COMMON PRACTICE TEST 5JEE MAINS

BATCH 12th TOP BATCH Date : 14/06/2015

MATHEAMTICS SOLUTION

1.

Sol. B

Let 2x 1 x t 2 1x 1 x

t

f (t) = 1 + 21t

range is (1, ).2.

Sol. A

Replacing f (x) by x, we f (x) = x2

ffff (x) = x16 also (f (x4))2 = x16.3.

Sol. Cf (x) = 7 + {8x} + |tan 2x + cot 2x|

period is LCM of 18

and 14

.

4.

Sol. C

If n(A) = n, n(B) = r then total number of functions = rn.

Total number of into function = rC1 (r 1)n rC2 (r 2)n +

Here r = 3, n = 4

rn = 34 = 81

X = 3C1 24 3C2 14 = 45

Y = 81 45 = 36

|X Y| = 9.

5.

Sol. A

Given equation can be written as

x2 3 = 3[sin x]

Case I: x2 3 = 3 when [sin x] = 1

x = 0 but sin x 1Case II: x2 3 = 0 when [sin x] = 0

x = 3 x = 3 gives [sin x] = 0Case III: x2 3 = 3 when [sin x] = 1

x = 6 but [sin x] 1.

Number of solution is one i.e. x = 3 .

6.

Sol. A

The Given function is f (x) = sin3 x sin 3x

f (x) = 3sin x sin3x4 sin 3x

f (x) = 38

(cos 2x cos 4x) 18

(1 cos 6x)

period of f (x) is .7.

Sol. A

Let g (x) be the inverse of f, so f (g (x)) = x

4(g(x))3 3g (x) = x

Let g (x) = cos cos 3 = x = xcos31 1

So g (x) = cos

xcos31 1

8.

Sol. D

f (x) 1 + f (1 x) 1 = 0.

So, g (x) + g (1 x) = 0.

Putting x = x + 12

we get, g 1 x2

+ g1 x2

= 0.

So it is symmetrical about 1, 02

.

9.

Sol. D

f (x) = 1 x3, so f1 (x) = (1 x)1/3.

Clearly f (x) = f1 (x) meet at (0, 1) and (1, 0) other then the line y = x.

10.

Sol. C

. 4x2 + 4x + 4 + sin (x) = (2x + 1)2 + sin (x) + 3 2.Now as 1 < 2 < e, the required value of n is 3.

11.

Sol. B

y = |sin1 2x2 1|

y =1 2

1 2 2

sin (2x 1) 4xsin (2x 1) | 2x | 1 x

x 0 and sin1 (2x2 1) 0 and |2x2 1| 1 |x| 1

x 0, x 12

.

12.

Sol. A

Put = tan x = 2 sin 2 y = 2 cos 2 z = x2 + y2 xy = 4 2 sin 4 z [2, 6].

13.

Sol. B

1 1 xf (x)x

21 x x 1 1 x 1 x 1 0x x 1 ( 1 as function is onto)Given that above equation has only one root

1 2 3 .

14.

Sol. D

Periods of sin 3xtanand

2xsin,x

4 are 8, 4 and 3 respectively

Period of the given function = L.C.M. (8, 4, 3) = 24.15.

Sol. B

1024

1r2 rlog =

102

1r2 rlog

=

12

2r2

2

rlog +

12

2r2

3

2

rlog +

12

2r2

4

3

rlog + ... +

12

2r2

10

9

rlog + 102 2log .

= 2.1 + 22 .2 +23 .3 +24.4 + . . . . + 29 .9 + 10

= 10r.29

1r

r

= 8204.

16.

Sol. C

F(x) = f(x).g(x).h(x)

F(x) = f(x).g(x).h(x) + f(x).g(x).h(x) + f(x).g(x).h(x) F(x1) = f( x1).g(x1).h(x1) + f(x1).g( x1).h(x1) + f(x1).g(x1).h( x1) 21F (x1) = 4 f ( x1).g(x1).h(x1) + (-7) f(x1).g ( x1).h(x1) + k f(x1).g(x1).h ( x1) 21 F(x1) = ( 4 - 7 + k) F(x1) k = 24 .

17.

Sol. (A)

Dividing the numerator and denominator by x2, the given integral becomes

x1xtan1

x1x

dxx11

12

2

Let x +x1 = tanv c|v|logvdv

= log cx

1xtan2

1

. Hence k = 1.

18.

Sol. B

x 22tanxe tan x dx1 tanx 4

x 2e tan x sec x dx

4 4

xe tan x c4 .

19.

Sol. (A)

Multiply Dr & Nr by cosec2x, then

I = 22

xcot3ecxcos4ecxcosxcot4xeccos3

dx

= cxcos34xsinc

xcot3ecxcos41dx

)x(f)x(f2

20.

Sol. A

Put t3x2x

I = c3x2x

58dtt

51 8

1

8/7

21

Sol. C

Put x9/2 = t then dtdxx29 2/7 , So given integral reduces to

c1xxIn92c1ttIn

92

1t

dt92 92/92

2

22.

Sol. (B)

Given integral = 222

sincossincos

d

= cossin sincos d = log|sin + cos| + c.23.

Sol. I = dx2xtanx)xcos1ln(= dx2xtanxdx1.)xcos1(ln(

= ln(1+ cosx). x + dx2xtanxdxx.2xcos2

2xcos

2xsin2

2

= ln(1+ cosx). x + dx2xtanxdx2xtanx + c= ln (1+ cosx). x + c.

24.

Sol. A

x 12 2 2 2

1 1 2xe tan x dx1 x 1 x (1 x )

x 1

21e tan x c

1 x .

25.

Sol. B

Let I = 2x tan x sec x dx(tan x x) = 22

2

x sin x dx(sin x xcos x)cos x

cos x

= 2x sinx dx

(sinx xcos x)Put sin x x cos x = t x sin x dx = dt

I = 2dt 1 1c c

t x cos x sin xt .

26.

Sol. C

Putting a6 + x8 = t2

I =2 3 3

2 6 3t a t adt t ln c

2t a t a .

27.

Sol. A

Let 3x = cos 3dx = - sin d

dcos3131dsinsin3cos

31 2

2

= c91sin

91 3 = cx3cos

91x91

91 312 .

28.

Sol. B

A is an idempotent matrix. Hence A2 = A An = A so thatAB = A(I A) = A A2 = A A = 0.

29.

Sol. B

We have A A = I, B B = I

A = A-1, B = B-1

Now (AB) = B A = B-1 A-1 = (AB)-1

30.

Sol. B

We have 4A A and A A A A A A A A A = 4 4 4A A A A 1 A A Hence (B) is the correct answer.