If for the solution curve y=f(x) of the differential equation d y ~d x +( x) y=(2+ x)/((1+2 x)^2) , x ∈ ((-π)/(2), (π)/(2)), f ((π)/(3))= 3 10 , then f((π)/(4)) is equal to :

Correct answer: D. 4- 2 14 aligned & If e^ ∫ xdx =e^ ( x) = x & y · x=∫\ 2+ x (1+2 x)^2 \ xdx & =∫ 2 x+1 ( x+2)^2 dx Let x= 1-t^2 1+t^2 & =∫ 2( 1-t^2 1+t^2 )+1 ( 1-t^2 1+t^2 +2)^2 2 dt aligned aligned & =∫ (2-2 t^2+1+t^2)/((1-t^2+2+2 t^2)^2) × 2 dt & =2 ∫ (3-t^2)/((t^2+3)^2) dt aligned Let t+ 3 t =u aligned & (1- 3 t^2 ) dt=du & =-2 ∫ du u^2 aligned aligned & y ·( x)=(2)/(u)+c & y · x=(2)/(t+(3)/(t))+c...(I) aligned At x=(π)/(3) , t= x 2 = 1 3 2 · 3 10 = 2 1 3 +3 3 +c aligned & 2. 3 10 = 2 3 10 +c C=0 & At x=(π)/(4), t= x 2 = 2 -1 & y · 2 = 2 2 -1+ 3 2 -1 & y · 2 = 2( 2 -1) 6-2 2 & y= 2 ( 2 -1) 2(3- 2 ) = 1 2 × 2 2 -1 7 & = 4- 2 14 aligned

JEE Main 202529 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 29 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2025 · 29 January · Evening Shift · Mathematics · Differential Equations

If for the solution curve y=f(x) of the differential equation d y...

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 :

Options

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 $\frac{4-\sqrt{2}}{14}$

Correct answer

D. $\frac{4-\sqrt{2}}{14}$

Step-by-step solution

\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 \mathrm{x}+1}{(\cos \mathrm{x}+2)^2} \mathrm{dx} \text{ Let } \cos \mathrm{x}=\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2} \\ & =\int \frac{2\left(\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2}\right)+1}{\left(\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2}+2\right)^2} 2 \mathrm{dt}\end{aligned}

\begin{aligned} & =\int \frac{2-2 t^2+1+t^2}{\left(1-t^2+2+2 t^2\right)^2} \times 2 \mathrm{dt} \\ & =2 \int \frac{3-t^2}{\left(t^2+3\right)^2} \mathrm{dt} \end{aligned}

Let

\mathrm{t}+\frac{3}{\mathrm{t}}=\mathrm{u}

\begin{aligned} & \left(1-\frac{3}{\mathrm{t}^2}\right) \mathrm{dt}=\mathrm{du} \\ & =-2 \int \frac{\mathrm{du}}{\mathrm{u}^2} \end{aligned}

\begin{aligned} & y \cdot(\sec x)=\frac{2}{u}+c \\ & y \cdot \sec x=\frac{2}{t+\frac{3}{t}}+c...(I) \end{aligned}

\text{ At } \mathrm{x}=\frac{\pi}{3}

,

\mathrm{t}=\tan \frac{\mathrm{x}}{2}=\frac{1}{\sqrt{3}} 2 \cdot \frac{\sqrt{3}}{10}=\frac{2}{\frac{1}{\sqrt{3}}+3 \sqrt{3}}+\mathrm{c}

\begin{aligned} & \text{ 2. } \frac{\sqrt{3}}{10}=\frac{2 \sqrt{3}}{10}+\mathrm{c} \Rightarrow \mathrm{C}=0 \\ & \text{ At } \mathrm{x}=\frac{\pi}{4}, \mathrm{t}=\tan \frac{\mathrm{x}}{2}=\sqrt{2}-1 \\ & \therefore \mathrm{y} \cdot \sqrt{2}=\frac{2}{\sqrt{2}-1+\frac{3}{\sqrt{2}-1}} \\ & \mathrm{y} \cdot \sqrt{2}=\frac{2(\sqrt{2}-1)}{6-2 \sqrt{2}} \\ & \mathrm{y}=\frac{\sqrt{2}(\sqrt{2}-1)}{2(3-\sqrt{2})}=\frac{1}{\sqrt{2}} \times \frac{2 \sqrt{2}-1}{7} \\ & =\frac{4-\sqrt{2}}{14}\end{aligned}

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