For the function $f(x)=x^3-6x^2-12x-3$,$x=2$ is a

The function $f(x)=2|x|+|x+2|-||x+2|-2|x||$ has a local minimum or a local maximum at $x=$

Let $f(x) = x \sin \pi x$,$x > 0$. Then for all natural numbers $n$,$f^{\prime}(x)$ vanishes at
$(A)$ a unique point in the interval $\left(n, n+\frac{1}{2}\right)$
$(B)$ a unique point in the interval $\left(n+\frac{1}{2}, n+1\right)$
$(C)$ a unique point in the interval $(n, n+1)$
$(D)$ two points in the interval $(n, n+1)$

For all real $x$,the minimum value of $\frac{1-x+x^2}{1+x+x^2}$ is

If $x = 1$ is a critical point of the function $f(x) = (3x^{2} + ax - 2 - a)e^{x}$,then

Let $f:[0,1] \rightarrow \mathbb{R}$ be a function. Suppose $f$ is twice differentiable,$f(0)=f(1)=0$ and satisfies $f^{\prime \prime}(x)-2 f^{\prime}(x)+f(x) \geq e^x$ for $x \in[0,1]$.
$1.$ Which of the following is true for $0 < x < 1$?
$(A)$ $0 < f(x) < \infty$
$(B)$ $-\frac{1}{2} < f(x) < \frac{1}{2}$
$(C)$ $-\frac{1}{4} < f(x) < 1$
$(D)$ $-\infty < f(x) < 0$
$2.$ If the function $g(x) = e^{-x} f(x)$ assumes its minimum in the interval $[0,1]$ at $x=\frac{1}{4}$,which of the following is true?
$(A)$ $f^{\prime}(x) < f(x)$ for $x \in (0, 1/4)$
$(B)$ $f^{\prime}(x) > f(x)$ for $x \in (0, 1/4)$
$(C)$ $f^{\prime}(x) < f(x)$ for $x \in (1/4, 1)$
$(D)$ $f^{\prime}(x) > f(x)$ for $x \in (1/4, 1)$