Let f(x) be a positive function and I_1=∫_ -(1)/(2) ^1 2 x f(2 x(1-2 x)) d x and I_2=∫_ -1 ^2 f(x(1-x)) d x . Then the value of (I_2)/(I_1) is equal to ________

Correct answer: D. 4 aligned & I_1=∫_ -(1)/(2) ^1 2 xf(2 x(1-2 x)) dx & 2 x=t 2 dx=dt I_1=(1)/(2) ∫_ -1 ^2 tf(t(1-t)) dt & 2 I_1=∫_ -1 ^2(1-t) f(1-t)(1-(1-t)) dt & 2 I_1=∫_ -1 ^2 f(t(1-t) dt-∫_ -1 ^2 tf(t(1-t) dt. & 2 I_1=I_2-2 I_1 & 4 I_1=I_2 & I_2 I_1 =4 aligned

JEE Main 20258 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 8 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Definite Integration

JEE Main 2025 · 8 April · Evening Shift · Mathematics · Definite Integration

Let f(x) be a positive function and I_1=∫_ -(1)/(2) ^1 2 x f(2...

Let

f(x)

be a positive function and

I_1=\int_{-\frac{1}{2}}^1 2 x f(2 x(1-2 x)) d x

and

I_2=\int_{-1}^2 f(x(1-x)) d x

. Then the value of

\frac{I_2}{I_1}

is equal to ________

Options

Correct answer

D. 4

Step-by-step solution

\begin{aligned} & \mathrm{I}_1=\int_{-\frac{1}{2}}^1 2 \mathrm{xf}(2 \mathrm{x}(1-2 \mathrm{x})) \mathrm{dx} \\ & \Rightarrow 2 \mathrm{x}=\mathrm{t} \Rightarrow 2 \mathrm{dx}=\mathrm{dt} \quad \Rightarrow \mathrm{I}_1=\frac{1}{2} \int_{-1}^2 \mathrm{tf}(\mathrm{t}(1-\mathrm{t})) \mathrm{dt} \\ & \Rightarrow 2 \mathrm{I}_1=\int_{-1}^2(1-\mathrm{t}) \mathrm{f}(1-\mathrm{t})(1-(1-\mathrm{t})) \mathrm{dt} \\ & \Rightarrow 2 \mathrm{I}_1=\int_{-1}^2 \mathrm{f}\left(\mathrm{t}(1-\mathrm{t}) \mathrm{dt}-\int_{-1}^2 \mathrm{tf}(\mathrm{t}(1-\mathrm{t}) \mathrm{dt}\right. \\ & \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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