JEE Main 2025 (Online) 29th January Evening ShiftQuestion 1
If for the solution curve $y=f(x)$ of the differential equation $\frac{dy}{dx}+(\tan x)y=\frac{2+\sec x}{(1+2\sec x)^{2}}$, $x\in\left(-\frac\pi2,\frac\pi2\right)$, $f\left(\frac\pi3\right)=\frac{\sqrt3}{10}$, then $f\left(\frac\pi4\right)$ is equal to :
Differential Equation
Solution
Concept. With IF $\sec x$ the right side becomes the derivative of $\frac{\sin x}{\cos x+2}$. Formulas used. (1) $\text{IF}=e^{\int\tan x\,dx}=\sec x$(2) $\left(\dfrac{\sin x}{\cos x+2}\right)'=\dfrac{1+2\cos x}{(\cos x+2)^2}$Step 1 - Multiply by the IF. Writing $\sec x=\frac1{\cos x}$: $$(y\sec x)'=\sec x\cdot\frac{2+\sec x}{(1+2\sec x)^2}=\frac{2\cos x+1}{(\cos x+2)^2}.$$ Step 2 - Integrate. By (2), $$y\sec x=\frac{\sin x}{\cos x+2}+C\ \Longrightarrow\ y=\frac{\sin x\cos x}{\cos x+2}+C\cos x.$$ Step 3 - Constant. At $\frac\pi3$: $\frac{\sqrt3/4}{5/2}+\frac C2=\frac{\sqrt3}{10}+\frac C2=\frac{\sqrt3}{10}$, so $C=0$. Step 4 - At $\frac\pi4$. $$f\left(\tfrac\pi4\right)=\frac{1/2}{\frac{\sqrt2}{2}+2}=\frac{1}{4+\sqrt2}=\frac{4-\sqrt2}{14}.$$ Answer: (D)