JEE Main 2025 (Online) 8th April Evening ShiftQuestion 1
Let $f(x)$ be a positive function and $I_1=\int_{-\frac{1}{2}}^{1}2xf(2x(1-2x))\,dx$ and $I_2=\int_{-1}^{2}f(x(1-x))\,dx$. Then the value of $\frac{I_2}{I_1}$ is equal to
Definite Integration
Solution
Concept. After $u=2x$, both integrals live on $[-1,2]$, an interval symmetric about $\frac12$, where $u(1-u)$ does not change under $u\to1-u$. Formulas used. (1) $\int_a^bg(u)\,du=\int_a^bg(a+b-u)\,du$Step 1 - Substitute $u=2x$. $dx=\frac{du}{2}$, $x:-\frac12\to1$ gives $u:-1\to2$: $$I_1=\frac12\int_{-1}^{2}u\,f\big(u(1-u)\big)\,du.$$ Step 2 - Symmetry. By (1) with $a+b=1$: $u\to1-u$ keeps $f(u(1-u))$, so $$\int_{-1}^2u\,f\,du=\int_{-1}^2(1-u)\,f\,du\ \Longrightarrow\ 2\int_{-1}^2u\,f\,du=\int_{-1}^2f\,du=I_2.$$ Step 3 - Ratio. $I_1=\frac12\cdot\frac{I_2}{2}=\frac{I_2}{4}$, so $\dfrac{I_2}{I_1}=4$. Answer: (D)