If $x=1$ and $x=2$ are extreme points of $f(x)=\alpha \log x+\beta x^2+x$,where $\alpha$ and $\beta$ are constants,then the value of $\alpha^2+2 \beta$ is

Let $f :[2,4] \rightarrow R$ be a differentiable function such that $(x \ln x) f'(x) + (\ln x + 1) f(x) \geq 1$ for all $x \in [2,4]$,with $f(2) = \frac{1}{2}$ and $f(4) = \frac{1}{4}$. Consider the following two statements:
$(A): f(x) \leq 1$ for all $x \in [2,4]$
$(B): f(x) \geq \frac{1}{8}$ for all $x \in [2,4]$
Then,

Statement-$I$: The sequence $a_n = \frac{n^2}{n^3 + 200}, n \in N$ has its $7^{th}$ term as the largest term.
Statement-$II$: The function $f(x) = \frac{x^2}{x^3 + 200}$ attains a local maximum at $x = 7$.

The minimum value of $\frac{\log x}{x}$ in the interval $(2, \infty)$ is

Let a function $f(x)$ be defined as $f(x) = \begin{cases} \cos^{-1}(\mu) + x^2, & 0 < x < 1 \\ 4x, & x \geqslant 1 \end{cases}$. The function $f(x)$ can have a local minimum at $x = 1$ if the value of $\mu$ lies in the interval:

An enemy Apache helicopter is flying along the curve given by $y = x^{2} + 7$. $A$ soldier,placed at $(3, 7)$,wants to shoot down the helicopter when it is nearest to him. Find the nearest distance.