The domain of definition of f(x)

(a) [4, ∞)

$f(x)=\sqrt{x-3-2 \sqrt{x-4}}-\sqrt{x-3+2 \sqrt{x-4}}$

For $f(x)$ to be defined, $x-4 \geq 0$

$\Rightarrow x-4 \geq 0$

$\Rightarrow x \geq 4$    .....(1)

Also, $x-3-2 \sqrt{x-4} \geq 0$

$\Rightarrow x-3-2 \sqrt{x-4} \geq 0$

$\Rightarrow x-3 \geq 2 \sqrt{x-4}$

$\Rightarrow(x-3)^{2} \geq(2 \sqrt{x-4})^{2}$

$\Rightarrow x^{2}+9-6 x \geq 4(x-4)$

$\Rightarrow x^{2}-10 x+25 \geq 0$

$\Rightarrow(x-5)^{2} \geq 0$, which is always true.

Similarly, $x-3+2 \sqrt{x-4} \geq 0$ is always true.

Thus, $\operatorname{dom}(\mathrm{f}(\mathrm{x}))=[4, \infty)$