If the maximum area bounded by $y^2=4x$ and the line $y=mx$ $\left(\forall m\in \left[1,3\right]\right)$ is $k$ square units, then the smallest prime number greater than $3k$ is

• A

  $3$
• B

  $5$
• C

  $7$
• Correct answer
  D

  $11$

Solution

The area covered is maximum if $‘m’$ is minimum, i.e. $m=1$

$\therefore$ Required area ${\int }_0^4\left(\sqrt{4x}-x\right)dx$
$={\left[2\cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^2}{2}\right]}_0^4$
$={\left[\frac{4}{3}{x}^{\frac{3}{2}}-\frac{x^2}{2}\right]}_0^4$
$=\frac{32}{3}-8=\frac{8}{3}$ sq. units
$\therefore k=\frac{8}{3}$ 
Thus required prime number $=11$

Hence, the correct answer is option: 'D'.