COMMON PRACTICE TEST 5JEE MAINS

BATCH 12th All Batch Date : 14/06/2015

MATHEAMTICS SOLUTION1.

Sol. A

y = 3 sin x 4 sin3 x + sin x = 4(sin x sin3 x)

now, dydx

= cos x(1 3 sin2 x) = 0

sin x = 13

, 1 we get max/minima at 13

.

2.

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

period is LCM of 18

and 14

.

3.

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.

4.

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 .

5.

Sol. A

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

f (x) = 3sinx sin3x4 sin 3x

f (x) = 38

(cos 2x cos 4x) 18

(1 cos 6x)

period of f (x) is .6.

Sol. B

Given that f (x) = x 2 sin x2

f (x) = 1 cos x2

0

f (x) is an increasing function. f (x) is oneone and onto.

7.

Sol. C

We should have [sin 2x] [cos 2x] we can have [sin 2x] = 1, [cos 2x] = 1, 0, 1[sin 2x] = 0, [cos 2x] = 0, 1

[sin 2x] = 1, [cos 2x] = 1

but [sin 2x] = 1, [cos 2x] = 1 and [sin 2x] = 1, [cos 2x] = 1 are not possible

So, range = {0, 1}.

8.

Sol. A

1 |sin x| + |cos x| 2 x R

ln 12 ln 1

| sinx | | cos x | 0

domain of f (x) is x = n2 , where n I

and range of f (x) is {0}

period is2 .

9.

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 about1, 02

.

10.

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

.

11.

So. D

. Period of sin (sin x) is 2 and e{3x} is 13

period of f (x) is LCM of 2 and 13

= 6.

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

The period of the given function =2

2 4 = 2.

14.

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.

15.

Sol. A

f(x) = log1/2log4log3[( x-4)2] log4( log3[ x 4)2] ) > 0 log3[ ( x 4)2] > 1

[(x 4)2] 4 (x 4)2 4 0 (x - 2)(x 6) 0 x (-, 2] [ 6, ).

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. (c) cxdxdxxx xxdxxxx 2)cos(sin cossin2sin1 cossin .

19. (d)xx

dxxf sinloglogsinlog)(

Differentiating both sides, we get.cot)(sinlog

cotsinlog)( xxfx

xx

xf

20. (c) dxxxx 2/12 )sectan2tan21( dxxxxx 2/122 )sectan2tan(sec

cxxxdxxx seclog)tanlog(sec)tan(seccxxx )tan(secseclog .

21.

Sol. A

Put t3x2x

I = c3x2x

58dtt

51 8

1

8/7

22.

Sol C

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

c1xxIn92c1ttIn

92

1t

dt92 92/92

2

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. (d).

11tan 21

abxbaxabx

baxdxd

abxbax

dxd

222222222

11xxbxaba

ba

.

25.

Sol. B

Let I = 2x tanx sec x dx(tanx 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. (a)11 222 yxyyxy

dxdyxxydx

dyy 22.2 222

xyxy

dxdy

.

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.