Question

$f\left(x\right)=\{\begin{array}{c}\frac{\sqrt{1+px}-\sqrt{1-px}}{x}\\ \frac{2x+1}{x-2},0\le x\le 1\end{array},-1\le x<0$ is continuous in the interval $[-1,1]$, then $p$ equals-

• A. $-1$
• B. $-\frac{1}{2}$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

The correct option is B $-\frac{1}{2}$

Given $f\left(x\right)=\{\begin{array}{c}\frac{\sqrt{1+px}-\sqrt{1-px}}{x}\\ \frac{2x+1}{x-2},0\le x\le 1\end{array},-1\le x<0$
Since, f(x) is continuous in $[-1,1]$, so f(x) is continuous at $x=0$
Now, $f\left(0\right)=\frac{-1}{2}$ (by def of f(x))
$LHL=\underset{x\to 0^{-}}{lim}f\left(x\right)$
$=\underset{h\to 0}{lim}f(0-h)$
$=\underset{h\to 0}{lim}\frac{\sqrt{1-ph}-\sqrt{1+ph}}{-h}$
$=\underset{h\to 0}{lim}\frac{\sqrt{1-ph}-\sqrt{1+ph}}{-h}\times \frac{\sqrt{1-ph}+\sqrt{1+ph}}{\sqrt{1-ph}+\sqrt{1+ph}}$
$=\underset{h\to 0}{lim}\frac{1-ph-1-ph}{-h(\sqrt{1-ph}+\sqrt{1+ph})}$
$=\underset{h\to 0}{lim}\frac{-2ph}{-h(\sqrt{1-ph}+\sqrt{1+ph})}$
$LHL=\frac{2p}{2}=p$
Since, $f\left(x\right)$ is continuous at $x=0$
$LHL=f\left(0\right)$
$\Rightarrow p=-\frac{1}{2}$