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  • If    <img alt="\mathrm{f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bf%28x%29%3D%5Csin%20%5E%7B-1%7D%5Cleft%28%5Cf

If    \mathrm{f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)}  , then f(x) is differentiable on

Option: 1

[-1,1]


Option: 2

 \mathrm{R-\mid-1,1\mid}


Option: 3

\mathrm{R-(-1,1)}


Option: 4

none of these


Answers (1)

best_answer

\mathrm{f^{\prime \prime}(x)=\frac{1}{\sqrt{1-\left(\frac{2 x}{1+z^2}\right)^t}} \frac{d}{d z}\left(\frac{2 z}{1-x^2}\right)}

\mathrm{ =\frac{1+z}{\sqrt{\left(1-z^2\right.}} \times \frac{2\left(1-z^2\right)}{(1+x)^2}=\frac{2}{1+z^2} \times \frac{1-x^2}{11-x^2 \mid} }
\mathrm{ =\left\{\begin{array}{l} \frac{2}{1+z^2} \text { if }|x|<1 \\ \frac{-2}{1+x^2}, \text { if }|x|>1 \end{array}\right. }
\mathrm{ \therefore } \mathrm{ f^{\prime}(z) }does not exist for |x|=1 ie, \mathrm{ z= \pm 1 }

Posted by

Ajit Kumar Dubey

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