# If f_1(x)=-((2x+7)/(x+3)), f_(n+1)(x)=f_1(f_n(x)) then f_2001(2002) is?

Jul 31, 2018

$2002$.

#### Explanation:

We have, ${f}_{1} \left(x\right) = - \left(\frac{2 x + 7}{x + 3}\right) , \mathmr{and} , {f}_{1} \left({f}_{n} \left(x\right)\right) = {f}_{n + 1} \left(x\right)$.

Observe that, ${f}_{1} \left(x\right) = - \left(\frac{2 x + 7}{x + 3}\right) = - \left(\frac{2 x + 6 + 1}{x + 3}\right)$

$= - \left(\frac{2 x + 6}{x + 3} + \frac{1}{x + 3}\right)$.

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

$\therefore {f}_{2} \left(x\right) = {f}_{1} \left({f}_{1} \left(x\right)\right)$

$= {f}_{1} \left(- 2 - \frac{1}{x + 3}\right) = {f}_{1} \left(y\right) , \text{ say, } y = - 2 - \frac{1}{x + 3}$,

$= - 2 - \frac{1}{y + 3} = - 2 - \frac{1}{- 2 - \frac{1}{x + 3} + 3}$,

$= - 2 - \frac{1}{1 - \frac{1}{x + 3}} = - 2 - \frac{x + 3}{x + 3 - 1}$,

$= - 2 - \frac{x + 3}{x + 2} = - 2 - \frac{x + 2 + 1}{x + 2}$,

$= - 2 - \left(\frac{x + 2}{x + 2} + \frac{1}{x + 2}\right)$

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

$\therefore {f}_{3} \left(x\right) = {f}_{1} \left({f}_{2} \left(x\right)\right)$,

$= {f}_{1} \left(t\right) , \text{ say, where, } t = {f}_{2} \left(x\right) = - 3 - \frac{1}{x + 2}$,

$= - 2 - \frac{1}{t + 3}$,

$= - 2 - \frac{1}{- 3 - \frac{1}{x + 2} + 3} = - 2 - \frac{1}{- \frac{1}{x + 2}} = - 2 + x + 2$.

$\Rightarrow {f}_{3} \left(x\right) = x$.

Clearly, then, ${f}_{6} \left(x\right) = {f}_{9} \left(x\right) = \ldots = {f}_{3 m} \left(x\right) = x , \forall m \in \mathbb{N}$.

Since, $2001 = 3 \left(667\right)$, we have,

f_2001(x)=x, & therefore, f_2001(2002)=2002.

$\textcolor{m a \ge n t a}{\text{Enjoy Maths.!}}$