मराठी

F ( X ) = { √ 1 + P X − √ 1 − P X X , − 1 ≤ X < 0 2 X + 1 X − 2 , 0 ≤ X ≤ 1 is Continuous in the Interval [−1, 1], Then P is Equal to - Mathematics

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प्रश्न

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

 

पर्याय

  • −1

  •  −1/2

  • 1/2

  • 1

MCQ
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उत्तर

 \[- \frac{1}{2}\]

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 

\[\lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^+} f\left( x \right)\]
\[\Rightarrow \lim_{h \to 0} f\left( - h \right) = \lim_{h \to 0} f\left( h \right) \]
\[ \Rightarrow \lim_{h \to 0} \left( \frac{\sqrt{1 - ph} - \sqrt{1 + ph}}{- h} \right) = \lim_{h \to 0} \left( \frac{2h + 1}{h - 2} \right)\]
\[ \Rightarrow \lim_{h \to 0} \left( \frac{\left( \sqrt{1 - ph} - \sqrt{1 + ph} \right)\left( \sqrt{1 - ph} + \sqrt{1 + ph} \right)}{- h\left( \sqrt{1 - ph} + \sqrt{1 + ph} \right)} \right) = \lim_{h \to 0} \left( \frac{2h + 1}{h - 2} \right)\]
\[ \Rightarrow \lim_{h \to 0} \left( \frac{\left( 1 - ph - 1 - ph \right)}{- h\left( \sqrt{1 - ph} + \sqrt{1 + ph} \right)} \right) = \lim_{h \to 0} \left( \frac{2h + 1}{h - 2} \right)\]
\[ \Rightarrow \lim_{h \to 0} \left( \frac{\left( - 2ph \right)}{- h\left( \sqrt{1 - ph} + \sqrt{1 + ph} \right)} \right) = \lim_{h \to 0} \left( \frac{2h + 1}{h - 2} \right)\]
\[ \Rightarrow \lim_{h \to 0} \left( \frac{\left( 2p \right)}{\left( \sqrt{1 - ph} + \sqrt{1 + ph} \right)} \right) = \lim_{h \to 0} \left( \frac{2h + 1}{h - 2} \right)\]
\[ \Rightarrow \left( \frac{\left( 2p \right)}{\left( 2 \right)} \right) = \left( \frac{1}{- 2} \right)\]
\[ \Rightarrow p = \frac{- 1}{2}\]
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पाठ 9: Continuity - Exercise 9.4 [पृष्ठ ४५]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 9 Continuity
Exercise 9.4 | Q 20 | पृष्ठ ४५

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