JEE Main 2025 — Mathematics Calculus

$$\begin{aligned} & x^2=2 y \& y=2 x+6 \\ & x^2=4 x+12\end{aligned}$$​x2=2y&y=2x+6x2=4x+12​

$$\left.x^2-4 x-12=0 \Rightarrow \begin{gathered}x=6 \\ y=18\end{gathered} \right\rvert\, \begin{gathered}\text { if } x=-2 \\ y=2\end{gathered}$$x2−4x−12=0⇒x=6y=18​​ if x=−2y=2​

$$\therefore(6,18) \&(-2,2)$$∴(6,18)&(−2,2)

Here $(6,18)$(6,18) Rejected because $(a, b)$(a,b) lies in $2^{\text {nd }}$2nd  quadrant

$$\begin{aligned} & \therefore \mathrm{a}=-2 \& \mathrm{~b}=2 \\ & \therefore \mathrm{I}=\int_{-2}^2 \frac{9 \mathrm{x}^2}{1+5^{\mathrm{x}}} \mathrm{dx}=\int_{-2}^2 \frac{9 \cdot 5^{\mathrm{x}} \cdot \mathrm{x}^2}{1+5^{\mathrm{x}}} \mathrm{dx} \\ & \therefore 2 \mathrm{I}=\int_{-2}^2 9 \mathrm{x}^2 \mathrm{dx}=18 \int_0^2 \mathrm{x}^2 \mathrm{dx}=18\left(\frac{\mathrm{x}^3}{3}\right)_0^2 \\ & 2 \mathrm{I}=48 \\ & \therefore \mathrm{I}=24\end{aligned}$$​∴a=−2& b=2∴I=∫−22​1+5x9x2​dx=∫−22​1+5x9⋅5x⋅x2​dx∴2I=∫−22​9x2dx=18∫02​x2dx=18(3x3​)02​2I=48∴I=24​