Let $a$ be a constant. Assume that the function $$f(x) = 2\sin^3 x + a\sin 2x + \frac{9}{2}\cos 2x - 9\cos x - 2ax + 6$$ takes a local extremum at $x = \frac{\pi}{3}$. We consider about the maximum and minimum values of $f(x)$ over the interval $0 \leqq x \leqq \frac{\pi}{2}$. (1) Since $f(x)$ takes a local extremum at $x = \frac{\pi}{3}$, it follows that $a = \frac{\mathbf{A}}{\mathbf{B}}$. Hence the derivative $f'(x)$ of $f(x)$ can be expressed as $$f'(x) = \mathbf{C}\cos x(\mathbf{D}\cos x - 1)(\sin x - \mathbf{E}).$$ (2) It can be seen from the result of (1) that $f(x)$ over $0 \leqq x \leqq \frac{\pi}{2}$ takes the maximum value at $x = \mathbf{F}$ and the minimum value at $x = \mathbf{G}$, where $\mathbf{F}$ and $\mathbf{G}$ are the appropriate expressions from among (0) $\sim$ (4) below. (0) $0$ (1) $\frac{\pi}{6}$ (2) $\frac{\pi}{4}$ (3) $\frac{\pi}{3}$ (4) $\frac{\pi}{2}$
Let $a$ be a constant. Assume that the function
$$f(x) = 2\sin^3 x + a\sin 2x + \frac{9}{2}\cos 2x - 9\cos x - 2ax + 6$$
takes a local extremum at $x = \frac{\pi}{3}$. We consider about the maximum and minimum values of $f(x)$ over the interval $0 \leqq x \leqq \frac{\pi}{2}$.
(1) Since $f(x)$ takes a local extremum at $x = \frac{\pi}{3}$, it follows that $a = \frac{\mathbf{A}}{\mathbf{B}}$.
Hence the derivative $f'(x)$ of $f(x)$ can be expressed as
$$f'(x) = \mathbf{C}\cos x(\mathbf{D}\cos x - 1)(\sin x - \mathbf{E}).$$
(2) It can be seen from the result of (1) that $f(x)$ over $0 \leqq x \leqq \frac{\pi}{2}$ takes the maximum value at $x = \mathbf{F}$ and the minimum value at $x = \mathbf{G}$, where $\mathbf{F}$ and $\mathbf{G}$ are the appropriate expressions from among (0) $\sim$ (4) below.\\
(0) $0$\\
(1) $\frac{\pi}{6}$\\
(2) $\frac{\pi}{4}$\\
(3) $\frac{\pi}{3}$\\
(4) $\frac{\pi}{2}$