# Free Objective Test 01 Practice Test - 11th and 12th

10 x71x4dx is equal to

A. 1
B. 13
C. 23
D. π3

#### SOLUTION

Solution : B

I=10x71x4dx=10x6x dx1x4
Put x2=sin θ2x dx=cos θ dθ
I=12 π20sin3 θ. cos θ dθcos θ=12 120 sin3 θ dθ
=12T2T(12)2.T(12)=T(12)4.32.12.T(12)=13

Let f be a positive function. Let
I1=k1k xf{x(1x)}dx,  I2=k1kf{x(1x)}dx
when 2k1>0. Then I1I2 is               [IIT 1997 Cancelled]

A.

2

B.

k

C.

12

D.

1

#### SOLUTION

Solution : C

I1=k1k xf{x(1x)}dx
=k1k(1k+kx)f[(1k+kx){1(1k+kx)}]dx
=k1k(1x)f{x(1x)}dx
=k1kf{x(1x)}dxk1kxf{x(1x)}dx=I2I1
2I1=I2I1I2=12

If I is the greatest of the definite integrals
I1=10excos2x dx,  I2=10ex2cos2 x dx
I3=10ex2dx, I4=10ex22dx, then

A.

I=I1

B.

I=I2

C.

I=I3

D.

I=I4

#### SOLUTION

Solution : D

For 0 < x < 1, we have 12x2<x2<x
x2>x, so that ex2<ex,
Hence     10ex2 cos2 x dx>10excos2x dx
Also cos2 x 1
Therefore 10ex2cos2 x dx10ex2dx<10ex22dx=I4
Hence I4 is the greatest integral

Area bounded by parabola y2=x and straight line 2y = x is

A. 43
B. 1
C.

23

D.

13

#### SOLUTION

Solution : A

y2=x and 2y=xy2=2yy=0,2
Required area=20(y22y)dy=(y33y2)20=43sq.unit

The correct evaluation of π0|sin4 x|dx is  [MP PET 1993]

A. 8π3
B. 2π3
C. 4π3
D. 3π8

#### SOLUTION

Solution : D

π0|sin4 x|dx=2π20 sin4 x dx
Applying gamma function,
2π20sin4 x dx=2T(52).T(12)2.T(62)=3π8

=limn[1n+1n2+n+1n2+2n++1n2+(n1)n] is equal to [RPET 2000]

A. 2+22
B. 222
C. 22
D. 2

#### SOLUTION

Solution : B

y=limn[1n+1n2+n++1n2+(n1)n]
y=limn1n+1n1+1n++1n1+(n1)n
y=1nlimn1+11+1n++11+(n1)n
y=limn1nk=1n11+(k1)n, Put k1n=x and 1n=dx
y=limnn1n0 dx1+x=limn2[1+x](n1n)0
y=2limn[2n1n1]=2limn2n1n2
y=2 limn 21n2=222

limn[1n+1n+1+1n+2++12n]=                           [Karnataka CET 1999]

A. 0
B. loge 4
C.

loge 3

D. loge 2

#### SOLUTION

Solution : D

limn[1n+1n+1+1n+2++12n]
=limn[1n+1n+1+1n+2++1n+n]
=1nlimn[1+11+1n+11+2n++11+nn]
=1nlimnnr=0[11+rn]=1011+xdx
=[loge (1+x)]10=loge 2loge 1=loge2

If f(x)=xa t3et dt, then ddxf(x)= [MP PET 1989]

A. ex(x3+3x2)
B. x3ex
C.

a3ex

D.

None of these

#### SOLUTION

Solution : B

f(x)=xa t3et dt=0at3.et dt+x0t3etdt
df(x)dx=ddx(0a t3.et dt)+ddx(x0t3.et dt)=x3ex

If f(x) is a continuous function defined on [a,b] such that f(x)0  x[a,b] then the area under the curve as the limit of a sum can be given as
(ba)limn1n[f(a)+f(a+h)...f(a+(n1)h)]
Where h=ban and h 0  as  n

A. True
B. False

#### SOLUTION

Solution : A

With all the information given let’s draw the graph of f(x) on cartesian coordinate system.

The area which we are interested in calculating is the area PRSQP. To calculate this area let’s divide this into rectangles of equal and infinitesimally small width. Now, what’s the purpose of this? Why are we doing this? Let’s discuss them one by one.
First why rectangle? Very simple, we do not know any approach to calculate the area of graphs like this. But we do know how to calculate the area of rectangle.
Second why infinitesimally small width? Just because we know how to calculate the area of rectangle we can’t calculate the area of this graph.
For example if we divide the stretch a to b in two parts, i.e. dividing it into two rectangles of equal width will give us area which is not equal to the area of this graph. The difference between the area of the rectangles and the area of this graph will be significant. So, to reduce this error of the difference we’ll divide this graph into infinite rectangles.
So, the the length a to b which is ba will be divided into n equal parts giving width as ban
Now consider regions ABCD, ABLC, ABDM.
By observation we can say -
Area of the rectangle ABLC < area of the region ABCD < Area of the rectangle ABDM
Actually when the width becomes infinitesimally small all these area become nearly equal to each other.
limnban   Area of the rectangle ABLC   Area of the region ABCD Area of the rectangle ABDM
Now we form the sums  -
sn=ban[f(a)+f(a+h)...f(a+(n1)h)]
Sn=ban[f(a+h)+f(a+2h)...f(a+nh)]
Where, sn  &  Sn denote the sum of the areas of lower rectangles and upper rectangles respectively.
And we know, sn<AreaPRSQP<Sn
As n strips becomes narrower and narrower and both the sum becomes equal.
So, limn  sn  = Area PRSQP = limn Sn
ban=h(given)
So, Area of the graph =
limn ban[f(a)+f(a+h)...f(a+(n1)h)]

Antiderivative of all the continuous functions can be written in terms of  elementary functions

A. False
B. True

#### SOLUTION

Solution : A

We have seen that there are some functions, antiderivatives of  which cannot be written in terms of elementary functions. Functions like ex2,ex2x,sin(1x), etc. are some of  the examples of such functions.

if a function f(x) is discontinuous in the interval (a,b) then baf(x)dx never exists.

A. False
B. True

#### SOLUTION

Solution : A

We have seen that there are functions which have discontinuity but still can be integrated. These functions have finite type discontinuities. Please try to understand one thing here, that definite integral gives you the area so area can be calculated by the curve into pieces as well.
For example -

We can see that the given function is discontinuous at x =1. So to integrate this function from (-3,3) we’ll have to integrate it into pieces. One piece will be (-3,1) and another will be (1, 3).
33f(x)dx=13f(x)dx+31f(x)dx
And given that for (3,1)f(x)=x2
And for (1,3) f(x) = 6-x
33f(x)dx=13(x2)dx+31(6x)dx
This way we can calculate the area even if the function is discontinuous.

11x2dx does not have a finite value

A. False
B. True

#### SOLUTION

Solution : A

Let’s calculate its value.
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.
11x2dx=limaa11x2dx
= lim a(1x)|a1 Since, 1x2dx=(1x)
=0(1)=1

Find the area enclosed by the curve x=y2+2, ordinates y = 0 & y = 3 and the Y - axis.

A. 5
B. 10
C. 15
D. 20

#### SOLUTION

Solution : C

We have seen that if g(y)0 for yϵ [c, d] then the area bounded by curve x = g(y) and y-axis between abscissa y = c and y = d is d(y=c)g(y)dy .
We'll use the same concept here. Here the curve given is
x=y2+2  and c = 0  &d=3.So,g(y)0xϵ(0,3)
Let the area enclosed be  A.
A=30(y2+2)dy.
A=(y33+2y)|30A=9+60A=15

In=π20cosnxcos(nx)dx,nϵN then I2001:I2002 can be the eccentricity of

A. Parabola
B. Ellipse
C. Circle
D. Hyperbola

#### SOLUTION

Solution : D

In+1=π20cosn+1xcos(n+1)xdx=π20cosn+1x(cosnxcosxsinnxsinx)dxIn+1=InIn+1             2In+1=InIn:In+1=2

In=10xntan1xdx. If anIn+2+bnIn=cn  nϵN,n1 then

A. a1,a2,a3.....are in A.P.
B. b1,b2,b3.....are in G.P.
C. c1,c2,c3.....are in H.P.
D. a1,a2,a3.....are in H.P.

#### SOLUTION

Solution : A

In=(xn+1n+1tan1x)1010xn+1n+1.11+x2dx(n+1)In=π410xn+11+x2dx(n+3)In+2=π410xn+31+x2dx(n+1)In+(n+3)In+2=π21n+2an=(n+3)a1,a2,a3.....are in A.P.
bn=(n+1)b1,b2..... are in A.P.
cn=π21n+2 not in any progression.