For the differentiable function $$f : \mathbb{R} - \{0\} - \mathbb{R}$$, let $$3f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$$, then $$\left|f(3) + f'\left(\frac{1}{4}\right)\right|$$ is equal to

Given: $$3f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$$ ... (1)

Replace $$x$$ by $$\frac{1}{x}$$:

$$3f\left(\frac{1}{x}\right) + 2f(x) = x - 10$$ ... (2)

From (1): $$3f(x) + 2f(1/x) = 1/x - 10$$

From (2): $$2f(x) + 3f(1/x) = x - 10$$

Multiply (1) by 3: $$9f(x) + 6f(1/x) = 3/x - 30$$

Multiply (2) by 2: $$4f(x) + 6f(1/x) = 2x - 20$$

Subtract: $$5f(x) = \frac{3}{x} - 30 - 2x + 20 = \frac{3}{x} - 2x - 10$$

$$f(x) = \frac{1}{5}\left(\frac{3}{x} - 2x - 10\right)$$

$$f(3) = \frac{1}{5}\left(1 - 6 - 10\right) = \frac{-15}{5} = -3$$

Finding $$f'(x)$$:

$$f'(x) = \frac{1}{5}\left(-\frac{3}{x^2} - 2\right)$$

$$f'\left(\frac{1}{4}\right) = \frac{1}{5}\left(-3 \times 16 - 2\right) = \frac{1}{5}(-50) = -10$$

$$|f(3) + f'(1/4)| = |-3 + (-10)| = |-13| = 13$$

This matches option 4: 13.