Let $f(x)$ be a polynomial of degree $6$ in $x$,in which the coefficient of $x^{6}$ is unity and it has extrema at $x=-1$ and $x=1$. If $\lim_{x \rightarrow 0} \frac{f(x)}{x^{3}}=1$,then $5 \cdot f(2)$ is equal to .............

The value of $f(4)-f(3)$ is

Let $f(x) = x|x|$,$g(x) = \sin x$ and $h(x) = (g \circ f)(x)$. Then

Let $f: ( -\infty, \infty ) \to ( -\infty, \infty )$ be defined by $f(x) = x^3 + 1$.
Statement $1$: The function $f$ has a local extremum at $x = 0$.
Statement $2$: The function $f$ is continuous and differentiable on $( -\infty, \infty )$ and $f'(0) = 0$.

Let $f(x) = \begin{cases} 3x, & x < 0 \\ \min \{1+x+[x], x+2[x]\}, & 0 \leq x \leq 2 \\ 5, & x > 2 \end{cases}$ where $[.]$ denotes the greatest integer function. If $\alpha$ and $\beta$ are the number of points where $f$ is not continuous and is not differentiable,respectively,then $\alpha + \beta$ equals.......

Let $f(x)$ be defined in $[-2, 2]$ by
$f(x) = \begin{cases} \max(4 - x^2, 1 + x^2), & -2 < x < 0 \\ \min(4 - x^2, 1 + x^2), & 0 < x < 2 \end{cases}$
Then $f(x)$: