JEE Mains · Maths · STD 12 - 8. Application and integration
Let a curve \(y = y ( x )\) pass through the point \((3,3)\) and the area of the region under this curve, above the \(x\)-axis and between the abscissae \(3\) and \(x(>3)\) be \(\left(\frac{y}{x}\right)^{3}\). If this curve also passes through the point \((\alpha, 6 \sqrt{10})\) in the first quadrant, then \(\alpha\) is equal to \(........\)
- A \(5\)
- B \(4\)
- C \(6\)
- D \(8\)
Answer & Solution
Correct Answer
(C) \(6\)
Step-by-step Solution
Detailed explanation
\(x^{4}=3 y x \cdot y^{\prime}-3 y^{2}\) \(3 x y \frac{d y}{d x}=3 y^{2}+x^{4}\) Put \(y ^{2}= t , y \frac{ dy }{ dx }=\frac{1}{2} \frac{ dt }{ dx }\) \(\frac{ dt }{ dx }-\frac{2}{ x } t =\frac{2}{3} x ^{3}\) \(\therefore \frac{ t }{ x ^{2}}=\frac{ x ^{2}}{3}+ C\)…
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