Let f(x) be a positive function and . Then the value of is equal to _______ [2025]

Question

Let f(x) be a positive function and $I_{1} = \int_{- 1 / 2}^{1} 2 x f (2 x (1 - 2 x)) d x and I_{2} = \int_{- 1}^{2} f (x (1 - x)) d x$ . Then the value of $\frac{I_{2}}{I_{1}}$ is equal to _______ [2025]

Smart Achievers · Accepted Answer

(3)

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

Let $2 x = t \Rightarrow 2 d x = d t$

when $x = \frac{- 1}{2}, t = - 1$ and

When for x = 1, t = 2

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

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

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

$\Rightarrow 2 I_{1} = I_{2} - 2 I_{1} \Rightarrow 4 I_{1} = I_{2}$

$\Rightarrow \frac{I_{1}}{I_{2}} = \frac{1}{4} i . e ., \frac{I_{2}}{I_{1}} = 4$

Solution

Q.
Let f(x) be a positive function and $I_{1} = \int_{- 1 / 2}^{1} 2 x f (2 x (1 - 2 x)) d x and I_{2} = \int_{- 1}^{2} f (x (1 - x)) d x$ . Then the value of $\frac{I_{2}}{I_{1}}$ is equal to _______ [2025]

1 12

2 6

3 4

4 9

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Solution

Q. Let f(x) be a positive function and I1=∫–1/212xf(2x(1–2x))dx and I2=∫–12f(x(1–x))dx. Then the value of I2I1 is equal to _______ [2025]

1 12

2 6

3 4

4 9

Ans. (3) I1=∫–1/212xf(2x(1–2x))dx Let 2x=t ⇒ 2dx=dt when x=–12, t=–1 and When for x = 1, t = 2 ∴ I1=12∫–12tf(t(1–t))dt ⇒ 2I1=∫–12(1–t)f((1–t)(1–(1–t)))dt ⇒ 2I1=∫–12f(t(1–t))dt–∫–12tf(t(1–t))dt ⇒ 2I1=I2–2I1 ⇒ 4I1=I2 ⇒ I1I2=14 i.e., I2I1=4

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Q.
Let f(x) be a positive function and $I_{1} = \int_{- 1 / 2}^{1} 2 x f (2 x (1 - 2 x)) d x and I_{2} = \int_{- 1}^{2} f (x (1 - x)) d x$ . Then the value of $\frac{I_{2}}{I_{1}}$ is equal to _______ [2025]