JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Definite Integrals

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

Answer: (D)

Solution

I_{1}

put

t = 2 x

, so

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

; the limits

x = - \frac{1}{2}, 1

become

t = - 1, 2

, and

2 x (1 - 2 x) = t (1 - t)

, giving

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

. Replacing

t \mapsto 1 - t

leaves

t (1 - t)

unchanged, so

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

; adding the two gives

2 \int_{- 1}^{2} t f d t = \int_{- 1}^{2} f d t = I_{2}

. Thus

\int_{- 1}^{2} t f d t = \frac{I _{2}}{2}

and

I_{1} = \frac{1}{2} \cdot \frac{I _{2}}{2} = \frac{I _{2}}{4}

, so

\frac{I _{2}}{I _{1}} = 4

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