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Question

Find the value of $$k$$, for which $$f(x)=\begin{Bmatrix}\dfrac{\sqrt{1+kx}-\sqrt{1-kx}}{x}&if -1\le x < 0\\ \dfrac{2x+1}{x-1}&if 0\le x < 1\end{Bmatrix}$$ is continuous at $$x = 0$$.


Solution

The given function f(x) is continuous at x = 0.

$$\displaystyle \lim_{x\rightarrow 0} f(x)$$

$$\dfrac{0+1}{0-1}=\displaystyle \lim_{x\rightarrow 0}\left(\dfrac{\sqrt{1+kx}-\sqrt{1-kx}}{x} \right )$$

$$\Rightarrow -1=\displaystyle \lim_{x\rightarrow 0}\left(\dfrac{\sqrt{1+kx}-\sqrt{1-kx}}{x} \right )\left(\dfrac{\sqrt{1+kx}+\sqrt{1-kx}}{\sqrt{1+kx}+\sqrt{1-kx}} \right )$$

$$\Rightarrow -1=\displaystyle \lim_{x\rightarrow 0}\left(\dfrac{1+kx-1+kx}{x[\sqrt{1+kx}+\sqrt{1-kx}]}\right )$$

$$\Rightarrow -1=\displaystyle \lim_{x\rightarrow 0}\left(\dfrac{2k}{\sqrt{1+kx}+\sqrt{1-kx}}\right )$$

When $$k = -1$$

$$\Rightarrow -1=\dfrac{2k}{2}$$

$$k = -1$$

Therefore the value of $$k$$ is $$-1.$$


Mathematics

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