JEE Mains · Maths · STD 12 - 9. differential equations
The slope of the tangent to a curve \(C : y = y ( x )\) at any point \([ x , y )\) on it is \(\frac{2 e ^{2 x }-6 e ^{- x }+9}{2+9 e ^{-2 x }}\). If \(C\) passes through the points \(\left(0, \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}\right)\) and \(\left(\alpha, \frac{1}{2} e ^{2 \alpha}\right)\) then \(e ^{\alpha}\) is equal to.
- A \(\frac{3+\sqrt{2}}{3-\sqrt{2}}\)
- B \(\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)\)
- C \(\frac{1}{\sqrt{2}}\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)\)
- D \(\frac{\sqrt{2}+1}{\sqrt{2}-1}\)
Answer & Solution
Correct Answer
(B) \(\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}=\frac{2 e^{2 x}-6 e^{-x}+9}{2+9 e^{-2 x}}\) \(\frac{d y}{d x}=e^{2 x}-\frac{6 e^{x}}{2 e^{2 x}+9}\) \(y=\frac{e^{2 x}}{2}-\tan ^{-1}\left(\frac{\sqrt{2} e^{x}}{3}\right)+c\) If \(C\) passes through the point…
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