Question

If $f:[-2,2]\to R$ is defined by $f\left(x\right)=\{\begin{array}{ccc}\frac{\sqrt{1+ex}-\sqrt{1-ex}}{x}& for& -2\le x<0\\ \frac{x+3}{x+1}& for& 0\le x\le 2\end{array}$
is continuous on$[-2,2]$, then $e=$

• A. $\frac{2}{\sqrt{3}}$
• B. $3$
• C. $\frac{3}{2}$
• D. $\frac{3}{\sqrt{2}}$

Answer 1

Answer: (B)

$3$

$\underset{x\to 0^{-}}{lim}f\left(x\right)$
$=limx\to 0^{-}\frac{\sqrt{1+ex}-\sqrt{1-ex}}{x}$
$=\underset{x\to 0^{-}}{lim}\frac{1+ex-(1-ex)}{(\sqrt{1+ex}+\sqrt{1-ex})x}$
$=\underset{x\to 0^{-}}{lim}\frac{2ex}{(\sqrt{1+ex}+\sqrt{1-ex})x}$
$=\underset{x\to 0^{-}}{lim}\frac{2e}{(\sqrt{1+ex}+\sqrt{1-ex})}$
$=\frac{2e}{2}$
$=e$ ............(i)
$\underset{x\to 0^{+}}{lim}f\left(x\right)$
$=\underset{x\to 0^{+}}{lim}\frac{x+3}{x+1}$
$=3$...............(ii)
For the function to be continuous
$LHL=RHL$
$\therefore e=3$.