A) \[-1\]
B) \[-1/2\]
C) 1/2
D) 1
Correct Answer: B
Solution :
\[f(x)=\left\{ \begin{matrix} \frac{\sqrt{(1+px)}-\sqrt{(1-px)}}{x}, & x<0 \\ \frac{2x+1}{x-2}, & 0\le x\le 1 \\ \end{matrix} \right.\] \[f(0+0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{2(0+h)+1}{0+h-2}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{2h+1}{h-2}=\frac{-1}{2}\] \[f(0-0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sqrt{1-ph}-\sqrt{1+ph}}{(0-h)}\] \[\times \frac{\sqrt{1-ph}+\sqrt{1+ph}}{\sqrt{1-ph}+\sqrt{1+ph}}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{1-ph-1-ph}{(\sqrt{1-ph}+\sqrt{1+ph})}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{2p}{\sqrt{1-ph}+\sqrt{1+ph}}\] \[=\frac{2p}{1+1}=p\] Since, at\[x=0,f(x)\]is continuous. \[\therefore \] \[f(0+0)=f(0-0)=f(0)\] \[\Rightarrow \] \[-\frac{1}{2}=p\] \[\Rightarrow \] \[p=-\frac{1}{2}\]
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