Solve this

Question:

$f(x)= \begin{cases}\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x}, & -1 \leq x<0 \\ \frac{2 x+1}{x-2} & , 0 \leq x \leq 1\end{cases}$

is continuous in the interval $[-1,1]$, then $p$ is equal to

(a) $-1$

(b) $-1 / 2$

(c) $1 / 2$

(d) 1

Solution:

(b) $-\frac{1}{2}$

Given: $f(x)=\left\{\begin{array}{c}\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x}, \text { if }-1 \leq x<0 \\ \frac{2 x+1}{x-2}, \text { if } 0 \leq x \leq 1\end{array}\right.$

If $f(x)$ is continuous at $x=0$, then

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)$

$\Rightarrow \lim _{h \rightarrow 0}\left(\frac{\sqrt{1-p h}-\sqrt{1+p h}}{-h}\right)=\lim _{h \rightarrow 0}\left(\frac{2 h+1}{h-2}\right)$

$\Rightarrow \lim _{h \rightarrow 0}\left(\frac{(\sqrt{1-p h}-\sqrt{1+p h})(\sqrt{1-p h}+\sqrt{1+p h})}{-h(\sqrt{1-p h}+\sqrt{1+p h})}\right)=\lim _{h \rightarrow 0}\left(\frac{2 h+1}{h-2}\right)$

$\Rightarrow \lim _{h \rightarrow 0}\left(\frac{(1-p h-1-p h)}{-h(\sqrt{1-p h}+\sqrt{1+p h})}\right)=\lim _{h \rightarrow 0}\left(\frac{2 h+1}{h-2}\right)$

$\Rightarrow \lim _{h \rightarrow 0}\left(\frac{(-2 p h)}{-h(\sqrt{1-p h}+\sqrt{1+p h})}\right)=\lim _{h \rightarrow 0}\left(\frac{2 h+1}{h-2}\right)$

$\Rightarrow \lim _{h \rightarrow 0}\left(\frac{(2 p)}{(\sqrt{1-p h}+\sqrt{1+p h})}\right)=\lim _{h \rightarrow 0}\left(\frac{2 h+1}{h-2}\right)$

$\Rightarrow\left(\frac{(2 p)}{(2)}\right)=\left(\frac{1}{-2}\right)$

$\Rightarrow p=\frac{-1}{2}$

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