JEE Main 2025 Apr 8 Shift 2, Mathematics Q8: Some Properties of Definite Integrals

JEE Main2025Apr 8, Shift 2Mathematics

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 ____

A
$9$
B
$6$
C
$12$
D
$4$

Solution

Substitute $t = 2 x$ in $I_{1}$ , so $d t = 2 d x$ and the limits change from $x \in [- 1/2, 1]$ to $t \in [- 1, 2]$ : $I_{1} = \frac{1}{2} \int_{- 1}^{2} t f (t (1 - t)) d t$ .

Apply the king reflection $t \to 1 - t$ on the interval $[- 1, 2]$ (which maps the interval to itself), giving $2 I_{1} = \int_{- 1}^{2} (1 - t) f ((1 - t) t) d t = \int_{- 1}^{2} (1 - t) f (t (1 - t)) d t$ .

Split the integrand: $2 I_{1} = \int_{- 1}^{2} f (t (1 - t)) d t - \int_{- 1}^{2} t f (t (1 - t)) d t = I_{2} - 2 I_{1}$ .

Solve: $4 I_{1} = I_{2}$ , so $\frac{I _{2}}{I _{1}} = 4$ , matching option (4).

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JEE Main 2025 Apr 8 Shift 2, Mathematics Q8: Some Properties of Definite Integrals

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