Question

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

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

Answer 1

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

Given: $f\left(x\right)$ is continuous
$\therefore \underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)$
and $f\left(0\right)=-\frac{1}{2}$

$\Rightarrow f\left(0\right)=\underset{x\to 0}{lim}\frac{\sqrt{1+\text{p}x}-\sqrt{1-\text{p}x}}{x}$
Applying L'hospital rule
$=\frac{\frac{p}{2\sqrt{1+0\cdot p}}-\frac{(-p)}{2\sqrt{1-0\cdot p}}}{1}$
$=p$
$\Rightarrow p=\frac{-1}{2}$