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Friday, February 3, 2023

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

Given that,

$f(x)=begin{cases} frac{sqrt{1+kx}-sqrt{1-kx}}{x}, text{if} -1le x<0 \ frac{2x+1}{x-1}, text{if} 0le x<1 end{cases}$

Since, f(x) is continuous at $x=0$,

$displaystylelim_{x to 0^-} f(x)=f(0)=displaystylelim_{x to 0^+} f(x)$ ………… (1)

Here, $f(0)=frac{2(0)+1}{0-1}$

$=frac{1}{-1}$

$f(0)=-1$ ………………… (2)

Now, $displaystylelim_{x to 0^-} f(x)=displaystylelim_{x to 0}frac{sqrt{1+kx}-sqrt{1-kx}}{x}$

$=displaystylelim_{x to 0}frac{sqrt{1+kx}-sqrt{1-kx}}{x}times frac{sqrt{1+kx}+sqrt{1-kx}}{sqrt{1+kx}+sqrt{1-kx}}$

$=displaystylelim_{x to 0}frac{{1+kx}-{1-kx}}{x(sqrt{1+kx}+sqrt{1-kx})}$

$=displaystylelim_{x to 0}frac{2kx}{x(sqrt{1+kx}+sqrt{1-kx})}$

$=displaystylelim_{x to 0}frac{2k}{sqrt{1+kx}+sqrt{1-kx}}$

$=frac{2k}{sqrt{1+0}+sqrt{1-0}}$

$=frac{2k}{2}$

$therefore displaystylelim_{x to 0^-} f(x)=k$ ………………… (3)

From (1), (2) and (3), we have

$k=-1$

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