Question

If $f\left(x\right)=p|\sin x|+q{e}^{|x|}+r|x{|}^3$ and if $f\left(x\right)$ 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 know that $\frac{d}{dx}\left|x\right|=\frac{\left|x\right|}{x}$

so, when $x⟶0+\frac{d}{dx}\left|x\right|=1x⟶0-\frac{d}{dx}\left|x\right|=-1$

now, $f^{\prime }\left(x\right)=p\frac{\left|\sin x\right|}{\sin x}\cos x+q{e}^{\left|x\right|}\frac{\left|x\right|}{x}+r\left(3{\left|x\right|}^2\right)\frac{\left|x\right|}{x}$

as $x⟶0+,\left|x\right|⟶xx⟶0-,\left|x\right|⟶-x$

Now, as $x⟶0+,f^{\prime }\left(x\right)=p\frac{\sin x}{\tan x}+q{e}^x+3r{x}^2{f}^{\prime }\left(x\right)=p\cos x+q{e}^x+3r{x}^{2}x⟶0-,f^{\prime }\left(x\right)=p\frac{-\sin x}{\tan x}-q{e}^{-x}-3r{x}^2{f}^{\prime }\left(x\right)=-p\cos x-q{e}^{-x}-3r{x}^2$

as $f\left(x\right)$ is differentiable at $x=0$ ,

${lim}_{x\to 0+}{f}^{\prime }\left(x\right)={lim}_{x\to 0-}{f}^{\prime }\left(x\right)$

$\Rightarrow p+q=-p-q\Rightarrow p+q=0$ ; r is any real number