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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY1 Mark
MCQ
$\text{f(x)}=\begin{cases}\frac{\sqrt{1+\text{px}}-\sqrt{1-\text{px}}}{\text{x}},&\text{if }0\leq\text{x}<0\\\frac{2\text{x}+1}{\text{x}-2},&\text{if }0\leq\text{x}\leq1\end{cases}$ is continuous in the interval [-1, 1], then p is equal to:
  • A
    $-1$
  • ✓
    $-\frac{1}{2}$
  • C
    $\frac{1}{2}$
  • D
    $1$
✓

Answer

Correct option: B.
$-\frac{1}{2}$
Given, $\text{f(x)}=\begin{cases}\frac{\sqrt{1+\text{px}}-\sqrt{1-\text{px}}}{\text{x}},&\text{if }0\leq\text{x}<0\\\frac{2\text{x}+1}{\text{x}-2},&\text{if }0\leq\text{x}\leq1\end{cases}$
If f(x) is continuous at x = 0,then
$\lim\limits_{\text{x}\rightarrow0^-}\text{f(x)}=\text{f(0)}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim\limits_{\text{h}\rightarrow0}\text{f(h)}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}}{-\text{h}}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\big(\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}\big)\big(\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}\big)}{-\text{h}\big(\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}\big)}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{(1-\text{ph}-1-\text{ph})}{-\text{h}\big(\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}\big)}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{(-2\text{ph})}{-\text{h}\big(\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}\big)}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{(2\text{p})}{\big(\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}\big)}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\Big(\frac{(2\text{p})}{(2)}\Big)=\Big(\frac{1}{-2}\Big)$
$\Rightarrow\text{p}=\frac{-1}{2}$

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