Question

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

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

Answer 1

Answer: (B)

$\frac{-1}{2}$

Given f(x) is continuous in [-1,1].
So, f(x) is continuous at x=0
${lim}_{x\to 0}f\left(x\right)=f\left(0\right)$
Here $f\left(0\right)=-\frac{1}{2}$
$LHL={lim}_{x\to 0^{-}}\frac{\sqrt{1+px}-\sqrt{1-px}}{x}$
$={lim}_{x\to 0^{-}}\frac{2px}{x(\sqrt{1+px}+\sqrt{1-px})}$
$=p$
Hence $p=-\frac{1}{2}$