JEE Mains · Maths · STD 12 - 9. differential equations
If for the solution curve \(y=f(x)\) of the differential equation \(\frac{\mathrm{d} y}{\mathrm{~d} x}+(\tan x) y=\frac{2+\sec x}{(1+2 \sec x)^2}\), \(x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right), f\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{10}\), then \(f\left(\frac{\pi}{4}\right)\) is equal to :
- A \(\frac{\sqrt{3}+1}{10(4+\sqrt{3})}\)
- B \(\frac{5-\sqrt{3}}{2 \sqrt{2}}\)
- C \(\frac{9 \sqrt{3}+3}{10(4+\sqrt{3})}\)
- D \(\frac{4-\sqrt{2}}{14}\)
Answer & Solution
Correct Answer
(D) \(\frac{4-\sqrt{2}}{14}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { If } \mathrm{e}^{\int \tan \mathrm{xdx}}=\mathrm{e}^{\ln (\sec \mathrm{x})}=\sec \mathrm{x} \\ & \therefore \mathrm{y} \cdot \sec \mathrm{x}=\int\left\{\frac{2+\sec \mathrm{x}}{(1+2 \sec \mathrm{x})^2}\right\} \sec \mathrm{xdx} \\ & =\int \frac{2 \cos…
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