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Question
Options
−1
−1/2
1/2
1
Solution
\[- \frac{1}{2}\]
Given:
If \[f\left( x \right)\] is continuous at x = 0, then
\[ \Rightarrow \lim_{h \to 0} \left( \frac{\sqrt{1 - ph} - \sqrt{1 + ph}}{- h} \right) = \lim_{h \to 0} \left( \frac{2h + 1}{h - 2} \right)\]
\[ \Rightarrow \lim_{h \to 0} \left( \frac{\left( \sqrt{1 - ph} - \sqrt{1 + ph} \right)\left( \sqrt{1 - ph} + \sqrt{1 + ph} \right)}{- h\left( \sqrt{1 - ph} + \sqrt{1 + ph} \right)} \right) = \lim_{h \to 0} \left( \frac{2h + 1}{h - 2} \right)\]
\[ \Rightarrow \lim_{h \to 0} \left( \frac{\left( 1 - ph - 1 - ph \right)}{- h\left( \sqrt{1 - ph} + \sqrt{1 + ph} \right)} \right) = \lim_{h \to 0} \left( \frac{2h + 1}{h - 2} \right)\]
\[ \Rightarrow \lim_{h \to 0} \left( \frac{\left( - 2ph \right)}{- h\left( \sqrt{1 - ph} + \sqrt{1 + ph} \right)} \right) = \lim_{h \to 0} \left( \frac{2h + 1}{h - 2} \right)\]
\[ \Rightarrow \lim_{h \to 0} \left( \frac{\left( 2p \right)}{\left( \sqrt{1 - ph} + \sqrt{1 + ph} \right)} \right) = \lim_{h \to 0} \left( \frac{2h + 1}{h - 2} \right)\]
\[ \Rightarrow \left( \frac{\left( 2p \right)}{\left( 2 \right)} \right) = \left( \frac{1}{- 2} \right)\]
\[ \Rightarrow p = \frac{- 1}{2}\]
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