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 .............

Let $f : R \to R$ be a differentiable function satisfying $f''(3) + f'(2) = 0$. Then $\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{1 + f\left( {3 + x} \right) - f\left( 3 \right)}}{{1 + f\left( {2 - x} \right) - f\left( 2 \right)}}} \right)^{\frac{1}{x}}}$ is equal to

If a function $f$ is defined by:
$\begin{cases} f(x) = x-1, & \text{when } -\infty < x < 1 \\ f(x) = 0, & \text{when } x=1 \\ f(x) = x^3-1, & \text{when } 1 < x < \infty \end{cases}$
then at $x=1$,$f$ is:

If $f(x) = \begin{cases} 3x^2 + 12x - 1, & -1 \le x \le 2 \\ 37 - x, & 2 < x \le 3 \end{cases}$,then:

If $f:R \to R$ and $f(x)$ is a polynomial function of degree $10$ such that $f(x)=0$ has all real and distinct roots,then the equation $(f'(x))^2 - f(x)f''(x) = 0$ has:

The line which is parallel to the $x$-axis and intersects the curve $y = \sqrt{x}$ at an angle of $\frac{\pi}{4}$ is: