If $x^2-4ax+5+a>0$ for all $x \in R$ whenever $a \in (\alpha, \beta)$,then $4\beta+\alpha=$

The set of all solutions of the inequation $x^2 - 2x + 5 \leq 0$ in $R$ is

Assertion $(A)$: $3x^2 - 16x + 4 > -16$ is satisfied for some values of real $x$ in $(0, \frac{10}{3})$.
Reason $(R)$: $ax^2 + bx + c$ and $a$ will have the same sign for some values of $x \in \mathbb{R}$ when $b^2 - 4ac > 0$.
The correct option among the following is

The set of values of $x$ for which the inequalities $x^2-3x-10 < 0$ and $10x-x^2-16 > 0$ hold simultaneously is:

For $x \in R-\{-6\}$,the value of $\frac{(x+2)(x+5)}{(x+6)}$ does not lie in the interval