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

We need to find f(5) - f(2) given the functional equation $$3f(x) + 2f\left(\frac{m}{19x}\right) = 5x$$.

First compute $$m = \sum_{i=1}^{9} i^2 = \frac{9 \times 10 \times 19}{6} = 285$$ so that $$\frac{m}{19} = \frac{285}{19} = 15$$.

With this value of m, the functional equation simplifies to $$3f(x) + 2f\left(\frac{15}{x}\right) = 5x\quad\cdots(1)$$.

Replacing x by $$\frac{15}{x}$$ in the same equation gives $$3f\left(\frac{15}{x}\right) + 2f(x) = \frac{75}{x}\quad\cdots(2)$$.

We now have the system of equations

$$3f(x) + 2f\left(\frac{15}{x}\right) = 5x$$

$$2f(x) + 3f\left(\frac{15}{x}\right) = \frac{75}{x}$$

Multiplying the first by 3 yields $$9f(x) + 6f\left(\frac{15}{x}\right) = 15x$$ and multiplying the second by 2 gives $$4f(x) + 6f\left(\frac{15}{x}\right) = \frac{150}{x}$$. Subtracting these equations eliminates $$f\left(\frac{15}{x}\right)$$, leading to $$5f(x) = 15x - \frac{150}{x}$$ and hence $$f(x) = 3x - \frac{30}{x}$$.

Substituting $$x=5$$ into this expression gives $$f(5) = 3(5) - \frac{30}{5} = 15 - 6 = 9$$, and for $$x=2$$ we find $$f(2) = 3(2) - \frac{30}{2} = 6 - 15 = -9$$. Therefore, $$f(5) - f(2) = 9 - (-9) = 18$$.

Therefore, f(5) - f(2) = Option 4: 18.