Question

Let $f\left(x\right)=a\sin |x|+b{e}^{|x|}$ is differentiable when

• A. $a=-b$
• B. $a=b$
• C. $a=0$
• D. $b=0$

Answer 1

Answer: (A)

$a=-b$

Given , $f\left(x\right)=a\sin |x|+b{e}^{|x|}$
$f\left(x\right)=\{\begin{array}{c}-a\sin x+b{e}^{-x}x<0\\ a\sin x+b{e}^{x}x\ge 0\end{array}$
Now, since, given $f\left(x\right)$ is differentiable function.
$f^{\prime }\left(x\right)=\{\begin{array}{c}-a\cos x-b{e}^{-x}x<0\\ a\cos x+b{e}^{x}x\ge 0\end{array}$
$f\left(x\right)$ is differentiable at $x=0$
$\Rightarrow f^{\prime }\left(0^{+}\right)=f^{\prime }\left(0^{-}\right)$
$\Rightarrow a+b=-a-b$
$\Rightarrow 2a=-2b$
$\Rightarrow a=-b$