JEE MAIN - Mathematics (2025 - 8th April Evening Shift - No. 15)
Let f(x) be a positive function and $I_{1} = \int\limits_{-\frac{1}{2}}^{1} 2x \, f(2x(1-2x)) \, dx$ and $I_{2} = \int\limits_{-1}^{2} f(x(1-x)) \, dx$. Then the value of $\frac{I_{2}}{I_{1}}$ is equal to ________
12
9
6
4
Explanation
$$\begin{aligned} & I_1=\int_{-\frac{1}{2}}^1 2 x f(2 x(1-2 x)) d x \\ & \Rightarrow 2 x=t \Rightarrow 2 d x=d t \quad \Rightarrow I_1=\frac{1}{2} \int_{-1}^2 t f(t(1-t)) d t \\ & \Rightarrow 2 I_1=\int_{-1}^2(1-t) f(1-t)(1-(1-t)) d t \end{aligned}$$
$$ \Rightarrow 2{I_1} = \int\limits_{ - 1}^2 {f(t(1 - t)dt - \int\limits_{ - 1}^2 {tf(t(1 - t)dt} } $$
$$\begin{aligned} & \Rightarrow 2 \mathrm{I}_1=\mathrm{I}_2-2 \mathrm{I}_1 \\ & \Rightarrow 4 \mathrm{I}_1=\mathrm{I}_2 \\ & \Rightarrow \frac{\mathrm{I}_2}{\mathrm{I}_1}=4 \end{aligned}$$
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