test - 5 (code-a) (answers) all india aakash test series ... · test - 5 (code-a) (hints &...

Test - 5 (Code-A) (Answers) All India Aakash Test Series for JEE (Main)-2019

1/15

1. (2)

2. (3)

3. (3)

4. (2)

5. (3)

6. (4)

7. (4)

8. (3)

9. (1)

10. (2)

11. (3)

12. (4)

13. (4)

14. (2)

15. (2)

16. (2)

17. (3)

18. (3)

19. (3)

20. (3)

21. (2)

22. (3)

23. (4)

24. (3)

25. (2)

26. (1)

27. (2)

28. (4)

29. (3)

30. (3)

PHYSICS CHEMISTRY MATHEMATICS

31. (1)

32. (2)

33. (3)

34. (3)

35. (2)

36. (1)

37. (3)

38. (4)

39. (4)

40. (4)

41. (2)

42. (2)

43. (4)

44. (3)

45. (3)

46. (3)

47. (2)

48. (3)

49. (1)

50. (2)

51. (4)

52. (2)

53. (3)

54. (2)

55. (1)

56. (2)

57. (3)

58. (1)

59. (1)

60. (3)

61. (3)

62. (4)

63. (3)

64. (3)

65. (1)

66. (3)

67. (4)

68. (3)

69. (2)

70. (2)

71. (3)

72. (3)

73. (4)

74. (2)

75. (4)

76. (1)

77. (1)

78. (3)

79. (2)

80. (1)

81. (1)

82. (1)

83. (3)

84. (2)

85. (3)

86. (2)

87. (2)

88. (2)

89. (2)

90. (3)

Test Date : 16/12/2018

ANSWERS

TEST - 5 - Code-A

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

All India Aakash Test Series for JEE (Main)-2019 Test - 5 (Code-A) (Hints & Solutions)

2/15

1. Answer (2)

Hint : Vc =

2

0

14

RG

r drr r

Sol. : M =

22

0

14 4

2

RR

r drr

Vc =

2

0

1 2(4 )

RG GM

r drr r R

2. Answer (3)

Hint : Two surfaces are formed in the capillary

Sol. : Since force due to surface tension becomes

twice

h = 2h

3. Answer (3)

Hint : Net force on particle = Centripetal force

Sol. :

2

3

mv

a

=

2

23Gm

a

v = Gm

a

T =

32

23

r a

v Gm

4. Answer (2)

Hint : p = T

r for cylindrical surface

Sol. : p = 2T T

r d

A = V m

d d

F = (p) × A = 2

2

Tm

d

5. Answer (3)

Hint : dr

dt =

2gR

r

Sol. : u = 2gR

21

2mv –

GMm

r =

21

2GMm

muR

PART - A (PHYSICS)

v = 2g

Rr

dr

dt =

2gR

r

4

R

R

r dr =

0

2 t

R g dt

t = 7 2

3

R

g

6. Answer (4)

Hint : Imagine hemisphere.

Sol. : Assume upper hemisphere

B = 32

3 R g

F2 = R2 × g(2R)

F1 = F

2 – B =

34

3 R g

7. Answer (4)

Hint : Torque = ( ) p dA r

Sol. : d = 45

( )3

g H x adx x ga

8. Answer (3)

Hint : dA

dt =

2

L ab

m T

Sol. : Areal speed, vA =

abT

Area(A B)

= 1

22 2

abb ae

t = ( )Area 1

12

A B

A

T

v

9. Answer (1)

Hint : Loss in PE = gain in KE

Sol. :21

2mv =

3 2 GMm GMm

R R

v = 3

GM

R

Test - 5 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019

3/15

10. Answer (2)

Hint : R = time of fall × speed of efflux then 0dR

dh

Sol. : hreq

= 2

2

HH

= 3

4

H

height from bottom of vessel = 3

4 2 4 H H H

11. Answer (3)

Hint : = p

B

Sol. : p = gh

= pB

� 2 kg/m3

12. Answer (4)

Hint : Force = Pav

× Projection area.

Sol. : Fx= P

av × Area

= ( )2

�gRR

=

2

2

�g R

13. Answer (4)

Hint : T = Vg, � = �T

YA

Sol. : T = × Vg = 34

3 R g

� =

34

3

� �T R g

YA YA

14. Answer (2)

Hint : Use equation of ellipse.

a = 1 2

2

r r

Sol. :E

S

r1r2

2 2

2 2x y

a b = 1

|x| = ae = a – r2,

a = 1 2

2

r r

y2 =

2

2

21

xb

a

y = 1 2

1 2

2

( )r r

r r

15. Answer (2)

Hint : W = 2( ) a x g dx

Sol. : F = (a2x)g

W =

0.1

2

0

( ) a

a x g dx

=

4

200

a g

16. Answer (2)

Hint : dv

dt = av

Sol. : av =

dv

dt

1 × 10–4 × 2gh =

330 10

60

h = 1.25 m

17. Answer (3)

Hint : E = U + K = 21

2kA

Sol. : v =

2 2

2 3

4 4 A A

A

v = 2v = 22

32

A

E = U + K =

2 2

21 1 34

2 2 2 4

A Ak m

=

221 1

32 4 2

Ak k A

2 2113

2 2 A

kA A


4/15

18. Answer (3)

Hint :

2

2

d

dt

= –2 for angular SHM

Sol. : = 6 6 3 3 L L L L

k k

2 2

2

2

2

3

mL dmL

dt = –

2 1 1

36 9

kL

2 2

2

5

3

mL d

dt =

25

36 kL

2

2

d

dt =

12 k

m

= 12

k

m

f = 1

2 12k

m =

1

4 3k

m

19. Answer (3)

Hint : 2R = R

R3R

Sol. : 2R = R

= 2

mg sinR =

22

2

7

5

dmRdt

=

22

2

14

5

dmRdt

2

2

d

dt =

5

14

g

R

= 5 14

, 214 5

g RT

R g

20. Answer (3)

Hint : fstring

= 2 µ

n T

L

fc =

4

V

L

Sol. : f0 =

4

V

L =

320100 Hz

4 0.8

100 = 2

50 cm2

TLL

L m

21. Answer (2)

Hint : g(x, t) = f((x – v(t – t0)), t)

Sol. : g(x, t) = f((x – v(t – t0)), t)

22. Answer (3)

Hint : Power, P = Area under E versus

Sol. : Power, P = Area under E versus

4

1/41 11 2 2

2 2

4 4 2

P TT T T

P T

23. Answer (4)

Hint : Work done by a gas in a cyclic process is

negative if P-V graph is in anticlockwise

sequence.

Sol. : Wby gas

=1

– 1 40 –20 J2

24. Answer (3)

Hint : For polytropic process, PVx = constant

C = 1

v

RC

x

Sol. : C = 5 2 1

13

R R =

3

2 2 R RR

Q = 3

,2 2 R R

n T U n T

W = –nRT

25. Answer (2)

Hint : For process PVx = constant

[ ]

1

i fnR T T

Wx

Sol. PT = constant P2V = constant

PV 1/2 = constant

W = [ ]

11

2

i fnR T T

= [600 300]

1

2

R = –600R


5/15

26. Answer (1)

Hint : n1CV1

+ n2CV2

= (n1 + n

2)CV

Sol. :3 3 5

2 2

R R = 4C

v

Cv =

18 9

2 4 4

R R

= 13

9

27. Answer (2)

Hint : 1 dV

V dT

Sol. :1 dV

V dT , PT2 = Constant

T3 = kV

2

2 33

dV T dVT k

dT k dT

2

3

3 1T dVk

V dTkT

28. Answer (4)

Hint : For maximum intensity, path difference = n

Sol. : Path difference = 2 × 8.5 cm = 17 cm

29. Answer (3)

Hint : For maximum, path difference = n

In a quadrant path difference varies continuously

from 3 to 0 4.

Sol. : For maximum, path difference = n


from 3 to 0 4.

30. Answer (3)

Hint : Work done by external pressure = Pex

[Vi – V

f]

Sol. : Work done by external pressure

= 2P0[V

0 – V] =

0

3[ ]

2R

n T T [–W = U]

Also 0 0 0 0

0 0

2 2PV PV VTT

T T V

P0V

0 = nRT

0

4[V0 – V] = 3[2V – V

0]

7V0 = 10V V =

07

10

V

31. Answer (1)

Hint : 3 2 2 2LiNO Li O NO O

Sol. : Lithium cannot form alkynide on reaction with

ethyne

32. Answer (2)

Hint : K form superoxide.

Sol. : KO2, Na

2O

2, Li

2O

33. Answer (3)

Hint : Concentrated solution is diamagnetic.

Sol. : Alkali metal dissolve in ammonia.

34. Answer (3)

Hint : Change in pH of buffer is minimum

PART - B (CHEMISTRY)

Sol. :

CH3COO– + H+ CH

3COOH

t = 0 1 0.1 1

t = eq 0.9 — 1.1

(pH)f = pK

a + log

0.9

1.1

(pH)i = pK

a + log

1

1

pH decreases

As concentration of buffer increase buffer

capacity increase

NaOH + H+ H2O

0.1 0.1

Salt is formed, pH = 7 and change in pH is

maximum


6/15

35. Answer (2)

Hint : Sulphate of 2nd group are white

Sol. : Solubility 2nd group sulphate or carbonate

decrease as atomic number increase

36. Answer (1)

Hint : Mg2+ has 6 or 8 molecule of water

Sol. : Ba(NO3)28H

2O does not exist.

37. Answer (3)

Hint : For ppt

Kip > K

sp

Sol. : Experiment I :

For saturated solution of AgCl

[Ag+] = [Cl–] = 10–5 M

If 10–3 mol NaBr is added it means,

[Br–] = 10–3 M.

(Kip)AgBr

= 10–5 × 10–3 = 10–8 > 10–14

(Ksp

of AgBr)

So AgBr form ppt.

Experiment II :

For saturated solution of AgBr

[Ag+] = [Br–] = 10–7

If 10–5 mol of NaCl is added in 100 mL solution so

[Cl–] = 10–4

Kip of AgCl = 10–7 × 10–4 = 10–11

Kip < K

sp so no ppt

38. Answer (4)

Hint : A = B3N

3H

3Cl

Y = B3N

3H

6

Y + 3HCl = B3N

3H

9Cl

3

Sol. :

YHCl

B N

B N

BN

—

—

Cl

H

H

H HH

H

H

HH

Cl

Cl

+

+

+

—

39. Answer (4)

Hint : Al2Cl

6 has 3c – 4e– as well as 2c – 2e– bond.

Sol. :

Cl

Cl

Cl

Cl

Cl

Cl

Al Al

40. Answer (4)

Hint : Buckminster fullerene is aromatic in nature.

Sol. : Stability: Graphite > Diamond > C60

41. Answer (2)

Hint : R3SiCl

Sol. : R3SiCl terminate the chain

42. Answer (2)

Hint :f

Distance moved by the substance

from base lineR

Distance moved by the solvent from

base line

Sol. :4

0.410

43. Answer (4)

Hint : H2O

2 is miscible with water

Sol. :3 2 2 2

Mg(HCO ) Mg(OH) 2CO

3 2 3 2 2Ca(HCO ) CaCO H O CO (g)

44. Answer (3)

Hint : Lassaigne's test for nitrogen

Sol. : Fe2+ + CN– [Fe(CN)6]4–

Blue coloured compound is prussian blue

45. Answer (3)

Hint : I effect act on O > M > P while hyperconjugation

effect act on O and P with equal intensity

Sol. : +I power CD3 > CH

3

+H power CH3 > CD

3

Stability order

A > D > C > B


7/15

46. Answer (3)

Hint : 2 mol

H H

O

O O

2 mol

H

O

H

Means P must contain 8 carbon

Sol. :

H

O

OH

O

H

(I) O 3

(II) Zn/H O2

+

1 mol

2 mol

H

O

2 mol

47. Answer (2)

Hint : A is 3° Alcohol

Sol. :

H+

Ring expansion

(+)

Ring expansion

(+)

H O2

OH

+

48. Answer (3)

Hint : In F – CH2 – CH

2 – OH, gauche form is more

stable than anti.

Sol. :H

H

F

H

H

O

H

Most stable

49. Answer (1)

Hint : 2Cr+6 2Cr+3

+6e–

2N3– N2 + 6e–

Sol. : So a total of 6e– are involved in the above

intramolecular redox

n factor = 6

50. Answer (2)

Hint : Let m moles of Na2CO

3 be x 2x meq

and NaOH be y y meq

Sol. : For phenolphthalein end point

2xy

2 = 30 × 0.5

x + y = 15

For methyl orange end point

2x

2 = 10 × 0.5

x = 5

y = 10

weight of Na2CO

3 = 5 × 10–3 × 106 = 0.53 g

weight of NaOH = 10 × 10–3 × 40 = 0.4 g

Impurity = 0.07 g (7% impurity is present)

51. Answer (4)

Hint : Zn + 4HNO3 4Zn(NO

3)2 + 2NO

2 + 2H

2O

Sol. : The H+ required for the conversion will be

provided by HNO3

Zn + 2NO + 4H Zn +2NO + 2H O3 2 2

– + 2++ 2NO3

–

+ 2NO3

–

Zn + 4HNO Zn(NO ) +2NO + 2H O3 3 2 2 2

(0.2)

0.8 moles of HNO3 will be consumed

1.6 litres of 0.5 M HNO3 is required

52. Answer (2)

Hint : M + H2SO

4 Ma+ + H

2

(8.96 L)

Ma+ + KMnO4 Mb+ + Mn2+

Sol. : Reaction (1)

g meq of H2 = g meq of metal


8/15

8.962

22.4

= 0.4 × (x)

0.8 = 0.4 × x

x = 2 a 2

Reaction (2)

eq of KMnO4 = eq of metal

40 × 2 × 5 × 10–3 = 0.4 × x

40 × 10 × 10–3 = 0.4x

x = 1

b – a = 1

b = 3 a + b = 5

53. Answer (3)

Hint : Conc. of solid is constant

Sol. : Addition of inert at constant pressure move

towards more no. of gaseous mole.

54. Answer (2)

Hint : Nitro benzene is formed which is m-directing

Sol. : 3Zn, CH OH

2 2 2 2 2Cl CH CHCl C H 2ZnCl

CH2 2

Red hot

Cu

HNO3

H SO2 4

NO2

Br /Fe2

Br

NO2

55. Answer (1)

Hint : As for polluted water B.O.D is higher.

Sol. : B.O.D: A < C < B < D

Pollution level in water A < C < B < D

56. Answer (2)

Hint : E2 elimination

Sol. : Anti H is removed

57. Answer (3)

Hint : AI

Sol. : AI has highest value of oxidation potential

58. Answer (1)

Hint : [H+] = 1 2

a 1 a 2K C K C

Sol. : Case of simultaneous equilibrium

4O

4O

HA A H , 1.7 10

(0.1 x) x (x y)

HB B H , 4.7 10

(0.1 y) y (y x)

��⇀↽��

��⇀↽��

x(x + y) = 1.7 × 10–5

y(x + y) = 4.7 × 10–5

(x + y)2 = 6.4 × 10–5

(x + y)2 = 64 × 10–6

x + y = 8 × 10–3

[H+] = 8 × 10–3

59. Answer (1)

Hint : Correct order is a > b > c

Sol. : As s character increases bond length decreases

60. Answer (3)

Hint :

AB(g) A(g) B(g)

1 p p p' p p ''

��⇀↽��

A(g) C(g) D(g)

p p' 1 p' p ' p ''

��⇀↽��

D B E

p' p '' p p '' p ''

��⇀↽��

Given [A][B] p p

0.15[AB] 1 p

Sol. : Given p = 0.2

1 – p = p – p

1 p''

2

= p, p = 0.6 atm


9/15

PART - C (MATHEMATICS)

61. Answer (3)

Hint :1

loglog

a

b

ba

Sol. :cos

cos

3log sin 4

log sin

logcossin = 1 or log

cossin = 3

There are two solutions.

62. Answer (4)

Hint : Max. value of sinnx = 1

Sol. : sinx + sin4x = 2

sinx = 1 and sin4x = 1

(4 1)2

x n and 4 (4 1)

2x m

(4 1)2

x n and (4 1)

8x m

There is no common solution.

63. Answer (3)

Hint : Max. value of sinx + cosx is 2 at x = 4

Sol. : 5 + 7sin2x + 3 3

sin cos2 22 2

x x

Max. value of 3sin 4cos2 2

x x is 3 at 2

x

Max. value of given expression = 15

64. Answer (3)

Hint : In ax2 + bx + c = 0

sum of roots = –b

a

Product of roots = c

a

From equilibrium of reaction I

(0.6 p')(0.4)

0.4

=

15

100

p = 0.45

pAB

= 0.4 atm

pA = 0.15 atm

pB = 0.4 atm

pC = 0.55

pD = 0.25

pE = 0.2

Sol. : cosx1 + cosx

2 = –a and sinx

1 + sinx

2 = –c

1 2 1 2–

2cos cos –2 2

x x x xa

and1 2 1 2

–2sin cos –

2 2

x x x xc

1 2tan

2

x x c

a

sec(x1 + x

2) =

2

2 22

2 2 2

2

1

–1–

c

a ca

c a c

a

65. Answer (1)

Hint : In a triangle

cot cot cot2 2 2

A B C = cot .cot .cot2 2 2

A B C

Sol. : In a triangle we know that

cot cot cot2 2 2

A B C = cot cot cot

2 2 2

A B C

cot cot cot 3cot2 2 2 2

A B C B

cot cot 32 2

A C

cot cot 2 cot cot

2 2 2 2

A C A B

2cot 2 32

B cot 32

B

1tan

2 3B


10/15

66. Answer (3)

Hint :

–1

–1

sin2

cos2 .cos2

r

r r rT

Sol. :

–1

–1

1

sin2

cos2 .cos2

rn

r r

r

=

–1

–1

1

sin(2 – 2 )

cos2 .cos2

r rn

r r

r

= –1

1

(tan2 – tan2 )n

r r

r

= (tan2n – tan)

67. Answer (4)

Hint : 2sinA sinB = cos(A – B) – cos(A + B)

Sol. :–

4 sin .sin2 2

A C A CR

= 2R(cosC – cosA)

= 2 2 2 2

2 –a c

R

a c a c

= a – c

68. Answer (3)

Hint :

2 2 2–

cos2

a c bB

ac

Sol. :

2 2 2 2 2 2– cos –

cos ,2 2

a c b b B bB

ac ac

=

2 2(cos –1)0

2

b B

ac

cosB < 0 2

B

69. Answer (2)

Hint : Speed = distance

time

Sol. :30°

45°

x

xy

1200 m

A B

P

1200tan45

y y = 1200 ...(1)

1200tan30

x y

1200 1

3x y

1200 3 = x + y

x = 1200 3 – 1200 x = 1200( 3 – 1)

Speed = 1200( 3 – 1)

5 = 240 3 –1 m/s

70. Answer (2)

Hint : Lines are y = x + 3; y = 1 – x, x = 0, y = 0

Sol. : Let the equation of tangent be

y = 25 5

3 2mx m

2t passes through (1, 2)

2 = 25 5

3 2m m

4m2 + 24m – 9 = 0

tan = 12

5

71. Answer (3)

Hint : Use property-vertex of parabola bisects the

subtangent and median through the point

passes through the vertex.

Sol. :

V(1, 2)

P(2, 1)

x

y

NT

∵ Vertex bisects the subtangent TN

Median through the point passes through

the vertex V(1, 2)


11/15

Equation of median of PTN, through P(2, 1)

is y – 1 = 1– 2

( – 2)2 –1

x x + y = 3

72. Answer (3)

Hint : Equation of normal at 1

,A tt

to the hyperbola

xy = 1.

is 1

–yt

= t2(x – t)

Sol. : Equations of normals at 1

1

1,A tt

and

2

2

1,B tt

to the hyperbola xy = 1 are

2

1 1

1

1– – ,y t x tt ...(i)

and 2

2 2

2

1– –y t x tt

...(ii) respectively.

(i) and (ii) will represent the same normal if

2 2

1 2t t and

3 3

1 2

1 2

1 1– –t t

t t

t1 = t

2 (Rejected) 3 3

1 2

1 2

1 1– –t t

t t

2 1

1 2

1 2

2 2

1 2 1 2

( )

( )

t tt t

t t

t t t t

∵ t

2 = –t

1

or t1 = –t

2 ...(iii) 4 2

1 11 1t t

t1 = ± 1 ...(iv)

From (iii) and (iv)

t1

= +1, t2

= –1 or t1 = –1, t

2 = 1

Points A and B are (1, 1) and (–1, –1)

2 2AB [AB] = 2

73. Answer (4)

Hint : Concept— If the normal at t1, again cuts the

parabola y2 = 4x at t2, then t

2 =

1

1

2– –t

t

Sol. : Camparing it with (y – m, x)(y – m, x) = 0 we

get

2

1

sin =

2 2

1 2

1 2

( ) tan cos

2tan

m m

mm

m1 + m

2 = 2

2tan 2

sin cossin

m1 + m

2 =

2 2 2 4

2 2 2

tan cos sin cos

sin sin (cos )

(m1 – m

2)2 = (m

1 + m

2)2 – 4m

1m2

2 2

4

sin cos =

2 2

2 2

4cos sin

cos sin

m1 – m

2 = ±2

tan1 – tan

2 = 2

74. Answer (2)

Hint : Solve equations of conics S1 and S

2, we find

values of x2 and y2 and non-substitute in the

equation of conic S3 = 0

Sol. : Solving conics (1) and (2), we get

2 2 2

2 1 1 1 1

2 2 2 2

1 1

( )a a a b a bx

a b a b

and

2 2 2

2 1 1 1 1

2 2 2 2

1 1

( – )b b a b a by

a b a b

Putting these values of x2 and y2 in the

equation of conic S3 = 0.

We get the requried result as in option (2).

75. Answer (4)

Hint : If points O(0, 0) and P(, 1) will lie on opposite

sides of line x + y = 2

(0 + 0 – 2) ( + 1 – 2) < 0 > 1 and

P(, 1) lies inside of circumcentre

< 2

Sol. :

O

(0, 0)(2, 0)

(0, 2)

Y

P( , 1)(0, 1)

(1, 1)(1 2,1)

(1 2,1)Y = 1


12/15

76. Answer (1)

Hint : Equation of circle touching the line x – 3y – 4 = 0

at the point (1, –1) is (x – 1)2 + (y + 1)2 +

(x – 3y – 4) = 0

Sol. : The equation of circle touching the line

x – 3y – 4 = 0 at the point (1, –1) is

y

O

x y – 3 – 4 = 0

(1, –1)

2 – 3 – 2,

2 2

y

(x – 1)2 + (y + 1)2 + (x – 3y – 4) = 0

x2 + y2 + x( – 2) + y(–3 + 2) + 2 – 4 = 0

whose centre is

C 2 – 3 – 2

,2 2

From figure, 2 –

2

> 0,

3 – 2

2

< 0 and

2 – 2 – 3

2 2

= 0

(If the circle touches both axes)

Circle is (x – 1)2 + (y + 1)2 = 0

Here, this circle does not touch the coordinate

axes.

Number of required circles is zero.

77. Answer (1)

Hint : If the line atx + 2ay + 1 = 0 touches the

parabola x2 = 4ay then at2x + 2atx

2

04

xt

a

will have equal roots and then D = 0

Sol. : For intersection point of the parabola

x2 = 4ay and the line atx + 2ay + 1 = 0,

Here equation (i) must have only one solution

for touching case

(, ) 1

– 2,2a

12 4 2 – 2 4 .

2a a

a

= –2 + 2 = 0

78. Answer (3)

Hint : (bsin, acos) lies on the ellipse

2 2

2 21

x y

a b ,

2 2 2 2

2 2

sin cos1

b a

a b

2 2 2 2

sin , cosa b

a b a b

Sol. : (bsin, acos) lies on the ellipse

2 2

2 21

x y

a b

2 2 2 2

2 2

sin cos1

b a

a b

2 2 2 2

sin , cosa b

a b a b

sin2 = 2sincos = 2 2 2

2

22

1

b

ab a

a b b

a

=

2 2

2 2

2 1– 2 1–

1 1– 2 –

e e

e e

2

2

21–

be

a

∵

79. Answer (2)

Hint :

P

A BD

C

CP = r3, CA = r

2, CD = r

1 = 2

2 1

3 2

sinr r

r r

2

2 1 3r r r

Sol. :

P

A BD

C(1, –2)

r2

r1


13/15

Here, CP = r3, CA = r

2,

CD = r1 = 2 2(–1) 2 – 3 = 2

From PAC, 2

3

sinr

r

and from ACD, 1

2

sinr

r

2 1

3 2

r r

r r

2

2 1 3r r r

where r2 = CA = 1 4 – ,

2 2

3(–1) 2 –1r = 2

(1 + 4 – ) = 2 2

= 5 – 2 2

80. Answer (1)

Hint :1 1 1 1

2 2

– – –2( )x y ax by c

a b a b

P x y( , )1 1

Q( , ) image of point

ax by c + + = 0

Sol.:

Let ( )M h, k

x y n + =

1,t

t

1

–(–1) –1

–

kt

h t and

1

2 2

kh t t

n

1– –t h kt ...(i)

12 – –t n h k

t ...(ii)

From (i) and (ii)

t = (n – k) and 1

t = (n – h) and

11t

t

(n – k)(n – h) = 1, locus of (h, k) is

(x – n)(y – n) = 1 whose centre is (n, n)

Sum of coordinates of

Ci = (1 + 1) + (2 + 2) + ... + (20 + 20)

= 20(21)

2 4202

81. Answer (1)

Hint : Equation of tangent at (acos, bsin) to the

ellipse

2 2

2 21

x y

a b is cos sin 1

x y

a b

Sol. : The equation of tangents at the points

(acos, bsin) and (asin, –bcos) to the given

ellipse are

cos sin 1x y

a b ...(i)

and sin – cos 1x y

a b ...(ii)

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

(i) and (ii)

cos sinh k

a b = 1 ...(iii)

sin – cosh k

a b = 1 ...(iv)

Now squaring and adding (iii) and (iv), we get

2 2

2 22

h k

a b

locus of P(h, k) is

2 2

2 21

2 2

x y

a b

...(v)

Let e be the eccentricity of ellipse (v)

2b2 = 2a2(1 – e2)

2

2

21–

be

a

1 – e2 = 1 – e2

e = e.


14/15

82. Answer (1)

Hint : G denotes OH in the ratio 1 : 2 intenally.

Centroid

O (Circumcentre)(3, 0)

6,

3 3

h kG

H h k ( , )orthocentre

Sol. : Centroid is

G 11 2sin 2cos 2sin 2cos

,3 3

Clearly circumcentre is O = (3, 0) so that

OA = OB = OC = 2. Let orthocentre be

H(h, k).

1

2

G

O

(3, 0)

H

h k( , )

6,

3 3

h k

G divides OH in the ratio 1 : 2.

G 1 2 1 2 0

,3 3

h b k

11 2sin 2cos 6

3 3

h ,

2sin 2cos

3 3

k

h – k = 5

locus of orthocentre H(h, k) is x – y = 5

(a straight line)

83. Answer (3)

Hint : Equation of required circle is s + 2 = 0

Sol. : Here the equation of the required circle is

x2 + y2 – 2x + 2y + 1 + (x + y – 1) = 0 ...(i)

Centre C 1– , – 1 –2 2

of circle (i) lies

on the line x + y = 1

1– –1– 12 2

= –1

Equation of circle is x2 + y2 – 3x + y + 2 = 0

84. Answer (2)

Hint : Tangents are y – 1 = ±x and y + 1 = ±x

Centre is C (0, 0)

Sol. : Tangents are

x2 – y2 + 2y – 1 = 0 and x2 – y2 – 2y – 1 = 0

x2 = y2 – 2y + 1 y + 1 = ±x

(y – 1)2 = x2

y – 1 = ±x

Equation of tangents are y = ±x + 1 and

y = ±x – 1

Radius is r = 2 1

2 2 , C (0, 0) centre

of circle

(∵ diameter of circle = 2 2

–1–12

1 1

= Distance between parallel tangents).

85. Answer (3)

Hint : Tangents at the ends of the minor axis of the

ellipse S1 = 0 are the chords of minimum

length of ellipse S2 = 0

Sol. : S1 =

2 2

2 2–1 0,

x y

a b

S2 =

2 2

2 2 2 2–1 0

4( )

x y

a b a b

Y

X

AB

X

Y

y b =

2 2

2 21

x y

a b

2 2

2 2 2 21

4( )

x y

a b a b

O

Tangents at the ends of the minor axis

of the ellipse

2 2

2 21

x y

a b are the

chords of maximum length of ellipse

2 2

2 2 2 21

4( )

x y

a b a b


15/15

��

Points A and B are intersection points of line

y = b and ellipse S2 = 0

A(2a, b) and B(–2a, b)

max.

= AB = 4a

86. Answer (2)

Hint : For common tangent, C1C

2 = |r

1 – r

2|

Sol. : For 1st circle

C1 (0, 0), r

1 = 1

A O

y

x(1, 0)

C2

(2, 0)

For 2nd circle,

C2 (2, 0), r

2 = 4 –

for exactly one common tangent

C1C

2 = |r

1 – r

2|

= –5

87. Answer (2)

Hint : Take images of points due to given line.

Sol. : Here, P(1, 1)

Q(1, 1) mirror image of P due to line y = x

1 7,

5 5R

17 31,

25 25S

lies on nx – y = 0 31

17n

88. Answer (2)

Hint : Make truth table

Sol. :

p q p q ~ p q( ) p q ~ p q( ) ~ p q ~p q( ) ( ) p p q ( ) ( ) p q p q p q ( )

T T T F T F T T T T

T F F T F T T F T T

F T T F F T F T F T

F F T F F T F T F T

89. Answer (2)

Hint : ~(p q) = p ~q

Sol. : ~((p q) (p q)) = (p q) (~(p q))

= (p q) (~p ~q)

90. Answer (3)


Sol. :

~ ~ ~ ~ ~ ~ ~ ~

T T T F F T T T F

T F F F T T T F T

F T T T F F T T T

F F T T T T F T T

p q p q p q p q p q q p q p

Test - 5 (Code-B) (Answers) All India Aakash Test Series for JEE (Main)-2019

1/15

1. (3)

2. (3)

3. (4)

4. (2)

5. (1)

6. (2)

7. (3)

8. (4)

9. (3)

10. (2)

11. (3)

12. (3)

13. (3)

14. (3)

15. (2)

16. (2)

17. (2)

18. (4)

19. (4)

20. (3)

21. (2)

22. (1)

23. (3)

24. (4)

25. (4)

26. (3)

27. (2)

28. (3)

29. (3)

30. (2)

PHYSICS CHEMISTRY MATHEMATICS

31. (3)

32. (1)

33. (1)

34. (3)

35. (2)

36. (1)

37. (2)

38. (3)

39. (2)

40. (4)

41. (2)

42. (1)

43. (3)

44. (2)

45. (3)

46. (3)

47. (3)

48. (4)

49. (2)

50. (2)

51. (4)

52. (4)

53. (4)

54. (3)

55. (1)

56. (2)

57. (3)

58. (3)

59. (2)

60. (1)

61. (3)

62. (2)

63. (2)

64. (2)

65. (2)

66. (3)

67. (2)

68. (3)

69. (1)

70. (1)

71. (1)

72. (2)

73. (3)

74. (1)

75. (1)

76. (4)

77. (2)

78. (4)

79. (3)

80. (3)

81. (2)

82. (2)

83. (3)

84. (4)

85. (3)

86. (1)

87. (3)

88. (3)

89. (4)

90. (3)

Test Date : 16/12/2018

ANSWERS

TEST - 5 - Code-B

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

All India Aakash Test Series for JEE (Main)-2019 Test - 5 (Code-B) (Hints & Solutions)

2/15

1. Answer (3)

Hint : Work done by external pressure = Pex

[Vi – V

f]

Sol. : Work done by external pressure

= 2P0[V

0 – V] =

0

3[ ]

2R

n T T [–W = U]

Also 0 0 0 0

0 0

2 2PV PV VTT

T T V

P0V

0 = nRT

0

4[V0 – V] = 3[2V – V

0]

7V0 = 10V V =

07

10

V

2. Answer (3)

Hint : For maximum, path difference = n


from 3 to 0 4.

Sol. : For maximum, path difference = n


from 3 to 0 4.

3. Answer (4)

Hint : For maximum intensity, path difference = n

Sol. : Path difference = 2 × 8.5 cm = 17 cm

4. Answer (2)

Hint : 1 dV

V dT

Sol. :1 dV

V dT , PT2 = Constant

T3 = kV

2

2 33

dV T dVT k

dT k dT

2

3

3 1T dVk

V dTkT

5. Answer (1)

Hint : n1CV

1 + n

2CV

2 = (n

1 + n

2)CV

PART - A (PHYSICS)

Sol. :3 3 5

2 2

R R = 4C

v

Cv =

18 9

2 4 4

R R

= 13

9

6. Answer (2)

Hint : For process PVx = constant

[ ]

1

i fnR T T

Wx

Sol. : PT = constant P2V = constant

PV1/2 = constant

W = [ ]

11

2

i fnR T T

= [600 300]

1

2

R = –600R

7. Answer (3)

Hint : For polytropic process, PVx = constant

C = 1

v

RC

x

Sol. : C = 5 2 1

13

R R =

3

2 2 R RR

Q = 3

,2 2 R R

n T U n T

W = –nRT

8. Answer (4)

Hint : Work done by a gas in a cyclic process is

negative if P-V graph is in anticlockwise

sequence.

Sol. : Wby gas

=1

– 1 40 –20 J2

9. Answer (3)

Hint : Power, P = Area under E versus

Sol. : Power, P = Area under E versus

4

1/41 11 2 2

2 2

4 4 2

P TT T T

P T

Test - 5 (Code-B) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2019

3/15

10. Answer (2)

Hint : g(x, t) = f((x – v(t – t0)), t)

Sol. : g(x, t) = f((x – v(t – t0)), t)

11. Answer (3)

Hint : fstring

= 2 µ

n T

L

fc =

4

V

L

Sol. : f0 =

4

V

L =

320100 Hz

4 0.8

100 = 2

50 cm2

TLL

L m

12. Answer (3)

Hint : 2R = R

R3R

Sol. : 2R = R

= 2

mg sinR =

22

2

7

5

dmRdt

=

22

2

14

5

dmRdt

2

2

d

dt =

5

14

g

R

= 5 14

, 214 5

g RT

R g

13. Answer (3)

Hint :

2

2

d

dt

= –2 for angular SHM

Sol. : = 6 6 3 3 L L L L

k k

2 2

2

2

2

3

mL dmL

dt = –

2 1 1

36 9

kL

2 2

2

5

3

mL d

dt =

25

36 kL

2

2

d

dt =

12 k

m

= 12

k

m

f = 1

2 12k

m =

1

4 3k

m

14. Answer (3)

Hint : E = U + K = 21

2kA

Sol. : v =

2 2

2 3

4 4 A A

A

v = 2v = 22

32

A

E = U + K =

2 2

21 1 34

2 2 2 4

A Ak m

=

221 1

32 4 2

Ak k A

2 2113

2 2 A

kA A

15. Answer (2)

Hint : dv

dt = av

Sol. : av =

dv

dt

1 × 10–4 × 2gh =

330 10

60

h = 1.25 m

16. Answer (2)

Hint : W = 2( ) a x g dx

Sol. : F = (a2x)g

W =

0.1

2

0

( ) a

a x g dx

=

4

200

a g


4/15

17. Answer (2)

Hint : Use equation of ellipse.

a = 1 2

2

r r

Sol. :E

S

r1r2

2 2

2 2x y

a b = 1

|x| = ae = a – r2,

a = 1 2

2

r r

y2 =

2

2

21

xb

a

y = 1 2

1 2

2

( )r r

r r

18. Answer (4)

Hint : T = Vg, � = �T

YA

Sol. : T = × Vg = 34

3 R g

� =

34

3

� �T R g

YA YA

19. Answer (4)

Hint : Force = Pav

× Projection area.

Sol. : Fx= P

av × Area

= ( )2

�gRR

=

2

2

�g R

20. Answer (3)

Hint : = p

B

Sol. : p = gh

= pB

� 2 kg/m3

21. Answer (2)

Hint : R = time of fall × speed of efflux then 0dR

dh

Sol. : hreq

= 2

2

HH

= 3

4

H

height from bottom of vessel = 3

4 2 4 H H H

22. Answer (1)

Hint : Loss in PE = gain in KE

Sol. :21

2mv =

3 2 GMm GMm

R R

v = 3

GM

R

23. Answer (3)

Hint : dA

dt =

2

L ab

m T

Sol. : Areal speed, vA =

abT

Area(A B)

= 1

22 2

abb ae

t = ( )Area 1

12

A B

A

T

v

24. Answer (4)

Hint : Torque = ( ) p dA r

Sol. : d = 45

( )3

g H x adx x ga

25. Answer (4)

Hint : Imagine hemisphere.

Sol. : Assume upper hemisphere

B = 32

3 R g

F2 = R2 × g(2R)

F1 = F

2 – B =

34

3 R g

26. Answer (3)

Hint : dr

dt =

2gR

r


5/15

31. Answer (3)

Hint : AB(g) A(g) B(g)

1 p p p' p p ''

��⇀↽��

A(g) C(g) D(g)

p p' 1 p' p ' p ''

��⇀↽��

D B E

p' p '' p p '' p ''

��⇀↽��

Given [A][B] p p

0.15[AB] 1 p

Sol. : Given p = 0.2

1 – p = p – p

1 p''

2

= p, p = 0.6 atm

PART - B (CHEMISTRY)

From equilibrium of reaction I

(0.6 p')(0.4)

0.4

=

15

100

p = 0.45

pAB

= 0.4 atm

pA = 0.15 atm

pB = 0.4 atm

pC = 0.55

pD = 0.25

pE = 0.2

32. Answer (1)

Hint : Correct order is a > b > c

Sol. : As s character increases bond length decreases

Sol. : u = 2gR

21

2mv –

GMm

r =

21

2GMm

muR

v = 2g

Rr

dr

dt =

2gR

r

4

R

R

r dr =

0

2 t

R g dt

t = 7 2

3

R

g

27. Answer (2)

Hint : p = T

r for cylindrical surface

Sol. : p = 2T T

r d

A = V m

d d

F = (p) × A = 2

2

Tm

d

28. Answer (3)

Hint : Net force on particle = Centripetal force

Sol. :

2

3

mv

a

=

2

23Gm

a

v = Gm

a

T =

32

23

r a

v Gm

29. Answer (3)

Hint : Two surfaces are formed in the capillary

Sol. : Since force due to surface tension becomes twice

h = 2h

30. Answer (2)

Hint : Vc =

2

0

14

RG

r drr r

Sol. : M =

22

0

14 4

2

RR

r drr

Vc =

2

0

1 2(4 )

RG GM

r drr r R


6/15

33. Answer (1)

Hint : [H+] = 1 2

a 1 a 2K C K C

Sol. : Case of simultaneous equilibrium

4O

4O

HA A H , 1.7 10

(0.1 x) x (x y)

HB B H , 4.7 10

(0.1 y) y (y x)

��⇀↽��

��⇀↽��

x(x + y) = 1.7 × 10–5

y(x + y) = 4.7 × 10–5

(x + y)2 = 6.4 × 10–5

(x + y)2 = 64 × 10–6

x + y = 8 × 10–3

[H+] = 8 × 10–3

34. Answer (3)

Hint : AI

Sol. : AI has highest value of oxidation potential

35. Answer (2)

Hint : E2 elimination

Sol. : Anti H is removed

36. Answer (1)

Hint : As for polluted water B.O.D is higher.

Sol. : B.O.D: A < C < B < D

Pollution level in water A < C < B < D

37. Answer (2)

Hint : Nitro benzene is formed which is m-directing

Sol. : 3Zn, CH OH

2 2 2 2 2Cl CH CHCl C H 2ZnCl

CH2 2

Red hot

Cu

HNO3

H SO2 4

NO2

Br /Fe2

Br

NO2

38. Answer (3)

Hint : Conc. of solid is constant

Sol. : Addition of inert at constant pressure move

towards more no. of gaseous mole.

39. Answer (2)

Hint : M + H2SO

4 Ma+ + H

2

(8.96 L)

Ma+ + KMnO4 Mb+ + Mn2+

Sol. : Reaction (1)

g meq of H2 = g meq of metal

8.962

22.4

= 0.4 × (x)

0.8 = 0.4 × x

x = 2 a 2

Reaction (2)

eq of KMnO4 = eq of metal

40 × 2 × 5 × 10–3 = 0.4 × x

40 × 10 × 10–3 = 0.4x

x = 1

b – a = 1

b = 3 a + b = 5

40. Answer (4)

Hint : Zn + 4HNO3 4Zn(NO

3)2 + 2NO

2 + 2H

2O

Sol. : The H+ required for the conversion will be

provided by HNO3

Zn + 2NO + 4H Zn +2NO + 2H O3 2 2

– + 2++ 2NO3

–

+ 2NO3

–

Zn + 4HNO Zn(NO ) +2NO + 2H O3 3 2 2 2

(0.2)

0.8 moles of HNO3 will be consumed

1.6 litres of 0.5 M HNO3 is required

41. Answer (2)

Hint : Let m moles of Na2CO

3 be x 2x meq

and NaOH be y y meq

Sol. : For phenolphthalein end point

2xy

2 = 30 × 0.5

x + y = 15


7/15

For methyl orange end point

2x

2 = 10 × 0.5

x = 5

y = 10

Weight of Na2CO

3 = 5 × 10–3 × 106 = 0.53 g

Weight of NaOH = 10 × 10–3 × 40 = 0.4 g

Impurity = 0.07 g (7% impurity is present)

42. Answer (1)

Hint : 2Cr+6 2Cr+3

+6e–

2N3– N2 + 6e–

Sol. : So a total of 6e– are involved in the above

intramolecular redox

n factor = 6

43. Answer (3)

Hint : In F – CH2 – CH

2 – OH, gauche form is more

stable than anti.

Sol. :H

H

F

H

H

O

H

Most stable

44. Answer (2)

Hint : A is 3° Alcohol

Sol. :

H+

Ring expansion

(+)

Ring expansion

(+)

H O2

OH

+

45. Answer (3)

Hint : 2 mol

H H

O

O O

2 mol

H

O

H

Means P must contain 8 carbon

Sol. :

H

O

OH

O

H

(I) O 3

(II) Zn/H O2

+

1 mol

2 mol

H

O

2 mol

46. Answer (3)

Hint : I effect act on O > M > P while hyperconjugation

effect act on O and P with equal intensity

Sol. : +I power CD3 > CH

3

+H power CH3 > CD

3

Stability order

A > D > C > B

47. Answer (3)

Hint : Lassaigne's test for nitrogen

Sol. : Fe2+ + CN– [Fe(CN)6]4–

Blue coloured compound is prussian blue

48. Answer (4)

Hint : H2O

2 is miscible with water

Sol. :3 2 2 2

Mg(HCO ) Mg(OH) 2CO

3 2 3 2 2Ca(HCO ) CaCO H O CO (g)

49. Answer (2)

Hint : f

Distance moved by the substance

from base lineR

Distance moved by the solvent from

base line

Sol. :4

0.410


8/15

50. Answer (2)

Hint : R3SiCl

Sol. : R3SiCl terminate the chain

51. Answer (4)

Hint : Buckminster fullerene is aromatic in nature.

Sol. : Stability: Graphite > Diamond > C60

52. Answer (4)

Hint : Al2Cl

6 has 3c – 4e– as well as 2c – 2e– bond.

Sol. :

Cl

Cl

Cl

Cl

Cl

Cl

Al Al

53. Answer (4)

Hint : A = B3N

3H

3Cl

Y = B3N

3H

6

Y + 3HCl = B3N

3H

9Cl

3

Sol. :

YHCl

B N

B N

BN

—

—

Cl

H

H

H HH

H

H

HH

Cl

Cl

+

+

+

—

54. Answer (3)

Hint : For ppt

Kip > K

sp

Sol. : Experiment I :

For saturated solution of AgCl

[Ag+] = [Cl–] = 10–5 M

If 10–3 mol NaBr is added it means,

[Br–] = 10–3 M.

(Kip)AgBr

= 10–5 × 10–3 = 10–8 > 10–14

(Ksp

of AgBr)

So AgBr form ppt.

Experiment II :

For saturated solution of AgBr

[Ag+] = [Br–] = 10–7

If 10–5 mol of NaCl is added in 100 mL solution so

[Cl–] = 10–4

Kip of AgCl = 10–7 × 10–4 = 10–11

Kip < K

sp so no ppt

55. Answer (1)

Hint : Mg2+ has 6 or 8 molecule of water

Sol. : Ba(NO3)28H

2O does not exist.

56. Answer (2)

Hint : Sulphate of 2nd group are white

Sol. : Solubility 2nd group sulphate or carbonate

decrease as atomic number increase

57. Answer (3)

Hint : Change in pH of buffer is minimum

Sol. :

CH3COO– + H+ CH

3COOH

t = 0 1 0.1 1

t = eq 0.9 — 1.1

(pH)f = pK

a + lo g

0.9

1.1

(pH)i = pK

a + log

1

1

pH decreases

As concentration of buffer increase buffer

capacity increase

NaOH + H+ H2O

0.1 0.1

Salt is formed, pH = 7 and change in pH is

maximum

58. Answer (3)

Hint : Concentrated solution is diamagnetic.

Sol. : Alkali metal dissolve in ammonia.

59. Answer (2)

Hint : K form superoxide.

Sol. : KO2, Na

2O

2, Li

2O

60. Answer (1)

Hint : 3 2 2 2LiNO Li O NO + O

Sol. : Lithium cannot form alkynide on reaction with

ethyne


9/15

PART - C (MATHEMATICS)

61. Answer (3)


Sol. :

~ ~ ~ ~ ~ ~ ~ ~

T T T F F T T T F

T F F F T T T F T

F T T T F F T T T

F F T T T T F T T

p q p q p q p q p q q p q p

62. Answer (2)

Hint : ~(p q) = p ~q

Sol. : ~((p q) (p q)) = (p q) (~(p q))

= (p q) (~p ~q)

63. Answer (2)


Sol. :

p q p q ~ p q( ) p q ~ p q( ) ~ p q ~p q( ) ( ) p p q ( ) ( ) p q p q p q ( )

T T T F T F T T T T

T F F T F T T F T T

F T T F F T F T F T

F F T F F T F T F T

64. Answer (2)

Hint : Take images of points due to given line.

Sol. : Here, P(1, 1)

Q(1, 1) mirror image of P due to line y = x

1 7,

5 5R

17 31,

25 25S

lies on nx – y = 0 31

17n

65. Answer (2)

Hint : For common tangent, C1C

2 = |r

1 – r

2|

Sol. : For 1st circle

C1 (0, 0), r

1 = 1

A O

y

x(1, 0)

C2 (2, 0)

For 2nd circle,

C2 (2, 0), r

2 = 4 –

for exactly one common tangent

C1C

2 = |r

1 – r

2|

= –5

66 .Answer (3)

Hint : Tangents at the ends of the minor axis of the

ellipse S1 = 0 are the chords of minimum

length of ellipse S2 = 0

Sol. : S1 =

2 2

2 2–1 0,

x y

a b

S2 =

2 2

2 2 2 2–1 0

4( )

x y

a b a b

Y

X

AB

X

Y

y b =

2 2

2 21

x y

a b

2 2

2 2 2 21

4( )

x y

a b a b

O

Tangents at the ends of the minor axis of the

ellipse

2 2

2 21

x y

a b are the chords

of maximum length of ellipse

2 2

2 2 2 21

4( )

x y

a b a b

Points A and B are intersection points of line

y = b and ellipse S2 = 0

A(2a, b) and B(–2a, b)

max.

= AB = 4a


10/15

67 Answer (2)

Hint : Tangents are y – 1 = ±x and y + 1 = ±x

Centre is C (0, 0)

Sol. : Tangents are

x2 – y2 + 2y – 1 = 0 and x2 – y2 – 2y – 1 = 0

x2 = y2 – 2y + 1 y + 1 = ±x

(y – 1)2 = x2

y – 1 = ±x

Equation of tangents are y = ±x + 1 and

y = ±x – 1

Radius is r = 2 1

2 2

, C (0, 0) centre

of circle

(∵ diameter of circle = 2 2

–1–12

1 1

= Distance between parallel tangents).

68. Answer (3)

Hint : Equation of required circle is s + 2 = 0

Sol. : Here the equation of the required circle is

x2 + y2 – 2x + 2y + 1 + (x + y – 1) = 0 ...(i)

Centre C 1– , – 1 –2 2

of circle (i) lies

on the line x + y = 1

1– –1– 12 2

= –1

Equation of circle is x2 + y2 – 3x + y + 2 = 0

69. Answer (1)

Hint : G denotes OH in the ratio 1 : 2 intenally.

Centroid

O (Circumcentre)(3, 0)

6,

3 3

h kG

H h k ( , )orthocentre

Sol. : Centroid is

G 11 2sin 2cos 2sin 2cos

,3 3

Clearly circumcentre is O = (3, 0) so that

OA = OB = OC = 2. Let orthocentre be

H(h, k).

1

2

G

O

(3, 0)

H

h k( , )

6,

3 3

h k

G divides OH in the ratio 1 : 2.

G 1 2 1 2 0

,3 3

h b k

11 2sin 2cos 6

3 3

h ,

2sin 2cos

3 3

k

h – k = 5

locus of orthocentre H(h, k) is x – y = 5

(a straight line)

70. Answer (1)

Hint : Equation of tangent at (acos, bsin) to the

ellipse

2 2

2 21

x y

a b is cos sin 1

x y

a b

Sol. : The equation of tangents at the points

(acos, bsin) and (asin, –bcos) to the given

ellipse are

cos sin 1x y

a b ...(i)

and sin – cos 1x y

a b ...(ii)

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

(i) and (ii)

cos sinh k

a b = 1 ...(iii)

sin – cosh k

a b = 1 ...(iv)

Now squaring and adding (iii) and (iv), we get

2 2

2 22

h k

a b

locus of P(h, k) is

2 2

2 21

2 2

x y

a b ...(v)


11/15

Let e be the eccentricity of ellipse (v)

2b2 = 2a2(1 – e2)

2

2

21–

be

a

1 – e2 = 1 – e2

e = e.

71. Answer (1)

Hint :1 1 1 1

2 2

– – –2( )x y ax by c

a b a b

P x y( , )1 1

Q( , ) image of point

ax by c + + = 0

Sol.:

Let ( )M h, k

x y n + =

1,t

t

1

–(–1) –1

–

kt

h t and

1

2 2

kh t t

n

1– –t h kt ...(i)

12 – –t n h k

t ...(ii)

From (i) and (ii)

t = (n – k) and 1

t = (n – h) and

11t

t

(n – k)(n – h) = 1, locus of (h, k) is

(x – n)(y – n) = 1 whose centre is (n, n)

Sum of coordinates of

Ci = (1 + 1) + (2 + 2) + ... + (20 + 20)

= 20(21)

2 4202

72. Answer (2)

Hint :

P

A BD

C

CP = r3, CA = r

2, CD = r

1 = 2

2 1

3 2

sinr r

r r

2

2 1 3r r r

Sol. :

P

A BD

C(1, –2)

r2

r1

Here, CP = r3, CA = r

2,

CD = r1 = 2 2(–1) 2 – 3 = 2

From PAC, 2

3

sinr

r

and from ACD, 1

2

sinr

r

2 1

3 2

r r

r r

2

2 1 3r r r

where r2 = CA = 1 4 – ,

2 2

3(–1) 2 –1r = 2

(1 + 4 – ) = 2 2

= 5 – 2 2


12/15

73. Answer (3)

Hint : (bsin, acos) lies on the ellipse

2 2

2 21

x y

a b ,

2 2 2 2

2 2

sin cos1

b a

a b

2 2 2 2

sin , cosa b

a b a b

Sol. : (bsin, acos) lies on the ellipse

2 2

2 21

x y

a b

2 2 2 2

2 2

sin cos1

b a

a b

2 2 2 2

sin , cosa b

a b a b

sin2 = 2sincos = 2 2 2

2

22

1

b

ab a

a b b

a

=

2 2

2 2

2 1– 2 1–

1 1– 2 –

e e

e e

2

2

21–

be

a

∵

74. Answer (1)

Hint : If the line atx + 2ay + 1 = 0 touches the

parabola x2 = 4ay then at2x + 2atx

2

04

xt

a

will have equal roots and then D = 0

Sol. : For intersection point of the parabola

x2 = 4ay and the line atx + 2ay + 1 = 0,

Here equation (i) must have only one solution

for touching case

(, ) 1

– 2,2a

12 4 2 – 2 4 .

2a a

a

= –2 + 2 = 0

75. Answer (1)

Hint : Equation of circle touching the line x – 3y – 4 = 0

at the point (1, –1) is (x – 1)2 + (y + 1)2 +

(x – 3y – 4) = 0

Sol. : The equation of circle touching the line

x – 3y – 4 = 0 at the point (1, –1) is

y

O

x y – 3 – 4 = 0

(1, –1)

2 – 3 – 2,

2 2

y

(x – 1)2 + (y + 1)2 + (x – 3y – 4) = 0

x2 + y2 + x( – 2) + y(–3 + 2) + 2 – 4 = 0

whose centre is

C 2 – 3 – 2

,2 2

From figure, 2 –

2

> 0,

3 – 2

2

< 0 and

2 – 2 – 3

2 2

= 0

(If the circle touches both axes)

Circle is (x – 1)2 + (y + 1)2 = 0

Here, this circle does not touch the coordinate

axes.

Number of required circles is zero.

76. Answer (4)

Hint : If points O(0, 0) and P(, 1) will lie on opposite

sides of line x + y = 2

(0 + 0 – 2) ( + 1 – 2) < 0 > 1 and

P(, 1) lies inside of circumcentre

< 2

Sol. :

O

(0, 0)(2, 0)

(0, 2)

Y

P( , 1)(0, 1)

(1, 1)(1 2,1)

(1 2,1)Y = 1

77. Answer (2)

Hint : Solve equations of conics S1 and S

2, we find

values of x2 and y2 and non-substitute in the

equation of conic S3 = 0


13/15

Sol. : Solving conics (1) and (2), we get

2 2 2

2 1 1 1 1

2 2 2 2

1 1

( )a a a b a bx

a b a b

and

2 2 2

2 1 1 1 1

2 2 2 2

1 1

( – )b b a b a by

a b a b

Putting these values of x2 and y2 in the

equation of conic S3 = 0.

We get the requried result as in option (2).

78. Answer (4)

Hint : Concept— If the normal at t1, again cuts the

parabola y2 = 4x at t2, then t

2 =

1

1

2– –t

t

Sol. : Camparing it with (y – m, x)(y – m, x) = 0 we

get

2

1

sin =

2 2

1 2

1 2

( ) tan cos

2tan

m m

mm

m1 + m

2 = 2

2tan 2

sin cossin

m1 + m

2 =

2 2 2 4

2 2 2

tan cos sin cos

sin sin (cos )

(m1 – m

2)2 = (m

1 + m

2)2 – 4m

1m2

2 2

4

sin cos =

2 2

2 2

4cos sin

cos sin

m1 – m

2 = ±2

tan1 – tan

2 = 2

79. Answer (3)

Hint : Equation of normal at 1

,A tt

to the hyperbola

xy = 1.

is 1

–yt

= t2(x – t)

Sol. : Equations of normals at 1

1

1,A tt

and

2

2

1,B tt

to the hyperbola xy = 1 are

2

1 1

1

1– – ,y t x tt ...(i)

and 2

2 2

2

1– –y t x tt

...(ii) respectively.

(i) and (ii) will represent the same normal if

2 2

1 2t t and

3 3

1 2

1 2

1 1– –t t

t t

t1 = t

2 (Rejected) 3 3

1 2

1 2

1 1– –t t

t t

2 1

1 2

1 2

2 2

1 2 1 2

( )

( )

t tt t

t t

t t t t

∵ t

2 = –t

1

or t1 = –t

2 ...(iii) 4 2

1 11 1t t

t1 = ± 1 ...(iv)

From (iii) and (iv)

t1

= +1, t2

= –1 or t1 = –1, t

2 = 1

Points A and B are (1, 1) and (–1, –1)

2 2AB [AB] = 2

80. Answer (3)

Hint : Use property-vertex of parabola bisects the

subtangent and median through the point

passes through the vertex.

Sol. :

V(1, 2)

P(2, 1)

x

y

NT

∵ Vertex bisects the subtangent TN

Median through the point passes through

the vertex V(1, 2)


14/15

Equation of median of PTN, through P(2, 1)

is y – 1 = 1– 2

( – 2)2 –1

x x + y = 3

81. Answer (2)

Hint : Lines are y = x + 3; y = 1 – x, x = 0, y = 0

Sol. : Let the equation of tangent be

y = 25 5

3 2mx m

2t passes through (1, 2)

2 = 25 5

3 2m m

4m2 + 24m – 9 = 0

tan = 12

5

82. Answer (2)

Hint : Speed = distance

time

Sol. :30°

45°

x

xy

1200 m

A B

P

1200tan45

y y = 1200 ...(1)

1200tan30

x y

1200 1

3x y

1200 3 = x + y

x = 1200 3 – 1200 x = 1200( 3 – 1)

Speed = 1200( 3 – 1)

5 = 240 3 –1 m/s

83. Answer (3)

Hint :

2 2 2–

cos2

a c bB

ac

Sol. :

2 2 2 2 2 2– cos –

cos ,2 2

a c b b B bB

ac ac

=

2 2(cos –1)0

2

b B

ac

cosB < 0 2

B

84. Answer (4)

Hint : 2sinA sinB = cos(A – B) – cos(A + B)

Sol. :–

4 sin .sin2 2

A C A CR

= 2R(cosC – cosA)

= 2 2 2 2

2 –a c

R

a c a c

= a – c

85. Answer (3)

Hint :

–1

–1

sin2

cos2 .cos2

r

r r rT

Sol. :

–1

–1

1

sin2

cos2 .cos2

rn

r r

r

=

–1

–1

1

sin(2 – 2 )

cos2 .cos2

r rn

r r

r

= –1

1

(tan2 – tan2 )n

r r

r

= (tan2n – tan)

86. Answer (1)

Hint : In a triangle

cot cot cot2 2 2

A B C = cot .cot .cot2 2 2

A B C

Sol. : In a triangle we know that

cot cot cot2 2 2

A B C = cot cot cot

2 2 2

A B C

cot cot cot 3cot2 2 2 2

A B C B

cot cot 32 2

A C


15/15

��

cot cot 2 cot cot

2 2 2 2

A C A B

2cot 2 32

B cot 32

B

1tan

2 3B

87. Answer (3)

Hint : In ax2 + bx + c = 0

sum of roots = –b

a

Product of roots = c

a

Sol. : cosx1 + cosx

2 = –a and sinx

1 + sinx

2 = –c

1 2 1 2–

2cos cos –2 2

x x x xa

and1 2 1 2

–2sin cos –

2 2

x x x xc

1 2tan

2

x x c

a

sec(x1 + x

2) =

2

2 22

2 2 2

2

1

–1–

c

a ca

c a c

a

88. Answer (3)

Hint : Max. value of sinx + cosx is 2 at x = 4

Sol. : 5 + 7sin2x + 3 3

sin cos2 22 2

x x

Max. value of 3sin 4cos2 2

x x is 3 at 2

x

Max. value of given expression = 15

89. Answer (4)

Hint : Max. value of sinnx = 1

Sol. : sinx + sin4x = 2

sinx = 1 and sin4x = 1

(4 1)2

x n and 4 (4 1)

2x m

(4 1)2

x n and (4 1)

8x m

There is no common solution.

90. Answer (3)

Hint :1

loglog

a

b

ba

Sol. :cos

cos

3log sin 4

log sin

logcossin = 1 or log

cossin = 3

There are two solutions.

test - 5 (code-a) (answers) all india aakash test series ... · test - 5 (code-a) (hints &...

Documents