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 what values of $x$ is $\frac{8x^2 + 16x - 51}{(2x - 3)(x + 4)} > 3$?

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

Let $f(x) = x^2 + 4x + 1$. Then

The solution of the equation $2x^2 + 3x - 9 \le 0$ is given by