Question

The function $f\left(x\right)=a\sin \left|x\right|+b{e}^{\left|x\right|}$ is differential at $x=0$ when

• A. $3a+b=0$
• B. $3a-b=0$
• C. $a+b=0$
• D. $a-b=0$

Answer 1

Answer: (C)

$a+b=0$

$f\left(x\right)=\{\begin{array}{c}a\sin x+b{e}^{x}x>0\\ bx=0\\ -a\sin x+b{e}^{-x}x<0\end{array}$

$f^{\prime }\left(x\right)=\{\begin{array}{c}a\cos x+b{e}^{x}x>0\\ 0x=0\\ -a\cos x-b{e}^{-x}x<0\end{array}$

at $x=0$

$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}(a\cos x+b{e}^x)$

$=a+b$

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}(-a\cos x-b{e}^{-x})$

$=-a-b$

Given that it is differentiable at $x=0$

LHL$=f^{\prime }\left(0\right)=$RHL

$\Rightarrow a+b=0$