JEE Mains · Maths · STD 12 - 9. differential equations
Let \(\mathrm{Y}=\mathrm{Y}(\mathrm{X})\) be a curve lying in the first quadrant such that the area enclosed by the line \(Y-y=Y^{\prime}(x)(X-x)\) and the co-ordinate axes, where \((\mathrm{x}, \mathrm{y})\) is any point on the curve, is always \(\frac{-y^2}{2 Y^{\prime}(x)}+1, Y^{\prime}(x) \neq 0\). If \(Y(1)=1\), then \(12 Y(2)\) equals
- A \(20\)
- B \(10\)
- C \(11\)
- D \(15\)
Answer & Solution
Correct Answer
(A) \(20\)
Step-by-step Solution
Detailed explanation
\(\mathrm{A}=\frac{1}{2}\left(\frac{-\mathrm{y}}{\mathrm{Y}^{\prime}(\mathrm{x})}+\mathrm{x}\right)(\mathrm{y}-\mathrm{xY} / \mathrm{x})=\frac{-\mathrm{y}^2}{2 \mathrm{Y}^{\prime}(\mathrm{x})}+1\)…
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