For the differentiable function f: ℝ - 0 - ℝ, let 3f(x) + 2f...

Given:

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

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

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

Eliminating $f\left(\dfrac{1}{x}\right)$ from \left(1\right) \text{ and } \left(2\right)$, we get</p><p>$⇒ 5f\left(x\right) = \dfrac{3}{x} - 2x - 10$</p><p>Let us replace$x$with$3$</p><p>$⇒ 5f\left(3\right) = \dfrac{3}{3} - 2\left(3\right) - 10$</p><p>$⇒ f\left(3\right) = - 3$</p><p>$⇒ 5{f}^{'}\left(x\right) = \dfrac{- 3}{{x}^{2}} - 2$</p><p>$⇒ 5{f}^{'}\left(\dfrac{1}{4}\right) = - 48 - 2$</p><p>$⇒ {f}^{'}\left(\dfrac{1}{4}\right) = - 10$</p><p>$⇒ \left|f ( 3 ) + {f}^{'}\left(\dfrac{1}{4}\right)\right| = 13

Hence this is the correct option.