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

Answer: (B)

$+1$

Answer 2

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