Question

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

• A. $0$
• B. $1$
• C. $i$
• D. $-i$

Answer 1

Answer: (B)

$1$

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

$\Rightarrow {(1-x)}^2=0$

$\Rightarrow x=1$