SATHEE: Area Question 6

Area Question 6

Question 6

Let $Missing or unrecognized delimiter for \left$ and $A (α)$ is area of the region $S (α)$ . If for $λ, 0 < λ < 4, A (λ) : A (4) = 2 : 5$ , then $λ$ 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, $Missing or unrecognized delimiter for \left$ and $A (α)$ is area of the region $S (α)$

Clearly, $A (λ) = 2 \int_{0}^{λ} \sqrt{x} d x = 2 \frac{x^{3 / 2}}{3 / 2}_{0}^{λ} = \frac{4}{3} λ^{3 / 2}$

Since, $\frac{A (λ)}{A (4)} = \frac{2}{5}, (0 < λ < 4)$

$\Rightarrow \frac{λ^{3 / 2}}{4^{3 / 2}} = \frac{2}{5} \Rightarrow \frac{λ^{3}}{4} = {\frac{2}{5}}^{2}$

$\Rightarrow \frac{λ}{4} = {\frac{4}{25}}^{1 / 3} \Rightarrow λ = 4 {\frac{4}{25}}^{1 / 3}$

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