Question

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

 

पर्याय

• −1

•  −1/2

• 1/2

• 1

Answer 1

 \[- \frac{1}{2}\]

Given:

\[f\left( x \right) = \begin{cases}\frac{\sqrt{1 + px} - \sqrt{1 - px}}{x}, if - 1 \leq x < 0 \\ \frac{2x + 1}{x - 2}, if 0 \leq x \leq 1 \\ \end{cases}\]

If  \[f\left( x \right)\]  is continuous at x = 0,  then 

\[\lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^+} f\left( x \right)\]

\[\Rightarrow \lim_{h \to 0} f\left( - h \right) = \lim_{h \to 0} f\left( h \right) \]
\[ \Rightarrow \lim_{h \to 0} \left( \frac{\sqrt{1 - ph} - \sqrt{1 + ph}}{- h} \right) = \lim_{h \to 0} \left( \frac{2h + 1}{h - 2} \right)\]
\[ \Rightarrow \lim_{h \to 0} \left( \frac{\left( \sqrt{1 - ph} - \sqrt{1 + ph} \right)\left( \sqrt{1 - ph} + \sqrt{1 + ph} \right)}{- h\left( \sqrt{1 - ph} + \sqrt{1 + ph} \right)} \right) = \lim_{h \to 0} \left( \frac{2h + 1}{h - 2} \right)\]
\[ \Rightarrow \lim_{h \to 0} \left( \frac{\left( 1 - ph - 1 - ph \right)}{- h\left( \sqrt{1 - ph} + \sqrt{1 + ph} \right)} \right) = \lim_{h \to 0} \left( \frac{2h + 1}{h - 2} \right)\]
\[ \Rightarrow \lim_{h \to 0} \left( \frac{\left( - 2ph \right)}{- h\left( \sqrt{1 - ph} + \sqrt{1 + ph} \right)} \right) = \lim_{h \to 0} \left( \frac{2h + 1}{h - 2} \right)\]
\[ \Rightarrow \lim_{h \to 0} \left( \frac{\left( 2p \right)}{\left( \sqrt{1 - ph} + \sqrt{1 + ph} \right)} \right) = \lim_{h \to 0} \left( \frac{2h + 1}{h - 2} \right)\]
\[ \Rightarrow \left( \frac{\left( 2p \right)}{\left( 2 \right)} \right) = \left( \frac{1}{- 2} \right)\]
\[ \Rightarrow p = \frac{- 1}{2}\]