JEE Mains · Maths · STD 12 - 9. differential equations
Let the tangent at any point \(P\) on a curve passing through the points \((1,1)\) and \(\left(\frac{1}{10}, 100\right)\), intersect positive \(x\)-axis and \(y\)-axis at the points \(A\) and \(B\) respectively. If \(P A: P B=1: k\) and \(y=y(x)\) is the solution of the differential equation \(e^{\frac{d y}{d x}}=k x+\frac{k}{2}\), \(y(0)=k\), then \(4 y(1)-5 \log _e 3\) is equal to \(.........\).
- A \(4\)
- B \(3\)
- C \(5\)
- D \(2\)
Answer & Solution
Correct Answer
(C) \(5\)
Step-by-step Solution
Detailed explanation
equation of tangent at \(P(x, y)\) \(Y-y=\frac{d y}{d x}(X-x)\) \(Y=0\) \(X=\frac{-y d x}{d y}+x\) \(\frac{ k \alpha+0}{ k +1}= x , \alpha=\frac{ k +1}{ k } x\) \(\frac{ k +1}{ k } x =- y \frac{ dx }{ dy }+ x\) \(x +\frac{ x }{ k }=- y \frac{ dx }{ dy }+ x\)…
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