MOCK TEST JEE (MAIN) 2015 HINTS & SOLUTIONS · 2019-07-04 · MOCK TEST - (JEE-MAIN)-2015_Solution...

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MOCK TESTJEE (MAIN) 2015

HINTS & SOLUTIONS


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13

Section - III : MATHEMATICS 61. (A)

We have 2sin 8cos 1x x =

1

sin cos2 2

x x⇒ =

Case I. When cos 0>x

In this case sin cos 1/ 2 2x x =

sin 2 1/ 2x⇒ =

2 / 4, 3 /4, 9 / 4, 19 / 4x⇒ = π π π π

/ 8, 3 / 8, 9 / 8, 19 / 8x⇒ = π π π π

As x lies between 0 and 2π and cos 0,x > therefore, / 8, 3 / 8.x = π π

Case II. When cos 0x <

In this case, sin cos 1 / 2 2x x =

sin cos 1/ 2 2x x⇒ = −

sin 2 1 / 2x⇒ = −

2 5 / 4, 7 / 4, 13 / 4, 15 / 4x⇒ = π π π π

5 / 8, 7 / 8, 13 / 8, 15 / 8x⇒ = π π π π

5 / 8, 7 / 8x⇒ = π π

Thus, the values of x satisfying the given equation which lie between 0 and 2π are π/8, 3π/8, 5π/8, 7π/8. These are in A.P. with common difference π/4.

62. (B)

2 0ax bx c+ + =

2 0b c

x xa a

⇒ + + =

The correct equation is 2 13 0c

x xa

+ + = ….(i)

The wrong equation is 2 17 0c

x xa

+ + = ….(ii)

Eq. (ii) has roots (−2) and (−15)

∴ ( )( )2 15c

a− − =

From Eq. (i), the correct equation is

2 13 30 0x x+ + =

⇒ ( )( )3 10 0x x+ + =

∴ 3, 10.x = − − 63. (B)

LHL = ( )

( ) ( )/2 0

lim lim / 2x n h

f x f n h→ π+π − →

= π + π −

( )( ) [ ]/ cos /2

0lim 1 cos / 2

ab n h

hn h

π+π −

→= + π + π −

13

( )( ) ( )/ sin

0lim 1 sin

ab n h

hn h

π−

→= + π −

( )( ) ( )1 1 sin

0lim 1 1 sin h

nab hn

h

−

→= + − ( ) /sin

0lim 1 sin

ab h ab

hh e

→= + =

( ) /

0lim 1h

h eη λ

→ + = ∵

V.F. ( )/ 2 a bf n e e= π + π = ⋅

RHL = ( )

( )/2 0

lim lim2x n h

f x f n h→ π+π + →

π = π + +

( ) ( )cot 2 2 /cot 8 4 8

0lim n h n h

he π+π+ π+ π+

→=

( )

0lim cot 2 /cot 8

hh h

e →= 0lim tan 8 / tan 2

hh h

e →= 0 04 lim tan 8 /8 lim 2 / tan 2

h xh h h h

e → →= 4 11 4e e⋅ ⋅= =

( )f x∵ is continuous in ( )( ), 1n nπ + π

∴ LHL = RHL = V.F.

4ab a be e e += = 4ab a b⇒ = = + 2a b⇒ = =

64. (A) Given ( ) ( ) ( )1 2cot . cot ..... cot 1nα α α =

1 1

cos sinn n

i ii i= =

⇒ α = α∏ ∏

2

1 1 1

sin 2 1cos sin cos

2 2

n n ni

i i i ni i i= = =

α⇒ α = α α = ≤∏ ∏ ∏

/2

1

1cos

2

n

i ni=

⇒ α ≤∏

Hence, maximum value of 1

cosn

ii=

α∏ is /2

1

2n

65. (B) Since, the given function is continuous but no differentiable at 0x = and

( ) ( )0 2, 0 2Lf Rf′ ′= − =

0x⇒ = is point of local minima and ( )0 1.f =

Also, there is no other critical point. Now, ( ) ( )2 11, 2 4 cos2f f− = = +

On comparing values of function at 2,0,2x = −

global maximum occurs at ( )2 2 11x f= − ⇒ − =

Hence, global maximum and local minimum values are 11,1. 66. (B)

We have 2 2 56m n= +

( )2 2 56 2 2 1 56m n n m n−⇒ − = ⇒ − =

( ) ( )2 2 1 8 8 1 3, 3n m n n m n−⇒ − = − ⇒ = − =

3 6 and 3.m n n⇒ = + = =

λ

�

m

14

67. (D)

The lines 3 1 3

1 2

x y z− − −= =−λ

and 1 2 1

3 4

x y z− − −= =λ

are coplanar

if

3 1 1 2 3 1

1 2 0

3 4

− − −−λ =

λ

i.e. if

2 1 2

1 2 0

3 4

−−λ =

λ

i.e. if ( ) ( ) ( )22 8 3 1 4 2 3 2 0+ λ + + λ + − λ =

i.e. if 2 26 0λ + λ + = which does not give any real value of λ. 68. (B)

Domain of 1 2cos

4

x− −

is given by 2

1 1 6 24

xx

−− ≤ ≤ ⇔ − ≤ − ≤

[ ]2 6 6 6,6 .x x x⇔ − ≤ ≤ ⇔ ≤ ⇔ ∈ −

Domain of ( )1

log 3 x−is the set of those real numbers for which 3 0,3 1x x− > − ≠

i.e., 3x < and 2,x ≠ i.e., ( ) ( ),2 2,3 .x∈ −∞ ∪

Domain of x is [0, ).∞

Hence domain of ( )f x is [ ] ( ) ( )( ) ( ) ( )6,6 ,2 2,3 [0, ) 0,2 2,3 .− ∩ −∞ ∪ ∩ ∞ = ∪

69. (C)

( )2 33

12 2

2 6.2 22 2 6lim lim

2 2 2 2

x xx x

x x xx x

−

− −→ →

− ++ − =− −

(Multiplying Num, and Den. by 2x)

( )( )( )

( )( )2

2 4 2 2 2 2lim

2 2 2 2

x x x

x x x→

− − +

− +

( )( )( )

2

2 4 2 2 2 2lim

2 4

x x x

xx→

− − +

−

( )( )2

lim 2 2 2 2x x

x→− + ( )( )2 22 2 2 2 8.= − + =

70. (A) As given for 1 1 1z x iy= + and 2 2 2,z x iy= + the symbol 1 2z z∩ is used if 1 2x x≤ and 1 2.y y≤

Therefore for 1 ,z∩ we have 1 x≤ and 0 y≤ as 1 = 1 + 0⋅i and z x iy= +

Now, ( )( )

( )( )( )2 2

1 1 11

1 1 1

x iy x iy x iyz

z x iy x y

− + − − + −− = =+ + + + +

( )( ) ( )

2 2

2 22 2

1 2

1 1

x y iy

x y x y

− −= −

+ + + +

But 1x ≥ and 0y ≥

2

15

∴ 2 2 1x y+ ≥ and 2 0y− ≤

⇒ ( )

2 2

2 2

10

1

x y

x y

− − ≤+ +

and ( )2 2

20

1

y

x y

− ≤+ +

∴ 1

0.1

z

z

− ∩+

71. (A)

( )4 1n + terms are

, 2, 4, 6, ,arein APa a a a+ + + …

( ) ( )1 14 , 4 , 4 ,....,arein G.P.

2 4a n a n a n+ + +

Middle term of AP ( )1n= + th term

( )2a n= + 2a n= +

and middle term GP ( )2 1 1

14

2

n

a n+ −

= +

( )

2

4

2 n

a n+=

According to the question

( )

2

42

2 n

a na n

++ =

( )2 2 12 1 4 2 4 0n na n n−− + ⋅ − =

( ) ( )2 2 12 1 4 2 1 0n na n −− + − =

∴ Middle term of the sequence 4a n= +

( )

( )2 1

2

4 2 14

2 1

n

n

nn

− −= − +

−

2 1

2

2 14 1

2 1

n

nn

− −= − −

2 2 1

2

2 24

2 1

n n

nn

− −= −

( )( )

2 1

2

4 2 2 1

2 1

n

n

n −⋅ −=

−

2 2 1

2 2

2 2 2

2 1 2 1

n n

n n

n n +⋅ ⋅= =− −

.

72. (B)

The number of matches in first round 6 62 2C C= + = 30

The number of matches in next round = 62 15C = and the number of matches in the semifinal

round 42 6C= =

Number of matches in the final round = 2 Hence, required number of matches 30 15 6 2 53= + + + = 73. (A) ∴ Number of intersection points

37 13 112 2 2 2C C C= − − + (∵ two points A and B)

= 535

A B

13 pass through A 11 pass through B

16

74. (D)

( ) ( )2 2 4 20 1 21 1

n n nnx C C x C x C x− = − + − + − ⋅∵ …

( ) ( )1 2 1 200

1 13 5 2 1

n n nCC Cx dx C

n⇒ − = − + − + −

+∫ … ( ) 11 2

01

nk x x dx

−= −∫ (given)

∴ ( ) ( ) 11 12 2

0 01 1

n nx dx k x x dx

−− = −∫ ∫

Put, sinx = θ in left integral ∴ cosdx d= θ θ

/2 2 1

0cos

2n k

dn

π + θ θ =∫

( )( )

( )( )( )2 2 2 2 4 .....4 2

2 1 2 1 2 3 .....5 3 1 2

n n n k

n n n n

− − ⋅=

+ − − ⋅ ⋅

⇒ ( )( ){ }

( )

22 2 2 2 4 .....4 2

2 1 ! 2

n n n k

n n

− − ⋅=

+

⇒ ( )

( )

22{2 !

2 1 ! 2

n n k

n n

⋅=

+ ∴

( )( )

22 12 !

2 1 !

n n nk

n

+ ⋅ ⋅=

+

75. (A) ( ) ( )1 3f f=∵

⇒ 11 6 27 9 33 6a b a b+ + − = + + − ⇒ 13 4 11a b+ = − …..(i)

and ( ) 23 2 11f x ax bx′ = + +

⇒ 2

1 1 12 3 2 2 2 11 0

3 3 3f a b ′ + = + + + + =

⇒ 1 4 1

3 4 2 2 11 03 3 3

a b + + + + + =

…..(ii)

From Eqs. (i) and (ii), we get 1, 6a b= = − 76. (B)

We have ,α β are the roots of 2 0,ax bx c+ + = then / , /b a c aα + β = − αβ =

∴ n nnS = α + β

∴ 1 2

1 2 3

2 3 4

3 1 1

1 1 1

1 1 1

S S

S S S

S S S

+ +∆ = + + +

+ + +

2 2

2 2 3 3

2 2 3 3 4 4

3 1 1

1 1 1

1 1 1

+ α + β + α + β

= + α + β + α + β + α + β

+ α + β + α + β + α + β

2 2 2 2

1 1 1 1 1 1

1 1

1 1

= α β × α β

α β α β

2

2 2

1 1 1

1

1

= α β

α β

17

( ){ } ( ){ }2 21 4= αβ − α + β + α + β − αβ

2 2

2

41

c b b c

a a aa

= + + −

( ) ( )2 2

4

4a b c b ac

a

+ + −=

77. (B)

Let ( )

12 9

35 3

2 5

1

x xI dx

x x

+=+ +

∫3 6

3

2 5

2 5

1 11

x x dx

x x

+ =

+ +

∫

Put 2 5

1 11 t

x x+ + = ∴

3 6

2 5dx dt

x x

− − =

Then, 3 2

1

2

dtI c

t t= − = +∫ 2

2 5

1

1 12 1

c

x x

= + + +

( )10

25 32 1

xc

x x= +

+ +

78. (B)

2 3 2 3

3 2 3 2

dy x d xf

dx x dx x

+ + ′= ⋅ − −

( ) ( )( )

( )2

3 2 2 2 3 2sin .

3 2

x x x

x

− − + −=

− ( )2

12sin .

3 2x=

−

79. (B) Apply basic rules of inverse, negation etc. 80. (D)

Put 2tan secx dx d= θ⇒ = θ θ

∴ 11 /4 2

0 0

tansec

tan

xdx d

x

− π θ= θ θθ∫ ∫

/4

0 sin cosd

π θ= θθ θ∫

/4 /2

0 0

12

sin 2 2 sin

td dt

t

π πθ= θ =θ∫ ∫ where 2 tθ = . Hence

1.

2k =

81. (C) Let E1(E2) denote the event that P1 and P2 are paired (not paired) together and let A denote the

event that one of two players P1 and P2 is among the winners. Since, P1 can be paired with any of the remaining 15 players.

We have, ( )11

15P E = and ( ) ( )2 1

1 141 1

15 15P E P E= − = − =

In case E1 occurs, it is certain that one of P1 and P2 will be among the winners. In case E2 occurs, the probability that exactly one of P1 and P2 is among the winners is

( ) ( ){ } ( ) ( )1 2 1 2 1 2 1 2P P P P P P P P P P P∩ ∪ ∩ = ∩ + ∩

( ) ( ) ( ) ( )1 2 1 2P P P P P P P P= + 1 1 1 1 1 1 1

1 12 2 2 2 4 4 2

= − + − = + =

ie, ( )1/ 1P A E = and ( )2/ 1 / 2=P A E

By the total probability Rule,

( ) ( ) ( )1 21 2

. .A A

P A P E P P E PE E

= +

( )1 14 1

115 15 2

= +

8

15=

18

82. (D)

1 3

2

i− +ω =∵ and 2 1 3

2

i− −ω =

Also, 3 1ω = and 2 1ω + ω = −

Then,

2

2

i iA

i i

ω ω = ω ω− −

1

1i

ω ω= −ω −

∴ 2

22

1 1

1 1A

i

ω ω ω= −ω − −ω −

2

2

2

1 0

0 1

− ω= −ω − ω

2 4

2 4

0

0

−ω + ω= −ω + ω

2

2

0

0

−ω + ω= −ω + ω

∵ ( ) 2 2f x x= +

∴ ( )2

2

2

0 2 02

0 20f A A I

−ω + ω = + = + −ω + ω

( ) ( )2

2

2

2 0 1 0 1 02 3 2

0 1 0 10 2

−ω + ω + = = −ω + ω + = + ω −ω + ω +

1 01 3

3 20 12

i − += + ( ) 1 02 3 .

0 1i

= +

83. (C)

( ) ( )2 2f x x bx b f x x b′= + − ⇒ = +

∴ ( )1 2f b′ = +

∴ Equations of tangent at (1, 1) is ( )( )1 2 1y b x− = + −

⇒ ( ) ( )2 1 0b x y b+ − − + =

∴ Length of x-intercept 1

2

b

b

+=+

Length of y-intercept ( )1b= − +

∴ Area of triangle ( )1 11 2

2 2

bb

b

+ = − + = + (given)

⇒ 2 6 9 0 3.b b b+ + = ⇒ = − 84. (D)

( ) ( ) ( )2 xaf x x f x f x e′′ ′+ + = and ( ) ( ) ( )2 xag x x g x g x e′′ ′+ + =

Subtracting, we get

( ) ( )( ) ( ) ( )( ) ( ) ( )2 0a f x g x x f x g x f x g x′′ ′′ ′ ′− + − + − =

⇒ ( ) ( )f x g x− is a solution of

2

22

0d y dy

a x ydxdx

+ + =

19

85. (B)

9 6

; ;20 5

OP OQ= =

9 6

: : 3: 420 5

OP OQ = =

O

P

Q

86. (C) An end of the latus rectum ( ),2 .a a= The equation of the tangent at ( ),2a a is

( ). 2 2 ,y a a x a= + i.e. .y x a= + The normal at ( ),2a a is 2 ,y x a a+ = + i.e. 3 .y x a+ = Solving

0y = and ,y x a= + we get , 0.x a y= − = Solving 0, 3 ,y y x a= + = we get 3 , 0.x a y= = The

area of the triangle with vertices ( ) ( ) ( ),2 , ,0 , 3 ,0a a a a− is 24 .a

87. (A) Let the roots are α and 2α

2

1 32

5 3

a

a a

−⇒ α + α =

− + and

2

2.2

5 3a aα α =

− +

2

2 2 2

1 (1 3 ) 22

9 ( 5 3) 5 3

a

a a a a

−⇒ = − + − +

2

2 22

(1 3 )9 9 6 1 9 45 27

( 5 3)

aa a a a

a a

−⇒ = ⇒ − + = − +

− +

2

39 263

a a⇒ = ⇒ = .

88. (C) [ ] ( )u v w u v w u w v w= ⋅ × ≤ × = ×� � � � � � � � � � �

But ˆˆ ˆ3 7 59v w i j k v w× = − − ⇒ × =� � � �

∴ [ ] 59.u v w ≤� � �

89. (B)

Let aX b aX b

Y Yc c

+ += ⇒ =

⇒ ( )aY Y X X

c− = −

⇒ ( ) ( )2

2 2

2

1 1aY Y X X

N NcΣ − = Σ −

⇒ 2

2 22

.y ya a

acc

= σ ⇒ σ = σ

90. (B)

We know that 1 1sin cos ,2

− − πθ + θ = then the given equation implies that

2 4 6

2..... .....2 4 2 4

x x x xx x− + − = − + −

⇒ 2

22

1 12 2

x xx x

x x= ⇒ =

+ + ⇒ ( )1 0 0x x x− = ⇒ = or 1.x =

v

3

MOCK TEST JEE (MAIN) 2015 HINTS & SOLUTIONS · 2019-07-04 · MOCK TEST - (JEE-MAIN)-2015_Solution...

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Transcript of MOCK TEST JEE (MAIN) 2015 HINTS & SOLUTIONS · 2019-07-04 · MOCK TEST - (JEE-MAIN)-2015_Solution...