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If $f(x)\left\{\begin{array}{cl}\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x}, & \text { for }-1 \leq x < 0 \\ 2 x^2+3 x-2, & \text { for } 0 \leq x \leq 1\end{array}\right.$ is continuous at $x=0$, then $k$ is equal to
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The correct answer is:
-2
Since, $f(x)$ is continuous at $x=0$
$\therefore \quad \lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(0+h)$
$\Rightarrow \lim _{h \rightarrow 0} \frac{\sqrt{1-k h}-\sqrt{1+k h}}{-h}=\lim _{h \rightarrow 0} 2 h^2+3 h-2$
$\Rightarrow \lim _{h \rightarrow 0} \frac{\frac{1}{2}(1-k h)^{-\frac{k}{2}}+\frac{1}{2}(1+k)^{-\frac{k}{2}}}{-1}=-2$
$\Rightarrow \quad \frac{k}{2}+\frac{k}{2}=-2$
$\therefore \quad k=-2$
$\therefore \quad \lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(0+h)$
$\Rightarrow \lim _{h \rightarrow 0} \frac{\sqrt{1-k h}-\sqrt{1+k h}}{-h}=\lim _{h \rightarrow 0} 2 h^2+3 h-2$
$\Rightarrow \lim _{h \rightarrow 0} \frac{\frac{1}{2}(1-k h)^{-\frac{k}{2}}+\frac{1}{2}(1+k)^{-\frac{k}{2}}}{-1}=-2$
$\Rightarrow \quad \frac{k}{2}+\frac{k}{2}=-2$
$\therefore \quad k=-2$
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