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

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

The set of all real numbers satisfying the inequation $x^2-|x+2|+x>0$ 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

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