The function defined by $f(x) = \begin{cases} |x - 3|, & x \ge 1 \\ \frac{1}{4}x^2 - \frac{3}{2}x + \frac{13}{4}, & x < 1 \end{cases}$ is

If $\operatorname{Lt}_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=e^x(x+1)$ and $f(0)=0$,then $\frac{d}{d x}\left(f(x) e^{-x}\right)+\frac{d}{d x}\left(\frac{f(x)}{x}\right)=$

Let $f: R \to R$ be a twice differentiable function such that the quadratic equation $f(x)m^{2}-2f^{\prime}(x)m+f^{\prime\prime}(x)=0$ in $m$ has two equal roots for every $x \in R$. If $f(0)=1$,$f^{\prime}(0)=2$ and $(\alpha, \beta)$ is the largest interval in which the function $g(x) = f(\log_{e}x-x)$ is increasing,then $\alpha+\beta$ is equal to:

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

Let $f(x) = \begin{cases} \int_{0}^{x} |1-t| dt, & x > 1 \\ x - \frac{1}{2}, & x \leq 1 \end{cases}$. Then:

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