Let 5f (x) + 4f ((1)/(x)) = (1)/(x) + 3, x > 0 . Then 18 ∫ _ 1 ^ 2...

Given:

$5f\left(x\right) + 4f\left(\dfrac{1}{x}\right) = \dfrac{1}{x} + 3 . . . . \left(1\right)$

Replace $x \to \dfrac{1}{x}$

$5f\left(\dfrac{1}{x}\right) + 4f\left(x\right) = x + 3 . . . . \left(2\right)$

Eliminating $f\left(\dfrac{1}{x}\right)$ from \left(1\right) \text{ and } \left(2\right)$, we get</p><p>$9f\left(x\right) = \dfrac{5}{x} + 15 - 4x - 12$</p><p>$⇒ 9f\left(x\right) = \dfrac{5}{x} - 4x + 3$</p><p>$⇒ 18f\left(x\right) = \dfrac{10}{x} - 8x + 6$</p><p>$⇒ 18{∫}_{1}^{2}f\left(x\right)dx = {∫}_{1}^{2}\left(\dfrac{10}{x} - 8x + 6\right)dx$</p><p>$⇒ 18{∫}_{1}^{2}f\left(x\right)dx = {\left[10\log x - 4{x}^{2} + 6x\right]}_{1}^{2}$</p><p>$⇒ 18{∫}_{1}^{2}f\left(x\right)dx = \left[\left(10\log 2 - 4\right) - \left(10\log 1 + 2\right)\right]$</p><p>$⇒ 18{∫}_{1}^{2}f\left(x\right)dx = 10\log_{e}2 - 6

Hence this is the correct option.