Free Objective Test 02 Practice Test - 11th and 12th
Question 1
∫π20 cos x1+sin xdx=
log 2
log e
12 log 3
0
SOLUTION
Solution : A
put 1+ sin x =t
Then ∫π20cos x1+sin xdx=[log|1+sin x|]π20=log 2
Question 2
If area bounded by the curves y2=4ax and y=mx is a23 then the value of m is
12
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 ax−mx)dx
∴∫4am20(√4ax−mx)dx=a23
⇒83a2m3=a23⇒m3=8⇒m=2
Question 3
∫π2−π2 sin2x dx=
π2
SOLUTION
Solution : B
∫π2−π2 sin2x dx=2∫π20 sin2x dx=2r(32)r(12)2r(2+22)=π2
Question 4
∫π20 sin2x cos3x dx= [RPET 1984, 2003]
0
None of these
SOLUTION
Solution : B
Using gamma function,
∫π20 sin2x cos3x dx=r(32)r22r(72)=215
Question 5
∫∞0 log(1+x2)1+x2dx=
π log 12
SOLUTION
Solution : B
Let I=∫∞0 log(1+x2)1+x2dx
Put x=tan θ⇒dx=sec2 θ dθ,
∴I=∫n20 log (sec θ)2dθ=2∫n20 log sec θ dθ
=−2∫n20log cos θ dθ =−2. π2log12=−π log 12=π log 2
Question 6
limπ→∞199+299+399+⋯⋯n99n100= [EAMCET 1994]
9100
1100
199
1101
SOLUTION
Solution : B
limπ→∞199+299+399+⋯⋯n99n100=limπ→∞∑nr=1(r99n100)
=limπ→∞1n∑nr=1(rn)99=∫10 x99 dx=[x100100]10=1100
Question 7
limπ→∞∑nr=1 1nern is [AIEEE 2004]
SOLUTION
Solution : B
limπ→∞∑nr=1 1nern=∫10exdx=[ex]10=e−1
Question 8
The value of the definite integral ∫10x dxx3+16 lies in the interval [a, b]. The smallest such interval is.
[0,1]
None of these
SOLUTION
Solution : A
f(x)=xx3+16⇒f′(x)=16−2x3(x3+16)2∴f′(x)=0⇒x=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(b−a)≤∫baf(x)dx≤M(b−a)
(where m and M are the smallest and greatest values of function)
⇒0≤∫10xx3+16dx≤117
Question 9
If ∫x0f(t)dt=x+∫1xt f(t) dt, then the value of f(1) is [IIT 1998; AMU 2005]
0
−12
SOLUTION
Solution : A
∫x0 dt=x+∫1xt f(t) dt⇒∫1xt f(t) dt=x−∫x1t f(t) dt
Differentiating w.r.t x, we get f(x)=1+{0−x f(x)}
⇒f(x)=1−x f(x)⇒(1+x)f(x)=1⇒f(x)=11+x
∴f(1)=11+1=12
Question 10
Calculate ∫3−2f(x)dx where
f(x)={6ifx>13x2ifx≤1
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.
∫3−2f(x)dx=∫1−23x2dx+∫316dx
Now we can integrate these integrands and put limits.
∫1−23x2dx+∫316dx=(x3)|1−2+(6x)|31=1−(−8)+18−6
= 21
Question 11
Find the integral ∞∫0e−xdx.
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.
∞∫0e−xdx=lima→∞a∫0e−xdx
=lima→∞(−e−x)a0
=lima→∞(−e−a−(−e0))
=0−(−1)
=1
Question 12
Area enclosed by curve y3−9y+x=0 and Y - axis is -
SOLUTION
Solution : C
The given equation of curve can be written as x=f(y)=9y−y3. 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(9−y2)
f(y)=y.(3+y)(3−y)
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=∫0−3|f(y)|dy+∫30f(y)dyA=∫0−3|9y−y3|dy+∫30(9y−y3)dyA=−∫0−3(9y−y3)dy+∫309y−y3dyA=−[9y22−y44]0−3+[9y22−y44]30A=−[0−0−812+814]+[812−814]A=81−812A=812
Question 13
∫10π0|sin x|dx is
SOLUTION
Solution : A
∴∫10π0|sin x|dx=10∫π0|sin x|dx
∴ |sin x| is positive in I & II quadrant and has period π
=10∫π0 sin x dx=10[−cos x]x0=20
Question 14
∫20[x2]dx is (where [.] is greastest integral function
SOLUTION
Solution : D
∫20[x2]dx
=∫10[x2]dx+∫√20[x2]dx+∫√3√2[x2]dx+∫2√3[x2]dx
=∫100dx+∫√201dx+∫√3√22dx+∫2√33dx
=√2−1+2√3−2√2+6−3√3
=5−√3−√2
Question 15
∫π20 log(tan x+cot x)dx=
SOLUTION
Solution : A
∫π20 log(tan x+cot x)dx=∫π20 log[2sin 2x]dx
=∫π20(log2−log sin 2x)dx=log 2(π2)+(π2)log 2=π log 2