Functions – Functional Equations – JEE Main 24 Jan 2026 Shift 2

Solution:

$$m = \sum_{i=1}^{9} i^2 = \frac{9(9+1)(2 \cdot 9+1)}{6} = \frac{9 \cdot 10 \cdot 19}{6} = 15 \cdot 19 = 285$$

The given equation is:
$$3f(x) + 2f\left(\frac{285}{19x}\right) = 5x$$
$$3f(x) + 2f\left(\frac{15}{x}\right) = 5x \quad \dots(1)$$

Replace $x$ with $\frac{15}{x}$ in (1):
$$3f\left(\frac{15}{x}\right) + 2f(x) = 5\left(\frac{15}{x}\right) = \frac{75}{x} \quad \dots(2)$$

Multiply (1) by 3 and (2) by 2:
$$9f(x) + 6f\left(\frac{15}{x}\right) = 15x$$
$$4f(x) + 6f\left(\frac{15}{x}\right) = \frac{150}{x}$$

Subtract the second from the first:
$$5f(x) = 15x – \frac{150}{x}$$
$$f(x) = 3x – \frac{30}{x}$$

Calculate $f(5)$ and $f(2)$:
$$f(5) = 3(5) – \frac{30}{5} = 15 – 6 = 9$$
$$f(2) = 3(2) – \frac{30}{2} = 6 – 15 = -9$$

$$f(5) – f(2) = 9 – (-9) = 18$$

Ans. (3)