xii class jee shikhar solution 20-12-2015

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JEE TEST SERIES / PART TEST # 03 / PAPER – 1 & 2 / SOLUTION / PAGE - 1

India’s First ‘Rank Improving’ All India JEE Test Series

Website : jeeshikhar.catalyser.in

JEE TEST SERIES PART TEST # 03

(XII + DRP) PAPER – 1 (ADVANCED) SOLUTIONS

PART I : PHYSICS

SECTION – 1 : (One Integer Value Correct Type)

Sol.1. (B) N = mg 2N mr

2mg m.2sin g

2sin

0.1 10

2 sin30

1rad / s.

Sol.2. (AC)

Sol.3. (C) B must be appearing at infinity

2 1 2 1h

2R

Solving 1

2

2

Sol.4. (AC) P = 0.5 W Energy delivered in 5 sec

E P t E 0.5 5 2.5J

This E will be in form of KE21

E mv2

212.5 0.2 v

2 v 5m / s

Sol.5. (BC) For pure rolling

2 R 5 R 7 3 7 73

X '7'

Sol.6. (AC) A ' 30t,100 2 40t B' 10t, 0

222

A 'B' 10t 30t 100 2 40t 0

222s 40t 100 2 40t

sds

2 2 40t .40 2 100 2 40t . 40

dt

For min

dss 0

dt

Solving this

20 5t 2 2

16 4

X 2

4

5 2

4 X '5'



JEE TEST SERIES PAPER – 1 & 2 / SOLUTION / PAGE - 2



60°

O

60°

B

A C

B

A

C

90 + r

20

)(2

r

310

y

A

D x

B

v t 0

Sol.7. (C) Using law of sine in ABC

10 3

sin=

20

sin(90 ) cos( )

AC

r r

... (i)

Snell’s law, 1 × sin 60 =3

sin2

r ... (ii)

AC = 4.1 cm

–102.07 10'

AC t

C s

Sol.8. (BD) 00 sin

v t x v t r

r

0cosv t

y r r r

Differentiating above equations

00 1 cos x

v t v v

r

00 siny

v t v v

r

2 2 002 sin

2 x y

v t v v v v

r

SECTION – 2 : (One or More Than One Option Correct Type)

Sol.9. (AC) Impulse = Change in momentum

= 2 12 v v ˆ ˆ2 3i j

As impulse is in the normal directing of colliding surface1

tan3

1 1tan

3

0 1 1

90 tan3

Sol.10. (ABC) v a

Sol.11. (BD)

Sol.12. (AC)

Sol.13. (ABCD) Maximum elongation =2F

k and equilibrium is at

F x

k

Amplitude =2F F F

k k k

Sol.14. (ABC) By Center of Linear Momentum

9 4 3 8 4 1 8V

2 /V m s






2 1 3 1

9 3 12 4e

2 21 1

4 1 8 2 182 2

f E

2 21 1

4 9 8 3 198

2 2

i E

loss = 18 198 180f i E E J

Sol.15. (ABD) 2 2v ax bx c (equation for parabola)

On differentiating

2 2vdv

ax bdx

2

vdv bax

dx

acceleration

0

at2

b x

a

Which is the vertex of parabola2 x x

After1 x velocity is increasing, hence v and a are in same direction.

After2 x velocity is decreasing, hence v and a are in opposite direction.

Sol.16. (AB) If x = a sin (t + )

Then V =dx

dt

= a cos (t + )

a =dv

dt = – a

2sin (t + )

Time period, T = const.

(B)aT

x = –

2T = const.

(A) a2T

2 + 4

2v

2= a

2

4 sin

2(t + ).

2

2

4

+ 4

2a

2

2 cos

2 (t + )

= a2

2 4

2[sin

2(t + ) + cos

2(t + )]

= a2

24

2= const.






Sol.17. (ABC) The path difference at C is = t ( – 1)

It may or may not be equal to n. Hence maxima may or may not occur at C .

(i) If t ( – 1) = n, the maxima will occur at C .

(ii) If ( 1) (2 1)2

t n

, the minima will occur at C .

(iii) The position on the screen where path difference = 0 is called central maximum

position.

(iv) The intensity at the point of minima will be zero only when 1 2I I .

Here I 1 I 2

At the minima, intensity will not be completely zero.

Hence (C) is the incorrect option.

Sol.18. (ACD) Use Snell’s law

SECTION – 3 : (Matrix Match Type)

Sol.19. (B) (P) The body is in pure rolling & there is no external force on it hence there is no friction(A), (B), (C), (D).

(Q) There is no torque about P and collision is elastic.

(B), (C), (D)

(R) There is no torque about P and collision is inelastic.

(B)

(S) Force acting is Mg (conservative). It will have a torque about P.

(T) Gravitational force is conservative, the force between these particles is internal force.

(A), (C), (B).

Sol.20. (C) For no emergence

c A 2 c

A / 2 c1sin

A 1sin2

2

For no emergence

c c A c csin A sin c

1sin A

csin A 1

Total internal reflection at 2x

2 cminr






PART II : CHEMISTRY


Sol.21. (AB) 2NO + O2 2NO2

(moles before reaction) 1 0.5 0

after reaction 1 –.5 0.5 1

= 1.5

n = |1.5 – 2.5| = 1

Change in pressure

P =nRT

V

=1 ×

1

12×

300

6.25 atm = 4 atm

Sol.22. (BD) Balanced equation is –

3I2(s) + 6 OH – 5I

– + –

3IO + 3H2O()

G° = 5×( –50) + ( – 123.5) + 3×( –233)

–0 – 6 × ( –150) = – 172.5 KJ/mol

Now

G° = – RTnk

– 172.5 =25

–

3 × 300 × 2.3 × 10

–3 log K

log K = 30

1030

= – 5 – –5 –1

3

– 6 – 6

[I ] [IO ] 10 10

[OH ] [OH ]

[OH –] = 10

–6

pOH = 6

pH = 8

Sol.23. (B) 2eq. of K2Cr 2O7 and eq. of SO2 = 1 mol of SO2

2SO2(g) + O2(g) 2 SO3(g)

1 mol 1 mol 0 mol

1 mol 0.5 mol 1 mol

2SO 1P

2.5 × 5 = 2 Atm,

2O

0.5 5P

2.5

= 1 Atm

3SO

1P

2.5 × 5 = 2Atm

Kp =

3

2

22

SO

2 2

SO

P 21.0

P 2 1

Sol.24. (AC) AB2(g) AB(g) + B(g)

Initial 100 0 0

At equ. 100 – x x x At eq. P = 120

120 = 100 – x + x + x

= 100 + x = 120x = 20

Kp =x x 20 20

100 – x 100 – 20

= 5






Sol.25. (C)

Initially :

At equilibrium :

CH3COOH + C2H5OHK C

CH3 COO C2H5 + H2O

0.1 mol 0.1 mol 0 0

(0.1 – x) mol (0.1 – x) mol x x

In 10 ml solution at equilibrium moles of

CH3COOH present = 0.1 (0.1 – x) 0.1 (0.1 – x)

moles of NaOH 0.1 (0.1 – x) = 80 × 10

–4 = 8 × 10

–3

or 0.01 – 0.1 x = 0.008

or x = 0.02

KC =2 2

2 2 2

x (0.02) 1 1

16(0.1 – x) (0.08) 4

= 32 KC = 32 ×1

16 = 2

Sol.26. (AC)

Sol.27. (C) Anhydride formation

Sol.28. (AC) W A B = – 10 J

W AB = PV ; WBC = 0

q = E – W

E = 0

q = 5 = 0 – W

W = – 5J = W A B + WB C + W C A

or – 5 J = – 10 + 0 + WC A

WC A = 5 J


Sol.29. (BC) Sol.30. (C) Sol.31. (AD) Sol.32. (BC)

Sol.33. (AD) Sol.34. (AD) Sol.35. (AC) Sol.36. (BC)

Sol.37. (AB) Sol.38. (ABC)SECTION – 3 : (Matrix Match Type)

Sol.39. (D) Sol.40. (C)


Sol.41. (D) f x x 1 x 2 x 3 x 4

2 2x 5x 4 x 5x 6

Now let 2x 5x t and t t 4 t 6

2t 10t 24

min

t 1 for t 5

and 2x 5x 5 has real roots

M 1 M 4 3.

PART III : MATHEMATICS






Sol.42. (C) 2 2 2 3

2 2

dy dx 1 1x y x y dy 0

y xy x

1 1d

x y1 1 1 1d y dy 0 ydy

1 1x y y x

x y

21 1 yln c k 2.

x y 2

Sol.43. (A) sinxcosy f 2x 2y f 2x 2y cosx.siny f 2x 2y f 2x 2y

f 2x 2y sin x y

f 2x 2y sin x y

f f k

sin sin2 2

x

f x k sin 2

4f ' x f x 0.

Sol.44. (AD) Since 2 x xsin cos

4 4 lies between 1 and 2 for x 2, 2

2 x xsin cos 1

4 4

Thus the problem reduces to finding the area bounded by 2 2 2x y 4, y x x 1 and line

y = 1

Now required area 2 0

2 2

13

2 3 2 4 x dx 1 x x 1 dx

2 13 k 6.

3 6

Sol.45. (D) /2

/2

1 2f x sinx sinx tcosx sint cost dt

k k

1 2 3 4sinx l l l l

Where,/2

1

/2

sinxl sin tdt 0

k

/2

2

/2

2sinx 4sinxl costdtk k

/2

3

/2

cosx 2cosxl t sin tdt

k k

/2

4

/2

2cosxl tcostdt 0

k

t cost is odd






1 2 4sinx 2cosx

sinx cosx sinxk k k k

1 41 k 3.

k k

Sol.46. (D) 1 1f x tan tan x cos cos x for3

x ,2 2

when x x x 02 2 2

x 02

tan x tan x tanx

1 1f x tan tanx cos cosx x x 2x

When 3 3

x x x 0 x2 2 2

0 x2

and x 2 x2

and cos 2 x cos x

f x x 2 x

So graph of f(x) and required area is shown in figure

2 22 3

area .4 4

Sol.47. (B) tanx 0, tanx 1

3

3 3

3

33

log 3 log 3

log tanx 1log tanxlog 3

Let3

log tanx 1 0

2t 3 1

t 2 2

t 3 1t

23t 2t 1 0

t 1 and1

t3

(not possible)

3log tan x 1

1

tanx x 63

only in 0, .






Sol.48. (B)

2 x 12

2

x e , 0 x 1f x a b 1

acos 2x 2 bx , 1 x 2

2 x 1 2 x 122xe 2x e , 0 x 1f ' x b 2

2bx 2asin 2x 2 , 1 x 2


Sol.49. (ACD) Let f ' x 0

1 1 1 1 3 3f , f , f

2 2 4 4 4 4

and so on

f x x

Let f ' x 0

1 1 1 3 3 1

f , f , f 2 2 4 4 4 4

And so on

f x 1 x

Sol.50. (AD) sinx cosx

sin x cos xe cos x e sinxf ' x e e

cosx sinx

f g x x f ' g x g' x 1

1g' x

f ' g x

1/ 21 1 eg' 0

5 2f ' g 0f '

4

Sol.51. (BC) Given curve is3 2 23 12 3 6 12

dy dy y x y y x

dx dx

23 4 6dy

y x dx

2

2

42

24

y dy x dx

dx dy x y

Here, 0dx

dy

2 40 2

2

y y

x

But 2y gives 2 28 3 24 3 16 x x

2 2 162 3 12 2 8 16

3

y x x

Required point is4

, 23

Sol.52. (AD) 4 3 3 2 4 3 21 4 3 4 2 3 4 A x x Bx C x x x x x

4, 2, 1 A B C






3 2

3/ 24 3 4 3

2 1 4 34

1 1

x x x dx f x

x x x x

1/2 1/2 1/24 3 4 3 4 3 4 3

2 2 1 2 2 144

1 1 1 1

x x dx dx c

x x x x x x x x

f 1 6

Sol.53. (AB) 2x

2 2

0

f x f t1 dt

1 x 1 t

2 2

2 22

1 x f ' x 2xf x f x

1 x1 x

2

2

dy 2xy y

dx 1 x

Let1

ty

2

dt 2xt 1

dx 1 x

solution is 3

21 x1 x x c

y 3

But 2

3

3 1 xf 0 1 c 1 y f x

x 3x 3

f 1 6

Sol.54. (ABC) The graph of g(x)

Sol.55. (BD) Area =

3 32 2 26 5 4 3 2

1 1x ax bx cx dx ex f mx n dx x 1 x 2 x 3 dx

1 1 1

22 22 2 2 2 4 2

1 1 0

t t 1 t 1 dt t t 1 dt 2 t t 1 2t dt

= 1 1 2 15 35 42 16

2 27 3 5 105 105






Sol.56. (AD) We have 2y x 2 2y 2 xf x 2y f x e f 2y e x 1 e 4y 1 e 4xy

Putting x y 0, we get f 0 2f 0 f 0 0

Now

y 0

f x 2y f xf ' x lim

2y

2y x 2 2y 2 x

y 0

f x e e f 2y x 1 e 4y 1 e 4xy f xf ' x lim

2y

2y 2y

x 2

y 0 y 0 y 0

e 1 e 1f 2y f 0f ' x f x lim e lim x lim 2x

2y 2y 2y

x 2f ' x f x e f ' 0 x 2x

x x x 2 xe f ' x e f x f ' 0 2xe x e

x x x 2 xe f ' x e f x 1 2xe x e

x 2 xd de f x x x e

dx dx

x 2 xe f x x x e c f 0 0 c 0

x 2f x xe x g x x and xh x e .

Sol.57. (AC) Equality is true if f(t) g(h(t)) = f(h(t)) g(t) or f(t).g(h(t)) = f(h(a + b – t)) g(a + b – t)

Sol.58. (ABC) Hence,

x 3 2 x 2 x 1 , x 3

x 3 2 x 2 x 1 , 3 x 2f x

x 3 2 x 2 x 1 , 2 x 1

x 3 2 x 2 x 1 , x 1

i.e.

2, x 3

2x 8, 3 x 2

f x 2x, 2 x 1

2, x 1

maxf 4 at x 2 and minf 2 at x 1 2f x m 3m will have no solution if

2m 3m 4 or 2

i.e. if 2m 3m 4 0 or 2m 3m 2 0 i.e. if m 4 m 1 0 or m 2 m 1 0

i.e. if m 4 or m 1 or 2 m 1

SECTION –

3 : (Matrix Match Type)Sol.59. (C) (A) Let x = sinθ and y = cosθ

x y sin cos

2sin4

minimum value = 2 .






(B)1

y acosx cos3x3

y ' 06

asin sin 06 2

a 2.

(C) f ' x 1 2 cos x 1

f ' x 0 cosx2

5x ,

3 3

a b 2.

(D) At x 0, y 1 also 1

y ' 02

equation of tangent 1

y 1 x 02

x y1 p q 3.

2 1

Sol.60. (B) (A) 2 3 21 sin x 24 sin x cosec x cosecx 24 cosec x 3

(B) 2 2f ' x 6x 2bx 4 cos x

f ' x 0 x R

2 26x 2bx 4 cos x 0 x R

26x 2bx 4 0 x R

23x bx 2 0 2b 24 0

24 b 24 b 0, 1, 2, 3, 4

(C) If 1 1sin sinx cos cosx , then 0 / 2.

Consider 4 different cases.

1

2

3

4

C 1 sinx 0

C 0 sinx 1

C sinx 0

C sinx 1

The solution set is 2n , 2n , n z2

(D) Let

2x

2xL lim 1 ,

x x

it is of form 1

2

22x 1 1 2

2x xx

x xL lim e lim e e

1






JEE TEST SERIES PART TEST # 03

(XII + DRP) PAPER – 2 (ADVANCED) SOLUTIONS

PART I : PHYSICS


Sol.1. (AC)

2

1 2

2

1 2

I I 49

9I I

Solve we get

1

2

I 5

2I 1

2

I 25

I 4

25I

4

4 4 25I 5

5 5 4

Sol.2. (AC)

Sol.3. (AB) T mg cos 2

P oscI w mgx 2

2

osc

mlw mgR

12

cos 1

For small and x R for small

osc 2

12gRw

l

2 2 l lT

w 12gR 3gR

T

3

12 12T 4

3

Sol.4. (AC)

Sol.5. (BD) Image is real

v

7u

7

77 u

1 7 u

1 7 u

u 8

Sol.6. (B) This would happen if there is no collision with the wall

on reaching RS = 0

2 21 1kx mv

2 2

2

21100 1 v

100

v 1/ ms

Sol.7. (AC) Particle is released at x 2

max at

min.u i f

TE TE 2110 15 mv

2

125

2 2 2

5 m / s Sol.8. (B) First reflection from (1) then

1 1 1

V 15 10

v 30

x

mg






1

30m

15

Next reflection from (2) from

1 1 1

v ' 10 10

v ' 5

2

5m

10

Total magnification

1 2m m m

30 5

15 10

m 1

Image is real & inverted.


Sol.9. (CD)

2

x Asinwt

v Awcoswt

a Aw sin wt

2 2 2 2v A w cos wt

22

2 2

vcos wt

A w …… (1)

2 2 4 2a A w sin wt

22

2 4

asin wt

A w ……. (2)

Add equation (1) and (2)2 2

2 2 2 4

v a1

A w A w

(A) 2v y a x

2

2 2 2 4

y x1

A w A w (Parabola)

(B) 2v y 2a x

2 2 2 4

y x1

A w A w (Straight line)

(C) v y a x

2 2

2 2 2 4

y x1

A w A w (Ellipse)

(D) v y 2a x

2

2 2 2 4

y x1

A w A w (Parabola)






Sol.10. (D)

Sol.11. (BCD)zF n mg 20 0 N mg 20 N 50 20 N 30

maxf N

0.4 30

12

Eternal force in the plane of surface =ˆ ˆ

6i 8 j =10 Nf 10N

F will be opposite to external force

Contact force

2 2

2 2

N f

30 10

10 10

Sol.12. (ABC)1 1 1

f v u

1 1 1

v f u f constant

1 1

kv u

1 1

versusv u is linear 1 1 1

f v u u u

1f v

u u1

v f

uku 1

v straight line.

Sol.13. (ABC) There is no external force acting on the system

a 0 p 0

constant

Sol.14. (BD)

Sol.15. (ABD) T 40N 2a 10 m / s KE 40J 21mv 40

2

1

22 2v 40 2v 40

2 2v a 2as 40 0 2 10 s

s 2m

mgW Mg h

2 10 2

40 J

TW 40 2

80 J

Sol.16. (ABCD)

A

B1 kg

1 kg 4 m/s

4 m/s

1 kg C

After collision linear momentum of (A + C) system should be conservable in horizontal

direction.

2 kg

40

20






A2 v 1 4

Av 2m / s

No horizontal force on B therefore no change in i t’s velocity.

Bv 4m / s

1 1 2 2 3 3com

1 2 3

m v m v m v ...... 1 4 1 4 1 0v

m m m 1 1 1

com

8v

3

2 m/s

4 m/s

8/3com

1/3

2/3

1 8w 2

3 3

w 2 rod / s

SECTION – 3 : (Comprehension Type)

Sol.17. (AB) Net force on system along horizontal directionxF 0

Hence linear momentum of system in horizontal direction will be conserved.

There is no non-conservative force on system. Therefore mechanical energy will be

conserved.

Sol.18. (A)

m

m2v

M 1v

initial

Initial Pi = 0 Pf = Mv1 + Mv2

TEi = mgh Pi = Pf 2 2

f 2 1

1 1TE mv Mv

2 2

1 2O Mv mv

2 1mv Mv ……. (1)

1

2

Mvv

m

i f TE TE

2 2

2 1

1 1mgh mv Mv

2 2






2

211

Mv1 1mgh m Mv

2 m 2

Solving this 1

2ghv m

M m M

Sol.19. (ABD)

f

R

F

netF F ma

Fa

m (accelerate in forward direction)

F f ma

2f F R MR

a

R

f F R 2MR a

R

f F ma

F f

ma

Fa ; f 0

2m

netF F ma

Fa

m (Accelerate in forward direction)

nett FR (In clockwise direction)

Sol.20. (ABC)

f

F

F f ma






fR I a

R

f R2MR

a

2 R

maf

2

maF ma

2

3maF

2

2Fa

3m

mf

2

2

F

3m

Ff

3

nett fR

PART II : CHEMISTRY


Sol.21. (AC)

Sol.22. (BC) H mass heatcapacity T

500 4.18 20T.

This increase in temperature is balanced by addition of ice cubes of one molecule. Thus

number of moles of ice cubes.

1

500 4.18 2077

6.02 1000Jmol

Sol.23. (BC)

Sol.24. (C) using the expression

a

saltpH pK log

acid

We get 5 Salt

4.75 log 1.34 10 log0.02M

which gives salt

4.75 4.87 log0.02M

Or salt

0.02M=0.76 or 2salt 1.52 10 M

Hence, Amount of sodium propanoate to be added = 21.52 10 mol

The addition of 0.01 mol of hydrogen chloride convert the equivalent amount of sodium

propanoate into propanoic acid. Hence, We will have

1

1

0.0152 0.01 molLpH 4.87 log 4.87 log 0.173 4.87 0.76 4.11

0.02 0.01 molL

The pH of 0.01 molar HCl solution would be

pH log 0.01 2 .

Sol.25. (AC)

Sol.26. (B) Use G 2.303RTlogK

Sol.27. (BD)






Sol.28. (BD)

0

eq

AB g A g B g

t 1 0 0

1 1 1t 1

3 3 3

2 1 1 4

n3 3 3 3

gn

A Bp

AB

n n PKn n

1

p

1 1

P 1 3P P P3 3 82 4 / 3 6 4 8 K

3


Sol.29. (B)

Sol.30. (AC) The solubility of 2

Zn OH will be given as

2 2sp2

c24

Ks Zn Zn OH K OH

OH

The minimum solubility will be obtained by settingds

0d OH

Hencesp

c3

2K2K OH 0

OH

1/4 1/417

sp 5

c

2K 1.2 10OH 9.8 10

2K 0.13

pH 14 4.01 9.99

172

5 9

25

1.2 10s 0.13 9.8 10 2.5 10 M.

9.8 10

Sol.31. (A)

As p-derivative give only one product whereas o & m give more than one nitro

product.






Sol.32. (ACD)

(A) In isothermal process, temperature is constant, therefore,

1 2T T

(C) w wisothermal adiabatic

Area under curve is greater for isothermal process, than that for adiabatic process.

(D)isothermal

U 0

For adiabatic process:

1TV constant

1 1

1 1 3 3T V T V

1

3 1

1 3

T V

T V

3 1T T hence internal energy will decrease.

i.e.,isothermal adiabaticU U

Sol.33. (ABD) Sol.34. (BCD) Sol.35. (ACD)

Sol.36. (ABD) dH dU PdV VdP


Sol.37. (D) Sol.38. (C)Sol.39. (AC) Equilibrium is achieved at point – X. In graph A, at point X, the free energy of both reactants

and products are equal and G 0 which is the characteristic of reaction at equilibrium.

(B) Is not correct because equilibrium is achieved

(C) Is correct because free energy is minimum at X.

(D) Cannot be correct.

Sol.40. (BD) The reaction is reversible. The graph B tells that the reactants concentration are decreasing

while products concentration are increasing and after certain time, both the concentrations

are becoming constant which is a characteristic of reversible reaction which attains

equilibrium.

(A) is not correct because conc. Of reactant is not reaching zero

For (C) and (D) as

PK

R then K is less than 1. So (D) is correct.






PART III : MATHEMATICS


Sol.41. (C) y = (tanx sinx) y Now z = 3x z

y2 = (tan x – sin x) + y z

2 = x

3 + z

y =1 1 4(tanx sinx)

2

z =

31 1 4x

2

y > 0 z > 0

y =1 1 4(tanx sinx)

2

z =

31 1 4x

2

x 0lim

3

1 1 4(tanx sinx)1

2

1 1 4x1

2

=x 0lim

3

1 1 4(tanx sinx)

1 1 4x

Rationalization

=x 0lim

3

4(tanx sinx)

4x

×31 1 4x

1 1 4(tanx sinx)

=x 0lim

3

tanx sinx

x

×2

2

=x 0lim

sinx

x ×

2

1 cosx

x

= 1 × 1/2 = ½

Sol.42. (C) Let A = cosec –1

2 13x 1

4

+ sec

–1 2 13x

4

= cosec –1

2 11 3x

4

+ sec

–1 2 13x

4

where 3x2

+

1

4 1

Now A will be minimum when 2 13x

4

is minimum

2 13x

4

= 1

Amin = cosec –1

2 + sec –1

1 =6

hence12

A = 2

Sol.43. (AC) y = f(x) = 4x

3

– x

2

– 2x + 1

3

1

2

1

4

1,

2

1






g(x) =

1f(x) ; 0 x

2

1f(1/ 2) ; x 1

2

3 x ; 1 x 2

So1

g4

=1

2 ;

3g

4

=1

4 ;

5g

4

= 3 – 5

4.

Sol.44. (AC)

–

–

– –

From graph it is clear that m 0,2

= 5

Sol.45. (CD)(a, 0)

(0, a)

( – a, 0)

|x| + |y| = a

x2 + y2 = a2

a|y||x|

(0, – a)

A1 area included between the line and circle in first quadrant

=2a 1

4 2

a

2 =

2( 2)a

4

A2 = area included between the line and

| x | + | y | = a

=1

2a

2 –

a2

0( a x ) dx

= (9 – 2)2a

12

Now 2

1

A 1

A 3

9 2

2

, A = 9, B = 1

A + B – 1 = 9






y = 3 /16

Y

O X

Sol.46. (BD) Minimum (x1 – x2)2 + x1 (0, 2 )

This can be translate to finding minimum distance between two curves.

y = & y =

Minimum distance his along the common normal.

The normal passes through (0, 0) [as y = is a semicirle]

y1 = – mN = +

Equation of normal y – y2 = (x – x2)

Satisfies (0, 0)

9y2 = x23 9 × = x2

3 x2

4 = 81

x2 = ± 3 as x2 R+

x2 = 3 y2 = 3

Hence minimum distance = = 3 = 2

Given expression is the equation of minimum distance

(2)2 = 8

Sol.47. (D) Since given differential equation is

y (x + y3) dx = x (y3 – x) dy

(xy dx + x2 dy) + y4 dx – y3x dy = 0

x (y dx + x dy) + y3 (y dx – x dy) = 0

xd (xy) = y3 (x dy – y dx)

xd (xy) = x2y3 d

d

On integrating – + c

at (4, – 2)

+ c c = 0 y3 = –2x y = ( – 2x)1/3

Since y = g(x) =

= g (x) = x . sin 2x + x . ( – sin 2x) = 0

y = c1

(constant)

put sin x = cos x =

c1 = (sin –1 + cos –1) dt =

y = g (x) =

Hence area between y = and y = ( – 2x)1/3

= x dy

=

= sq. units.

2

2

21

x

9x2

21x2

1x

9

22x

9

9

x22

2x

9

233 22

2

xy

21

xy1

xsin

8/1

xcos

8/1

112 2

dttcosdttsin

dy2

y316/3

0

4

16

3.

8

1






Sol.48. (AB) f() = .....(1)

x = dx = – dt

f() = =

....(2)

Equation (1) + (2)

2f() = = = 2 l n

Now g()

g() = dx = = 2 l n

f(200) – g(50) = l n(200) – l n(50) = l n 4 = 3 · l n 4

a+ b = 7


Sol.49. (ACD) Diff. we get 2f(x) f (x) = f(x)

22sec x

4 tanx

f(x) 0, so f (x) =2sec x

4 tanx

f(x) = n (4 + tan x) + c

Given f(0) = n 4. So f(x) = n (4 + tan x).

Sol.50. (ABCD) (A) sgn (e –x)

e –x > 0 for every real x

every constant function is a periodic function.

(C) f(x) =8 8

1 cosx 1 cosx

=2

16

1 cos x =

4

|sinx|

= 4|cosec x|

this is a periodic function.

(D) f(x) =1

x2

+1

x2

+ 2[ – x]

dxx23

1xx

x3x2

1xxcot

x

1

1

2

2

21

t

12t

1

dt

t

1

t2t3

1tt

3t2

1ttcott

2

1

2

221

dt

t2t3

1tt

3t2

1ttcot

t

1

1 2

221

dtt3t2

1tt

t23

1ttcot

t

1

12

221

1 t

1

nn l l

dx1

|1x||1x|

|)2x)(1x(||2x3x|n

1n)x(f )x(f .e.ifunctionOdd

2

l

l

ln

1ln

dx.1

1nn l l






–1, –3

64

6

(1,0)

=1

x2

+

1x

2

+ 2 [ – x]

=1 1 1 1

x x x x2 2 2 2

+ 2 ( – x – { – x})

= – 1

x2

–

1x

2

– 2 { – x}

Since { } is a periodic function hence this function is periodic. Sol.51. (AB) f(x) = [x]2 + [x + 1] – 3 = {[x] + 2} {[x] – 1}

So, x = 1, 1.1, 1.2, .... f(x) = 0 f(x) is many one

only integral values will be attained f(x) is into

Sol.52. (BD) f(x) =

2

2

x 2 , 1 x 2

x 2, 2 x 3

2

11, x 3

3

f (x) =2x , 1 x 2

x , 2 x 3

function is increasing in each of the two intervals (1, 2) and (2, 3) but not in [1, 3]

Clearly least value of f(x) is 3 and greatest value of f(x) does not exist.

Sol.53. (AB) Domain of sin –1

x and cos –1

x, each is [ –1, 1] and that of sec –1

x and cosec –1

x,

each is ( –, –1] [1, )

Domain of f(x) must be { –1, 1}

Range of f(x) will be {f( –1), f(1)}

where f( –1) = sin –1

( –1) · cos –1

( –1) · tan –1

( –1) · cot –1

( –1) · sec –1

( –1) · cosec –1

( –1)

= 3

· · · · ·2 4 4 2

=

63

64

and f(1) = 0 {as cos

–1 1 = 0}

(i) Thus, the graph of f(x) is a two point graph which

doesn't lie above x - axis (A) is correct statement

(ii) f(x)max= 0 and f(x)min =63

64

6

max min

3f(x) f(x)

64

(B) is correct statement

(iii) f(x) is one-one hence injective

(C) is incorrect statement

(iv) Domain is { –1, 1}

Number of non-negative integers in the domain of f(x) is one (D) is incorrect statement

Sol.54. (ACD)

Sol.55. (BD) (A) We have

x

1x 1 3

(t 1) sin t dtLim

(x 1)

0form

0






=x 1 2

(x 1) sin xLim

3(x 1)

= x 1

sin x 0Lim

3(x 1) 0

= x 1

sin( x)Lim

3(x 1)

=

x 1

sin(1 x)Lim

3(x 1)

=x 1

sin (x 1)Lim

3 (x 1)

= (1)3

=

3

(B) We have 2x + y = 0, 4y + 3z = 24 and 12x – 7z = 24.

On solving, we get the point of intersection as ( – 12, 24, – 24) whose

distance from (0, 0, 0) is 36.

(C) We have1

2

sint

x g(x) dx = (1 – sin t) ........... (i)

Differentiating both the sides of (1) with respect to 't', we get

0 – sin2 t · g(sin t) · cos t = – cos t

g(sin t) =2

1

sin t ........... (ii)

Putting t =4

in (2),

We get g1

2

= 2

(D) We have

x2 y

2

3

12 3x

y2

33

3

2x y

4 x2y 4

Maximum value = 4

Sol.56. (AC) We have

I =a 1 x 1 x x

1 a 1 x 2x

sin e cos e edx

tan e tan e e 1

=

a x

1 a 1 x 2x

1 e. dx

2 (tan e tan e ) (e 1)

Put tan –1ex = t x

2x

edx

e 1 = dt

I =

1 atan e

1 a

0

dt

2 (t tan e )

=1 atan e

1 a

0ln(t tan e )

2

= 1 a 1 aln 2 tan e ln tan e2

= ln22


Sol.57. & 58 (B & C)

f(x) =

n

nxlim cosn

=

n

nxlim 1 cos 1n

=n

xlim cos 1 n

ne

=

2

n

1 xlim 2sin n

2 ne




=

2

n

1 x

2 n2 lim

1

ne

= n

x2 lim

2

x1 ne e

14

n

y = f(x) = e –x/2, x 0 range = (0, 1]

g(x) =xlim

(1 – x + x n e )n

=

1

n

n

( e 1)lim x

1/ne

= ex x R

h(x) = tan –1 (g –1 (f –1(x))

x

2 = ny x = 2n

1

y f –1(x) = 2n

1

x 0 < x 1

y = g(x) = ex

x = ny g –1(x) = nx

g –1 1

2 nx

= n 1

2 nx

for 0 < x < 1

h(x) = tan –1 2

1n n

x

for 0 < x < 1

Sol.59. (D) Let 0 < < < < 1, and , are the roots of

f(x) = x3 – 3x + k = 0 f() = f() = 0

f(x) satisfies RMVT f (c) = 0 3c2 = 0 c = ± 1

but c must be lies between & .

Hence k

Sol.60. (A) Let f(x) = tan –1x

then for some a (x, y), f () =1 1tan y tan x

y x

(LMVJ)

2

1

1 =

1 1tan x tan y

x y

1

|tan –1x – tan –1y| |x – y|

END OF Solution

xii class jee shikhar solution 20-12-2015

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