JEE Main 2025 Apr 3 Shift 1, Mathematics Q17: Definite Integral as Limit of a Sum

JEE Main2025Apr 3, Shift 1Mathematics

Let the domain of the function $f (x) = lo g_{2} lo g_{4} lo g_{6} (3 + 4 x - x^{2})$ be $(a, b)$ . If $\int_{0}^{b - a} [x^{2}] d x = p - q - r$ , $p, q, r \in N$ , $g cd (p, q, r) = 1$ , where $[\cdot]$ is the greatest integer function, then $p + q + r$ is equal to

A
10
B
8
C
11
D
9

Solution

Domain demands $lo g_{4} lo g_{6} (3 + 4 x - x^{2}) > 0$ , equivalently $lo g_{6} (3 + 4 x - x^{2}) > 1$ , i.e. $3 + 4 x - x^{2} > 6$ , i.e. $x^{2} - 4 x + 3 < 0$ , so $x \in (1, 3)$ .

Hence $a = 1, b = 3, b - a = 2.$

Evaluate $\int_{0}^{2} [x^{2}] d x .$ The integrand jumps at $x = 1, 2, 3, 2.$

$\int_{0}^{1} 0 d x + \int_{1}^{2} 1 d x + \int_{2}^{3} 2 d x + \int_{3}^{2} 3 d x$

$= 0 + (2 - 1) + 2 (3 - 2) + 3 (2 - 3)$

$= 2 - 1 + 23 - 22 + 6 - 33$

$= 5 - 2 - 3 .$

Comparing with $p - q - r$ : $p = 5, q = 2, r = 3$ (or $q = 3, r = 2$ ). $g cd (5, 2, 3) = 1.$

$p + q + r = 5 + 2 + 3 = 10.$

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JEE Main 2025 Apr 3 Shift 1, Mathematics Q17: Definite Integral as Limit of a Sum

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