Key Concepts and Formulas

• First Derivative Test: A critical point $c$ of a function $f(x)$ is a local maximum if $f'(x)$ changes its sign from positive to negative as $x$ increases through $c$.
• Derivatives of Trigonometric Functions: $\frac{d}{dx}(\cos x) = -\sin x$.
• Trigonometric Identities: $\sin 3x = 3 \sin x - 4 \sin^3 x$ and $\cos 3x = 4\cos^3 x - 3\cos x$.

Step-by-Step Solution

Step 1: Find the first derivative of f(x)

We are given $f(x) = 4 \cos^3 x + 3 \sqrt{3} \cos^2 x - 10$. We need to find $f'(x)$. $f'(x) = \frac{d}{dx}(4 \cos^3 x + 3 \sqrt{3} \cos^2 x - 10)$ $f'(x) = 4 \cdot 3 \cos^2 x (-\sin x) + 3 \sqrt{3} \cdot 2 \cos x (-\sin x) - 0$ $f'(x) = -12 \cos^2 x \sin x - 6 \sqrt{3} \cos x \sin x$ $f'(x) = -6 \cos x \sin x (2 \cos x + \sqrt{3})$

Step 2: Find the critical points by setting f'(x) = 0

To find the critical points, we set $f'(x) = 0$. $-6 \cos x \sin x (2 \cos x + \sqrt{3}) = 0$ This gives us three possibilities:

1. $\cos x = 0$
2. $\sin x = 0$
3. $2 \cos x + \sqrt{3} = 0 \implies \cos x = -\frac{\sqrt{3}}{2}$

Now, we solve for $x$ in the interval $(0, 2\pi)$:

1. $\cos x = 0 \implies x = \frac{\pi}{2}, \frac{3\pi}{2}$
2. $\sin x = 0 \implies x = \pi$
3. $\cos x = -\frac{\sqrt{3}}{2} \implies x = \frac{5\pi}{6}, \frac{7\pi}{6}$

Thus, the critical points are $x = \frac{\pi}{2}, \pi, \frac{3\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}$. Arranging in ascending order, we have $\frac{\pi}{2}, \frac{5\pi}{6}, \pi, \frac{7\pi}{6}, \frac{3\pi}{2}$.

Step 3: Analyze the sign of f'(x) around each critical point

We will use the first derivative test to find the local maxima. Recall that $f'(x) = -6 \cos x \sin x (2 \cos x + \sqrt{3}) = -3 \sin(2x) (2 \cos x + \sqrt{3})$.

• Around $x = \frac{\pi}{2}$:

  • For $x < \frac{\pi}{2}$ (e.g., $x = \frac{\pi}{3}$), $f'(\frac{\pi}{3}) = -3 \sin(\frac{2\pi}{3}) (2 \cos(\frac{\pi}{3}) + \sqrt{3}) = -3 (\frac{\sqrt{3}}{2}) (1 + \sqrt{3}) < 0$.
  • For $x > \frac{\pi}{2}$ (e.g., $x = \frac{2\pi}{3}$), $f'(\frac{2\pi}{3}) = -3 \sin(\frac{4\pi}{3}) (2 \cos(\frac{2\pi}{3}) + \sqrt{3}) = -3 (-\frac{\sqrt{3}}{2}) (-1 + \sqrt{3}) = -3 (\frac{\sqrt{3}}{2}) (\sqrt{3} - 1) < 0$. Since the sign does not change, $x = \frac{\pi}{2}$ is neither a local maximum nor a local minimum.

• Around $x = \frac{5\pi}{6}$:

  • For $x < \frac{5\pi}{6}$ (e.g., $x = \pi$), $f'(\pi) = -3 \sin(2\pi) (2 \cos(\pi) + \sqrt{3}) = 0$. Let's take $x = \frac{4\pi}{5}$. Then $f'(\frac{4\pi}{5}) = -3 \sin(\frac{8\pi}{5})(2\cos(\frac{4\pi}{5})+\sqrt{3})$. We know that $\cos(\frac{4\pi}{5}) \approx -0.8$, so $2\cos(\frac{4\pi}{5}) + \sqrt{3} \approx -1.6 + 1.73 > 0$. Also, $\sin(\frac{8\pi}{5}) < 0$, so $f'(\frac{4\pi}{5}) < 0$.
  • For $x > \frac{5\pi}{6}$ (e.g., $x = \frac{11\pi}{12}$), $f'(\frac{11\pi}{12}) = -3 \sin(\frac{11\pi}{6})(2\cos(\frac{11\pi}{12})+\sqrt{3}) = -3 (-\frac{1}{2}) (2(-\frac{\sqrt{2}+\sqrt{6}}{4}) + \sqrt{3}) = \frac{3}{2} (-\frac{\sqrt{2}+\sqrt{6}}{2} + \sqrt{3}) = \frac{3}{4} (2\sqrt{3} - \sqrt{2} - \sqrt{6})$. Since $2\sqrt{3} \approx 3.46$, $\sqrt{2} \approx 1.41$, and $\sqrt{6} \approx 2.45$, we have $3.46 - 1.41 - 2.45 = -0.4 < 0$. So $f'(\frac{11\pi}{12}) < 0$. Since the sign does not change, $x = \frac{5\pi}{6}$ is neither a local maximum nor a local minimum.

• Around $x = \pi$:

  • For $x < \pi$ (e.g., $x = \frac{5\pi}{6}$), $f'(\frac{5\pi}{6}) < 0$ as shown above.
  • For $x > \pi$ (e.g., $x = \frac{7\pi}{6}$), $f'(\frac{7\pi}{6}) = -3 \sin(\frac{7\pi}{3}) (2 \cos(\frac{7\pi}{6}) + \sqrt{3}) = -3 \sin(\frac{\pi}{3}) (2(-\frac{\sqrt{3}}{2}) + \sqrt{3}) = -3 (\frac{\sqrt{3}}{2}) (-\sqrt{3} + \sqrt{3}) = 0$. Let's take $x = \frac{6\pi}{5}$. Then $f'(\frac{6\pi}{5}) = -3 \sin(\frac{12\pi}{5})(2\cos(\frac{6\pi}{5})+\sqrt{3})$. We know that $\cos(\frac{6\pi}{5}) \approx -0.3$, so $2\cos(\frac{6\pi}{5}) + \sqrt{3} \approx -0.6 + 1.73 > 0$. Also, $\sin(\frac{12\pi}{5}) > 0$, so $f'(\frac{6\pi}{5}) < 0$. Since the sign does not change, $x = \pi$ is neither a local maximum nor a local minimum.

• Around $x = \frac{7\pi}{6}$:

  • For $x < \frac{7\pi}{6}$ (e.g., $x = \pi$), $f'(\pi) = 0$. Let's take $x = \frac{6\pi}{7}$. Then $f'(x) < 0$.
  • For $x > \frac{7\pi}{6}$ (e.g., $x = \frac{4\pi}{3}$), $f'(\frac{4\pi}{3}) = -3 \sin(\frac{8\pi}{3}) (2 \cos(\frac{4\pi}{3}) + \sqrt{3}) = -3 \sin(\frac{2\pi}{3}) (2 (-\frac{1}{2}) + \sqrt{3}) = -3 (\frac{\sqrt{3}}{2}) (-1 + \sqrt{3}) = -3 (\frac{\sqrt{3}}{2}) (\sqrt{3} - 1) < 0$. Since the sign does not change, $x = \frac{7\pi}{6}$ is neither a local maximum nor a local minimum.

• Around $x = \frac{3\pi}{2}$:

  • For $x < \frac{3\pi}{2}$ (e.g., $x = \frac{5\pi}{4}$), $f'(\frac{5\pi}{4}) = -3 \sin(\frac{5\pi}{2}) (2 \cos(\frac{5\pi}{4}) + \sqrt{3}) = -3 (1) (2 (-\frac{\sqrt{2}}{2}) + \sqrt{3}) = -3 (-\sqrt{2} + \sqrt{3}) = -3 (\sqrt{3} - \sqrt{2}) < 0$.
  • For $x > \frac{3\pi}{2}$ (e.g., $x = \frac{7\pi}{4}$), $f'(\frac{7\pi}{4}) = -3 \sin(\frac{7\pi}{2}) (2 \cos(\frac{7\pi}{4}) + \sqrt{3}) = -3 (-1) (2 (\frac{\sqrt{2}}{2}) + \sqrt{3}) = 3 (\sqrt{2} + \sqrt{3}) > 0$. Since the sign changes from negative to positive, $x = \frac{3\pi}{2}$ is a local minimum.

Let's reconsider $f'(x) = -6\cos x \sin x (2\cos x + \sqrt{3})$. The critical points are $\cos x = 0$, $\sin x = 0$, and $\cos x = -\frac{\sqrt{3}}{2}$. In $(0, 2\pi)$, these are $x = \frac{\pi}{2}, \frac{3\pi}{2}, \pi, \frac{5\pi}{6}, \frac{7\pi}{6}$.

Consider the interval $(0, \frac{\pi}{2})$. Let's take $x = \frac{\pi}{3}$. Then $f'(\frac{\pi}{3}) = -6 (\frac{1}{2}) (\frac{\sqrt{3}}{2}) (1 + \sqrt{3}) = -\frac{3\sqrt{3}}{2}(1+\sqrt{3}) < 0$. Consider the interval $(\frac{\pi}{2}, \frac{5\pi}{6})$. Let's take $x = \frac{2\pi}{3}$. Then $f'(\frac{2\pi}{3}) = -6 (-\frac{1}{2}) (\frac{\sqrt{3}}{2}) (-1 + \sqrt{3}) = \frac{3\sqrt{3}}{2}(\sqrt{3}-1) > 0$. Since $f'(x)$ changes from negative to positive at $x=\frac{\pi}{2}$, it's a local minimum.

Consider the interval $(\frac{5\pi}{6}, \pi)$. Let's take $x = \frac{11\pi}{12}$. Then $f'(\frac{11\pi}{12}) = -3\sin(\frac{11\pi}{6})(2\cos(\frac{11\pi}{12})+\sqrt{3}) = -3(-\frac{1}{2})(2(-\frac{\sqrt{6}+\sqrt{2}}{4}) + \sqrt{3}) = \frac{3}{2}(-\frac{\sqrt{6}+\sqrt{2}}{2}+\sqrt{3}) = \frac{3}{4}(2\sqrt{3}-\sqrt{6}-\sqrt{2})$. $2\sqrt{3} \approx 3.46$, $\sqrt{6} \approx 2.45$, $\sqrt{2} \approx 1.41$. $3.46 - 2.45 - 1.41 = -0.4 < 0$. So $f'(\frac{11\pi}{12}) < 0$. Since $f'(x)$ changes from positive to negative at $x = \frac{5\pi}{6}$, it's a local maximum.

Consider the interval $(\pi, \frac{7\pi}{6})$. Let's take $x = \frac{6\pi}{7}$. Then $f'(x) < 0$. Consider the interval $(\frac{7\pi}{6}, \frac{3\pi}{2})$. Let's take $x = \frac{5\pi}{4}$. Then $f'(\frac{5\pi}{4}) = -3(1)(2(-\frac{\sqrt{2}}{2}) + \sqrt{3}) = -3(\sqrt{3}-\sqrt{2}) < 0$. Since $f'(x)$ doesn't change sign at $\pi$ or $\frac{7\pi}{6}$, they're neither local max nor min. Consider the interval $(\frac{3\pi}{2}, 2\pi)$. Let's take $x = \frac{7\pi}{4}$. Then $f'(\frac{7\pi}{4}) = -3(-1)(2(\frac{\sqrt{2}}{2}) + \sqrt{3}) = 3(\sqrt{2}+\sqrt{3}) > 0$. Since $f'(x)$ changes from negative to positive at $x = \frac{3\pi}{2}$, it's a local minimum.

So, we have a local maximum at $x = \frac{5\pi}{6}$. Thus, there is only 1 local maximum.

Common Mistakes & Tips

• Carefully calculate the derivatives and simplify the expression. Sign errors are very common.
• When using the first derivative test, make sure to check the sign of the derivative in the intervals immediately to the left and right of the critical point.
• Don't assume that every critical point is a local maximum or minimum. It could be a point of inflection.

Summary

We found the first derivative of the given function and set it to zero to find the critical points. Then, we analyzed the sign of the first derivative in the intervals around each critical point to determine whether it is a local maximum, a local minimum, or neither. We found that there is only one point of local maxima in the given interval $(0, 2\pi)$.

Final Answer

The final answer is $\boxed{1}$, which corresponds to option (A).