Let f : R $\to$ R be a continuous function such that $f(3x) - f(x) = x$. If $f(8) = 7$, then $f(14)$ is equal to :

<p>$f(3x) - f(x) = x$ ...... (1)</p> <p>$x \to {x \over 3}$</p> <p>$f(x) - f\left( {{x \over 3}} \right) = {x \over 3}$ ....... (2)</p> <p>Again $x \to {x \over 3}$</p> <p>$$f\left( {{x \over 3}} \right) - f\left( {{x \over 9}} \right) = {x \over {{3^2}}}$$ ...... (3)</p> <p>Similarly</p> <p>$$f\left( {{x \over {{3^{n - 2}}}}} \right) - f\left( {{x \over {{3^{n - 1}}}}} \right) = {x \over {{3^{n - 1}}}}\,.....\,(n)$$</p-> <p>Adding all these and applying $n \to \infty$</p> <p>$$\mathop {\lim }\limits_{n \to \infty } \left( {f(3x) - f\left( {{x \over {{3^{n - 1}}}}} \right)} \right) = x\left( {1 + {1 \over 3} + {1 \over {{3^2}}}\, + \,....} \right)$$</p> <p>$f(3x) - f(0) = {{3x} \over 2}$</p> <p>Putting $x = {8 \over 3}$</p> <p>$f(8) - f(0) = 4$</p> <p>$\Rightarrow f(0) = 3$</p> <p>Putting $x = {{14} \over 3}$</p> <p>$f(14) - 3 = 7 \Rightarrow f(14) = 10$</p>