Question

Let $f\left(x\right)=\begin{array}{cc}(x-e){.2}^{\frac{-2}{e-x}},& x\ne e\\ 0,& x=e\end{array}$ then -

• A. f is continuous and differentiable at $x=e$
• B. f is continuous but not differentiable at $x=e$
• C. f is neither continuous nor differentiable at $x=e$
• D. geometrically f has sharp corner at $x=e$

Answer 1

Answer: (B), (D)

geometrically f has sharp corner at $x=e$
f is continuous but not differentiable at $x=e$

Given: $f\left(x\right)=(x-e).2^{\frac{-2}{e-x}},x\ne e=0x=e$

To find: Nature of continuity and differentiability of $f\left(x\right)$ at x=e

Sol: ${lim}_{x\to e}(x-e).2^{\frac{-2}{e-x}}$

let $x-e=t⟹{lim}_{t\to 0}t.2^{\frac{-2}{-t}}$

$⟹{lim}_{t\to 0}\frac{t}{2^{\frac{-2}{t}}}=0$ (using l'Hospital rule)

$⟹f\left(x\right)$ is continous at x=e

Now, $f^{{}^{\prime }}\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$f^{{}^{\prime }}\left(e\right)={lim}_{h\to 0}\frac{f(e+h)-f\left(e\right)}{h}=\frac{f(e+h)}{h}$

$={lim}_{h\to 0}\frac{h.2^{\frac{2}{h}}}{h}={lim}_{h\to 0}{2}^{\frac{2}{h}}=\infty$

$⟹$ limit does not exist

Hence, f is continous but not differentiable at $x=e$