Question

If $f\left(x\right)=p\left|\sin x\right|+q{e}^{\left|x\right|}+r{\left|x\right|}^3$ and $f\left(x\right)$ is differentiable at $x=0$, then

• A. $q+r=0$; $p$ is any real number
• B. $p+q=0$; $r$ is any real number
• C. $q=0;r=0$; $p$ is any real number
• D. $r=0,p=0$; $q$ is any real number

Answer 1

The correct option is B $p+q=0$; $r$ is any real number

We have

$f\left(x\right)=p|\sin x|+q{e}^{|x|}+r|x{|}^3$

By chain rule of differentiation,

$\frac{d}{dx}\left(p|\sin x|\right)=p\frac{|\sin x|}{\sin x}\cos x$

$\frac{d}{dx}\left(q{e}^{|x|}\right)=q{e}^{|x|}\frac{|x|}{x}$

$\frac{d}{dx}\left(r|x{|}^3\right)=3r|x{|}^2\frac{|x|}{x}=3rx|x|$

$\therefore f^{\prime }\left(x\right)=$ $p\frac{|\sin x|}{\sin x}\cos x+q{e}^{|x|}\frac{|x|}{x}+3rx|x|$

$f\left(x\right)$ is differentiable at $x=0$ if $f^{\prime }\left(x\right)$ exists at $x=0$.

At $x=0$, $\cos x=1$, $\frac{|\sin x|}{\sin x}=\frac{|x|}{x}$, $e^{|x|}=1$ and $x|x|=0$

Hence, at $x=0,f^{\prime }\left(x\right)$ reduces to

$f^{\prime }\left(0\right)=(p+q)\frac{|x|}{x}+3r\times 0$

$f^{\prime }\left(x\right)$ exists for any real $r$ and $p+q=0$