Let S() = (x,y) : y^2 x, 0 x and A() be the area of the region S().

Question

Let $S(\alpha) = \{(x,y) : y^2\le x,\ 0\le x\le\alpha\}$ and $A(\alpha)$ be the area of the region $S(\alpha)$. If for $\lambda$, $0<\alpha<4$, $A(\lambda):A(\alpha) = 2:5$... *(as stated in the source: $A(\lambda):A(4)=2:5$)*, then $\lambda$ equals:

*[Online May 12, 2012]* *(Note: source metadata for this JEE Main question shows "[April 08, 2019 (II)]" style; retaining as printed.)*

Accepted Answer

$$A(\alpha) = 2\int_0^\alpha\sqrt x\,dx = \frac43\alpha^{3/2}$$

Given $\dfrac{A(\lambda)}{A(4)}=\dfrac25$:

$$\frac{\lambda^{3/2}}{4^{3/2}} = \frac25 \implies \lambda^{3/2} = \frac{2}{5}\cdot8 = \frac{16}{5} \implies \lambda = \left(\frac{16}{5}\right)^{2/3} = 4\left(\frac{4}{25}\right)^{1/3}$$

Solutions · 1