If $\quad f:[-2,2] \rightarrow R \quad$ is defined by $f(x)=\left\{\begin{array}{cc}\frac{\sqrt{1+c…

If $\quad f:[-2,2] \rightarrow R \quad$ is defined by $f(x)=\left\{\begin{array}{cc}\frac{\sqrt{1+c x}-\sqrt{1-c x}}{x} & \text { for }-2 \leq x < 0 \\ \frac{x+3}{x+1} & \text { for } 0 \leq x \leq 2\end{array}\right.$ continuous on $[-2,2]$, then $c$ is equal to
  1. $\frac{2}{\sqrt{3}}$
  2. $3$
  3. $\frac{3}{2}$
  4. $\frac{3}{\sqrt{2}}$

Solution

Given, $f:[-2,2] \rightarrow R$ $ f(x)=\left\{\begin{array}{cc} \frac{\sqrt{1+c x}-\sqrt{1-c x}}{x}, & -2 \leq x < 0 \\ \frac{x+3}{x+1}, & 0 \leq x \leq 2 \end{array}\right. $ Now, $\quad$ LHL $=\lim _{x \rightarrow 0^{-}} f(x)$ $ \begin{aligned} & =\lim _{h \rightarrow 0} \frac{\sqrt{1-c h}-\sqrt{1+c h}}{-h} \\ & \times \frac{\sqrt{1-c h}+\sqrt{1+c h}}{\sqrt{1-c h}+\sqrt{1+c h}} \\ & =\lim _{h \rightarrow 0} \frac{(1-c h)-(1+c h)}{-h(\sqrt{1-0}+\sqrt{1+0})} \\ & =\lim _{h \rightarrow 0} \frac{-2 c h}{-h(1+1)}=c \end{aligned} $ and $ \mathrm{RHL}=\lim _{x \rightarrow 0^{+}} f(x) $ $ \begin{aligned} & =\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} \frac{0+h+3}{0+h+1} \\ & =\lim _{h \rightarrow 0} \frac{h+3}{h+1}=\frac{0+3}{0+1}=3 \end{aligned} $ Since, $f$ is continuous at $x=0$. $ \therefore \mathrm{LHL}=\mathrm{RHL} \Rightarrow \mathrm{c}=3 $

Asked in: AP EAMCET 2014

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