⨍ (x)=1+px-1-pxxi⨍ -1≤x<0 2x-1x-2i⨍ 0≤x≤1 continuous on -1, 1 then p is

LHL=limx→0− 1+px−1−pxx=limx→0− 2pxx(1+px+1−px)=pRHL=limx→0+2x-1x-2=-12 since f is a continuous function LHL=RHL ⇒p=-12

⨍ (x)=1+px-1-pxxi⨍ -1≤x<0 2x-1x-2i⨍ 0≤x≤1 continuous on -1, 1 then p is

LHL=limx→0− 1+px−1−pxx=limx→0− 2pxx(1+px+1−px)=pRHL=limx→0+2x-1x-2=-12 since f is a continuous function LHL=RHL ⇒p=-12

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$⨍ (x)= \{\begin{cases} \frac{\sqrt{1 + px} - \sqrt{1 - px}}{x} & i ⨍ -1≤x<0 \\ \frac{2 x - 1}{x - 2} & i ⨍ 0≤x≤1 \end{cases}$
continuous on $[- 1, 1]$ then $p$ is

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-1/2

-1

1/2

answer is D.

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Detailed Solution

$\begin{matrix} LHL = lim_{x \to 0^{-}} \frac{\sqrt{1 + p x} - \sqrt{1 - p x}}{x} \\ = lim_{x \to 0^{-}} \frac{2 p x}{x (\sqrt{1 + p x} + \sqrt{1 - p x})} = p \end{matrix}$

$R H L = \lim_{x \to 0^{+}} \frac{2 x - 1}{x - 2} = \frac{- 1}{2} \sin c e f i s a c o n t i n u o u s f u n c t i o n L H L = R H L \Rightarrow p = \frac{- 1}{2}$