Area Question 6
Question 6
- Let $S(\alpha)=\left{(x, y): y^{2} \leq x, 0 \leq x \leq \alpha\right}$ and $A(\alpha)$ is area of the region $S(\alpha)$. If for $\lambda, 0<\lambda<4, A(\lambda): A(4)=2: 5$, then $\lambda$ equals
(2019 Main, 8 April II) (a) $2 \frac{4}{25}^{\frac{1}{3}}$ (b) $4 \frac{2}{5}^{\frac{1}{3}}$ (c) $4 \frac{4}{25}^{\frac{1}{3}}$ (d) $2 \frac{2}{5}^{\frac{1}{3}}$
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Solution:
- Given, $S(\alpha)=\left{(x, y): y^{2} \leq x, 0 \leq x \leq \alpha\right}$ and $A(\alpha)$ is area of the region $S(\alpha)$
Clearly, $A(\lambda)=2 \int_{0}^{\lambda} \sqrt{x} d x=2 \frac{x^{3 / 2}}{3 / 2}{ }_{0}^{\lambda}=\frac{4}{3} \lambda^{3 / 2}$
Since, $\frac{A(\lambda)}{A(4)}=\frac{2}{5},(0<\lambda<4)$
$\Rightarrow \quad \frac{\lambda^{3 / 2}}{4^{3 / 2}}=\frac{2}{5} \quad \Rightarrow \quad \frac{\lambda^{3}}{4}=\frac{2}{5}^{2}$
$\Rightarrow \quad \frac{\lambda}{4}=\frac{4}{25}^{1 / 3} \Rightarrow \lambda=4 \frac{4}{25}^{1 / 3}$
