JEE Main 2025 Jan 28 Shift 2, Mathematics Q7: Definite Integration And Fundamental Theorem

JEE Main2025Jan 28, Shift 2Mathematics

Let $f$ be a real valued continuous function defined on the positive real axis such that $g (x) = \int_{0}^{x} t f (t) d t$ . If $g (x^{3}) = x^{6} + x^{7}$ , then the value of $r = 1 \sum 15 f (r^{3})$ is :

A
$320$
B
$340$
C
$270$
D
$310$

Solution

Substitute $x \to x^{1/3}$ to get $g (x) = x^{2} + x^{7/3}$ . Then $g^{'} (x) = 2 x + \frac{7}{3} x^{4/3}$ .

From the definition of $g$ , $g^{'} (x) = x f (x)$ , so $f (x) = \frac{g ^{'} ( x )}{x} = 2 + \frac{7}{3} x^{1/3}$ .

Hence $f (r^{3}) = 2 + \frac{7 r}{3}$ , and

$r = 1 \sum 15 f (r^{3}) = r = 1 \sum 15 (2 + \frac{7 r}{3}) = 2 (15) + \frac{7}{3} \cdot \frac{15 \cdot 16}{2} = 30 + 280 = 310$ .

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JEE Main 2025 Jan 28 Shift 2, Mathematics Q7: Definite Integration And Fundamental Theorem

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