Test Index

Functions Part 2

Section: Mathematics

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Question : 13 of 73

Marks: +1, -0

Let

f

be a function such that

3 f(x) + 2 f\left(\,\frac{m}{19 x}\right) = 5x, x \neq 0

m = \sum\limits_{i=1}^{9(i)}{}^{2}

f(5)-f(2)

[24 Jan 2026 Shift 2]

36
9
-9
18

Solution:

Concept:

The problem uses a functional equation and the sum of squares formula.

Explanation:

First, compute

m = \sum_{i=1}^{9} i^2

.

Use the formula:

\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}

.

So

m = \frac{9 \times 10 \times 19}{6} = 285

.

The given equation is

3f(x) + 2f\left(\frac{m}{19x}\right) = 5x

.

Substitute

m = 285

:

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

. (1)

Replace

x

by

\frac{15}{x}

in (1):

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

. (2)

From (1):

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

.

Substitute into (2):

3\left(\frac{5x - 3f(x)}{2}\right) + 2f(x) = \frac{75}{x}

.

Multiply by 2:

15x - 9f(x) + 4f(x) = \frac{150}{x}

.

Simplify:

15x - 5f(x) = \frac{150}{x}

.

Thus

5f(x) = 15x - \frac{150}{x}

, so

f(x) = 3x - \frac{30}{x}

.

Now compute

f(5) - f(2)

.

f(5) = 3(5) - \frac{30}{5} = 15 - 6 = 9

.

f(2) = 3(2) - \frac{30}{2} = 6 - 15 = -9

.

Therefore

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

.

Answer:

f(5) - f(2) = 18

, which corresponds to option D.

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