Area Ques 67
Question
- 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) $2 (\frac{2}{5}^){\frac{1}{3}}$
Show Answer
Answer:
Correct Answer: 67.(c)
Solution:
Formula:
- 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, $\frac{A(\lambda)}{A(4)}=\frac{2}{5},(0<\lambda<4)$
$\Rightarrow \quad \frac{\lambda^{3 / 2}}{4^{3 / 2}}=\frac{2}{5} \quad \Rightarrow \quad (\frac{\lambda}{4})^3=(\frac{2}{5})^{2}$
$\Rightarrow \quad \frac{\lambda}{4}=(\frac{4}{25})^{1 / 3} \Rightarrow \lambda=4 (\frac{4}{25})^{1 / 3}$
BETA
