Definite Integration — Page 11 | JEE Mathematics

Q101

Let

f:R \to R

be a function defined by :

f(x) = \left\{ \begin{array}{lll}{\max \,\{ {t^3} - 3t\} \,t \le x} & ; & {x \le 2} \\ {{x^2} + 2x - 6} & ; & {2 5} \end{array} \right.

where [t] is the greatest integer less than or equal to t. Let m be the number of points where f is not differentiable and

I = \int\limits_{ - 2}^2 {f(x)\,dx}

. Then the ordered pair (m, I) is equal to :

A

\left( {3,\,{{27} \over 4}} \right)

B

\left( {3,\,{{23} \over 4}} \right)

C

\left( {4,\,{{27} \over 4}} \right)

D

\left( {4,\,{{23} \over 4}} \right)

Correct Answer

Option C

Solution

\left\{\begin{array}{l} f(x)=x^3-3 x, x \leq-1 \\\\ 2,-1 5 \end{array}\right.

Clearly $\mathrm{f}(\mathrm{x})$ is not differentiable at

\begin{aligned} & x=2,3,4,5 \Rightarrow m=4 \\\\ & I=\int_{-2}^{-1}\left(x^3-3 x\right) d x+\int_{-1}^2 2 \cdot d x=\frac{27}{4} \end{aligned}

Q102

\int_0^5 {\cos \left( {\pi \left( {x - \left[ {{x \over 2}} \right]} \right)} \right)dx}

, where [t] denotes greatest integer less than or equal to t, is equal to:

A

-

3

B

-

2

C 2

D 0

Correct Answer

Option D

Solution

We know,

\left[ {{x \over 2}} \right]

is discontinuous at 1, 2, 3, 4 ........ $\therefore$ [ x ] is discontinuous at 2, 4, 6, 8 .....

In between 0 to 5 it is discontinuous at 2 and 4.

Break the integration into 3 parts (1) 0 to 2 (2) 2 to 4 (3) 4 to 5 $\therefore$

\int\limits_0^5 {\cos \left( {\pi \left( {x - \left[ {{x \over 2}} \right]} \right)} \right)dx}

= \int\limits_0^2 {\cos \left( {\pi (x - 0)} \right)dx + \int\limits_2^4 {\cos \left( {\pi (x - 1)} \right)dx + \int\limits_4^5 {\cos \left( {\pi (x - 2)} \right)dx} } }

= \int\limits_0^2 {\cos \pi x\,dx + \int\limits_2^4 {\cos (\pi x - \pi )dx + \int\limits_4^5 {\cos (\pi x - 2\pi )dx} } }

= \int\limits_0^2 {\cos \pi dx - \int\limits_2^4 {\cos \pi x\,dx + \int\limits_4^5 {\cos \pi x\,dx} } }

= \left[ {{{\sin \pi x} \over \pi }} \right]_0^2 - \left[ {{{\sin \pi x} \over \pi }} \right]_2^4 + \left[ {{{\sin \pi x} \over \pi }} \right]_4^5

= 0 - 0 + 0

= 0

Q103

Let f be a real valued continuous function on [0, 1] and

f(x) = x + \int\limits_0^1 {(x - t)f(t)dt}

. Then, which of the following points (x, y) lies on the curve y = f(x) ?

A (2, 4)

B (1, 2)

C (4, 17)

D (6, 8)

Correct Answer

Option D

Solution

Given,

f(x) = x + \int_0^1 {(x - t)f(t)dt}

= x + x\int_0^1 {f(t)dt - \int_0^1 {tf(t)dt} }

= x\left( {1 + \int_0^1 {f(t)dt} } \right) - \int_0^1 {tf(t)dt}

Now, let

A = 1 + \int_0^1 {f(t)dt}

and

B = \int_0^1 {tf(t)dt}

$\therefore$

f(x) = Ax - B

\Rightarrow f(t) = At - B

So,

A = 1 + \int_0^1 {f(t)dt}

= 1 + \int_0^1 {(At - B)dt}

= 1 + \left[ {{{A{t^2}} \over 2} - Bt} \right]_0^1

= 1 + {A \over 2} - B

\Rightarrow {A \over 2} = 1 - B

...... (1)

B = \int_0^1 {tf(t)dt}

= \int_0^1 {t(At - B)dt}

= \int_0^1 {(A{t^2} - Bt)dt}

= \left[ {{{A{t^3}} \over 3} - {{B{t^2}} \over 2}} \right]_0^1

= {A \over 3} - {B \over 2}

\Rightarrow {{3B} \over 2} = {A \over 3}

....... (2) Solving equation (1) and (2) we get,

A = {{18} \over {13}}

and

B = {4 \over {13}}

$\therefore$

f(x) = {{18} \over {13}}x - {4 \over {13}}

By checking all the options you can see when x = 6 we get

y = f(x) = {{18} \over {13}} \times 6 - {4 \over {13}}

= {{108 - 4} \over {13}} = 8

$\therefore$ Point (6, 8) lies on the curve.

Q104

If

\int\limits_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx = \int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy + \int\limits_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy + I} } }

, then I equals

A

\int\limits_0^1 {\left( {1 + \sqrt {1 - {y^2}} } \right)dy}

B

\int\limits_0^1 {\left( {{{{y^2}} \over 2} - \sqrt {1 - {y^2}} + 1} \right)dy}

C

\int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} } \right)dy}

D

\int\limits_0^1 {\left( {{{{y^2}} \over 2} + \sqrt {1 - {y^2}} + 1} \right)dy}

Correct Answer

Option C

Solution

Given,

\int_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx = \int_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy + \int_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy + I} } }

Now,

L.H.S. = \int_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx}

= \sqrt 2 \int_0^2 {{x^{1/2}}dx - \int_0^2 {\sqrt { - ({x^2} - 2x)} dx} }

= \sqrt 2 \left[ {{{{x^{3/2}}} \over {{3 \over 2}}}} \right]_0^2 - \int_0^2 {\sqrt { - \left[ {{{(x - 1)}^2} - 1} \right]} dx}

= \sqrt 2 \times {2 \over 3}\left[ {{2^{{3 \over 2}}}} \right] - \int_0^2 {\sqrt {1 - {{(x - 1)}^2}} dx}

[We know,

\sqrt {{a^2} - {x^2}} = {x \over 2}\sqrt {{a^2} - {x^2}} + {{{a^2}} \over 2}{\sin ^{ - 1}}\left( {{x \over a}} \right) + C

]

= {{2\sqrt 2 \times 2\sqrt 2 } \over 3} - \left[ {{{x - 1} \over 2}\sqrt {1 - {{(x - 1)}^2}} + {1 \over 2}{{\sin }^{ - 1}}\left( {{{x - 1} \over 1}} \right)} \right]_0^2

= {8 \over 3} - \left[ {{1 \over 2}\sqrt {1 - 1} + {1 \over 2}{{\sin }^{ - 1}}(1) - \left( {\left( {{1 \over 2}} \right)\sqrt {1 - 1} + {1 \over 2}{{\sin }^{ - 1}}( - 1)} \right)} \right]

= {8 \over 3} - {1 \over 2}{\sin ^{ - 1}}\left( {\sin {\pi \over 2}} \right) + {1 \over 2}{\sin ^{ - 1}}\left( {\sin {{3\pi } \over 2}} \right)

= {8 \over 3} - {1 \over 2}\,.\,{\pi \over 2} + {1 \over 2}\,.\,{{3\pi } \over 2}

= {8 \over 3} - {\pi \over 4} + {{3\pi } \over 4}

= {8 \over 3} - {{2\pi } \over 4}

= {8 \over 3} - {\pi \over 2}

Now in R.H.S.,

\int_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy}

= \int_0^1 {dy - \int_0^1 {\sqrt {1 - {y^2}} - \int_0^1 {{{{y^2}} \over 2}dy} } }

= \left[ y \right]_0^1 - \left[ {{y \over 2}\sqrt {1 - {y^2}} + {1 \over 2}{{\sin }^{ - 1}}\left( {{y \over 1}} \right)} \right]_0^1 - \left[ {{{{y^3}} \over 6}} \right]_0^1

= 1 - \left[ {\left( {{1 \over 2}\sqrt {1 - 0} + {1 \over 2}{{\sin }^{ - 1}}(1)} \right) - \left( {0 + {{\sin }^{ - 1}}(0)} \right)} \right] - {1 \over 6}

= 1 - {1 \over 2}{\sin ^{ - 1}}\left( {\sin {\pi \over 2}} \right) - {1 \over 6}

= 1 - {1 \over 2} \times {\pi \over 2} - {1 \over 6}

= {5 \over 6} - {\pi \over 4}

Now,

\int_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy}

= \left[ {2y - {{{y^3}} \over 6}} \right]_1^2

= \left[ {\left( {4 - {8 \over 6}} \right) - \left( {2 - {1 \over 6}} \right)} \right]

= 4 - {8 \over 6} - 2 + {1 \over 6}

= 2 - {7 \over 6}

= {5 \over 6}

$\therefore$

R.H.S. = {5 \over 6} - {\pi \over 4} + {5 \over 6} + I

As

L.H.S. = R.H.S.

\Rightarrow {8 \over 3} - {\pi \over 2} = {{10} \over 6} - {\pi \over 4} + I

\Rightarrow {{10} \over 6} - {8 \over 3} + I = - {\pi \over 2} + {\pi \over 4}

\Rightarrow {{ - 6} \over 6} + I = - {\pi \over 4}

\Rightarrow I = 1 - {\pi \over 4}

From option (c)

\int_0^1 {(1 - \sqrt {1 - {y^2}} )dy}

= \left[ y \right]_0^1 - \int_0^1 {(\sqrt {1 - {y^2}} )dy}

= 1 - \left[ {{y \over 2}\sqrt {1 - {y^2}} + {1 \over 2}{{\sin }^{ - 1}}\left( {{y \over 1}} \right)} \right]_0^1

= 1 - \left[ {{1 \over 2}\sqrt {1 - 1} + {1 \over 2}{{\sin }^{ - 1}}(1) - \left( {0 + {1 \over 2}{{\sin }^{ - 1}}(0)} \right)} \right]

= 1 - \left[ {{1 \over 2} \times {\pi \over 2} - 0} \right]

= 1 - {\pi \over 4} = I

$\therefore$ Option (c) is the right answer.

Note : There is no way to guess which option is correct.

You have to check all the options to see which give value equal to I.

Q105

Let f : R

\to

R be a differentiable function such that

f\left( {{\pi \over 4}} \right) = \sqrt 2 ,\,f\left( {{\pi \over 2}} \right) = 0

and

f'\left( {{\pi \over 2}} \right) = 1

and let

g(x) = \int_x^{\pi /4} {(f'(t)\sec t + \tan t\sec t\,f(t))\,dt}

for

x \in \left[ {{\pi \over 4},{\pi \over 2}} \right)

. Then

\mathop {\lim }\limits_{x \to {{\left( {{\pi \over 2}} \right)}^ - }} g(x)

is equal to :

A 2

B 3

C 4

D

-

3

Correct Answer

Option B

Solution

Given : $f\left(\dfrac{\pi}{4}\right)=\sqrt{2}, f\left(\dfrac{\pi}{2}\right)=0$ and $f^{\prime}\left(\dfrac{\pi}{2}\right)=1$

\begin{aligned} &g(x)=\int_{x}^{\frac{\pi}{4}}\left(f^{\prime}(t) \sec t+\tan t \sec t f(t)\right) d t \\\\ &=[\sec t+f(t)]_{x}^{\frac{\pi}{4}}=2-\sec x f(x) \end{aligned}

Now, $\lim \limits_{x \rightarrow \dfrac{\pi^{-}}{2}} g(x)=\lim \limits_{h \rightarrow 0} g\left(\dfrac{\pi}{2}-h\right)$

=\lim \limits_{h \rightarrow 0} 2-(\operatorname{cosec} h) f\left(\frac{\pi}{2}-h\right)

$=\lim \limits_{h \rightarrow 0}\left[2-\dfrac{f\left(\dfrac{\pi}{2}-h\right)}{\sin h}\right]$ $=\lim \limits_{h \rightarrow 0}\left[2+\dfrac{f^{\prime}\left(\dfrac{\pi}{2}-h\right)}{\cos h}\right]$

=3

Q106

Let f : R

\to

R be a continuous function satisfying f(x) + f(x + k) = n, for all x

\in

R where k > 0 and n is a positive integer. If

{I_1} = \int\limits_0^{4nk} {f(x)dx}

and

{I_2} = \int\limits_{ - k}^{3k} {f(x)dx}

, then :

A

{I_1} + 2{I_2} = 4nk

B

{I_1} + 2{I_2} = 2nk

C

{I_1} + n{I_2} = 4{n^2}k

D

{I_1} + n{I_2} = 6{n^2}k

Correct Answer

Option C

Solution

$f: R \rightarrow R$ and $f(x)+f(x+k)=n \quad \forall x \in R$

\begin{aligned} &x \rightarrow x+k \\\\ &f(x+k)+f(x+2 k)=n \\\\ &\therefore \quad f(x+2 k)=f(x) \end{aligned}

So, period of $f(x)$ is $2 k$

\begin{aligned} &\text{ Now, } I_{1}=\int_{0}^{4 n k} f(x) d x = 2 n \int_{0}^{2 k} f(x) d x \\\\ &=2 n\left[\int_{0}^{k} f(x) d x+\int_{k}^{2 k} f(x) d x\right] \\\\ &x=t+k \Rightarrow d x=d t \text{ (in second integral) } \\\\ &=2 n\left[\int_{0}^{k} f(x) d x+\int_{0}^{k} f(t+k) d t\right] \\\\ &=2 n^{2} k \end{aligned}

Now, $I_2=\int_{-k}^{3 k} f(x) d x=2 \int_{0}^{2 k} f(x) d x$

I_{2}=2(n k)

$\therefore \quad l_{1}+n l_{2}=4 n^{2} k$

Q107

Let [t] denote the greatest integer less than or equal to t. Then, the value of the integral

\int\limits_0^1 {[ - 8{x^2} + 6x - 1]dx}

is equal to :

A

-

1

B

{{ - 5} \over 4}

C

{{\sqrt {17} - 13} \over 8}

D

{{\sqrt {17} - 16} \over 8}

Correct Answer

Option C

Solution

$\int_{0}^{1}\left[-8 x^{2}+6 x-1\right] d x$ $=\int_{0}^{\dfrac{1}{4}}(-1) d x+\int_{\dfrac{1}{4}}^{\dfrac{3}{4}} 0 d x+\int_{\dfrac{1}{2}}^{\dfrac{3}{4}}-1 d x+\int_{\dfrac{3}{4}}^{8}-2 d x+\int_{\dfrac{3+\sqrt{17}}{8}}^{1}-3 d x$ $=-\dfrac{1}{4}-\dfrac{1}{4}-2\left(\dfrac{3+\sqrt{17}}{8}-\dfrac{3}{4}\right)-3\left(1-\dfrac{3+\sqrt{17}}{8}\right)$ $=\dfrac{\sqrt{17}-13}{8}$

Q108

If m and n respectively are the number of local maximum and local minimum points of the function

f(x) = \int\limits_0^{{x^2}} {{{{t^2} - 5t + 4} \over {2 + {e^t}}}dt}

, then the ordered pair (m, n) is equal to

A (3, 2)

B (2, 3)

C (2, 2)

D (3, 4)

Correct Answer

Option B

Solution

f(x) = \int_0^{{x^2}} {{{{t^2} - 5t + 4} \over {2 + {e^t}}}dt}

f'(x) = 2x\left( {{{{x^4} - 5{x^2} + 4} \over {2 + {e^{{x^2}}}}}} \right) = 0

x = 0

, or

({x^2} - 4)({x^2} - 1) = 0

x = 0,

x = \pm 2,\, \pm 1

Now,

f'(x) = {{2x(x + 1)(x - 1)(x + 2)(x - 2)} \over {\left( {{e^{{x^2}}} + 2} \right)}}

f'(x) changes sign from positive to negative at x = $-$ 1, 1 So, number of local maximum points = 2 f'(x) changes sign from negative to positive at x = $-$ 2, 0, 2 So, number of local minimum points = 3 $\therefore$ m = 2, n = 3

Q109

Let f be a differentiable function in

\left( {0,{\pi \over 2}} \right)

. If

\int\limits_{\cos x}^1 {{t^2}\,f(t)dt = {{\sin }^3}x + \cos x}

, then

{1 \over {\sqrt 3 }}f'\left( {{1 \over {\sqrt 3 }}} \right)

is equal to

A

6 - 9\sqrt 2

B

6 - {9 \over {\sqrt 2 }}

C

{9 \over 2} - 6\sqrt 2

D

{9 \over {\sqrt 2 }} - 6

Correct Answer

Option B

Solution

\int\limits_{\cos x}^1 {{t^2}f(t)dt = {{\sin }^3}x + \cos x}

\Rightarrow \sin x{\cos ^2}x\,f(\cos x) = 3{\sin ^2}x\cos x - \sin x

\Rightarrow f(\cos x) = 3\tan x - {\sec ^2}x

\Rightarrow f'(\cos x).\,( - \sin x) = 3{\sec ^2}x - 2{\sec ^2}x\tan x

Put

\cos x = {1 \over {\sqrt 3 }}

, $\therefore$

f'\left( {{1 \over {\sqrt 3 }}} \right)\left( { - {{\sqrt 2 } \over {\sqrt 3 }}} \right) = 9 - 6\sqrt 2

{1 \over {\sqrt 3 }}f'\left( {{1 \over {\sqrt 3 }}} \right) = 6 - {9 \over {\sqrt 2 }}

Q110

The integral

\int\limits_0^1 {{1 \over {{7^{\left[ {{1 \over x}} \right]}}}}dx}

, where [ . ] denotes the greatest integer function, is equal to

A

1 + 6{\log _e}\left( {{6 \over 7}} \right)

B

1 - 6{\log _e}\left( {{6 \over 7}} \right)

C

{\log _e}\left( {{7 \over 6}} \right)

D

1 - 7{\log _e}\left( {{6 \over 7}} \right)

Correct Answer

Option A

Solution

\int_0^1 {{1 \over {{7^{\left[ {{1 \over x}} \right]}}}}dx}

, let

{1 \over x} = t

{{ - 1} \over {{x^2}}}dx = dt

= \int_\infty ^1 {{1 \over { - {t^2}{7^{[t]}}}}dt = \int_1^\infty {{1 \over {{t^2}{7^{[t]}}}}dt} }

= \int_1^2 {{1 \over {7{t^2}}}dt + \int_2^3 {{1 \over {{7^2}{t^2}}}dt + \,\,....} }

= {1 \over 7}\left[ { - {1 \over t}} \right]_1^2 + {1 \over {{7^2}}}\left[ {{{ - 1} \over t}} \right]_2^3 + {1 \over {{7^3}}}\left[ { - {1 \over t}} \right]_2^3 + \,\,....

= \sum\limits_{n = 1}^\infty {{1 \over {{7^n}}}\left( {{1 \over n} - {1 \over {n + 1}}} \right)}

= \sum\limits_{n = 1}^\infty {{{{{\left( {{1 \over 7}} \right)}^n}} \over n} - 7\sum\limits_{n = 1}^\infty {{{{{\left( {{1 \over 7}} \right)}^{n + 1}}} \over {n + 1}}} }

= - \log \left( {1 - {1 \over 7}} \right) + 7\log \left( {1 - {1 \over 7}} \right) + 1

= 1 + 6\log {6 \over 7}