Question

If $f(x) = \begin{cases} \frac{\sqrt{1+ kx }-\sqrt{1- kx }}{ X } & \text{for $-1 \leq x<0$} \\[2ex] 2 x^{2}+3 x-2 & \text{for $0 \leq x \leq 1$} \end{cases}$
continuous at $x =0$, then $k$ is equal to

• A. -4
• B. -3
• C. -2
• D. -1

Answer 1

Answer: (C)

$LHL =\displaystyle\lim _{ x \rightarrow 0^{-}} \frac{\sqrt{1+ kx }-\sqrt{1- kx }}{ x }$
$=\displaystyle\lim _{x \rightarrow 0^{-}} \frac{2 kx }{x(\sqrt{1+ kx }+\sqrt{1- kx })}= k$
$RHL =\displaystyle\lim _{ x \rightarrow 0^{+}}\left(2 x ^{2}+3 x -2\right)=-2 \& f (0)=-2$
$\because$ It is given that $f ( x )$ is continuous
$\therefore LHL = RHL = f (0)$
$\Rightarrow k =-2$