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AP EAPCET 06-Jul-2022 Shift 1 Paper

Section: Mathematics

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Question : 64 of 160

Marks: +1, -0

If

\;\frac{d}{dx}\left(A\log\left(\frac{\sqrt{1-x^{3}}+B}{\sqrt{1-x^{3}}+1}\right)\right)=\frac{1}{x\sqrt{1-x^{3}}}

AB=

$\frac{1}{3}$
$\frac{-1}{3}$
$\frac{-2}{3}$
$\frac{2}{3}$

Solution: 👈: Video Solution

We have

\;\frac{d}{dx}\left[A\log\;\frac{\sqrt{1-x^{3}}+B}{\sqrt{1-x^{3}+1}}\right]\because\frac{1}{x\sqrt{1-x^{3}}}

\Rightarrow A\log\left(\:\frac{\sqrt{1-x^{3}}+B}{\sqrt{1-x^{3}}+1}\right)=\int\frac{1}{x\sqrt{1-x^{3}}}dx

.....(i)

Let

I=\int\frac{1}{x\sqrt{1-x^{3}}}dx

Let

x^{3}=\sin^{2}t

3x^{2}dx=2\sin t\cos t

I=\frac{2}{3}\int\frac{1\times\sin t\cos t}{\sin^{2}t\cos t}dt

=\frac{2}{3}\int\csc t\,dt=\frac{2}{3}\log\left|\csc t-\cot t\right|

=\frac{1}{3}\log\left|\frac{1-\cos t\,t^{2}}{\sin t}\right|^{x^{3}}\mid

=\frac{1}{3}\log\left|\frac{(1-\sqrt{1-x^{3}})^{2}}{x^{3}}\right|

=\frac{1}{3}\log\left|\frac{(1-\sqrt{1-x^{3}})^{2}(1+\sqrt{1-x^{3}})}{x^{3}(1+\sqrt{1-x^{3}})}\right|.

=\frac{1}{3}\log\left|\frac{1-\sqrt{1-x^{3}}}{1+\sqrt{1-x^{3}}}\right|.

From Eq. (i), we get

A=\frac{1}{3}

and

B=-1

\Rightarrow AB=-\frac{1}{3}

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