QBManager

Integral Calculus

86375 $\int_{- π / 2}^{π / 2} \log (\frac{2 - \sin x}{2 + \sin x}) d x$

1 1

2 3

3 2

4 0

Explanation:

(D) : Given,
$\int_{- π / 2}^{π / 2} \log (\frac{2 - \sin x}{2 + \sin x}) d x$
Let, $f (x) = \log (\frac{2 - \sin x}{2 + \sin x})$
Then, $f (- x) = \log (\frac{2 - \sin (- x)}{2 + \sin (- x)})$
$= \log (\frac{2 + \sin x}{2 - \sin x}) = - \log (\frac{2 - \sin x}{2 + \sin x})$
$= - f (x)$
$∴ f (x)$ is an odd function.
So, $\int_{- π / 2}^{π / 2} f (x) d x = 0$

MHT CET-2016

Integral Calculus

86437 If $\int^{b} x^{3} d x = 0$ and if $\int^{b} x^{2} d x = \frac{2}{3}$ , then $a$ and $b$ are respectively

1 1,1

2 -1.1

3

1, - 1

4 1

Explanation:

(B) : Given,
$\int_{a}^{b} x^{3} d x = 0$
${[\frac{x^{4}}{4}]}_{a}^{b} = 0$
$\frac{b^{4} - a^{4}}{4} = 0$
$b^{4} - a^{4} = 0$
$b^{4} = a^{4}$
$b = \pm a$
and, $\int_{a}^{b} x^{2} d x = \frac{2}{3}$
${[\frac{x^{3}}{3}]}_{a}^{b} = \frac{2}{3}$
$\frac{b^{3} - a^{3}}{3} = \frac{2}{3}$
$b^{3} - a^{3} = 2$
Let, b = a
Then, $b^{3} - a^{3} = a^{3} - a^{3} = 0 \neq 2$
Now, let $b = - a$
Then, $b^{β} - a^{3}$
$\Rightarrow (- a)^{3} - a^{3} = 2$
$\Rightarrow - 2 a^{3} = 2$
$\Rightarrow a^{3} = - 1$
$a = - 1$
$b = - a = - (- 1) = 1$
So, $(a, b) = (- 1, 1)$

MHT CET-2022

Integral Calculus

86379 $\int_{0}^{π / 2} \sin 2 x \cdot \log \tan x d x$ is equal to

1 0

2 2

3 4

4 7

Explanation:

(A) : Let $I = \int_{0}^{π / 2} \sin 2 x \cdot \log \tan x d x$
$I = \int_{0}^{π / 2} (\log \tan x) \sin 2 x d x$
$I = \int_{0}^{π / 2} \log \tan (\frac{π}{2} - x) \sin 2 (\frac{π}{2} - x) d x$
$[∵ \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x]$
$I = \int_{0}^{π / 2} \log \cot x \cdot \sin 2 x d x$
$[∵ \sin (π - 2 x) = \sin 2 x]$
On adding equation (i) and (ii), we get
$2 I = \int_{0}^{π / 2} (\log \tan x) \sin 2 x d x + \int_{0}^{π / 2} (\log \cot x) \sin 2 x d x$
$2 I = \int_{0}^{π / 2} [(\log \tan x) \sin 2 x + (\log \cot x) \sin 2 x] d x$
$∵ \log m + \log n = \log m . n$
$2 I = \int_{0}^{π / 2} \sin 2 x \log (\tan x \cdot \cot x) d x$
$∵ \tan x \cdot \cot x = 1$
$2 I = \int_{0}^{π / 2} \sin 2 x \log (1) d x$
$2 I = 0 ∵ \log 1 = 0$
$I = 0$

VITEEE-2015

Integral Calculus

86380 If $y + \sqrt{1 + y^{2}} = e^{x}$ , then the value of $y$ is

1

e^{x} - e^{- x}

2

e^{x} + e^{- x}

3

\frac{e^{x} + e^{- x}}{2}

4 None of these

Explanation:

(D) : Given,
$y + \sqrt{1 + y^{2}} = e^{x}$
$\sqrt{1 + y^{2}} = e^{x} - y$
Squaring on both side, we get -
$1 + y^{2} = {(e^{x} - y)}^{2}$
$1 + y^{2} = e^{2 x} + y^{2} - 2 e^{x} y$
$e^{2 x} - 1 = 2 e^{x} y$
$y = \frac{e^{2 x} - 1}{2 e^{x}} = \frac{1}{2} [e^{x} - e^{- x}]$

UPSEE-2018

Integral Calculus

86381 If $\int_{0}^{t^{2}} x f (x) d x = \frac{2}{5} t^{5}, t > 0$ , then $f (\frac{4}{25})$ is

1

\frac{2}{5}

2

\frac{5}{2}

3

- \frac{2}{5}

4 None of these

Explanation:

(A) : Given,
$\int_{0}^{t^{2}} x . f (x) d x = \frac{2}{5} t^{5}; t > 0$
Differentiating both side, we get -
$t^{2} f (t^{2}) (2 t) = 2 t^{4} (By Leibnitz theorem)$
$f (t^{2}) = t$
So, $f (\frac{4}{25}) = \frac{2}{5}$

UPSEE-2012

Integral Calculus

86375 $\int_{- π / 2}^{π / 2} \log (\frac{2 - \sin x}{2 + \sin x}) d x$

1 1

2 3

3 2

4 0

Explanation:

(D) : Given,
$\int_{- π / 2}^{π / 2} \log (\frac{2 - \sin x}{2 + \sin x}) d x$
Let, $f (x) = \log (\frac{2 - \sin x}{2 + \sin x})$
Then, $f (- x) = \log (\frac{2 - \sin (- x)}{2 + \sin (- x)})$
$= \log (\frac{2 + \sin x}{2 - \sin x}) = - \log (\frac{2 - \sin x}{2 + \sin x})$
$= - f (x)$
$∴ f (x)$ is an odd function.
So, $\int_{- π / 2}^{π / 2} f (x) d x = 0$

MHT CET-2016

Integral Calculus

86437 If $\int^{b} x^{3} d x = 0$ and if $\int^{b} x^{2} d x = \frac{2}{3}$ , then $a$ and $b$ are respectively

1 1,1

2 -1.1

3

1, - 1

4 1

Explanation:

(B) : Given,
$\int_{a}^{b} x^{3} d x = 0$
${[\frac{x^{4}}{4}]}_{a}^{b} = 0$
$\frac{b^{4} - a^{4}}{4} = 0$
$b^{4} - a^{4} = 0$
$b^{4} = a^{4}$
$b = \pm a$
and, $\int_{a}^{b} x^{2} d x = \frac{2}{3}$
${[\frac{x^{3}}{3}]}_{a}^{b} = \frac{2}{3}$
$\frac{b^{3} - a^{3}}{3} = \frac{2}{3}$
$b^{3} - a^{3} = 2$
Let, b = a
Then, $b^{3} - a^{3} = a^{3} - a^{3} = 0 \neq 2$
Now, let $b = - a$
Then, $b^{β} - a^{3}$
$\Rightarrow (- a)^{3} - a^{3} = 2$
$\Rightarrow - 2 a^{3} = 2$
$\Rightarrow a^{3} = - 1$
$a = - 1$
$b = - a = - (- 1) = 1$
So, $(a, b) = (- 1, 1)$

MHT CET-2022

Integral Calculus

86379 $\int_{0}^{π / 2} \sin 2 x \cdot \log \tan x d x$ is equal to

1 0

2 2

3 4

4 7

Explanation:

(A) : Let $I = \int_{0}^{π / 2} \sin 2 x \cdot \log \tan x d x$
$I = \int_{0}^{π / 2} (\log \tan x) \sin 2 x d x$
$I = \int_{0}^{π / 2} \log \tan (\frac{π}{2} - x) \sin 2 (\frac{π}{2} - x) d x$
$[∵ \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x]$
$I = \int_{0}^{π / 2} \log \cot x \cdot \sin 2 x d x$
$[∵ \sin (π - 2 x) = \sin 2 x]$
On adding equation (i) and (ii), we get
$2 I = \int_{0}^{π / 2} (\log \tan x) \sin 2 x d x + \int_{0}^{π / 2} (\log \cot x) \sin 2 x d x$
$2 I = \int_{0}^{π / 2} [(\log \tan x) \sin 2 x + (\log \cot x) \sin 2 x] d x$
$∵ \log m + \log n = \log m . n$
$2 I = \int_{0}^{π / 2} \sin 2 x \log (\tan x \cdot \cot x) d x$
$∵ \tan x \cdot \cot x = 1$
$2 I = \int_{0}^{π / 2} \sin 2 x \log (1) d x$
$2 I = 0 ∵ \log 1 = 0$
$I = 0$

VITEEE-2015

Integral Calculus

86380 If $y + \sqrt{1 + y^{2}} = e^{x}$ , then the value of $y$ is

1

e^{x} - e^{- x}

2

e^{x} + e^{- x}

3

\frac{e^{x} + e^{- x}}{2}

4 None of these

Explanation:

(D) : Given,
$y + \sqrt{1 + y^{2}} = e^{x}$
$\sqrt{1 + y^{2}} = e^{x} - y$
Squaring on both side, we get -
$1 + y^{2} = {(e^{x} - y)}^{2}$
$1 + y^{2} = e^{2 x} + y^{2} - 2 e^{x} y$
$e^{2 x} - 1 = 2 e^{x} y$
$y = \frac{e^{2 x} - 1}{2 e^{x}} = \frac{1}{2} [e^{x} - e^{- x}]$

UPSEE-2018

Integral Calculus

86381 If $\int_{0}^{t^{2}} x f (x) d x = \frac{2}{5} t^{5}, t > 0$ , then $f (\frac{4}{25})$ is

1

\frac{2}{5}

2

\frac{5}{2}

3

- \frac{2}{5}

4 None of these

Explanation:

(A) : Given,
$\int_{0}^{t^{2}} x . f (x) d x = \frac{2}{5} t^{5}; t > 0$
Differentiating both side, we get -
$t^{2} f (t^{2}) (2 t) = 2 t^{4} (By Leibnitz theorem)$
$f (t^{2}) = t$
So, $f (\frac{4}{25}) = \frac{2}{5}$

UPSEE-2012

Integral Calculus

86375 $\int_{- π / 2}^{π / 2} \log (\frac{2 - \sin x}{2 + \sin x}) d x$

1 1

2 3

3 2

4 0

Explanation:

(D) : Given,
$\int_{- π / 2}^{π / 2} \log (\frac{2 - \sin x}{2 + \sin x}) d x$
Let, $f (x) = \log (\frac{2 - \sin x}{2 + \sin x})$
Then, $f (- x) = \log (\frac{2 - \sin (- x)}{2 + \sin (- x)})$
$= \log (\frac{2 + \sin x}{2 - \sin x}) = - \log (\frac{2 - \sin x}{2 + \sin x})$
$= - f (x)$
$∴ f (x)$ is an odd function.
So, $\int_{- π / 2}^{π / 2} f (x) d x = 0$

MHT CET-2016

Integral Calculus

86437 If $\int^{b} x^{3} d x = 0$ and if $\int^{b} x^{2} d x = \frac{2}{3}$ , then $a$ and $b$ are respectively

1 1,1

2 -1.1

3

1, - 1

4 1

Explanation:

(B) : Given,
$\int_{a}^{b} x^{3} d x = 0$
${[\frac{x^{4}}{4}]}_{a}^{b} = 0$
$\frac{b^{4} - a^{4}}{4} = 0$
$b^{4} - a^{4} = 0$
$b^{4} = a^{4}$
$b = \pm a$
and, $\int_{a}^{b} x^{2} d x = \frac{2}{3}$
${[\frac{x^{3}}{3}]}_{a}^{b} = \frac{2}{3}$
$\frac{b^{3} - a^{3}}{3} = \frac{2}{3}$
$b^{3} - a^{3} = 2$
Let, b = a
Then, $b^{3} - a^{3} = a^{3} - a^{3} = 0 \neq 2$
Now, let $b = - a$
Then, $b^{β} - a^{3}$
$\Rightarrow (- a)^{3} - a^{3} = 2$
$\Rightarrow - 2 a^{3} = 2$
$\Rightarrow a^{3} = - 1$
$a = - 1$
$b = - a = - (- 1) = 1$
So, $(a, b) = (- 1, 1)$

MHT CET-2022

Integral Calculus

86379 $\int_{0}^{π / 2} \sin 2 x \cdot \log \tan x d x$ is equal to

1 0

2 2

3 4

4 7

Explanation:

(A) : Let $I = \int_{0}^{π / 2} \sin 2 x \cdot \log \tan x d x$
$I = \int_{0}^{π / 2} (\log \tan x) \sin 2 x d x$
$I = \int_{0}^{π / 2} \log \tan (\frac{π}{2} - x) \sin 2 (\frac{π}{2} - x) d x$
$[∵ \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x]$
$I = \int_{0}^{π / 2} \log \cot x \cdot \sin 2 x d x$
$[∵ \sin (π - 2 x) = \sin 2 x]$
On adding equation (i) and (ii), we get
$2 I = \int_{0}^{π / 2} (\log \tan x) \sin 2 x d x + \int_{0}^{π / 2} (\log \cot x) \sin 2 x d x$
$2 I = \int_{0}^{π / 2} [(\log \tan x) \sin 2 x + (\log \cot x) \sin 2 x] d x$
$∵ \log m + \log n = \log m . n$
$2 I = \int_{0}^{π / 2} \sin 2 x \log (\tan x \cdot \cot x) d x$
$∵ \tan x \cdot \cot x = 1$
$2 I = \int_{0}^{π / 2} \sin 2 x \log (1) d x$
$2 I = 0 ∵ \log 1 = 0$
$I = 0$

VITEEE-2015

Integral Calculus

86380 If $y + \sqrt{1 + y^{2}} = e^{x}$ , then the value of $y$ is

1

e^{x} - e^{- x}

2

e^{x} + e^{- x}

3

\frac{e^{x} + e^{- x}}{2}

4 None of these

Explanation:

(D) : Given,
$y + \sqrt{1 + y^{2}} = e^{x}$
$\sqrt{1 + y^{2}} = e^{x} - y$
Squaring on both side, we get -
$1 + y^{2} = {(e^{x} - y)}^{2}$
$1 + y^{2} = e^{2 x} + y^{2} - 2 e^{x} y$
$e^{2 x} - 1 = 2 e^{x} y$
$y = \frac{e^{2 x} - 1}{2 e^{x}} = \frac{1}{2} [e^{x} - e^{- x}]$

UPSEE-2018

Integral Calculus

86381 If $\int_{0}^{t^{2}} x f (x) d x = \frac{2}{5} t^{5}, t > 0$ , then $f (\frac{4}{25})$ is

1

\frac{2}{5}

2

\frac{5}{2}

3

- \frac{2}{5}

4 None of these

Explanation:

(A) : Given,
$\int_{0}^{t^{2}} x . f (x) d x = \frac{2}{5} t^{5}; t > 0$
Differentiating both side, we get -
$t^{2} f (t^{2}) (2 t) = 2 t^{4} (By Leibnitz theorem)$
$f (t^{2}) = t$
So, $f (\frac{4}{25}) = \frac{2}{5}$

UPSEE-2012

Integral Calculus

86375 $\int_{- π / 2}^{π / 2} \log (\frac{2 - \sin x}{2 + \sin x}) d x$

1 1

2 3

3 2

4 0

Explanation:

(D) : Given,
$\int_{- π / 2}^{π / 2} \log (\frac{2 - \sin x}{2 + \sin x}) d x$
Let, $f (x) = \log (\frac{2 - \sin x}{2 + \sin x})$
Then, $f (- x) = \log (\frac{2 - \sin (- x)}{2 + \sin (- x)})$
$= \log (\frac{2 + \sin x}{2 - \sin x}) = - \log (\frac{2 - \sin x}{2 + \sin x})$
$= - f (x)$
$∴ f (x)$ is an odd function.
So, $\int_{- π / 2}^{π / 2} f (x) d x = 0$

MHT CET-2016

Integral Calculus

86437 If $\int^{b} x^{3} d x = 0$ and if $\int^{b} x^{2} d x = \frac{2}{3}$ , then $a$ and $b$ are respectively

1 1,1

2 -1.1

3

1, - 1

4 1

Explanation:

(B) : Given,
$\int_{a}^{b} x^{3} d x = 0$
${[\frac{x^{4}}{4}]}_{a}^{b} = 0$
$\frac{b^{4} - a^{4}}{4} = 0$
$b^{4} - a^{4} = 0$
$b^{4} = a^{4}$
$b = \pm a$
and, $\int_{a}^{b} x^{2} d x = \frac{2}{3}$
${[\frac{x^{3}}{3}]}_{a}^{b} = \frac{2}{3}$
$\frac{b^{3} - a^{3}}{3} = \frac{2}{3}$
$b^{3} - a^{3} = 2$
Let, b = a
Then, $b^{3} - a^{3} = a^{3} - a^{3} = 0 \neq 2$
Now, let $b = - a$
Then, $b^{β} - a^{3}$
$\Rightarrow (- a)^{3} - a^{3} = 2$
$\Rightarrow - 2 a^{3} = 2$
$\Rightarrow a^{3} = - 1$
$a = - 1$
$b = - a = - (- 1) = 1$
So, $(a, b) = (- 1, 1)$

MHT CET-2022

Integral Calculus

86379 $\int_{0}^{π / 2} \sin 2 x \cdot \log \tan x d x$ is equal to

1 0

2 2

3 4

4 7

Explanation:

(A) : Let $I = \int_{0}^{π / 2} \sin 2 x \cdot \log \tan x d x$
$I = \int_{0}^{π / 2} (\log \tan x) \sin 2 x d x$
$I = \int_{0}^{π / 2} \log \tan (\frac{π}{2} - x) \sin 2 (\frac{π}{2} - x) d x$
$[∵ \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x]$
$I = \int_{0}^{π / 2} \log \cot x \cdot \sin 2 x d x$
$[∵ \sin (π - 2 x) = \sin 2 x]$
On adding equation (i) and (ii), we get
$2 I = \int_{0}^{π / 2} (\log \tan x) \sin 2 x d x + \int_{0}^{π / 2} (\log \cot x) \sin 2 x d x$
$2 I = \int_{0}^{π / 2} [(\log \tan x) \sin 2 x + (\log \cot x) \sin 2 x] d x$
$∵ \log m + \log n = \log m . n$
$2 I = \int_{0}^{π / 2} \sin 2 x \log (\tan x \cdot \cot x) d x$
$∵ \tan x \cdot \cot x = 1$
$2 I = \int_{0}^{π / 2} \sin 2 x \log (1) d x$
$2 I = 0 ∵ \log 1 = 0$
$I = 0$

VITEEE-2015

Integral Calculus

86380 If $y + \sqrt{1 + y^{2}} = e^{x}$ , then the value of $y$ is

1

e^{x} - e^{- x}

2

e^{x} + e^{- x}

3

\frac{e^{x} + e^{- x}}{2}

4 None of these

Explanation:

(D) : Given,
$y + \sqrt{1 + y^{2}} = e^{x}$
$\sqrt{1 + y^{2}} = e^{x} - y$
Squaring on both side, we get -
$1 + y^{2} = {(e^{x} - y)}^{2}$
$1 + y^{2} = e^{2 x} + y^{2} - 2 e^{x} y$
$e^{2 x} - 1 = 2 e^{x} y$
$y = \frac{e^{2 x} - 1}{2 e^{x}} = \frac{1}{2} [e^{x} - e^{- x}]$

UPSEE-2018

Integral Calculus

86381 If $\int_{0}^{t^{2}} x f (x) d x = \frac{2}{5} t^{5}, t > 0$ , then $f (\frac{4}{25})$ is

1

\frac{2}{5}

2

\frac{5}{2}

3

- \frac{2}{5}

4 None of these

Explanation:

(A) : Given,
$\int_{0}^{t^{2}} x . f (x) d x = \frac{2}{5} t^{5}; t > 0$
Differentiating both side, we get -
$t^{2} f (t^{2}) (2 t) = 2 t^{4} (By Leibnitz theorem)$
$f (t^{2}) = t$
So, $f (\frac{4}{25}) = \frac{2}{5}$

UPSEE-2012

Integral Calculus

86375 $\int_{- π / 2}^{π / 2} \log (\frac{2 - \sin x}{2 + \sin x}) d x$

1 1

2 3

3 2

4 0

Explanation:

(D) : Given,
$\int_{- π / 2}^{π / 2} \log (\frac{2 - \sin x}{2 + \sin x}) d x$
Let, $f (x) = \log (\frac{2 - \sin x}{2 + \sin x})$
Then, $f (- x) = \log (\frac{2 - \sin (- x)}{2 + \sin (- x)})$
$= \log (\frac{2 + \sin x}{2 - \sin x}) = - \log (\frac{2 - \sin x}{2 + \sin x})$
$= - f (x)$
$∴ f (x)$ is an odd function.
So, $\int_{- π / 2}^{π / 2} f (x) d x = 0$

MHT CET-2016

Integral Calculus

86437 If $\int^{b} x^{3} d x = 0$ and if $\int^{b} x^{2} d x = \frac{2}{3}$ , then $a$ and $b$ are respectively

1 1,1

2 -1.1

3

1, - 1

4 1

Explanation:

(B) : Given,
$\int_{a}^{b} x^{3} d x = 0$
${[\frac{x^{4}}{4}]}_{a}^{b} = 0$
$\frac{b^{4} - a^{4}}{4} = 0$
$b^{4} - a^{4} = 0$
$b^{4} = a^{4}$
$b = \pm a$
and, $\int_{a}^{b} x^{2} d x = \frac{2}{3}$
${[\frac{x^{3}}{3}]}_{a}^{b} = \frac{2}{3}$
$\frac{b^{3} - a^{3}}{3} = \frac{2}{3}$
$b^{3} - a^{3} = 2$
Let, b = a
Then, $b^{3} - a^{3} = a^{3} - a^{3} = 0 \neq 2$
Now, let $b = - a$
Then, $b^{β} - a^{3}$
$\Rightarrow (- a)^{3} - a^{3} = 2$
$\Rightarrow - 2 a^{3} = 2$
$\Rightarrow a^{3} = - 1$
$a = - 1$
$b = - a = - (- 1) = 1$
So, $(a, b) = (- 1, 1)$

MHT CET-2022

Integral Calculus

86379 $\int_{0}^{π / 2} \sin 2 x \cdot \log \tan x d x$ is equal to

1 0

2 2

3 4

4 7

Explanation:

(A) : Let $I = \int_{0}^{π / 2} \sin 2 x \cdot \log \tan x d x$
$I = \int_{0}^{π / 2} (\log \tan x) \sin 2 x d x$
$I = \int_{0}^{π / 2} \log \tan (\frac{π}{2} - x) \sin 2 (\frac{π}{2} - x) d x$
$[∵ \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x]$
$I = \int_{0}^{π / 2} \log \cot x \cdot \sin 2 x d x$
$[∵ \sin (π - 2 x) = \sin 2 x]$
On adding equation (i) and (ii), we get
$2 I = \int_{0}^{π / 2} (\log \tan x) \sin 2 x d x + \int_{0}^{π / 2} (\log \cot x) \sin 2 x d x$
$2 I = \int_{0}^{π / 2} [(\log \tan x) \sin 2 x + (\log \cot x) \sin 2 x] d x$
$∵ \log m + \log n = \log m . n$
$2 I = \int_{0}^{π / 2} \sin 2 x \log (\tan x \cdot \cot x) d x$
$∵ \tan x \cdot \cot x = 1$
$2 I = \int_{0}^{π / 2} \sin 2 x \log (1) d x$
$2 I = 0 ∵ \log 1 = 0$
$I = 0$

VITEEE-2015

Integral Calculus

86380 If $y + \sqrt{1 + y^{2}} = e^{x}$ , then the value of $y$ is

1

e^{x} - e^{- x}

2

e^{x} + e^{- x}

3

\frac{e^{x} + e^{- x}}{2}

4 None of these

Explanation:

(D) : Given,
$y + \sqrt{1 + y^{2}} = e^{x}$
$\sqrt{1 + y^{2}} = e^{x} - y$
Squaring on both side, we get -
$1 + y^{2} = {(e^{x} - y)}^{2}$
$1 + y^{2} = e^{2 x} + y^{2} - 2 e^{x} y$
$e^{2 x} - 1 = 2 e^{x} y$
$y = \frac{e^{2 x} - 1}{2 e^{x}} = \frac{1}{2} [e^{x} - e^{- x}]$

UPSEE-2018

Integral Calculus

86381 If $\int_{0}^{t^{2}} x f (x) d x = \frac{2}{5} t^{5}, t > 0$ , then $f (\frac{4}{25})$ is

1

\frac{2}{5}

2

\frac{5}{2}

3

- \frac{2}{5}

4 None of these

Explanation:

(A) : Given,
$\int_{0}^{t^{2}} x . f (x) d x = \frac{2}{5} t^{5}; t > 0$
Differentiating both side, we get -
$t^{2} f (t^{2}) (2 t) = 2 t^{4} (By Leibnitz theorem)$
$f (t^{2}) = t$
So, $f (\frac{4}{25}) = \frac{2}{5}$

UPSEE-2012