# Free Objective Test 02 Practice Test - 11th and 12th

π20 cos x1+sin xdx=

A.

log 2

B.

log e

C.

12 log 3

D.

0

#### SOLUTION

Solution : A

put 1+ sin x =t
Then π20cos x1+sin xdx=[log|1+sin x|]π20=log 2

If area bounded by the curves y2=4ax and y=mx is a23  then the value of m is

A. 2
B. -2
C.

12

D. None of these

#### SOLUTION

Solution : A

The two curves y2 = 4ax and y = mx intersect at (4am2,4am) and the area enclosed by the two curves is given by 4am20(4 axmx)dx
4am20(4axmx)dx=a23
83a2m3=a23m3=8m=2

π2π2 sin2x dx=

A. π
B.

π2

C. π212
D. π1

#### SOLUTION

Solution : B

π2π2 sin2x dx=2π20 sin2x dx=2r(32)r(12)2r(2+22)=π2

π20 sin2x cos3x dx= [RPET 1984, 2003]

A.

0

B. 215
C. 415
D.

None of these

#### SOLUTION

Solution : B

Using gamma function,
π20 sin2x cos3x dx=r(32)r22r(72)=215

0 log(1+x2)1+x2dx=

A.

π log 12

B. π log 2
C. 2π log 12
D. 2π log 2

#### SOLUTION

Solution : B

Let I=0 log(1+x2)1+x2dx
Put x=tan θdx=sec2 θ dθ,
I=n20 log (sec θ)2dθ=2n20 log sec θ dθ
=2n20log cos θ dθ =2. π2log12=π log 12=π log 2

limπ199+299+399+n99n100= [EAMCET 1994]

A.

9100

B.

1100

C.

199

D.

1101

#### SOLUTION

Solution : B

limπ199+299+399+n99n100=limπnr=1(r99n100)
=limπ1nnr=1(rn)99=10 x99 dx=[x100100]10=1100

limπnr=1 1nern is [AIEEE 2004]

A. e + 1
B. e - 1
C. 1 - e
D. e

#### SOLUTION

Solution : B

limπnr=1 1nern=10exdx=[ex]10=e1

The value of the definite integral 10x dxx3+16 lies in the interval [a, b]. The smallest such interval is.

A. [0,117]
B.

[0,1]

C. [0,127]
D.

None of these

#### SOLUTION

Solution : A

f(x)=xx3+16f(x)=162x3(x3+16)2f(x)=0x=2Also f′′(2)<0 The function f(x)=xx3+16 is an increasing function in [0,1], so Min f(x) = f(0) = 0 and Max f(x) = f(1) = 117
Therefore by the property m(ba)baf(x)dxM(ba)
(where m and M are the smallest and greatest values of function)
010xx3+16dx117

If x0f(t)dt=x+1xt f(t) dt, then the value of f(1) is      [IIT 1998; AMU 2005]

A. 12
B.

0

C. 1
D.

12

#### SOLUTION

Solution : A

x0 dt=x+1xt f(t) dt1xt f(t) dt=xx1t f(t) dt
Differentiating w.r.t x, we get f(x)=1+{0x f(x)}
f(x)=1x f(x)(1+x)f(x)=1f(x)=11+x
f(1)=11+1=12

Calculate 32f(x)dx where
f(x)={6ifx>13x2ifx1

A. 19
B. 21
C. 17
D. 15

#### SOLUTION

Solution : B

Here, we can see that the integrand f(x) in not continuous in the interval (-2, 3) as it has a discontinuity at x = 1. So we can’t just integrate f(x) and put limits.We’ll have to break this integral into integrals which have limits such that integrands is continuous in those limits.
32f(x)dx=123x2dx+316dx
Now we can integrate these integrands and put limits.
123x2dx+316dx=(x3)|12+(6x)|31=1(8)+186
= 21

Find the integral 0exdx.

A. 0
B. 1
C. 2
D.

#### SOLUTION

Solution : B

We can see that the given integral is an improper integral as one of its limits is not finite.
In such cases where we have to deal with infinity as the limits of definite integral, we’ll  change the limit which is not finite to a variable and then put the limits.
0exdx=limaa0exdx
=lima(ex)a0
=lima(ea(e0))
=0(1)
=1

Area enclosed by curve y39y+x=0 and Y - axis is -

A. 92
B. 9
C. 812
D. 81

#### SOLUTION

Solution : C

The given equation of curve can be written as x=f(y)=9yy3. Now to calculate the area we need to find the boundaries of this curve i.e ordinates or the point where this curve is meeting Y - axis.
f(y)=y(9y2)
f(y)=y.(3+y)(3y)
So, the points where f(y) is meeting y - axis are y = -3,  y = 0 & y = 3.
Important thing to note here is that the function is changing its signs.
I.e. from y = -3 to y = 0  f(y) is negative.
& from y = 0 to y = 3  f(y) is positive.
Since, the function in negative in the interval (-3, 0 ) we’ll take absolute value of it. Because we are interested in the area enclosed and not the algebraic sum of area.
Let the area enclosed be A.
A=03|f(y)|dy+30f(y)dyA=03|9yy3|dy+30(9yy3)dyA=03(9yy3)dy+309yy3dyA=[9y22y44]03+[9y22y44]30A=[00812+814]+[812814]A=81812A=812

10π0|sin x|dx is

A. 20
B. 8
C. 10
D. 18

#### SOLUTION

Solution : A

10π0|sin x|dx=10π0|sin x|dx
|sin x| is positive in I & II quadrant and has period π
=10π0 sin x dx=10[cos x]x0=20

20[x2]dx is (where [.] is greastest integral function

A. 22
B. 2+2
C. 21
D. 23+5

#### SOLUTION

Solution : D

20[x2]dx
=10[x2]dx+20[x2]dx+32[x2]dx+23[x2]dx
=100dx+201dx+322dx+233dx
=21+2322+633
=532

π20 log(tan x+cot x)dx=

A. π log 2
B. π log 2
C. π2 log 2
D. π2 log 2

#### SOLUTION

Solution : A

π20 log(tan x+cot x)dx=π20 log[2sin 2x]dx
=π20(log2log sin 2x)dx=log 2(π2)+(π2)log 2=π log 2