Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{1-\cos x}{x^2},& x<0\\ \frac{1}{2}{e}^x,& x\ge 0\end{array},$
then which among the following statements is correct

• A. $f\left(x\right)$ is not a continuous function
• B. $f\left(x\right)$ is differentiable function
• C. $f\left(x\right)$ is continuous and differentiable function
• D. $f\left(x\right)$ is continuous but not differentiable function

Answer 1

The correct option is D $f\left(x\right)$ is continuous but not differentiable function

Given : $f\left(x\right)=\{\begin{array}{cc}\frac{1-\cos x}{x^2},& x<0\\ \frac{1}{2}{e}^x,& x\ge 0\end{array}$
$f\left(0\right)=\frac{1}{2}R.H.L.=\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{1}{2}{e}^h=\frac{1}{2}L.H.L.=\underset{x\to 0^{-}}{lim}=\underset{h\to 0}{lim}\frac{1-\cos h}{h^2}=\frac{1}{2}$
Hence, $f\left(x\right)$ is continuous function.

Now,
$R.H.D.=\underset{h\to 0^{+}}{lim}\frac{f(0+h)-\frac{1}{2}}{h}=\underset{h\to 0^{+}}{lim}\frac{1}{2}\left(\frac{e^h-1}{h}\right)=\frac{1}{2}L.H.D.=\underset{h\to 0^{+}}{lim}\frac{f(0-h)-\frac{1}{2}}{-h}=\underset{h\to 0^{+}}{lim}\frac{-1}{h}(\frac{1-\cos h}{h^2}-\frac{1}{2})$
Now, using expansion series
$=\underset{h\to 0}{lim}\frac{-1}{h}[\frac{1-(1-\frac{h^2}{2!}+\frac{h^4}{4!}-\cdots )}{h^2}-\frac{1}{2}]=0\Rightarrow L.H.D.\ne R.H.D.$

Hence, $f\left(x\right)$ is not differentiable function.