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1

JEE Main 2019 (Online) 11th January Evening Slot

MCQ (Single Correct Answer)
If   $$\int {{{x + 1} \over {\sqrt {2x - 1} }}} \,dx$$ = f(x) $$\sqrt {2x - 1} $$ + C, where C is a constant of integration, then f(x) is equal to :
A
$${2 \over 3}$$ (x $$-$$ 4)
B
$${1 \over 3}$$ (x + 4)
C
$${1 \over 3}$$ (x + 1)
D
$${2 \over 3}$$ (x + 2)

Explanation

$$\sqrt {2x - 1} = t \Rightarrow 2x - 1 = {t^2} \Rightarrow 2dx = 2t.dt$$

$$\int {{{x + 1} \over {\sqrt {2x - 1} }}dx = \int {{{{{{t^2} + 1} \over 2} + 1} \over t}tdt = \int {{{{t^2} + 3} \over 2}dt} } } $$

$$ = {1 \over 2}\left( {{{{t^3}} \over 3} + 3t} \right) = {t \over 6}\left( {{t^2} + 9} \right) + c$$

$$ = \sqrt {2x - 1} \left( {{{2x - 1 + 9} \over 6}} \right) + c = \sqrt {2x - 1} \left( {{{x + 4} \over 3}} \right) + c$$

$$ \Rightarrow f\left( x \right) = {{x + 4} \over 3}$$
2

JEE Main 2019 (Online) 10th January Evening Slot

MCQ (Single Correct Answer)
If  $$\int \, $$x5.e$$-$$4x3 dx = $${1 \over {48}}$$e$$-$$4x3 f(x) + C, where C is a constant of inegration, then f(x) is equal to -
A
$$-$$2x3 $$-$$ 1
B
$$-$$ 2x3 + 1
C
4x3 + 1
D
$$-$$4x3 $$-$$ 1

Explanation

$$\int {{x^5}} .{e^{ - 4{x^3}}}\,dx = {1 \over {48}}{e^{ - 4{x^3}}}f\left( x \right) + c$$

Put  $${x^3} = t$$

$$3{x^2}\,dx = dt$$

$$\int {{x^3}.{e^{ - 4{x^3}}}.\,{x^2}} dx$$

$${1 \over 3}\int {t.{e^{ - 4t}}dt} $$

$${1 \over 3}\left[ {t.{{{e^{ - 4t}}} \over { - 4}} - \int {{{{e^{ - 4t}}} \over { - 4}}dt} } \right]$$

$$ - {{{e^{ - 4t}}} \over {48}}\left[ {4t + 1} \right] + c$$

$${{ - {e^{ - 4{x^3}}}} \over {48}}\left[ {4{x^3} + 1} \right] + c$$

$$ \therefore $$  $$f(x) = - 1 - 4{x^3}$$
3

JEE Main 2019 (Online) 10th January Morning Slot

MCQ (Single Correct Answer)
Let n $$ \ge $$ 2 be a natural number and $$0 < \theta < {\pi \over 2}.$$ Then $$\int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta $$ is equal to - (where C is a constant of integration)
A
$${n \over {{n^2} - 1}}{\left( {1 + {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C$$
B
$${n \over {{n^2} - 1}}{\left( {1 - {1 \over {{{\sin }^{n + 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C$$
C
$${n \over {{n^2} - 1}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C$$
D
$${n \over {{n^2} + 1}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C$$

Explanation

$$\int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta $$

$$ = \int {{{\sin \theta {{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)}^{1/n}}} \over {{{\sin }^{n + 1}}\theta }}} \,d\theta $$

Put $$1 - {1 \over {{{\sin }^{n - 1}}\theta }} = t$$

So $${{\left( {n - 1} \right)} \over {{{\sin }^n}\theta }}\cos \theta d\theta = dt$$

Now  $${1 \over {n - 1}}\int {{{\left( t \right)}^{{1 \over n} + 1}}dt} $$

$$ = {1 \over {\left( {n - 1} \right)}}{{{{\left( t \right)}^{{1 \over n} + 1}}} \over {{1 \over n} + 1}} + C$$

$$ = {n \over {\left( {n^2 - 1} \right)}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{1 \over n} + 1}} + C$$
4

JEE Main 2019 (Online) 9th January Evening Slot

MCQ (Single Correct Answer)
English
Hindi
If   $$f\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{{\left( {{x^2} + 1 + 2{x^7}} \right)}^2}}}} \,dx,\,\left( {x \ge 0} \right),$$

$$f\left( 0 \right) = 0,$$    then the value of $$f(1)$$ is :
A
$$ - $$ $${1 \over 2}$$
B
$$ - $$ $${1 \over 4}$$
C
$${1 \over 2}$$
D
$${1 \over 4}$$

Explanation

$$f\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{{\left( {{x^2} + 1 + 2{x^7}} \right)}^2}}}} \,dx$$

$$f\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{x^{14}}{{\left( {{x^{ - 5}} + {x^{ - 7}} + 2} \right)}^2}}}} \,dx$$

$$f\left( x \right) = \int {{{5{x^{ - 6}} + 7{x^{ - 8}}} \over {{{\left( {{x^{ - 5}} + {x^{ - 7}} + 2} \right)}^2}}}} \,dx$$

Let $${x^{ - 5}} + {x^{ - 7}} + 2 = t$$

$$\left( { - 5{x^{ - 6}} - 7{x^{ - 8}}} \right)dx = dt$$

$$\left( {5{x^{ - 6}} + 7{x^{ - 8}}} \right)dx = - dt$$

$$f(x) = \int {{{ - dt} \over {{t^2}}}} = {1 \over t} + c$$

$$f\left( x \right) = {1 \over {{x^{ - 5}} + {x^{ - 7}} + 2}} + c$$

$$f\left( x \right) = {{{x^7}} \over {{x^2} + 1 + 2{x^7}}} + c$$

$$f\left( 0 \right) = 0$$   

$$ \therefore $$ $$c = 0$$

$$f\left( x \right) = {{{x^7}} \over {\left( {{x^2} + 1 + 2{x^7}} \right)}}$$

$$f(1) = {1 \over {1 + 1 + 2}} = {1 \over 4}$$

यदि $$f(x)=\int \frac{5 x^{8}+7 x^{6}}{\left(x^{2}+1+2 x^{7}\right)^{2}} d x,(x \geq 0)$$, तथा $$f(0)=0$$ है, तो $$f(1)$$ का मान है-

A
$$-\frac{1}{2}$$
B
$$-\frac{1}{4}$$
C
$$\frac{1}{2}$$
D
$$\frac{1}{4}$$

Questions Asked from Indefinite Integrals

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