Let S(α) = {(x, y) : y2 ≤ x, 0 ≤ x ≤ α} and A(α) is area of the region S(α). If for aλ, 0 < λ < 4, A(λ) : A(4) = 2 : 5, then λ equals:

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

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

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

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

Let S(α) = {(x, y) : y2 ≤ x, 0 ≤ x ≤ α} and A(α) is area of t

1.	Let S(α) = {(x, y) : y2 ≤ x, 0 ≤ x ≤ α} and A(α) is area of the region S(α). If for aλ, 0 < λ < 4, A(λ) : A(4) = 2 : 5, then λ equals:
A.	\(2{\left( {\frac{4}{{25}}} \right)^{\frac{1}{3}}}\)
B.	\(2{\left( {\frac{2}{5}} \right)^{\frac{1}{3}}}\)
C.	\(4{\left( {\frac{2}{5}} \right)^{\frac{1}{3}}}\)
D.	\(4{\left( {\frac{4}{{25}}} \right)^{\frac{1}{3}}}\)
Answer» E.

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Let S(α) = {(x, y) : y2 ≤ x, 0 ≤ x ≤ α} and A(α) is area of the region S(α). If for aλ, 0 < λ < 4, A(λ) : A(4) = 2 : 5, then λ equals:

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