Question

$f\left(x\right)=\{\begin{array}{cc}\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x},& if-1\le x<0\\ \frac{2x+1}{x-1},& if0\le x\le 1\end{array}atx=0$

What is the value of k if the function is continuous at all points?

Answer 1

We have, $f\left(x\right)=\{\begin{array}{cc}\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x},& if-1\le x<0\\ \frac{2x+1}{x-1},& if0\le x\le 1\end{array}atx=0$

$\therefore$ LHL $=\underset{x\to 0^{-}}{lim}\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}=\underset{x\to 0^{-}}{lim}\left(\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}\right).\left(\frac{\sqrt{1+kx}+\sqrt{1-kx}}{\sqrt{1+kx}+\sqrt{1-kx}}\right)=\underset{x\to 0^{-}}{lim}\frac{1+kx-1+kx}{x[\sqrt{1+kx}+\sqrt{1-kx}]}\underset{h\to 0}{lim}\frac{2k}{\sqrt{1+k(0-h)}+\sqrt{1-k(0-h)}}=\underset{h\to 0}{lim}\frac{2k}{\sqrt{1-kh}+\sqrt{1+kh}}=\frac{2k}{2}=k$

$f\left(0\right)=\frac{2\times 0+1}{0-1}=-1$

$\Rightarrow k=-1$ [$\because$ LHL=RHL = f(0)]