JEE Main 2025 — Mathematics Calculus

$$\begin{aligned} \text { I.F. } & =\mathrm{e}^{\int 2 \sec ^2 x d x} \\ & =\mathrm{e}^{2 \tan x}\end{aligned}$$ I.F. ​=e∫2sec2xdx=e2tanx​

Solution of diff. eq.

$$\begin{gathered}y \cdot e^{2 \tan x}=\int e^{2 \tan x}\left(2 \sec ^2 x+3 \tan x \cdot \sec ^2 x\right) d x \\ y \cdot e^{2 \tan x}=\int e^{2 \tan x} \cdot\left(2 \sec ^2 x\right) d x+\int e^{2 \tan x} \cdot\left(3 \tan x \cdot \sec ^2 x\right) d x\end{gathered}$$y⋅e2tanx=∫e2tanx(2sec2x+3tanx⋅sec2x)dxy⋅e2tanx=∫e2tanx⋅(2sec2x)dx+∫e2tanx⋅(3tanx⋅sec2x)dx​

y. $\displaystyle \mathrm{e}^{2 \tan x}=\mathrm{e}^{2 \tan x} \cdot 2 \tan x-\int \mathrm{e}^{2 \tan x} \cdot 2 \sec ^2 x \times 2 \tan x d x+\int \mathrm{e}^{2 \tan x} \cdot 3 \tan x \cdot \sec ^2 x d x$e2tanx=e2tanx⋅2tanx−∫e2tanx⋅2sec2x×2tanxdx+∫e2tanx⋅3tanx⋅sec2xdx

$$\begin{aligned} & y \cdot e^{2 \tan x}=2 \tan x \cdot e^{2 \tan x}-\int \mathrm{e}^{2 \tan x} \cdot \tan x \sec ^2 x d x \\ & y . e^{2 \tan x}=2 \tan x \cdot e^{2 \tan x}-\frac{\tan x \cdot e^{2 \tan x}}{2}+\frac{e^{2 \tan x}}{4}+C\end{aligned}$$​y⋅e2tanx=2tanx⋅e2tanx−∫e2tanx⋅tanxsec2xdxy.e2tanx=2tanx⋅e2tanx−2tanx⋅e2tanx​+4e2tanx​+C​

$$\begin{aligned} & y=2 \tan x-\frac{\tan x}{2}+\frac{1}{4}+\mathrm{Ce}^{-2 \tan x} \\ & x=0, y=\frac{5}{4} \\ & c=1\end{aligned}$$​y=2tanx−2tanx​+41​+Ce−2tanxx=0,y=45​c=1​

$$\mathrm{y}\left(\frac{\pi}{4}\right)=\frac{7}{4}+\mathrm{e}^{-2}$$y(4π​)=47​+e−2

Then $12\left(y\left(\frac{\pi}{4}\right)-\mathrm{e}^{-2}\right)=12\left(\frac{7}{4}\right)=21$12(y(4π​)−e−2)=12(47​)=21