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1

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

Let $$\int {{{{x^{{1 \over 2}}}} \over {\sqrt {1 - {x^3}} }}dx = {2 \over 3}g(f(x)) + c} $$ ; then

(c denotes constant of integration)

A
$$f(x) = \sqrt x ,g(x) = {x^{{3 \over 2}}}$$
B
$$f(x) = {x^{{3 \over 2}}},g(x) = {\sin ^{ - 1}}x$$
C
$$f(x) = \sqrt x ,g(x) = {\sin ^{ - 1}}x$$
D
$$f(x) = {\sin ^{ - 1}}x,g(x) = {x^{{3 \over 2}}}$$

Explanation

$$I = \int {{{{x^{1/2}}} \over {\sqrt {1 - {x^3}} }}dx} $$

Let $${x^{3/2}} = t$$

$$ \Rightarrow {3 \over 2}{x^{1/2}}dx = dt$$

$$\therefore$$ $$I = \int {{2 \over 3}\,.\,{{dt} \over {\sqrt {1 - {t^2}} }}} $$

$$ = {2 \over 3}{\sin ^{ - 1}}t + C$$

$$ = {2 \over 3}{\sin ^{ - 1}}({x^{3/2}}) + C$$

$$ = {2 \over 3}g(f(x)) + C$$

$$\therefore$$ $$f(x) = {x^{3/2}}$$ and $$g(x) = {\sin ^{ - 1}}(x)$$

মনে কর $$\int {{{{x^{{1 \over 2}}}} \over {\sqrt {1 - {x^3}} }}dx = {2 \over 3}g(f(x)) + c} $$ । সেক্ষেত্রে

(c সমাকলনের যদৃচ্ছ ধ্রুবক বুঝায়)

A
$$f(x) = \sqrt x ,g(x) = {x^{{3 \over 2}}}$$
B
$$f(x) = {x^{{3 \over 2}}},g(x) = {\sin ^{ - 1}}x$$
C
$$f(x) = \sqrt x ,g(x) = {\sin ^{ - 1}}x$$
D
$$f(x) = {\sin ^{ - 1}}x,g(x) = {x^{{3 \over 2}}}$$
2

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

$$I = \int {\cos (\ln x)dx} $$. Then I =

A
$${x \over 2}\{ \cos (\ln x) + \sin (\ln x)\} + c$$ (c denotes constant of integration)
B
$${x^2}\{ \cos (\ln x) - \sin (\ln x)\} + c$$ (c denotes constant of integration)
C
$${x^2}\sin (\ln x) + c$$ (c denotes constant of integration)
D
$$x\cos (\ln x) + c$$ (c denotes constant of integration)

Explanation

$$I = \int {\cos (\ln x)dx} $$

Let, $$\ln x = t$$

$$ \Rightarrow x = {e^t}$$

$$ \Rightarrow dx = {e^t}dt$$

$$\therefore$$ $$I = \int {\cos (t)\,.\,{e^t}dt} $$

Apply integration by parts theorem,

$$ = \cos t\,.\,\int {{e^t}dt - \int {\left( {{d \over {dt}}(\cos t)\,.\,\int {{e^t}dt} } \right)dt} } $$

$$ = \cos t\,.\,{e^t} + \int {\sin t\,.\,{e^t}dt} $$

$$I = \cos t\,.\,{e^t} + \sin t\,.\,{e^t} - \int {(\cos t\,.\,{e^t})\,dt} $$

$$ \Rightarrow I = \cos t\,.\,{e^t} + \sin t\,.\,{e^t} - I$$

$$ \Rightarrow 2I = \cos t\,.\,{e^t} + \sin t\,.\,{e^t} + C$$

$$ \Rightarrow 2I = {e^t}(\cos t + \sin t) + C$$

$$ \Rightarrow 2I = x\left( {\cos (\ln x) + \sin (\ln x)} \right) + C$$

$$ \Rightarrow I = {x \over 2}\left[ {\cos (\ln x) + \sin (\ln x)} \right] + C$$

$$I = \int {\cos (\ln x)dx} $$, সেক্ষেত্রে I =

A
$${x \over 2}\{ \cos (\ln x) + \sin (\ln x)\} + c$$ (c সমাকলনের যদৃচ্ছ ধ্রুবক বুঝায়)
B
$${x^2}\{ \cos (\ln x) - \sin (\ln x)\} + c$$ (c সমাকলনের যদৃচ্ছ ধ্রুবক বুঝায়)
C
$${x^2}\sin (\ln x) + c$$ (c সমাকলনের যদৃচ্ছ ধ্রুবক বুঝায়)
D
$$x\cos (\ln x) + c$$ (c সমাকলনের যদৃচ্ছ ধ্রুবক বুঝায়)
3

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
If $$\int {{{\sin 2x} \over {{{(a + b\cos x)}^2}}}dx} = \alpha \left[ {{{\log }_e}\left| {a + b\cos x} \right| + {a \over {a + b\cos x}}} \right] + c$$, then $$\alpha$$ is equal to
A
$${2 \over {{b^2}}}$$
B
$${2 \over {{a^2}}}$$
C
$$ - {2 \over {{b^2}}}$$
D
$$ - {2 \over {{a^2}}}$$

Explanation

Let $$I = \int {{{\sin 2x} \over {{{(a + b\cos x)}^2}}}dx} $$

$$ \Rightarrow I = \int {{{2\sin x\cos x} \over {{{(a + b\cos x)}^2}}}dx} $$

Put $$a + b\cos x = t$$

$$ \Rightarrow \cos x = {{t - a} \over b} \Rightarrow - b\sin xdx = dt$$

$$I = - {2 \over b}\int {\left( {{{t - a} \over b}} \right){1 \over {{t^2}}}dt} $$

$$ = - {2 \over {{b^2}}}\int {{{t - a} \over {{t^2}}}dt} $$

$$ = - {2 \over {{b^2}}}\left[ {\int {{1 \over t}dt - \int {{a \over {{t^2}}}dt} } } \right]$$

$$ = - {2 \over {{b^2}}}\int {\left[ {\log (t) - a{{{t^{ - 2 + 1}}} \over { - 2 + 1}}} \right] + C} $$

$$ = - {2 \over {{b^2}}}\left[ {\log \left| t \right| + {a \over t}} \right] + C$$

$$\therefore$$ $$I = - {2 \over {{b^2}}}\left[ {\log \left| {a + b\cos x} \right| + {a \over {a + b\cos x}}} \right] + C$$

$$ \Rightarrow \alpha = - {2 \over {{b^2}}}$$
যদি $$\int {{{\sin 2x} \over {{{(a + b\cos x)}^2}}}dx} = \alpha \left[ {{{\log }_e}\left| {a + b\cos x} \right| + {a \over {a + b\cos x}}} \right] + c$$ হয়, তবে $$\alpha$$ = ?
A
$${2 \over {{b^2}}}$$
B
$${2 \over {{a^2}}}$$
C
$$ - {2 \over {{b^2}}}$$
D
$$ - {2 \over {{a^2}}}$$

Explanation

ধরা যাক $$I = \int {{{\sin 2x} \over {{{(a + b\cos x)}^2}}}dx} $$

$$ \Rightarrow I = \int {{{2\sin x\cos x} \over {{{(a + b\cos x)}^2}}}dx} $$

বসিয়ে $$a + b\cos x = t$$

$$ \Rightarrow \cos x = {{t - a} \over b} \Rightarrow - b\sin xdx = dt$$

$$I = - {2 \over b}\int {\left( {{{t - a} \over b}} \right){1 \over {{t^2}}}dt} $$

$$ = - {2 \over {{b^2}}}\int {{{t - a} \over {{t^2}}}dt} $$

$$ = - {2 \over {{b^2}}}\left[ {\int {{1 \over t}dt - \int {{a \over {{t^2}}}dt} } } \right]$$

$$ = - {2 \over {{b^2}}}\int {\left[ {\log (t) - a{{{t^{ - 2 + 1}}} \over { - 2 + 1}}} \right] + C} $$

$$ = - {2 \over {{b^2}}}\left[ {\log \left| t \right| + {a \over t}} \right] + C$$

$$\therefore$$ $$I = - {2 \over {{b^2}}}\left[ {\log \left| {a + b\cos x} \right| + {a \over {a + b\cos x}}} \right] + C$$

$$ \Rightarrow \alpha = - {2 \over {{b^2}}}$$
4

WB JEE 2020

MCQ (Single Correct Answer)
English
Bengali
$$\int {{{f(x)\phi '(x) + \phi (x)f'(x)} \over {(f(x)\phi (x) + 1)\sqrt {f(x)\phi (x) - 1} }}dx = } $$
A
$${\sin ^{ - 1}} = \sqrt {{{f(x)} \over {\phi (x)}}} + c$$
B
$${\cos ^{ - 1}}\sqrt {{{(f(x))}^2} - {{(\phi (x))}^2}} + c$$
C
$$\sqrt 2 {\tan ^{ - 1}}\sqrt {{{f(x)\phi (x) - 1} \over 2}} + c$$
D
$$\sqrt 2 {\tan ^{ - 1}}\sqrt {{{f(x)\phi (x) + 1} \over 2}} + c$$

Explanation

$$I = \int {{{f(x)\phi '(x) + \phi (x)\,f'(x)} \over {(f(x)\phi (x) + 1)\sqrt {f(x)\phi (x) - 1} }}dx} $$

Put f(x)$$\phi $$(x) $$ - $$ 1 = t2

$$ \Rightarrow $$ (f(x)$$\phi $$'(x) + $$\phi $$(x) f'(x)) dx = 2t dt

$$ \therefore $$ $$I = \int {{{2tdt} \over {({t^2} + 2)t}}} $$

$$ \Rightarrow I = 2\int {{{dt} \over {{t^2} + 2}}} $$

= $${2 \over {\sqrt 2 }}{\tan ^{ - 1}}\left( {{t \over {\sqrt 2 }}} \right) + c$$

$$ \Rightarrow I = \sqrt 2 {\tan ^{ - 1}}\sqrt {{{f(x)\phi (x) - 1} \over 2}} + c$$

$$\int {{{f(x)\varphi '(x) + \varphi (x)f'(x)} \over {(f(x)\varphi (x) + 1)\sqrt {f(x)\varphi (x) - 1} }}dx = } $$

A
$${\sin ^{ - 1}}\sqrt {{{f(x)} \over {\varphi (x)}}} + c$$
B
$${\cos ^{ - 1}}\sqrt {{{(f(x))}^2} - {{(\varphi (x))}^2}} + c$$
C
$$\sqrt 2 {\tan ^{ - 1}}\sqrt {{{f(x)\varphi (x) - 1} \over 2}} + c$$
D
$$\sqrt 2 {\tan ^{ - 1}}\sqrt {{{f(x)\varphi (x) + 1} \over 2}} + c$$

Explanation

$$I = \int {{{f(x)\varphi '(x) + \varphi (x)f'(x)} \over {(f(x)\varphi (x) + 1)\sqrt {f(x)\varphi (x) - 1} }}dx} $$

$$ = \int {{{dz} \over {(z + 1)\sqrt {z - 1} }}} $$ এখানে $$z = f(x)\varphi (x)$$

$$ = \int {{{2pdp} \over {({p^2} + 2)\,.\,p}}} $$ [$$z - 1 = {p^2}$$ $$\therefore$$ $$dz = 2pdp$$]

$$ = 2\int {{{dp} \over {{p^2} + 2}}} $$

$$ = 2.{1 \over {\sqrt 2 }}{\tan ^{ - 1}}{p \over {\sqrt 2 }} + c$$

$$ = \sqrt 2 {\tan ^{ - 1}}{{\sqrt {f(x)\varphi (x) - 1} } \over {\sqrt 2 }} + c$$

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