The maximum value of ${e^{(2 + \sqrt 3 \cos x + \sin x)}}$ is:
A) ${e^2}$
B) ${e^{2 - \sqrt 3 }}$
C) ${e^4}$
D) $1$

Hint:Let $f$ be a function of variable $x$ . We say that $f$ has a local maximum at $a$ if $f(x) \leqslant f(a)$ for all points $x$ in the neighbourhood of $a$ . We say that $f$ has a local minimum at $a$ if $f(x) \geqslant f(a)$ for all points $x$ in the neighbourhood of $a$ . If $f$ has a local extremum (i.e., local maximum or minimum) at $a$ then, $\dfrac{{d(f(a))}}{{dx}} = 0$ .

Formula Used:
$\sin (a + b) = \sin a\cos b + \cos a\sin b$

Complete step by step Solution:
Let $f(x) = {e^{(2 + \sqrt 3 \cos + \sin x)}}$ be a function of variable $x$ .
Now, to find the maximum value of this function, we differentiate it with respect to $x$.
$\dfrac{d}{{dx}}(f(x)) = \dfrac{d}{{dx}}({e^{(2 + \sqrt 3 \cos + \sin x)}})$
As we know, $\dfrac{d}{{dx}}({e^{g(x)}}) = {e^{g(x)}}\dfrac{d}{{dx}}(g(x))$ .
Here, $g(x) = 2 + \sqrt 3 \cos x + \sin x$ so we get
$\dfrac{d}{{dx}}(f(x)) = f'(x) = {e^{(2 + \sqrt 3 \cos x + \sin x)}}\dfrac{d}{{dx}}(2 + \sqrt 3 \cos x + \sin x)$
$\dfrac{d}{{dx}}(f(x)) = {e^{(2 + \sqrt 3 \cos x + \sin x)}}( - \sqrt 3 \sin x + \cos x)$
As we know that the value of the exponential function is always greater than zero, hence we will take
$( - \sqrt 3 \sin x + \cos x) = 0$ to find the maximum value of the function $f(x)$
(We can divide the equation by $2$ on both sides)
$ \Rightarrow ( - \dfrac{{\sqrt 3 }}{2}\sin x + \dfrac{1}{2}\cos x) = 0$
$ \Rightarrow \cos ( - \dfrac{\pi }{6})\sin x + \sin ( - \dfrac{\pi }{6})\cos x = 0$
Using formula: $\sin (a + b) = \sin a\cos b + \cos a\sin b$
$ \Rightarrow \sin (x - \dfrac{\pi }{6}) = 0$
$ \Rightarrow \sin (x - \dfrac{\pi }{6}) = \sin 0$
So, we can equalize the angles on both sides, we will get,
$x = \dfrac{\pi }{6}$
Now, differentiating $\dfrac{d}{{dx}}f(x)$ again with respect to $x$
$\dfrac{d}{{dx}}\left( {\dfrac{d}{{dx}}(f(x))} \right) = {e^{(2 + \sqrt 3 \cos x + \sin x)}}(\sqrt 3 \sin x - \cos x)((3 - 2\sin x)\sin x + \sqrt 3 (2\sin x + 3)\cos x)$
If we substitute the value $x = \dfrac{\pi }{6}$ in the second derivative, we get the negative value of the function i.e., maximum value. Hence, the maximum value of $f(x) = {e^4}$ .

Therefore, the correct option is C.

Note: A point $a$ such that $d(f(a)) = 0$ is called a critical point of $f$ . At a critical point, a function could have a local maximum or a local minimum, or none. If $f$ is a continuous function of two variables then, the point $(a,b)$ is called a saddle point for the function $f(x,y)$ , if for each $r > 0$ , $\exists $ at least one point $(x,y)$ , within a distance $r$ of $(a,b)$ , for which $f(x,y) > f(a,b)$ or $f(x,y) < f(a,b)$ .