Let $f(x)$ be a polynomial of degree 5, and have extrema at $x = 1$ and $x = -1$. If $\lim\limits_{x \to 0} \left( \frac{f(x)}{x^3} \right) = -5$, then $f(2) - f(-2)$ is equal to:

The number of critical points of the function

$f(x) = \begin{cases} |\frac{\sin x}{x}|, & x \neq 0 \\ 1, & x = 0 \end{cases}$ in the interval $(-2\pi, 2\pi)$ is equal to :

Consider the following three statements for the function $f:(0, \infty) \rightarrow \mathbb{R}$ defined by $f(x)=\left|\log _e x\right|-|x-1|$ :

(I) $f$ is differentiable at all $x>0$.

(II) $f$ is increasing in $(0,1)$.

(III) $f$ is decreasing in $(1, \infty)$.

Then.

The least value of $\left(\cos ^2 \theta-6 \sin \theta \cos \theta+3 \sin ^2 \theta+2\right)$ is