# Let s(alpha)

Question:

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

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

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

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

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

Correct Option: , 4

Solution:

Area of the region $=2 \times \int_{0}^{\lambda} y d x=2 \int_{0}^{\lambda} \sqrt{x} d x$

$=2 \times \frac{2}{3} \pi^{\frac{3}{2}}$

$\mathrm{A}(\lambda)=2 \times \frac{2}{3}(\lambda \times \sqrt{\lambda})=\frac{4}{3} \lambda^{\frac{3}{2}}$

Given, $\frac{A(\lambda)}{A(4)}=\frac{2}{5} \Rightarrow \frac{\lambda^{\frac{3}{2}}}{8}=\frac{2}{5}$

$\lambda_{Q 23}=\left(\frac{16}{5}\right)^{\frac{2}{3}}=4 \cdot\left(\frac{4}{25}\right)^{\frac{1}{3}}$