JEE Mains · Maths · STD 12 - 8. Application and integration
Let \(S\left( \alpha \right) = \left\{ {\left( {x,y} \right):{y^2} \leq x,0 \leq \alpha } \right\}\) and \(A(\alpha )\) is area of the regions \(S(\alpha )\). If for a \(\lambda ,0 < \lambda < 4,A (\lambda ) : A\left( 4 \right)\,=\,2:5\) then \(\lambda \) equals
- A \(4{\left( {\frac{2}{5}} \right)^{\frac{1}{3}}}\)
- B \(2{\left( {\frac{2}{5}} \right)^{\frac{1}{3}}}\)
- C \(4{\left( {\frac{4}{25}} \right)^{\frac{1}{3}}}\)
- D \(2{\left( {\frac{4}{25}} \right)^{\frac{1}{3}}}\)
Answer & Solution
Correct Answer
(C) \(4{\left( {\frac{4}{25}} \right)^{\frac{1}{3}}}\)
Step-by-step Solution
Detailed explanation
\(S\left( \lambda \right) = 2\int\limits_0^\lambda {\sqrt x dx = \frac{4}{3}} {\lambda ^{3/2}}\) \(\frac{S(\lambda)}{S(4)}=\frac{2}{5} \Rightarrow \frac{\lambda^{3 / 2}}{4^{3 / 2}}=\frac{2}{5}\) \(\Rightarrow \lambda=4\left(\frac{4}{25}\right)^{1 / 3}\)
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