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

Question 1

$\int_{0}^{1} \frac{x^{7}}{\sqrt{1 - x^{4}}} d x$ is equal to

A. 1

\frac{1}{3}

\frac{2}{3}

\frac{π}{3}

SOLUTION

Solution : B

$I = \int_{0}^{1} \frac{x^{7}}{\sqrt{1 - x^{4}}} d x = \int_{0}^{1} \frac{x^{6} x d x}{\sqrt{1 - x^{4}}}$
$P u t x^{2} = s i n θ \Rightarrow 2 x d x = c o s θ d θ$
$I = \frac{1}{2}_{0}^{\frac{π}{2}} \frac{s i n^{3} θ . c o s θ d θ}{c o s θ} = \frac{1}{2} \int_{0}^{\frac{1}{2}} s i n^{3} θ d θ$
$= \frac{1}{2} \frac{T 2 T (\frac{1}{2})}{2. T (\frac{1}{2})} = \frac{T (\frac{1}{2})}{4. \frac{3}{2} . \frac{1}{2} . T (\frac{1}{2})} = \frac{1}{3}$

Question 2

Let f be a positive function. Let
$I_{1} = \int_{1 - k}^{k} x f {x (1 - x)} d x, I_{2} = \int_{1 - k}^{k} f {x (1 - x)} d x$
$w h e n 2 k - 1 > 0. T h e n \frac{I_{1}}{I_{2}} i s$ [IIT 1997 Cancelled]

$\frac{1}{2}$

SOLUTION

Solution : C

$I_{1} = \int_{1 - k}^{k} x f {x (1 - x)} d x$
$= \int_{1 - k}^{k} (1 - k + k - x) f [(1 - k + k - x) {1 - (1 - k + k - x)}] d x$
$= \int_{1 - k}^{k} (1 - x) f {x (1 - x)} d x$
$= \int_{1 - k}^{k} f {x (1 - x)} d x - \int_{1 - k}^{k} x f {x (1 - x)} d x = I_{2} - I_{1}$
$∴ 2 I_{1} = I_{2} \Rightarrow \frac{I_{1}}{I_{2}} = \frac{1}{2}$

Question 3

If I is the greatest of the definite integrals
$I_{1} = \int_{0}^{1} e^{- x} c o s^{2} x d x, I_{2} = \int_{0}^{1} e^{- x^{2}} c o s^{2} x d x$
$I_{3} = \int_{0}^{1} e^{- x^{2}} d x, I_{4} = \int_{0}^{1} e^{\frac{- x^{2}}{2}} d x, t h e n$

$I = I_{1}$

$I = I_{2}$

$I = I_{3}$

$I = I_{4}$

SOLUTION

Solution : D

For 0 < x < 1, we have $\frac{1}{2} x^{2} < x^{2} < x$
$\Rightarrow - x^{2} > - x, s o t h a t e^{- x^{2}} < e^{- x},$
$H e n c e \int_{0}^{1} e^{- x^{2}} c o s^{2} x d x > \int_{0}^{1} e^{- x} c o s^{2} x d x$
$A l s o c o s^{2} x \leq 1$
$T h e r e f o r e \int_{0}^{1} e^{- x^{2}} c o s^{2} x d x \leq \int_{0}^{1} e^{- x^{2}} d x < \int_{0}^{1} e^{\frac{- x^{2}}{2}} d x = I_{4}$
Hence $I_{4}$ is the greatest integral

Question 4

Area bounded by parabola $y^{2} = x$ and straight line 2y = x is

\frac{4}{3}

1

$\frac{2}{3}$

$\frac{1}{3}$

SOLUTION

Solution : A

$y^{2} = x a n d 2 y = x \Rightarrow y^{2} = 2 y \Rightarrow y = 0, 2$
$∴ R e q u i r e d a r e a = \int_{0}^{2} (y^{2} - 2 y) d y = {(\frac{y^{3}}{3} - y^{2})}_{0}^{2} = \frac{4}{3} s q . u n i t$

Question 5

The correct evaluation of $\int_{0}^{π} | s i n^{4} x | d x$ is [MP PET 1993]

\frac{8 π}{3}

\frac{2 π}{3}

\frac{4 π}{3}

\frac{3 π}{8}

SOLUTION

Solution : D

$\int_{0}^{π} | s i n^{4} x | d x = 2 \int_{0}^{\frac{π}{2}} s i n^{4} x d x$
Applying gamma function,
$2 \int_{0}^{\frac{π}{2}} s i n^{4} x d x = 2 \frac{T (\frac{5}{2}) . T (\frac{1}{2})}{2. T (\frac{6}{2})} = \frac{3 π}{8}$

Question 6

$= {lim}_{n \to \infty} [\frac{1}{n} + \frac{1}{\sqrt{n^{2} + n}} + \frac{1}{\sqrt{n^{2} + 2 n}} + \dots + \frac{1}{\sqrt{n^{2} + (n - 1) n}}]$ is equal to [RPET 2000]

2 + 2 \sqrt{2}

2 \sqrt{2} - 2

2 \sqrt{2}

2

SOLUTION

Solution : B

$y = {lim}_{n \to \infty} [\frac{1}{n} + \frac{1}{\sqrt{n^{2} + n}} + \dots + \frac{1}{\sqrt{n^{2} + (n - 1) n}}]$
$\Rightarrow y = {lim}_{n \to \infty} ⎡ ⎢ ⎣ \frac{1}{n} + \frac{1}{n \sqrt{1 + \frac{1}{n}}} + \dots + \frac{1}{n \sqrt{1 + \frac{(n - 1)}{n}}} ⎤ ⎥ ⎦$
$\Rightarrow y = \frac{1}{n} {lim}_{n \to \infty} ⎡ ⎢ ⎣ 1 + \frac{1}{\sqrt{1 + \frac{1}{n}}} + \dots + \frac{1}{\sqrt{1 + \frac{(n - 1)}{n}}} ⎤ ⎥ ⎦$
$y = {lim}_{n \to \infty} \frac{1}{n} \sum_{n}^{k = 1} \frac{1}{\sqrt{1 + \frac{(k - 1)}{n}}}, P u t \frac{k - 1}{n} = x a n d \frac{1}{n} = d x$
$\Rightarrow y = {lim}_{n \to \infty} \int_{0}^{\frac{n - 1}{n}} \frac{d x}{\sqrt{1 + x}} = {lim}_{n \to \infty} 2 {[\sqrt{1 + x}]}_{0}^{(\frac{n - 1}{n})}$
$\Rightarrow y = 2 {lim}_{n \to \infty} [\sqrt{\frac{2 n - 1}{n}} - 1] = 2 {lim}_{n \to \infty} \sqrt{\frac{2 n - 1}{n}} - 2$
$\Rightarrow y = 2 {lim}_{n \to \infty} \sqrt{2 - \frac{1}{n}} - 2 = 2 \sqrt{2} - 2$

Question 7

${lim}_{n \to \infty} [\frac{1}{n} + \frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{2 n}] =$ [Karnataka CET 1999]

0

l o g_{e} 4

$l o g_{e} 3$

l o g_{e} 2

SOLUTION

Solution : D

${lim}_{n \to \infty} [\frac{1}{n} + \frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{2 n}]$
$= {lim}_{n \to \infty} [\frac{1}{n} + \frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{n + n}]$
$= \frac{1}{n} {lim}_{n \to \infty} [1 + \frac{1}{1 + \frac{1}{n}} + \frac{1}{1 + \frac{2}{n}} + \dots + \frac{1}{1 + \frac{n}{n}}]$
$= \frac{1}{n} {lim}_{n \to \infty} \sum_{r = 0}^{n} [\frac{1}{1 + \frac{r}{n}}] = \int_{0}^{1} \frac{1}{1 + x} d x$
$= [l o g_{e} (1 + x)]_{0}^{1} = l o g_{e} 2 - l o g_{e} 1 = l o g_{e} 2$

Question 8

If $f (x) = \int_{a}^{x} t^{3} e^{t} d t, t h e n \frac{d}{d x} f (x) =$ [MP PET 1989]

e^{x} (x^{3} + 3 x^{2})

x^{3} e^{x}

$a^{3} e^{x}$

$N o n e o f t h e s e$

SOLUTION

Solution : B

$f (x) = \int_{a}^{x} t^{3} e^{t} d t = \int_{a}^{0} t^{3} . e^{t} d t + \int_{0}^{x} t^{3} e^{t} d t$
$\Rightarrow \frac{d f (x)}{d x} = \frac{d}{d x} (\int_{a}^{0} t^{3} . e^{t} d t) + \frac{d}{d x} (\int_{0}^{x} t^{3} . e^{t} d t) = x^{3} e^{x}$

Question 9

If $f (x)$ is a continuous function defined on $[a, b]$ such that $f (x) \geq 0$ $x \in [a, b]$ then the area under the curve as the limit of a sum can be given as
$(b - a) lim n \to \infty \frac{1}{n} [f (a) + f (a + h) . . . f (a + (n - 1) h)]$
Where $h = \frac{b - a}{n}$ and $h \to 0$ as $n \to \infty$

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 $P R S Q P .$ 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 $\frac{b - a}{n}$
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.
$lim n \to \infty \frac{b - a}{n}$ Area of the rectangle ABLC $\approx$ Area of the region ABCD $\approx$ Area of the rectangle ABDM
Now we form the sums -
$s_{n} = \frac{b - a}{n} [f (a) + f (a + h) . . . f (a + (n - 1) h)]$
$S_{n} = \frac{b - a}{n} [f (a + h) + f (a + 2 h) . . . f (a + n h)]$
Where, $s_{n} & S_{n}$ denote the sum of the areas of lower rectangles and upper rectangles respectively.
And we know, $s_{n} < A r e a P R S Q P < S_{n}$
As $n \to \infty$ strips becomes narrower and narrower and both the sum becomes equal.
So, $lim n \to \infty s_{n}$ = Area PRSQP = $lim n \to \infty S_{n}$
$\frac{b - a}{n} = h (g i v e n)$
So, Area of the graph =
$lim n \to \infty \frac{b - a}{n} [f (a) + f (a + h) . . . f (a + (n - 1) h)]$

Question 10

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

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 $e^{x^{2}}, \frac{e^{x^{2}}}{x}, s i n (\frac{1}{x}),$ etc. are some of the examples of such functions.

Question 11

if a function f(x) is discontinuous in the interval (a,b) then $\int_{a}^{b} f (x) d x$ 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).
$\int_{- 3}^{3} f (x) d x = \int_{- 3}^{1} f (x) d x + \int_{1}^{3} f (x) d x$
And given that for $(- 3, 1) f (x) = x^{2}$
And for (1,3) f(x) = 6-x
$\int_{- 3}^{3} f (x) d x = \int_{- 3}^{1} (x^{2}) d x + \int_{1}^{3} (6 - x) d x$
This way we can calculate the area even if the function is discontinuous.

Question 12

$\int_{1}^{\infty} \frac{1}{x^{2}} d x$ 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.
$\int_{1}^{\infty} \frac{1}{x^{2}} d x = {lim}_{a \to \infty} \int_{1}^{a} \frac{1}{x^{2}} d x$
= lim $a \to \infty (- \frac{1}{x}) |_{1}^{a}$ Since, $\int \frac{1}{x^{2}} d x = (- \frac{1}{x})$
$= 0 - (- 1) = 1$

Question 13

Find the area enclosed by the curve $x = y^{2} + 2$ , 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 $g (y) \geq 0$ for $y ϵ$ [c, d] then the area bounded by curve x = g(y) and y-axis between abscissa y = c and y = d is $\int_{(y = c)}^{d} g (y) d y$ .
We'll use the same concept here. Here the curve given is
$x = y^{2} + 2$ and c = 0 $& d = 3. S o, g (y) \geq 0 \forall x ϵ (0, 3)$
Let the area enclosed be A.
$A = \int_{0}^{3} (y^{2} + 2) d y .$
$A = (\frac{y^{3}}{3} + 2 y) |_{0}^{3} A = 9 + 6 - 0 A = 15$

Question 14

$I_{n} = \int_{0}^{\frac{π}{2}} {cos}^{n} x cos (n x) d x, n ϵ N$ then $\sqrt{I_{2001} : I_{2002}}$ can be the eccentricity of

A. Parabola

B. Ellipse

C. Circle

D. Hyperbola

SOLUTION

Solution : D

$I_{n + 1} = \int_{0}^{\frac{π}{2}} {cos}^{n + 1} x cos (n + 1) x d x = \int_{0}^{\frac{π}{2}} {cos}^{n + 1} x (cos n x cos x - sin n x sin x) d x ∴ I_{n + 1} = I_{n} - I_{n + 1} \Rightarrow 2 I_{n + 1} = I_{n} ∴ I_{n} : I_{n + 1} = 2$

Question 15

$I_{n} = \int_{0}^{1} x^{n} {tan}^{- 1} x d x . If a_{n} I_{n + 2} + b_{n} I_{n} = c_{n} \forall n ϵ N, n \geq 1$ then

a_{1}, a_{2}, a_{3}

.....are in A.P.

b_{1}, b_{2}, b_{3}

.....are in G.P.

c_{1}, c_{2}, c_{3}

.....are in H.P.

a_{1}, a_{2}, a_{3}

.....are in H.P.

SOLUTION

Solution : A

$I_{n} = {(\frac{x^{n + 1}}{n + 1} t a n^{- 1} x)}_{0}^{- 1} - \int_{0}^{1} \frac{x^{n + 1}}{n + 1} . \frac{1}{1 + x^{2}} d x (n + 1) I_{n} = \frac{π}{4} - \int_{0}^{1} \frac{x^{n + 1}}{1 + x^{2}} d x (n + 3) I_{n + 2} = \frac{π}{4} - \int_{0}^{1} \frac{x^{n + 3}}{1 + x^{2}} d x ∴ (n + 1) I_{n} + (n + 3) I_{n + 2} = \frac{π}{2} - \frac{1}{n + 2} ∴ a_{n} = (n + 3) \Rightarrow a_{1}, a_{2}, a_{3}$ .....are in A.P.
$b_{n} = (n + 1) \Rightarrow b_{1}, b_{2}$ ..... are in A.P.
$c_{n} = \frac{π}{2} - \frac{1}{n + 2}$ not in any progression.

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

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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