Question

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

• A. $4{\left(\frac{4}{25}\right)}^{1/3}$
• B. $4{\left(\frac{2}{5}\right)}^{1/3}$
• C. $2{\left(\frac{2}{5}\right)}^{1/3}$
• D. $2{\left(\frac{4}{25}\right)}^{1/3}$

Answer 1

The correct option is A $4{\left(\frac{4}{25}\right)}^{1/3}$

$A\left(\alpha \right)=2\underset{0}{\overset{\alpha }{\int }}\sqrt{x}dx=\frac{4}{3}{\alpha }^{3/2}$
$\frac{A\left(\lambda \right)}{A\left(4\right)}=\frac{2}{5}$
$\Rightarrow \frac{{\lambda }^{3/2}}{4^{3/2}}=\frac{2}{5}$
$\Rightarrow {\lambda }^{3/2}=\frac{16}{5}$
$\Rightarrow \lambda ={\left(\frac{16}{5}\right)}^{2/3}$
$\Rightarrow \lambda =4{\left(\frac{4}{25}\right)}^{1/3}$