We have,
$$
f(x)=\left\{\begin{array}{cc}
\frac{\sqrt{1+a x}-\sqrt{1-a x}}{x}, & -1 \leq x < 0 \\
\frac{x^2+2}{x-2}, & 0 \leq x \leq 1
\end{array}\right.
$$
Since, $f(x)$ is continuous on $[-1,1]$.

$$
\begin{aligned}
& \therefore f(x) \text { is continuous at } x=0 \\
& \therefore \text { LHL (at } x=0)=\text { RHL (at } x=0) \\
& \Rightarrow \quad \lim _{x \rightarrow 0} \frac{\sqrt{1+a x}-\sqrt{1-a x}}{x}=\lim _{x \rightarrow 0} \frac{x^2+2}{x-2} \\
& \Rightarrow \quad \lim _{x \rightarrow 0} \frac{(1+a x)-(1-a x)}{x[\sqrt{1+a x}+\sqrt{1-a x}]}=\lim _{x \rightarrow 0} \frac{x^2+2}{x-2} \\
& \Rightarrow \quad \lim _{x \rightarrow 0} \frac{2 a}{\sqrt{1+a x}+\sqrt{1-a x}}=\lim _{x \rightarrow 0} \frac{x^2+2}{x-2} \\
& \Rightarrow \quad a=-1
\end{aligned}
$$