JEE Main 2025 — 8 April, Evening Shift — Question 38

$I_{1}=\int_{-\frac{1}{2}}^{1} 2 x f(2 x(1-2 x) d x)$

$$\begin{aligned} & I_{1}=\int_{-\frac{1}{2}}^{1} 2\left(\frac{1}{2}-x\right) f\left(2\left(\frac{1}{2}-x\right)\left(1-2\left(\frac{1}{2}-x\right)\right)\right) d x \\& I_{1}=\int_{-\frac{1}{2}}^{1}(1-2 x) f((1-2 x)(2 x)) d x \end{aligned}$$

$I_{1}=\int_{-\frac{1}{2}}^{1} f((1-2 x)(2 x)) d x-\int_{-\frac{1}{2}}^{1} 2 x \underbrace{f((1-2 x)(2 x)) d x}_{I_{1}}$

$2 I_{1}=\int_{-\frac{1}{2}}^{1} f((1-2 x)(2 x)) d x$

Put $2 x=t$

$2 d x=d t$

$d x=\frac{d t}{2}$

$2 I_{1}=\frac{1}{2} \int_{-1}^{2} f((1-t)(t)) d t$

$I_{1}=\frac{1}{4} \int_{-1}^{2} f((1-x)(x)) d x$

$I_{1}=\frac{1}{4} I_{2}$

$\Rightarrow \frac{I_{2}}{l_{1}}=4$