JEE Mains · Maths · STD 12 - 8. Application and integration
If the area enclosed by the parabolas \(P_1: 2 y=5 x^2\) and \(P_2: x^2-y+6=0\) is equal to the area enclosed by \(P_1\) and \(y=\alpha x, \alpha > 0\), then \(\alpha^3\) is equal to \(......\).
- A \(559\)
- B \(600\)
- C \(601\)
- D \(602\)
Answer & Solution
Correct Answer
(B) \(600\)
Step-by-step Solution
Detailed explanation
Abscissa of point of intersection of \(2 y =5 x ^2\) and \(y=x^2+6\) is \(\pm 2\) \(\text { Area }=2 \int \limits_0^2\left(x^2+6-\frac{5 x^2}{2}\right) d x=\int \limits_0^{\frac{2 \alpha}{5}}\left(\alpha x-\frac{5 x^2}{2}\right) d x\)…
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