Let f: R → R be a continuous function such that f (3x) - f (x) = x...

Given $f\left(3x\right) - f\left(x\right) = x$

Now replance $x\text{by}\dfrac{x}{3}$ infinite times,

$f\left(x\right) - f\left(\dfrac{x}{3}\right) = \dfrac{x}{3}$

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

On adding, we get

$f\left(x\right) - \lim_{n \to \infty}f\left(\dfrac{x}{{3}^{n}}\right) = x\left(\dfrac{1}{3} + \dfrac{1}{{3}^{2}} + ⋯ + \infty\right)$

$f\left(x\right) - \lim_{n \to \infty}f\left(\dfrac{x}{{3}^{n}}\right) = x\left(\dfrac{\dfrac{1}{3}}{1 - \dfrac{1}{3}}\right) = \dfrac{x}{2}$

$⇒ f\left(x\right) - f\left(0\right) = \dfrac{x}{2}$

Given $f\left(8\right) = 7$

so $f\left(8\right) - f\left(0\right) = 4$

i.e. $f\left(0\right) = 3$

$∴ f\left(x\right) = \dfrac{x}{2} + 3$

Hence $f\left(14\right) = 10$