Let S α = x , y : y 2 ≤ x ,   0 ≤ x ≤ α   and A α is area of the…

1. $4 {\left(\frac{2}{5}\right)}^{\frac{1}{3}}$
2. $2 {\left(\frac{4}{25}\right)}^{\frac{1}{3}}$
3. $4 {\left(\frac{4}{25}\right)}^{\frac{1}{3}}$
4. $2 {\left(\frac{2}{5}\right)}^{\frac{1}{3}}$

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 $\frac{S\left(\lambda \right)}{S\left(4\right)}=\frac{2}{5}\Rightarrow \frac{{\lambda }^{\frac{3}{2}}}{4^{\frac{3}{2}}}=\frac{2}{5}$
$\Rightarrow \lambda =4{\left(\frac{4}{25}\right)}^{\frac{1}{3}}$

Asked in: JEE Main 2019 (08 Apr Shift 2)

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