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If $f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$, then $f(x)$ is
differentiable on
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differentiable on
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Verified Answer
The correct answer is:
$R-\{-1,1\}$
Given, $f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$
On differentiating w.r.t. $x$, we get $\begin{aligned} f^{\prime}(x) &=\frac{1}{\sqrt{1-\left(\frac{2 x}{1+x^{2}}\right)^{2}} \times \frac{d}{d x}\left(\frac{2 x}{1+x^{2}}\right)} \\ &=\frac{1+x^{2}}{\sqrt{\left(1-x^{2}\right)^{2}}} \times \frac{2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}} \end{aligned}$
$=\frac{2}{1+x^{2}} \times \frac{1-x^{2}}{\left|1-x^{2}\right|}$
$=\left\{\begin{array}{l}\frac{2}{1+x^{2}}, \text { if }|x| < 1 \\ -\frac{2}{1+x^{2}}, \text { if }|x|>1\end{array}\right.$
$\therefore f^{\prime}(x)$ does not exist for $|x|=1, i e, x=\pm 1$
Hence, $f(x)$ is differentiable on $R-\{-1,1\}$.
On differentiating w.r.t. $x$, we get $\begin{aligned} f^{\prime}(x) &=\frac{1}{\sqrt{1-\left(\frac{2 x}{1+x^{2}}\right)^{2}} \times \frac{d}{d x}\left(\frac{2 x}{1+x^{2}}\right)} \\ &=\frac{1+x^{2}}{\sqrt{\left(1-x^{2}\right)^{2}}} \times \frac{2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}} \end{aligned}$
$=\frac{2}{1+x^{2}} \times \frac{1-x^{2}}{\left|1-x^{2}\right|}$
$=\left\{\begin{array}{l}\frac{2}{1+x^{2}}, \text { if }|x| < 1 \\ -\frac{2}{1+x^{2}}, \text { if }|x|>1\end{array}\right.$
$\therefore f^{\prime}(x)$ does not exist for $|x|=1, i e, x=\pm 1$
Hence, $f(x)$ is differentiable on $R-\{-1,1\}$.
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