cpt-5 jee mains maths (top) held on 14-june-15
DESCRIPTION
Cpt-5 Jee Mains Maths (Top) Held on 14-June-15 IIT AshramTRANSCRIPT
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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.
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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.
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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 .
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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
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= 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
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= 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 .
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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.