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

### Question 1

∫10 x7√1−x4dx is equal to

#### SOLUTION

Solution :B

I=∫10x7√1−x4dx=∫10x6x dx√1−x4

Put x2=sin θ⇒2x 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

### Question 2

Let f be a positive function. Let

I1=∫k1−k xf{x(1−x)}dx, I2=∫k1−kf{x(1−x)}dx

when 2k−1>0. Then I1I2 is [IIT 1997 Cancelled]

2

k

12

1

#### SOLUTION

Solution :C

I1=∫k1−k xf{x(1−x)}dx

=∫k1−k(1−k+k−x)f[(1−k+k−x){1−(1−k+k−x)}]dx

=∫k1−k(1−x)f{x(1−x)}dx

=∫k1−kf{x(1−x)}dx−∫k1−kxf{x(1−x)}dx=I2−I1

∴2I1=I2⇒I1I2=12

### Question 3

If I is the greatest of the definite integrals

I1=∫10e−xcos2x dx, I2=∫10e−x2cos2 x dx

I3=∫10e−x2dx, I4=∫10e−x22dx, then

I=I1

I=I2

I=I3

I=I4

#### SOLUTION

Solution :D

For 0 < x < 1, we have 12x2<x2<x

⇒−x2>−x, so that e−x2<e−x,

Hence ∫10e−x2 cos2 x dx>∫10e−xcos2x dx

Also cos2 x ≤1

Therefore ∫10e−x2cos2 x dx≤∫10e−x2dx<∫10e−x22dx=I4

Hence I4 is the greatest integral

### Question 4

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

23

13

#### SOLUTION

Solution :A

y2=x and 2y=x⇒y2=2y⇒y=0,2

∴ Required area=∫20(y2−2y)dy=(y33−y2)20=43sq.unit

### Question 5

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

#### SOLUTION

Solution :D

∫π0|sin4 x|dx=2∫π20 sin4 x dx

Applying gamma function,

2∫π20sin4 x dx=2T(52).T(12)2.T(62)=3π8

### Question 6

=limn→∞[1n+1√n2+n+1√n2+2n+⋯+1√n2+(n−1)n] is equal to [RPET 2000]

#### SOLUTION

Solution :B

y=limn→∞[1n+1√n2+n+⋯+1√n2+(n−1)n]

⇒y=limn→∞⎡⎢⎣1n+1n√1+1n+⋯+1n√1+(n−1)n⎤⎥⎦

⇒y=1nlimn→∞⎡⎢⎣1+1√1+1n+⋯+1√1+(n−1)n⎤⎥⎦

y=limn→∞1n∑k=1n1√1+(k−1)n, Put k−1n=x and 1n=dx

⇒y=limn→∞∫n−1n0 dx√1+x=limn→∞2[√1+x](n−1n)0

⇒y=2limn→∞[√2n−1n−1]=2limn→∞√2n−1n−2

⇒y=2 limn→∞ √2−1n−2=2√2−2

### Question 7

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

loge 3

#### 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]

=1nlimn→∞∑nr=0[11+rn]=∫1011+xdx

=[loge (1+x)]10=loge 2−loge 1=loge2

### Question 8

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

a3ex

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

### Question 9

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

(b−a)limn→∞1n[f(a)+f(a+h)...f(a+(n−1)h)]

Where h=b−an and h →0 as n →∞

#### 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 b−a will be divided into n equal parts giving width as b−an

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.

limn→∞b−an Area of the rectangle ABLC ≈ Area of the region ABCD ≈ Area of the rectangle ABDM

Now we form the sums -

sn=b−an[f(a)+f(a+h)...f(a+(n−1)h)]

Sn=b−an[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

b−an=h(given)

So, Area of the graph =

limn→ ∞b−an[f(a)+f(a+h)...f(a+(n−1)h)]

### Question 10

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

#### 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.

### Question 11

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

#### 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).

∫3−3f(x)dx=∫1−3f(x)dx+∫31f(x)dx

And given that for (−3,1)f(x)=x2

And for (1,3) f(x) = 6-x

∫3−3f(x)dx=∫1−3(x2)dx+∫31(6−x)dx

This way we can calculate the area even if the function is discontinuous.

### Question 12

∫∞11x2dx does not have a finite value

#### 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=lima→∞∫a11x2dx

= lim a→∞(−1x)|a1 Since, ∫1x2dx=(−1x)

=0−(−1)=1

### Question 13

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

#### 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)≥0∀xϵ(0,3)

Let the area enclosed be A.

A=∫30(y2+2)dy.

A=(y33+2y)|30A=9+6−0A=15

### Question 14

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

#### SOLUTION

Solution :D

In+1=∫π20cosn+1xcos(n+1)xdx=∫π20cosn+1x(cosnxcosx−sinnxsinx)dx∴In+1=In−In+1 ⇒2In+1=In∴In:In+1=2

### Question 15

In=∫10xntan−1xdx. If anIn+2+bnIn=cn ∀ nϵN,n≥1 then

#### SOLUTION

Solution :A

In=(xn+1n+1tan−1x)10−∫10xn+1n+1.11+x2dx(n+1)In=π4−∫10xn+11+x2dx(n+3)In+2=π4−∫10xn+31+x2dx∴(n+1)In+(n+3)In+2=π2−1n+2∴an=(n+3)⇒a1,a2,a3.....are in A.P.

bn=(n+1)⇒b1,b2..... are in A.P.

cn=π2−1n+2 not in any progression.