Question

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

• A. $[-1,1]$
• B. $R-\{-1,1\}$
• C. $R-(-1,1)$
• D. None of these

Answer 1

Answer: (B)

Given, $f(x)=si{n}^{-1}\left(\frac{2x}{1+x^2}\right)$
On differentiating w.r.t. $x,$ we get
$f^{\prime }(x)=\frac{1}{\sqrt{1-{\left(\frac{2x}{1+x^2}\right)}^2}}\times \frac{d}{dx}\left(\frac{2x}{1+x^2}\right)$
$=\frac{1+x^2}{\sqrt{{\left(1-x^2\right)}^2}}\times \frac{2\left(1-x^2\right)}{{\left(1+x^2\right)}^2}$
$=\frac{2}{1+x^2}\times \frac{1-x^2}{|1-x^2|}$
$=\{\begin{array}{ll}\frac{2}{1+x^2}& \text{If ∣x∣<1}\\ -\frac{2}{1+x^2},& \text{If ∣x∣>1 }\end{array}$
$\therefore f^{\prime }(x)$ does not exist for $|x|=1$, i e, $x=\pm 1$
Hence, $f(x)$ is differentiable on $R-\{-1,1\}$