AP DSC School Assistant Math Model Test 1

Concept:The local maxima of a function occur at points where the first derivative is zero and the second derivative is negative (or the derivative changes sign from positive to negative). We apply this to the given trigonometric function.Explanation:Given $f(x) = 4\cos^3 x + 3\sqrt{3}\cos^2 x - 10$, for $x \in (0, 2\pi)$.Step 1: Differentiate $f(x)$ with respect to $x$:$$f'(x) = 4 \cdot 3\cos^2 x (-\sin x) + 3\sqrt{3} \cdot 2\cos x (-\sin x) = -12\cos^2 x \sin x - 6\sqrt{3}\cos x \sin x.$$Factor out $-6\sin x \cos x$:$$f'(x) = -6\sin x \cos x \left(2\cos x + \sqrt{3}\right).$$Step 2: Set $f'(x) = 0$ to find critical points in $(0, 2\pi)$:$-6\sin x \cos x (2\cos x + \sqrt{3}) = 0$ implies:$\sin x = 0$, $\cos x = 0$, or $2\cos x + \sqrt{3} = 0 \Rightarrow \cos x = -\frac{\sqrt{3}}{2}$.Step 3: Solve each equation in $(0, 2\pi)$:- $\sin x = 0$ gives $x = \pi$ (since $x=0,2\pi$ are excluded).- $\cos x = 0$ gives $x = \frac{\pi}{2}, \frac{3\pi}{2}$.- $\cos x = -\frac{\sqrt{3}}{2}$ gives $x = \frac{5\pi}{6}, \frac{7\pi}{6}$ (since cosine is negative in Quadrants II and III).Total critical points: $\pi, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}$.Step 4: Determine which are local maxima. Use the sign of $f'(x)$ around each point or the second derivative test. Compute $f''(x)$ or analyze sign changes. A simpler approach: examine the factors' sign patterns. The factor $2\cos x + \sqrt{3}$ changes sign at $\frac{5\pi}{6}$ and $\frac{7\pi}{6}$. The factor $\sin x$ is positive in $(0,\pi)$ and negative in $(\pi,2\pi)$ (except zeros). The factor $\cos x$ changes sign at $\frac{\pi}{2}$ and $\frac{3\pi}{2}$. By checking intervals, we find that $f'(x)$ changes from positive to negative at $x = \frac{5\pi}{6}$ and $x = \frac{7\pi}{6}$. At other points, the derivative changes from negative to positive (minima) or does not change. Therefore, the local maxima occur at $x = \frac{5\pi}{6}$ and $x = \frac{7\pi}{6}$.Step 5: Count the points of local maxima in $(0, 2\pi)$: two points.Answer:2 points of local maxima, corresponding to option B.