$f(x)=\frac{4 x^{3}-3 x^{2}}{6}-2 \sin x+(2 x-1) \cos x$
$f^{\prime}(x)=\left(2 x^{2}-x\right)-2 \cos x+2 \cos x-\sin x(2 x-1)$
$\quad=(2 x-1)(x-\sin x)$
for $x>0, x-\sin x>0$
$\quad x<0, x-\sin x<0$
for $x \in(-\infty, 0] \cup\left[\frac{1}{2}, \infty\right), f^{\prime}(x) \geq 0$
for $x \in\left[0, \frac{1}{2}\right], f^{\prime}(x) \leq 0$
$\Rightarrow \quad f(x)$ increases in $\left[\frac{1}{2}, \infty\right)$
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($A$) There is exactly one choice for such $\vec{v}$
($B$) There are infinitely many choices for such $\vec{v}$
($C$) If $\hat{u}$ lies in the $x y$-plane then $\left|u_1\right|=\left|u_2\right|$
($D$) If $\hat{u}$ lies in the $x z$-plane then $2\left|u_1\right|=\left|u_3\right|$