The solution set contained in $R^{+}$ of the inequation $3^x + 3^{1-x} - 4 < 0$ is

The solution set of $x^{2} \leq 4$ 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

For $0 \leq p \leq 1$ and for any positive $a, b$, let $I(p)=(a+b)^{p}$ and $J(p)=a^{p}+b^{p}$. Then:

If $x^2-5x-14 > 0$ implies $x$ lies outside $[\alpha, \beta]$,then find the value of $\frac{\alpha}{\beta}$.

The integral value of $k$ for which $x^2 - 2(4k - 1)x + 15k^2 - 2k - 7 > 0$ for all $x \in R$ is