Question 8

For $x \neq 0$, $f(x) = \frac{6x + \sin x}{2x + \sin x}$.

$\lim_{x \to 0} f(x) = \frac{7}{3} = f(0)$, so $f$ is continuous.

$f(x) - f(0) = \frac{18x + 3\sin x - 14x - 7\sin x}{3(2x + \sin x)} = \frac{4(x - \sin x)}{3(2x + \sin x)}$.

Near $x=0$, for $x > 0$, $x - \sin x > 0$ and $2x + \sin x > 0$, so $f(x) > f(0)$. For $x < 0$, $x - \sin x < 0$ and $2x + \sin x < 0$,

so $f(x) > f(0)$. Thus $x=0$ is a local minima (B).

$f'(x) = \frac{4(\sin x - x\cos x)}{(2x + \sin x)^2}$. $f'(x) = 0 \Rightarrow \tan x = x$.

Let $h(x) = \tan x - x$. In $(n\pi, n\pi + \pi/2)$, there is one root $x_n$.

Analysis of $f''$ or $h$ shows that local maxima occur for odd $n$ and local minima for even $n$. In $[\pi, 6\pi]$, roots occur in $(\pi, 3\pi/2), (2\pi, 5\pi/2), (3\pi, 7\pi/2), (4\pi, 9\pi/2), (5\pi, 11\pi/2)$.

Maxima are at $n=1, 3, 5$ (3 points), so (C) is true.

Minima in $[2\pi, 4\pi]$ is only at $n=2$ (near $2.5\pi$), so (D) is true.