Given that $x$ is a real number satisfying $\frac{5x^{2}-26x+5}{3x^{2}-10x+3} < 0$,then

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

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

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 number of integral values of $x$ satisfying $9 x-2 < (x+2)^2 < 12 x-3$ is

If the inequality $kx^2 - 2x + k \geq 0$ holds good for at least one real $x$,then the complete set of values of $k$ is