JEE Main 2026 — Jan 24, Shift 2 — Mathematics solution

JEE Main2026Jan 24, Shift 2Mathematics

Let $f$ be a function such that $3 f (x) + 2 f (\frac{m}{19 x}) = 5 x$ , $x \neq = 0$ , where $m = i = 1 \sum 9 i^{2}$ . Then $f (5) - f (2)$ is equal to

A
$- 9$
B
36
C
18
D
9

Solution

$m = \frac{9 \cdot 10 \cdot 19}{6} = 285 = 15 \cdot 19$ , so $\frac{m}{19 x} = \frac{15}{x}$ .

Equation (1): $3 f (x) + 2 f (\frac{15}{x}) = 5 x$ .

Replace $x$ by $\frac{15}{x}$ in (1): $3 f (\frac{15}{x}) + 2 f (x) = \frac{75}{x}$ ... (2)

Multiply (1) by 3 and (2) by 2 and subtract: $9 f (x) + 6 f (15/ x) - 4 f (x) - 6 f (15/ x) = 15 x - \frac{150}{x}$ , giving $5 f (x) = 15 x - \frac{150}{x}$ .

$f (x) = 3 x - \frac{30}{x}$ . Then $f (5) = 15 - 6 = 9$ and $f (2) = 6 - 15 = - 9$ .

$f (5) - f (2) = 9 - (- 9) = 18$ .

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