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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
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Verified Answer
The correct answer is:
$3$
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 $x=0$.
$$
\therefore \mathrm{LHL}=\mathrm{RHL} \Rightarrow \mathrm{c}=3
$$
$$
f(x)=\left\{\begin{array}{cc}
\frac{\sqrt{1+c x}-\sqrt{1-c x}}{x}, & -2 \leq x < 0 \\
\frac{x+3}{x+1}, & 0 \leq x \leq 2
\end{array}\right.
$$
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 $x=0$.
$$
\therefore \mathrm{LHL}=\mathrm{RHL} \Rightarrow \mathrm{c}=3
$$
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