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Mock Test - 2 (Code-A) (Answers) All India Aakash Test Series for JEE (Main)-2020

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

Test Date : 02/01/2020

ANSWERS

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

MOCK TEST - 2 - Code-A

1/14

PHYSICS CHEMISTRY MATHEMATICS

1. (1)

2. (3)

3. (2)

4. (3)

5. (3)

6. (3)

7. (3)

8. (1)

9. (3)

10. (4)

11. (4)

12. (2)

13. (3)

14. (2)

15. (1)

16. (2)

17. (2)

18. (2)

19. (3)

20. (3)

21. (17.00)

22. (41.00)

23. (19.50)

24. (10.00)

25. (18.00)

26. (3)

27. (2)

28. (4)

29. (2)

30. (2)

31. (3)

32. (3)

33. (4)

34. (4)

35. (1)

36. (4)

37. (2)

38. (4)

39. (1)

40. (3)

41. (2)

42. (2)

43. (1)

44. (2)

45. (1)

46. (01.00)

47. (18.00)

48. (30.18)

49. (06.00)

50. (11.00)

51. (3)

52. (2)

53. (3)

54. (2)

55. (3)

56. (3)

57. (1)

58. (3)

59. (2)

60. (3)

61. (2)

62. (4)

63. (4)

64. (1)

65. (2)

66. (3)

67. (1)

68. (2)

69. (4)

70. (2)

71. (05.00)

72. (12.00)

73. (00.25)

74. (00.50)

75. (-02.37)

All India Aakash Test Series for JEE (Main)-2020 Mock Test - 2 (Code-A) (Hints & Solutions)


2/14

1. Answer (1)

Hint : Itriangle = 2

6Mh

Sol. :

2

2

12

6 24

LM MLI

× = =

2

2 2

2

32

6 242 3

LM L MLI M

× = − × =

2

1 2 12cMLI I I∴ = + =

2. Answer (3)

Hint : vmin = ucosθ

Sol. : 2min

2

1(KE) ( cos )21 2 (20 cos ) 1442

m u= θ

= × × × θ =

12 3cos20 5

⇒ θ = =

22 sin cosRange u

gθ θ

∴ =

3 42 20 205 5

10

× × × ×=

32 12 38.4 m10×

= =

3. Answer (2)

Hint : 00

s

v vn nv v

−′ = ×−

Sol. : n1 = 300 Hz

2 0330 5 300 309 Hz330 5

c

c

v vn n

v v+ +

= × = × =− −

∴ Beats = 309 – 300 = 9 Hz.

4. Answer (3)

Hint : dE dldt

φ⋅ = −∫

Sol. : 2

2 B RE Rt

× π× π =

∆

2BRE

t⇒ =

∆

2( )E Q R t MR∴ × × × ∆ = × ω

2

2BR QR t MR

t⇒ × × ∆ = × ω

∆

2BQ

m⇒ ω =

5. Answer (3)

Hint : Linear momentum is not conserved but angular momentum is conserved.

Sol. : 2

20 3

mLmv L mL

× = + × ω

034vL

⇒ ω =

Now, 2

21 42 3 2

mL LmgL mg

× × ω = +

2202

92 33 216

vmL mgLL

⇒ × =

20 04 2v gL v gL⇒ = ⇒ =

6. Answer (3)

Hint : Theoretical

Sol. : Demodulation is carried out with rectifier and envelope detector.

7. Answer (3)

Hint : Capacitors come into parallel combination.

Sol. : Q1 = 1 × 100 = 100 µC

Q2 = 2 × 20 = 40 µC

∴ Charge on plates connected through

S1 = –100 + 40 = –60 µC

When switches are closed, the two capacitor becomes in parallel.

PART - A (PHYSICS)

Mock Test - 2 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020


3/14

11 ( 60) 20 C

1 2⇒ = × − = − µ

+q

2

2

40 C,( ) 20 100

qQ S

= − µ

∴ ∆ = − +

= 80 µC

8. Answer (1)

Hint : Time period = 2 mqBπ

Sol. : 2 , at ,m mT tqB qBπ π

= ∴ = velocity of

particle becomes

2ˆ ˆ ˆ2 3v i j k= − −

Now, 0m eF F+ =

2( ) 0q v B qE⇒ × + =

ˆ ˆ ˆ ˆ ˆ ˆ( 2 3 ) (4 ) 12 8E i j k i j k⇒ = − − × = − +

9. Answer (3)

Hint : Use 2

0 20

, ,4

qF i B B nI Fr

= = µ =πε

Sol. : 0 2[ ] Fi

µ =

2

0 2[ ] qFr

ε =

2 2

02 2

0

F ri q−

µ ∴ = ε

2 2 2

2 2(MLT ) L

A (AT)

− ×=

×

= M2L4T–6A–4

10. Answer (4)

Hint : Theoretical

Sol. : Size of collector plate is largest.

11. Answer (4)

Hint : The leakage process is slow so use equilibrium equations.

Sol. : Tcosθ = mg

2

20

sin4

qTx

θ =πε

2

20

tan4

qx mg

⇒ θ =πε ×

2

202 4

x qmgx

⇒ =πε

3/202 mg x qπε⇒ × =

0232

mgdq dxxdt dt

πε ∴ − = ⋅ ×

0232

mga πε=

12. Answer (2)

Hint : Use deviation graph

Sol. : i i A′δ = + −

90 (initial)i i ′+ = °

( ) 90 (final)i i ′+ < °

50 90i ′+ < °

40i ′⇒ < ° but 37i ′ > °

∴ Angle of emergence can be 38°

13. Answer (3)

Hint : Use conservation of energy

Sol. : 212 3

GMm GMmmvR R

−= +



4/14

21 22 3

GMmmvR

= ×

4 23 3GM gRvE

⇒ = =

14. Answer (2)

Hint : Speed of efflux = 2gh

Sol. : 1 2, 2 ,h h h v gh= − = 1 22dh dhA Adt dt

− × =

and 1dhA a vdt

− = ×

2 23

A Hta g

⇒ =

15. Answer (1)

Hint : Restoring force, ∆F = kx

Sol. : Restoring force

1 1 22 2

F k x kx ∆ = × × × × =

2dmm kxdt

∴ = −

2 mTk

∴ = π

16. Answer (2)

Hint : 1 2tan ; tanC LX XR R

φ = φ =

Sol. : tan60 3LL

X X RR

° = ⇒ =

tan60 3CC

X X RR

° = ⇒ =

2 2( )C Lz R X X R∴ = + − =

rms200 2 A100

I∴ = =

17. Answer (2)

Hint : Shift = ( 1)tDd

µ −

Sol. : Shift = ( 1)tDd

µ −

∴ Number of fringes = ( 1) ( )( )tD d

d Dµ − ×

× λ

( 1) 10tµ −= =

λ

18. Answer (2)

Hint : Use shift = 11 t − × µ

Sol. : 1Shift 1 1cmt = − × = µ

21 1 20 cmu∴ = − =

20 15 60 cm(20 15)

v ×∴ = =

−

final 60 1 61cmu∴ = + =

19. Answer (3)

Hint : 20 2 2

1 2

1 1E E zn n

∆ = −

Sol. : 21 0

11 40.84

E E z ∆ = − =

20 54.4 eVE z⇒ =

24 3 0

1 1 7 (54.4)9 16 9 16

E E z→ ∆ = − = × ×

1240 9 16 470 nm7 54.4

× ×∴ λ =

×

20. Answer (3)

Hint : KEmax = hν – φ

Sol. : h kν = φ +

21. Answer (17.00)

Hint : 2

cc

vra

=

Sol. : ˆ ˆ ˆ ˆ4 , At 1 s, 4v t i j t u i j= + = = +

ˆ4a i=

2

2 2, cc

vr a a aa τ= = −

2

16 16 11716 1

tat

τ× ×

= =+



5/14

216 16 1 416

17 17 17ca ×∴ = − = =

16 1 17 17 m4c

ra

+∴ = =

22. Answer (41.00)

Hint : distancetimeavv =

Sol. : 2 0 0,1 sv t t t= − = ⇒ =

1 1

2

0 0

( )x

dx t t dt= −∫ ∫

13 2

10

1 m3 2 6t tx

⇒ = − = −

And 2 4

2

0 0

40( ) m3

x

dx t t dt= − =∫ ∫

total40 1 412 m3 6 3

S∴ = + × =

avg41 41 m/s

3 4 12V∴ = =

×

23. Answer (19.50)

Hint : ∆θ = ∆U + ∆W

Sol. : A B vQ nC T→∆ = ∆

Let at 0, ,AA T T= then 02BT T=

0 0 05 5( )2 2ABQ n R T P V∴ ∆ = × × =

0(2 )ln2BC BCQ nR T∆ = ∆ω = ×

0 0 0 02 0.7 1.4P V P V= × =

0 0 0 0 0 02.5 1.4 3.9ABCQ P V P V P V∴ ∆ = + =

⇒ 5n = 19.50

24. Answer (10.00)

Hint : dN Ndt

= α − λ

Sol. : dN Ndt

= α − λ

0 0

N tdN dtN

⇒ =α − λ∫ ∫

11Ne

α ⇒ = − λ

25. Answer (18.00)

Hint : dN Ndt

− = λ

Sol. : (238)

(206)

102383

206

NN

=

0

00.75 1.33NN

N N∴ = ⇒

0 tNe

Nλ

∴ =

10 90.72.303log (1.33)

4.5 10t⇒ = ×

×

91.8 10 yearst⇒ = ×

26. Answer (3)

Hint: A buffer system has maximum buffer capacity when pH= pKa.

Sol. : For (3), pKa = 8 – log 6.2

= 7.2

27. Answer (2)

Hint: Geminal diacids on heating lose CO2.

Sol. :

PART - B (CHEMISTRY)



6/14

28. Answer (4)

Hint: Ammonia causes asymmetrical cleavage.

B2H6 + NH3 → [BH2(NH3)2]+ [ –4BH ]

Sol. : LiBH4 is a reducing Agent.

29. Answer (2)

Hint: 4 3 2 2NH NO N O H O∆→ +

Sol. : NH4ClO4, NH4NO2 and (NH4)2Cr2O7

produce N2 on heating.

30. Answer (2)

Hint: Pentagonal rings increase the deviation from planarity.

Sol. : Deviation from planarity will be most severe at (C) and least severe at (B).

31. Answer (3)

Hint: Greater the energy of incident light, greater is the energy of the photoelectron.

Sol. : Blue light has greater energy than Red light.

32. Answer (3)

Hint: Aromaticity based on Huckel’s Rule.

Sol. :

33. Answer (4)

Hint: Minute living organisms dispersed in

atmosphere are examples of viable

pollutant.

Sol. : Smoke, Mist, Fumes, and Dust are non-

viable pollutants.

34. Answer (4)

Hint: Functional isomers are not metamers.

Sol. : 2° amines and 3° amines are functional

isomers.

35. Answer (1)

Hint:

Sol. :

And

36. Answer (4)

Hint: Valium is

Sol. : Valium contains cyclic amide



7/14

37. Answer (2)

Hint: Loss of aromaticity is the rate determining step.

Sol. : The stabilisation of the carbanionic intermediate plays a major role in rate of ArSN2.

38. Answer (4)

Hint: Cu

3 2 3573 KCH — CH — OH CH — CHO→

Sol. : In 1° and 2° alcohols, Cu causes dehydrogenation. In 3° alcohols, it causes dehydration.

39. Answer (1)

Hint: Cannizzaro reaction followed by intramolecular esterification.

Sol. :

40. Answer (3)

Hint: The reaction is electrophilic aromatic substitution.

Sol. : Greater the electrophilic nature, greater is the rate of reaction.

41. Answer (2)

Hint:

( )2H O3 2

3 –3 3 6FeFe (OH) HCH COO CH COO + ++ + → +

Sol.:( )

( ) ( )3 2 2

2 3 3

3 6Fe (OH) H O

Fe O

CH C

H CH COO CH COOH

OO

H

+

+

+ →

+ +

42. Answer (2)

Hint : is a copolymer.

Sol. : It is urea formaldehyde resin.

43. Answer (1)

Hint: Pepsin and trypsin convert proteins into amino acids.

Sol. : Invertase acts on sucrose and maltase acts on Maltose.

44. Answer (2)

Hint: (N atoms in Y) × 3 = 3 × 2.

⇒ N atoms in Y = 2

Sol. : N3H has O.S of N as – 13

N2H4 has O.S of N as –2

KMnO4 cannot reduce N .

45. Answer (1)

Hint: Configuration of chiral carbon does not change as no bond is broken.

Sol. : L-Lactic acid has S configuration. So both the chiral centre in dilactide should have S configuration.

46. Answer (01.00)

Hint: Rate law ; [ ]2d N Odt

= [ ]2 2

3

NO NHK

H O+

Sol.: Step (ii) is RDS.

∴ Rate = 3 2K [NO NH ]−

and Keq = [ ] [ ]

–2 31

2 2 2 2

NO NH H OKK NO NH H O

+ =

[ ] –– 21 2

22 3

NO NHK H ONO NH

K H O+

∴ =

1 3 2 22

2 3

K K [NO NH ]Rate [H O]K [H O ]+

∴ =

2 2

3

[NO NH ]K[H O ]+

=

In a buffer solution [H3O+] = constant

So, order = 1



8/14

47. Answer (18.00)

Hint: ‘a’ is insignificant

P (Vm – b) = RT

Sol. : Vm = 10b

⇒ b = mV10

So, m9PV RT

10 =

mPV 10RT 9

⇒ =

R

i

V 10ZV 9

∴ = =

Ri

9VV10

⇒ =

⇒ Vi = 9b

48. Answer (30.18)

Hint: Csystem = Ccalorimeter+ Cwater

Sol. : Total heat liberated (Q)

= 0.6 3050 15 kJ122

× =

Q 15000Total heat capacity 6000T 2.5

∴ = = =∆

∴ 710 × 4.2 + C = 6000

C = 3018 JK–1

49. Answer (06.00)

Hint: Diacids can be geminal (1, 1), vicinal (1, 2) or isolated (1, 3)

Sol. :

50. Answer (11.00)

Hint: B is exocyclic alkene.

Sol. :

51. Answer (3)

Hint : Coefficient of x9 in 10 10( 1) (1 )dx xdx

+ ⋅ +

Sol. : 10 2 100 1 2 10(1 ) ...x C C x C x C x+ = + + + +

Differentiating w.r.t. x 9 2

1 2 39

10

10(1 ) 1 2 3 ...

10 ...(i)

x C C x C x

C x

+ = ⋅ + ⋅ + ⋅ +

+ ⋅

10 10 9 80 1 2 10( 1) ...x C x C x C x C+ = + + + +

…(ii)

Comparing the coefficient of x9 in (i) × (ii)

19 2 2 2 29 1 2 3 1010 1 2 3 ... 10C C C C C⋅ = ⋅ + ⋅ + ⋅ + + ⋅

52. Answer (2)

Hint : Favourable cases are (2, 2, 1), (1, 2, 3), (2, 2, 2), (1, 3, 3) and (2, 2, 3), (1, 1, 3)

Sol. : Total favourable cases are

(1, 2, 2), (1, 2, 3), (2, 2, 2), (1, 3, 3) and (2, 2, 3), (1, 1, 3)

PART - C (MATHEMATICS)



9/14

Required probability

1 3 3 1 3 2 3 3 33 66 6 6 6 6 6 6 6 6

= ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅

1 2 2 3 3 2 1 33 36 6 6 6 6 6 36 4

+ ⋅ ⋅ + ⋅ ⋅ + =

53. Answer (3)

Hint : 62 0, 14 1

x mm

= − ⇒ = −+

Sol. : x + 4y = 8m and y = mx + 1

x + 4mx + 4 = 8m

8 4 624 1 4 1mxm m

−= = −

+ +

624 1m

− +

is an integer then 4m + 1

= ±1, ±2, ±3, ±6 but in is also integer then m = 0, –1

54. Answer (2)

Hint : tan75° = 2 3+

Sol. : Let height of the flag pole be h

50 50tan603

xx

° = ⇒ =

Also, 50tan75 hx+

° =

502 3 503

h+⇒ + =

100 50 503

h⇒ + = +

1003

h =

55. Answer (3)

Hint : |z1 + z2| ≤ |z1| + |z2| and equality holds when z1, z2 and origin are collinear and z1, z2 lie on the same side of origin.

Sol. : |z1 – z2| = |z1 + z2| ⇒ OP ⊥ OQ because z1 lies on the perpendicular bisector of the line segment joining z2 and –z2.

|z1 + iz3| = |z1| + |iz3| ⇒ z1 and iz3 are collinear with origin.

⇒ OP ⊥ OR

So Q and R will be collinear with O.

56. Answer (3)

Hint : Use sine rule and cosine rule.

Sol. :

coscot sin

cos sin cos sincot cotsin sin

AA A

B C C BB CB C

=⋅ ++

2cos sin sin sin sin cossin( ) sin sin

A B C B C AB C A A

⋅ ⋅= = ⋅

+ ⋅

2 2 2 2 2 2

2 2 22 2

bc b c a b c abca a

+ − + −= ⋅ = =

57. Answer (1)

Hint : S1 must be independent of α.

Sol. : Let point P is (h, k)

Length of tangent = 1S

2 2 2( 2 ) (3 ) ( 4)h k h h k h k= + + + α + + α − −

3h + k = 0

and h – k = 4

h = 1

k = –3

Point P(1, –3)

2 21 ( 3) 10OP = + − =



10/14

58. Answer (3)

Hint : Chord with given mid-point T = S1

Sol. : Equation of chord T = S1

1 11 14 3 4 3x y

⇒ + − = + −

3 4 7x y⇒ + =

On solving this line with the equation of ellipse

22 1 3 7 14 3 4x x − + =

−

2 212 (9 42 49) 48x x x⇒ + − + =

221 42 1 0x x⇒ − + =

Let the point of intersections be (x1, y1) and (x2, y2)

2 21 2

1 80( ) (2) 421 21

x x − = − =

Also 1 2 1 23| | | |4

y y x x− = −

So 21 2

45( )21

y y− =

Length of chord

2 21 2 1 2

125( ) ( )21

x x y y= − + − =

59. Answer (2)

Hint : 2

2( )ixx

Nσ = −∑

Sol. : Incorrect 7x =

⇒ Incorrect 7 6 42ix = × =∑

So correct 42 9 6 39ix = − + =∑

⇒ correct 39 136 2

x = =

Incorrect σ = 1

22Incorrect

(7) 16

ix− =∑

Incorrect 2 300ix =∑

Correct 2 2 2300 9 6 255ix = − + =∑

So correct

22correct

(correct ( ))6

ixxσ = −∑

2255 13 0.5

6 2 = − =

60. Answer (3)

Hint : lim

lim

1lim ( )n

n

bnb

n r a an

rf f x dxn n

→∞

→∞

→∞ =

= ∑ ∫

Sol. : 3 31

n

nr

rIn r=

=+

∑

31

1lim lim1

n

nn n r

rnI

n rn

→∞ →∞ =

= +

∑

1

3/23

0

let1

x dx x tx

= =+∫

1

1/22

0

2 23 31

dt x dx dtt

= =+

∫

1

20

2 ln 13

t t = + +

( )2 ln 1 23

= +

61. Answer (2)

Hint : Check L.H.D. and R.H.D. at x = 1 and x = 3.

Sol. : f(x) = (x – 3)(x – 4)|(x – 1)(x –3)| + |x – 3|sinπx

Let f(x) = g(x) + h(x)

Where g(x) = (x – 3)(x – 4)|(x – 1)(x – 3)|

g(x) is not differentiable at x = 1 only.



11/14

( x = 3 is the repeated root of g(x), so it will be

differentiable at x = 3)

h(x) = |x – 3|sinπx is always differentiable

( x = 3 is repeated root of h(x) = 0)

So f(x) is not differentiable at x = 1 only.

62. Answer (4)

Hint : 2

(2) lim ( )x

f f x→

>

Sol. : For x > 2

f(x) is decreasing

So it cannot have a local minima at x = 2 for any value of k.

63. Answer (4)

Hint : A is skew symmetric matrix.

Sol. : 0ij jk kia a a+ + =

Put 3 0iii j k a= = =

0 1, 2, 3iia i⇒ = ∀ =

Put 0ij ji iii k a a a= + + =

ij jia a⇒ = −

So matrix A will be skew symmetric matrix. Then

sum of all elements 1 , 3

0iji j

a≤ ≤

=∑

Also Tr(A) = 0 and det(A) = 0.

64. Answer (1)

Hint : ( ) ( )a b d c b d× × = × ×

Sol. : a b c b× = ×

( ) ( )a b d c b d⇒ × × = × ×

( ) ( ) ( ) ( )a d b b d a c d b b d c⇒ ⋅ − ⋅ = ⋅ − ⋅

0a d⋅ =

c da c bb d

⋅⇒ = − ⋅

65. Answer (2)

Hint : ∆ABC is right angled isosceles triangle.

Sol. : AB = 2, BC = 2 and AC = 2

So, ∆ABC is right angled isosceles triangle.

12

ABR = = and

1 2 2 122 12 2 2

2

rS

⋅ ⋅∆= = =

++

66. Answer (3)

Hint : ( 1) let ( 2)1 ( 2)

xx

xe x dx x e t

x e

−−

−+

+ =+ +∫

Sol. : 1 ( 1)( 2) 1 ( 2)

x

x xx x edx dx

e x x e

−

−+ +

=+ + + +∫ ∫

Let (x + 2)e–x = t

(–(x + 2)e–x + e–x)dx = dt

(x + 1)e–x dx = –dt

1dt

t= −

+∫

= –ln(1 + t) + c

= –ln(1 + (x + 2)e–x) + c

= x – ln(ex + x + 2) + c

67. Answer (1)

Hint : Use 0 0

( ) ( )f x dx f x dxπ π

= π −∫ ∫

Sol. : Let 2

/20

sinx

xI dxe e

π

π=+∫ …(i)

2 2

/2 /2 /20 0

sin sin( )

x

x xx x eI dx dx

e e e e e

π π

π π− π π⋅

= =+ +∫ ∫ …(ii)

(i) + (ii)

2 /2

/2 /20

sin ( )2( )

x

xx e eI dx

e e e

π π

π π+

=+∫

2/2

0

12 sinI x dxe

π

π⇒ = ∫



12/14

/20

1 1 cos222

xI dxe

π

π− ⇒ =

∫

/222

Ieππ

⇒ =

/214

I e−π⇒ = π

68. Answer (2)

Hint : Draw graph

Sol. : , (2,8)x y ∈

Then [ ]3 3 1x y = =

So, curves are y2 = 2x and x2 = 2y

Area = 4 8 82

2 4 2

(12 ) 22x dx x dx x dx+ − −∫ ∫ ∫

8 843 2

3/2

2 24

2 2126 2 3x xx x

= + − − ⋅

56 2 2(48 24) (16 2 2 2)6 3

= + − − −

443

=

69. Answer (4)

Hint : Variable separable form

Sol. : 2 2( )x dy yd x ydx+ =

2

22 2

( ) 1ln( )d x y dx x y cxx y x

⇒ = ⇒ = − +

When x = 1, y = 1 ⇒ c = 1

When 1 12; ln(4 ) 12 2

x y= = − =

4ey⇒ =

70. Answer (2)

Hint : Truth table

Sol. :

71. Answer (05.00)

Hint : Find roots of both equations.

Sol. : 2 ( 3) 2 2 0x k x k− + + + =

2( 3) ( 6 9) 8 8

2k k k kx + ± + + − −

⇒ =

( 3) ( 1)2

k kx + ± −⇒ =

1, 2x k⇒ = +

2 2 2( 4) 3 3 0x k x k− + + + =

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

2k k kx + ± + − +

⇒ =

2 2 2( 4) ( 2)

2k kx + ± −

⇒ =

2 1, 3x k⇒ = +

21,{2, 3, 1}k kA B +∪ = +

A B∪ contains only 3 elements then

k + 1 = k2 + 1 ⇒ k = 0, 1

k + 1 = 2 ⇒ k = 1

k2 + 1 = 2 ⇒ k = 1, –1



13/14

k + 1 = 3 ⇒ k = 2

k2 + 1 = 3 ⇒ 2, 2k = −

But at k = 1, A ∪ B contains exactly 2 elements

So k = 0, 2, –1, 2, 2−

72. Answer (12.00)

Hint : loga ba b=

Sol. : ( ) 232 2

log 3log 5log 3 log 52 2 5a = = =

( ) 33 5 3

log 5log 5 log 7 log 73 3 7b = = =

73. Answer (00.25)

Hint : If normal to the parabola at P(t1) cuts it

again at Q(t2) then 2 11

2t tt

= − −

Sol. : 21 1( , 2 )L at at where t1 = 1

If normal cuts the parabola at 22 2( , 2 )M at at

Then 2 11

2 3t tt

= − − = −

Tangents at L and M intersect at N then

1 2 1 2( , ( ))N at t a t t+

So the vertices of ∆LMN are 1 1 9 3, , ,4 2 4 2

L M −

and 3 1,4 2

N − −

Area of ∆LMN =

1 1 14 2

1 9 3 1 22 4 2

3 1 14 2

− =

− −

74. Answer (00.50)

Hint : Let 1,P tt

then

: 2xT ytt

+ = and 221: yN xt t

t t− = −

Sol. : Let point P as 1,tt

Equation of tangent : T = 0

2x ytt

+ =

2(2 ,0) and 0,A t Bt

Equation of normal

221yxt t

t t− = −

4

31,0tC

t −

and 4 10, tDt

−−

Area of 4

31 1 122

tPAC ttt

− ∆ = ⋅ − ⋅

4

41 12

tt

+=

Area of 41 2 1

2tPBD t

t t −

∆ = + ⋅

4 12

t +=

1 1area of area ofPAC PBD

+∆ ∆

4

4 42 2 2

1 1t

t t= + =

+ +

75. Answer (-02.37)

Hint : 3( ) 1f x x= −



14/14

Sol. : If 1 1( ) ( )f x f f x fx x

+ = ⋅

1 1( ) ( ) 1 1f x f f x fx x

⋅ − − + =

1( ( ) 1) 1 1f x fx

− − =

Let ( ) 1 ( )f x g x− =

Then 1( ) 1g x gx

⋅ =

( ) ng x x⇒ = ±

So ( ) 1 nf x x= ±

( 2) 9f − =

Then 3( ) 1f x x= −

3 192 8

f − =

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