If the value of _ x → 0 (2 - x cos 2x ) ^ ( x + 2 x ^ 2 ) is equal...

We have,

$\lim_{x \to 0}{\left(2 - \cos x\sqrt{\cos 2x}\right)}^{\left(\dfrac{x + 2}{{x}^{2}}\right)} \to {1}^{\infty}$

Hence,
$= {e}^{\lim_{x \to 0}\left(\dfrac{1 - \cos x\sqrt{\cos 2x}}{{x}^{2}}\right) \times \left(x + 2\right)}$
Now,

$\lim_{x \to 0}\left(\dfrac{1 - \cos x\sqrt{\cos 2x}}{{x}^{2}}\right) \to \left(\dfrac{0}{0}\right)$

Using L'Hospital Rule

$\lim_{x \to 0}\left\{\dfrac{\sin x\sqrt{\cos 2x} - \cos x \times \dfrac{1}{2\sqrt{\cos 2x}} \times \left(- 2\sin 2x\right)}{2x}\right\}$

$= \lim_{x \to 0}\left\{\dfrac{\sin x \cos 2x + \sin 2x \cdot \cos x}{2x\sqrt{\cos 2x}}\right\}$

$= \lim_{x \to 0}\left\{\dfrac{\sin 3x}{2x\sqrt{\cos 2x}}\right\}$

$= \dfrac{3}{2}\lim_{x \to 0}\left\{\dfrac{\sin 3x}{3x}\right\} \times \lim_{x \to 0}\left\{\dfrac{1}{\sqrt{\cos 2x}}\right\}$

$= \dfrac{3}{2}\lim_{x \to 0}\left\{\dfrac{\sin 3x}{3x}\right\} \times 1$

$= \dfrac{3}{2}$

So,

${e}^{\lim_{x \to 0}\left(\dfrac{1 - \cos x\sqrt{\cos 2x}}{{x}^{2}}\right)\left(x + 2\right)}$

$= {e}^{\dfrac{3}{2} \times 2} = {e}^{3}$

$⇒ a = 3$