The value of f(0), so that the function

Solution:

(C) $-a^{\frac{1}{2}}$

Given: $f(x)=\frac{\sqrt{a^{2}-a x+x^{2}}-\sqrt{a^{2}+a x+x^{2}}}{\sqrt{a+x}-\sqrt{a-x}}$

$\Rightarrow f(x)=\frac{\left(\sqrt{a^{2}-a x+x^{2}}-\sqrt{a^{2}+a x+x^{2}}\right)\left(\sqrt{a^{2}-a x+x^{2}}+\sqrt{a^{2}+a x+x^{2}}\right)}{(\sqrt{a+x}-\sqrt{a-x})\left(\sqrt{a^{2}-a x+x^{2}}+\sqrt{a^{2}+a x+x^{2}}\right)}$

$\Rightarrow f(x)=\frac{\left(a^{2}-a x+x^{2}-\left(a^{2}+a x+x^{2}\right)\right)}{(\sqrt{a+x}-\sqrt{a-x})\left(\sqrt{a^{2}-a x+x^{2}}+\sqrt{a^{2}+a x+x^{2}}\right)}$

$\Rightarrow f(x)=\frac{(-2 a x)(\sqrt{a+x}+\sqrt{a-x})}{(\sqrt{a+x}-\sqrt{a-x})\left(\sqrt{a^{2}-a x+x^{2}}+\sqrt{a^{2}+a x+x^{2}}\right)(\sqrt{a+x}+\sqrt{a-x})}$

$\Rightarrow f(x)=\frac{(-2 a x)(\sqrt{a+x}+\sqrt{a-x})}{(a+x-a+x)\left(\sqrt{a^{2}-a x+x^{2}}+\sqrt{a^{2}+a x+x^{2}}\right)}$

$\Rightarrow f(x)=\frac{(-2 a x)(\sqrt{a+x}+\sqrt{a-x})}{(2 x)\left(\sqrt{a^{2}-a x+x^{2}}+\sqrt{a^{2}+a x+x^{2}}\right)}$

$\Rightarrow f(x)=\frac{-a(\sqrt{a+x}+\sqrt{a-x})}{\left(\sqrt{a^{2}-a x+x^{2}}+\sqrt{a^{2}+a x+x^{2}}\right)}$

If $f(x)$ is continuous for all $x$, then it will be continuous at $x=0$ as well.

So, if $f(x)$ is continuous at $x=0$, then

$\lim _{x \rightarrow 0} f(x)=f(0)$

$\Rightarrow \lim _{x \rightarrow 0}\left[\frac{-a(\sqrt{a+x}+\sqrt{a-x})}{\left(\sqrt{a^{2}-a x+x^{2}}+\sqrt{a^{2}+a x+x^{2}}\right)}\right]=f(0)$

$\Rightarrow\left[\frac{-2 a(\sqrt{a})}{\left(\sqrt{a^{2}}+\sqrt{a^{2}}\right)}\right]=f(0)$

$\Rightarrow\left[\frac{-2 a(\sqrt{a})}{(a+a)}\right]=f(0)$

$\Rightarrow f(0)=-\sqrt{a}$