Let f be a function such that 3f(x) + 2f(m/19x) = 5x

Q. 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

(A) 36

(B) 9

(D) 18

Correct Answer: 18

Explanation

First evaluate the value of $ m $:

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

So the given equation becomes:

$$ 3f(x) + 2f\!\left(\frac{285}{19x}\right) = 5x $$ $$ \Rightarrow 3f(x) + 2f\!\left(\frac{15}{x}\right) = 5x $$

Replace $ x $ by $ \frac{15}{x} $:

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

Now solve the two equations:

$$ \begin{cases} 3f(x) + 2f(15/x) = 5x \\ 2f(x) + 3f(15/x) = \frac{75}{x} \end{cases} $$

Multiply first by 3 and second by 2:

$$ 9f(x) + 6f(15/x) = 15x $$ $$ 4f(x) + 6f(15/x) = \frac{150}{x} $$

Subtracting:

$$ 5f(x) = 15x - \frac{150}{x} $$ $$ \Rightarrow f(x) = 3x - \frac{30}{x} $$

Now compute:

$$ f(5) = 15 - 6 = 9 $$ $$ f(2) = 6 - 15 = -9 $$ $$ f(5) - f(2) = 9 - (-9) = 18 $$

Hence, the correct answer is 18.

Explanation

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