At x = 5π6, f (x) = 2 sin 3x + 3 cos 3x is

The correct option is neither maximum nor minimum.

To determine the nature of the function at $x=\frac{5\pi }{6}$, we can analyze its derivatives.

Let the given function be:
$$f(x)=2\sin (3x)+3\cos (3x)$$

First, we find the first derivative of the function with respect to $x$ using the chain rule:
$$f^{\prime }(x)=\frac{d}{dx}[2\sin (3x)+3\cos (3x)]$$
$$f^{\prime }(x)=2\cdot 3\cos (3x)-3\cdot 3\sin (3x)$$
$$f^{\prime }(x)=6\cos (3x)-9\sin (3x)$$

Next, we evaluate this first derivative at the given point $x=\frac{5\pi }{6}$:
$$3x=3\left(\frac{5\pi }{6}\right)=\frac{5\pi }{2}$$

Substituting $3x=\frac{5\pi }{2}$ into the expression for $f^{\prime }(x)$:
$$f^{\prime }\left(\frac{5\pi }{6}\right)=6\cos \left(\frac{5\pi }{2}\right)-9\sin \left(\frac{5\pi }{2}\right)$$

We know the trigonometric values:
$$\cos \left(\frac{5\pi }{2}\right)=\cos (2\pi +\frac{\pi }{2})=\cos \left(\frac{\pi }{2}\right)=0$$
$$\sin \left(\frac{5\pi }{2}\right)=\sin (2\pi +\frac{\pi }{2})=\sin \left(\frac{\pi }{2}\right)=1$$

Thus:
$$f^{\prime }\left(\frac{5\pi }{6}\right)=6(0)-9(1)=-9$$

For a function to have a local maximum or a local minimum at a given point, the first derivative at that point must equal zero (i.e., it must be a critical point, assuming the function is differentiable).
Since $f^{\prime }\left(\frac{5\pi }{6}\right)=-9\ne 0$, the point is not a stationary/critical point. Consequently, the function can be neither a maximum nor a minimum at this value of $x$.