Question

If the function defined by
$f(x)=<br/>\{\begin{array}{ll}{\left(x^2+e^{\frac{1}{2-x}}\right)}^{-1},& \text{for}x\ne 2\\ k,& \text{for x = 2}\end{array}$
is right continuous at $x=2$, then $k=$

• A. $-\frac{1}{4}$
• B. 0
• C. $\frac{1}{4}$
• D. 1

Answer 1

Answer: (C)

We have
$f(x)=<br/><br/>\{\begin{array}{ll}{\left(x^2+e^{\frac{1}{2-x}}\right)}^{-1},& \text{for}x\ne 2\\ k,& \text{for x = 2}\end{array}$
is right continuous at right of $x=2$
$\therefore \underset{x\to 2^{+}}{\lim }{\left(x^2+e^{\frac{1}{2-x}}\right)}^{-1}=k$
$\Rightarrow k=\underset{h\to 0}{\lim }{\left[(2+h{)}^2+e^{\frac{1}{2-(2+h)}}\right]}^{-1}$
$=\underset{h\to 0}{\lim }{\left((2+h{)}^2+e^{-\frac{1}{h}}\right)}^{-1}$
$=\underset{h\to 0}{\lim }{\left((2+h{)}^2+\frac{1}{e^{\frac{1}{h}}}\right)}^{-1}$
$=(4{)}^{-1}=\frac{1}{4}$