Question

If $f\left(x\right)=a\left|\sin x\right|+b{e}^{\left|x\right|}+c{\left|x\right|}^3$ and if f (x) is differentiable at x = 0, then
(a) $a=b=c=0$
(b) $a=0,b=0;c\in R$
(c) $b=c=0,a\in R$
(d) $c=0,a=0,b\in R$

Answer 1

(b) $a=0,b=0;c\in R$

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ f\left(x\right)=a\left|\sin x\right|+b{e}^{\left|x\right|}+c{\left|x\right|}^3 \\ =\left\{\begin{array}{l}a\sin x+b{e}^x+c{x}^{3}0<x<\frac{\mathrm{\pi }}{2}\\ -a\sin x+b{e}^{-x}-c{x}^3-\frac{\mathrm{\pi }}{2}<x<0\end{array}\right. \\ \mathrm{Here},f\left(x\right)\mathrm{is}\mathrm{differentiable}\mathrm{at}x=0 \\ \mathrm{Therefore},\left(\mathrm{LHD}\mathrm{at}x=0\right)=\left(\mathrm{RHD}\mathrm{at}x=0\right) \\ \Rightarrow \underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0} \\ \Rightarrow \underset{x\to 0^{-}}{\lim }\frac{-a\sin x+b{e}^{-x}-c{x}^3-b}{x}=\underset{x\to 0^{+}}{\lim }\frac{a\sin x+b{e}^x+c{x}^3-b}{x} \\ \Rightarrow \underset{h\to 0}{\lim }\frac{-a\sin \left(0-h\right)+b{e}^{-\left(0-h\right)}-c{\left(0-h\right)}^3-b}{0-h}=\underset{h\to 0}{\lim }\frac{a\sin \left(0+h\right)+b{e}^{\left(0+h\right)}+c{\left(0+h\right)}^3-b}{0+h} \\ \Rightarrow \underset{h\to 0}{\lim }\frac{a\sin h+be{}^h+c{h}^3-b}{-h}=\underset{h\to 0}{\lim }\frac{a\sin h+b{e}^h+c{h}^3-b}{h} \\ \Rightarrow \underset{h\to 0}{\lim }\frac{a\cos h+b{e}^h+3c{h}^2}{-1}=\underset{h\to 0}{\lim }\frac{a\cos h+b{e}^h+3c{h}^2}{1}\left(\mathrm{By}\mathrm{L}\text{'}\mathrm{Hospital}\mathrm{rule}\right) \\ \Rightarrow -\left(a+b\right)=a+b \\ \Rightarrow -2\left(a+b\right)=0 \\ \Rightarrow a+b=0 \\ \mathrm{This}\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{all}\mathrm{value}\mathrm{of}c \\ \therefore c\in \mathrm{R} \\ \mathrm{In}\mathrm{the}\mathrm{given}\mathrm{options},\mathrm{option}\left(\mathrm{b}\right)\mathrm{satisfies}a+b=0\mathrm{and}\mathit{}c\in \mathrm{R} \end{gathered}$$