) x xf x x e e

0 0(0) (0 1) 2f e e

f(0) + f (0) = –1 + 2 = 1

Option (B) and (C) are correct

54. Answer (A, C)

Hint: Reciprocal inequality.


x3 – x2 + (a4 + 4a2 + 1)x – a2 = 0

= 1 ...(1)

+ + = a4 + 4a2 + 1 ...(2)

= a2 ...(3)

Now,

1

1 1 1

, from (1)

1 1 13

3

4 2

2

4 13

a a

a

2

2

11a

a

3.

Minimum value of

1

3

Now, 3 2

2 2log 8 log 2 3log2 3

4

2 2 2log 16 log 2 4log 2 4

3

3 3 3log 27 log 3 3log 3 3

and 3

2 5 5log 125 log 5 3log 5 3


Hint: Use the graph.

Solution :

(A) Here [ ] 1

( )[ ] 1

xf x

x x

0 1

0 1x

, if 0 x < 1

1 1

1 1x

, 1 x < 2

2 1 5, 2

2 1 2

x

x

1, 0 1

1

2, 1 2

3 5, 2

1 2

x

x

x

x

x

x

10

y

x2 5/2

1/2

1

2

3

From graph it is discontinuous at the points x = 1, 2

f(x) is discontinuous as well as non-diff.

Also, 1, 3

2

fR = co-domain of f(x)

f(x) is surjective


15/16

Also, 01 1

1 1 1lim ( ) lim lim

( 1) 1 1 2hx x

f xx h

and 01 1

2 2 2lim ( ) lim lim 2

1 1 0hx x

f xx h

1 1

1min. lim ( ), lim ( ) max. ,2

2x x

f x f x

option (C) is correct 1

(1)2

f

56. Answer (A, C, D)

Hint: Every constant function is periodic.

Solution :

(A) 0x

e x R

f(x) sgn(e–x) = 1

and every constant function is periodic.

f(x) = sgn(e–x) is a periodic function.

(B) Let T > 0 be a rational number.

Then , if

( ) ( ),if

x x Qf T x f x

x x Q

f(x) is a non-periodic function

(C)2 2

4(1 sin 1 sin ) 8( )

1 sin cos

x xf x

x x

2 2 sec x which is a periodic function.

(D) f(x) = {x} + 1

Which is also a periodic function, since {x} is a

periodic function.


Hint: px2 + qy2 = –r represents different conics on

conditions of p, q and r.


(px2 +qy2 + r)(4x2 + 4y2 – 8x – 5) = 0

Either, 4x2 + 4y2 – 8x – 5 = 0 ...(1)

2 2 52 0

4x y x

2 2 9( 2 1)

4x x y

2 2 9( 1)

4x y

Which is a circle with centre (1, 0) and radius = 3

2

or, px2 + qy2 + r = 0 ...(2)

Here equation (2) will represent a pair of straight lines

when r = 0 and p and q will be opposite signs.

Also equation (2) will represent a circle if p = q and r

will be of sign opposite to that of p.

Again equation (3) will represent a hyperbola when p

and q are of opposite sign and r 0.

Also, equation (3) will represent an ellipse if p and q

are unequal and of same and r has sign opposite to

that of p.

58. Answer (A, B)

Hint: Use tree diagram to get result.

Solution :

C1

CoinsCoin C2

C3

,Fair

biased

H-Blue

T-Blue

T-White

H-Red

Tree diagram of the experiment is shown below:

C1

C2

C2C

3

C1C

2

C1

C3

C2

H B( )

H B( )

H R( )H R( )

H B( )

H R( )

T B( )

T W( )

T B( )

T W( )

T B( )

T B( )C

2

C1

C3

C3C

1

C3

Required probability (When both coins show up the

same colour)

1 1 1 1 1 1(1 ) ( ) (1 )(1 ) (1 )

3 2 3 3 3 2k k k k k k

2 22(1 ) 1[ (1 ) ]

3 2 3

kk k

2

2 21 2 3 2 29[1 2 1] (given)

3 3 96

k kk k k k

2 292 3 2

32k k

64k2 – 96k + 35 = 0

64k2 – 56k – 40k + 35 = 0

8k (8k – 7) – 5(8k – 7) = 0

5 7or

8 8k


16/16

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

D(R, S)

Hint:

lim

lim

1lim ( )

n

n

b

nb

nr a a

n

rf f x dx

n n

Solution : (A) Given f(x) = 2

1

2 (ln 2ln )

x

x t t dt

Differentiating both sides w.r.t . x,

(ln + 1)2 = 0

1 1

x e

e

= 1

0, 1a be a b

(B) Given 2 2

1 1

2

0 0

2 t tt e dt e dt

2 21 1

0 0

( )( 2 )t tt t e dt e dt

2 2 21 1

1

0

0 0

[( ) ] ( 1)t t tt e e dt e dt

2 21 1

1

0 0

[( 1) 0] t te e dt e dt

= e–1 = + e–1 (given)

= 0, = 1

–= 1 – 0 = 1

(C)

2

20

(cos 1)lim

2

0lim 2 cos 1

2

x

xx

x

x

xe

1 1

2 2

e p qe

p = 0, q = 1 and 1

2 2

p q

(D) tan

2

lim (cosec cos ) Form 1

x

x

l x x

2

cosec cos 1lim

cot

x

x x

x

e

2

2

cosec cot sinlim

cosec

x

x x x

x

e

1 5 q

e p e

p = 0, q = 5

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

Hint: Use graph of y = sin–1(sinx), y = cos–1(cosx),

y = tan–1(tanx)

Solution : (A) f2(x) = f

1(f

1(x)) = x

f3(x) = f

1(f

2(x)) = f

1(x) = x

From given equation

xxx – 25xx + 175x = 375

x3 – 25x2 + 175x – 375 = 0

x3 – 5x2 – 20x2 + 100x + 75x – 375 = 0

x2(x – 5) – 20x(x – 5) + 75(x – 5) = 0

(x – 5) (x2 – 20x + 75) = 0

(x – 5)2 (x – 15) = 0

x = 5, 15

(B) 2 2( 2) 2( 1)2 2 2 52

x x x

22 64 3

xx

(C) Here domain of f(x) is [–1, 1]

f(x) = 5x + 4x – x = 8x

f(x) [–8, 8]

(D) Here domain of f(x) is [–1, 1]

For positive roots, x [0, 1].

Then from given equation

21

22

xx x

1 + x2 = 2x

x = 1

|10a| = 10

[|10a|] = 10

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mock test - 1 (paper-1) - code-a...mock test - 1 (paper - 1) (code-a) (hints & solutions) all...

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