Answer:
Option A
Explanation:
We have ,
$f(x)=\begin{cases}\frac{\sqrt{1+ax}-\sqrt{1-ax}}{x} ,& -1\leq x <0\\ \frac{x^{2}+2}{x-2}, & 0\leq x \leq 1\end{cases} is $
Since, f(x) is continuous on [-1,1]
$\therefore$ f(x) is continuous at x=0
$\therefore$ LHS (at x=0)=RHL(at x=0)
$\Rightarrow \lim_{x \rightarrow 0}\frac{\sqrt{1+ax}-\sqrt{1-ax}}{x}=\lim_{x \rightarrow 0}\frac{x^{2}+2}{x-2}$
$\Rightarrow \lim_{x \rightarrow 0}\frac{(1+ax)-{(1-ax)}}{x[\sqrt{1+ax}+\sqrt{1-ax}]}=\lim_{x \rightarrow 0}\frac{x^{2}+2}{x-2}$
$\Rightarrow \lim_{x \rightarrow 0}\frac{2a}{\sqrt{1+ax}+\sqrt{1-ax}}=\lim_{x \rightarrow 0}\frac{x^{2}+2}{x-2}$
$\Rightarrow$ a=-1
