Question

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

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

$\underset{x\to 0^{+}}{lim}f\left(x\right)\underset{x\to 0^{+}}{lim}(p|sinx|+q{e}^{\left(x\right)}+r{x}^3)=\underset{x\to 0^{+}}{lim}p\sin x+\underset{x\to 0^{+}}{lim}q{e}^x+\underset{x\to 0}{lim}r{x}^3=0+q+0=q\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}(-p\sin x)+\underset{x\to 0^{-}}{lim}q{e}^{-x}+\underset{x\to 0}{lim}-r{x}^3=q$

$f^{\prime }\left(x\right)=p\cos x+q{e}^{|x|}+3r(x{)}^2$ for $x>0$

$f^{\prime }\left(x\right)=-(p\cos x+q{e}^x+3r{x}^2)$ for $x<0$

$\underset{x\to 0^{+}}{lim}{f}^{\prime }\left(x\right)=(p+q+0)\underset{x\to 0^{-}}{lim}{f}^{\prime }\left(x\right)=-(p+q+0)\Rightarrow p+q=0$

Hence, $p+q=0$ and $r$ is any real number.