Question

Assertion :If $f\left(x\right)={\cos }^{-1}\left(\frac{2x}{1+x^2}\right)$, then $f\left(x\right)$ is differentiable everywhere Reason: For $f\left(x\right)={\cos }^{-1}\left(\frac{2x}{1+x^2}\right)$,$f^{\prime }\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\frac{-2}{1+x^2},& \left|x\right|<1\end{array}\\ \begin{array}{cc}\frac{2}{1+x^2},& \left|x\right|>1\end{array}\end{array}$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is D Assertion is incorrect but Reason is correct

$f^{\prime }\left(x\right)=\frac{-1}{\sqrt{1-{\left(\frac{2x}{1+x^2}\right)}^2}}\times \frac{d}{dx}\left(\frac{2x}{1+x^2}\right)$

$=\frac{-(1+x^2)}{\sqrt{{(1+x^2)}^2-4x^2}}\times \frac{2(1-x^2)}{{(1+x^2)}^2}$

$=\frac{-2}{1+x^2}.\frac{1-x^2}{|1-x^2|}=\{\begin{array}{c}\begin{array}{cc}\frac{-2}{1+x^2},& \left|x\right|<1\end{array}\\ \begin{array}{cc}\frac{2}{1+x^2},& \left|x\right|>1\end{array}\end{array}$

Clearly, $f\left(x\right)$ is differentiable everywhere except at the points where $\left|x\right|=1$ i.e., $x=\pm 1.$

Hence, $f\left(x\right)$ is differentiable on $(-\infty ,\infty )-\{-1,1\}.$