Question

The function $f\left(x\right)$ is defined by $f\left(x\right)=\{\begin{array}{cc}{(x^2+e^{\frac{1}{2-x}})}^{-1},& x\ne 2\\ k,& x=2\end{array}$.
If it is continuous from right at the point $x=2$, then $k$ is equal to

• A. $0$
• B. $\frac{1}{4}$
• C. $-\frac{1}{2}$
• D. None of these

Answer 1

The correct option is B $\frac{1}{4}$

$f\left(2\right)=k=\underset{x\to 2^{+}}{\text{lim}}f\left(x\right)$ as $f\left(x\right)$ is right continuous at $x=2$
$\Rightarrow$ $k=\underset{x\to 2^{+}}{\text{lim}}{(x^2+e^{\frac{1}{2-\left(x\right)}})}^{-1}$
$=\underset{h\to 0}{\text{lim}}{({(2+h)}^2+e^{-\frac{1}{h}})}^{-1}$
$={(4+e^{-\infty })}^{-1}[\because e^{-\infty }\to 0]$
$\Rightarrow k=4^{-1}=\frac{1}{4}$