Question

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

• A. $0$
• B. $1$
• C. $i$
• D. $-i$

Answer 1

The correct option is B

$1$

Explanation for the correct option.

Find the value of $x$:

Given,

$f\left(x\right)={\tan }^{-1}x$ and $f\prime \left(x\right)+f\prime \prime \left(x\right)=0$.

Now differentiate with respect to $x$.

$f\text{'}\left(x\right)=\frac{1}{1+x^2}$

Again differentiate.

$f\text{'}\text{'}\left(x\right)=-\frac{2x}{{\left(1+x^2\right)}^2}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left[\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v{\displaystyle \frac{du}{dx}}-u{\displaystyle \frac{dv}{dx}}}{v^2}\right]$

Therefore,

$\begin{array}{rcl}f\text{'}\left(x\right)+f\text{'}\text{'}\left(x\right)& =& 0\\ & \Rightarrow & \frac{1}{1+x^2}-\frac{2x}{{\left(1+x^2\right)}^2}=0\\ & \Rightarrow & \frac{1+x^2-2x}{{\left(1+x^2\right)}^2}=0\\ & \Rightarrow & \frac{{\left(1-x\right)}^2}{{\left(1+x^2\right)}^2}=0\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{·}\mathbf{a}\mathbf{·}\mathbf{b}\mathbf{=}{\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}}^{\mathbf{2}}\mathbf{]}\\ & \Rightarrow & 1-x=0\\ & \Rightarrow & x=1\end{array}$

If $x=i$ then we will get denominator as zero.

Hence, the correct option is B.