The number of polynomials $p: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $p(0)=0$,$p(x) > x^2$ for all $x \neq 0$,and $p^{\prime \prime}(0) = \frac{1}{2}$ is

$\mathop {\lim }\limits_{x \to 0} \frac{d}{{dx}}\left( {\frac{{{e^{{e^{{x^2}}}}} - e}}{x}} \right)$ is

The function $f(x) = \sqrt{1 - \sqrt{1 - x^2}}$

Let $f(x)=x^2+a x+b$,where $a, b \in R$. If $f(x)=0$ has all its roots imaginary,then the roots of $f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=0$ are

Let $f, g: R \rightarrow R$ be two real-valued functions defined as $f(x)=\begin{cases} -|x+3| & , x < 0 \\ e^{x} & , x \geq 0 \end{cases}$ and $g(x)=\begin{cases} x^{2}+k_{1} x & , x < 0 \\ 4 x+k_{2} & , x \geq 0 \end{cases}$,where $k_{1}$ and $k_{2}$ are real constants. If $(g \circ f)$ is differentiable at $x=0$,then $(g \circ f)(-4)+(g \circ f)(4)$ is equal to

If $y=|\cos x-\sin x|+|\tan x-\cot x|$,then $\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{3}}+\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{6}}=$