Let $S\left(\alpha \right)=\left\{\left(x,y\right):y^2\le x, 0\le x\le \alpha \right\}$ and $\mathrm{A}\left(\alpha \right)$ is area of the region $S\left(\alpha \right).$ If for a $\lambda ,0<\lambda <4, \mathrm{A}\left(\lambda \right):\mathrm{A}\left(4\right)=2:5,$then $\lambda$ equals:

• A

  $4 {\left(\frac{2}{5}\right)}^{\frac{1}{3}}$
• B

  $2 {\left(\frac{4}{25}\right)}^{\frac{1}{3}}$
• Correct answer
  C

  $4 {\left(\frac{4}{25}\right)}^{\frac{1}{3}}$
• D

  $2 {\left(\frac{2}{5}\right)}^{\frac{1}{3}}$

Solution

$S\left(\lambda \right)=2\underset{0}{\overset{\lambda }{\int }}\sqrt{x}dx=2{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^{\lambda }$
$=\frac{4}{3}{\lambda }^{\frac{3}{2}}$
$\therefore S\left(4\right)=\frac{4}{3}{\left(4\right)}^{\frac{3}{2}}$
Given $\frac{S\left(\lambda \right)}{S\left(4\right)}=\frac{2}{5}\Rightarrow \frac{{\lambda }^{\frac{3}{2}}}{4^{\frac{3}{2}}}=\frac{2}{5}$
$\Rightarrow \lambda =4{\left(\frac{4}{25}\right)}^{\frac{1}{3}}$

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