Question

Let $f(x) = \tan^{-1}x$. Then $f'(x) + f''(x)$ is = $0$, when $x$ is equal to

• A. $0$
• B. $+1$
• C. $i$
• D. $-i$

Answer 1

The Correct Option is B

Given, \(f(x)=\tan ^{-1} x\)
\(\Rightarrow f^{\prime}(x)=\frac{1}{1+x^{2}}\)
\(\Rightarrow f^{\prime \prime}(x)=-\frac{1}{\left(1+x^{2}\right)^{2}}(2 x)=-\frac{2 x}{\left(1+x^{2}\right)^{2}}\)
\(\because f^{\prime}(x)+f^{\prime \prime}(x)=0\)
\(\therefore \frac{1}{1+x^{2}}-\frac{2 x}{\left(1+x^{2}\right)^{2}}=0\)
\(\Rightarrow \frac{\left(1+x^{2}\right)-2 x}{\left(1+x^{2}\right)^{2}}=0\)
\(\Rightarrow (1-x)^{2}=0 \Rightarrow x=1\)

So, the correct answer is (B): +1