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

Answer 1

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

$S\left(\alpha \right)=\left\{\right(x,y):y^2\le x,0\le x\le \alpha \}$
$A\left(\alpha \right)=2{\int }_0^{\alpha }\sqrt{x}dx=2{\alpha }^{\frac{3}{2}}$
$A\left(4\right)=2\times 4^{3/2}=16$
$A\left(\lambda \right)=2\times {\lambda }^{3/2}$
$\frac{A\left(\lambda \right)}{A\left(4\right)}=\frac{2}{5}\Rightarrow \lambda =4.{\left(\frac{4}{25}\right)}^{1/3}$