If $x=-2$ and $x=4$ are the extreme points of $y=x^3-\alpha x^2-\beta x+5$,then

Twenty metres of wire is available to fence off a flower bed in the form of a circular sector. What must the radius of the circle be, if the area of the flower bed is to be the greatest (in $m$)?

Let $a, b \in R$ be such that the function $f(x) = \ln|x| + bx^2 + ax, x \neq 0$ has extreme values at $x = -1$ and $x = 2$.
Statement-$1$: $f$ has a local maximum at $x = -1$ and $x = 2$.
Statement-$2$: $a = \frac{1}{2}$ and $b = -\frac{1}{4}$.

Let a function $f(x) = \begin{cases} -\ln(3x - [3x]) & ; 3x \neq n, n \in N \\ \ln(\operatorname{sgn}(3x)) & ; 3x = n, n \in N \end{cases}$,where $[.]$ and $\operatorname{sgn}(x)$ denote the greatest integer function and signum function respectively. Then the number of points where $f(x)$ is minimum in $x \in (0, 5)$ is:

Let $f: R \rightarrow R$ be given by $f(x) = |x^{2} - 1|$, $x \in R$. Then:

If the function $f$ given by $f(x) = x^3 - 3(a - 2)x^2 + 3ax + 7$ for some $a \in R$ is increasing in $(0, 1]$ and decreasing in $[1, 5)$,then a root of the equation $\frac{f(x) - 14}{(x - 1)^2} = 0$ $(x \neq 1)$ is