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

  • A
    $0$
  • B
    $4$
  • C
    $5$
  • D
    $8$

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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

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