If $f_{1} (x) = - (\frac{2 x + 7}{x + 3}), f_{n + 1} (x) = f_{1} (f_{n} (x))$ $f_1(x)=-((2x+7)/(x+3)), f_(n+1)(x)=f_1(f_n(x))$ then $f_{2001} (2002)$ $f_2001(2002)$ is?

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Answer 1 · 2018-07-31T18:47:41

$2002$ $2002$ .

We have, $f_{1} (x) = - (\frac{2 x + 7}{x + 3}), and, f_{1} (f_{n} (x)) = f_{n + 1} (x)$ $f_1(x)=-((2x+7)/(x+3)), and, f_1(f_n(x))=f_(n+1)(x)$ .

Observe that, $f_{1} (x) = - (\frac{2 x + 7}{x + 3}) = - (\frac{2 x + 6 + 1}{x + 3})$ $f_1(x)=-((2x+7)/(x+3))=-((2x+6+1)/(x+3))$

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

$\Rightarrow f_{1} (x) = - (2 + \frac{1}{x + 3}) = - 2 - \frac{1}{x + 3}$ $rArr f_1(x)=-(2+1/(x+3))=-2-1/(x+3)$

$:. f_2(x)=f_1(f_1(x))$

$=f_1(-2-1/(x+3))=f_1(y)," say, "y=-2-1/(x+3)$ ,

$=-2-1/(y+3)=-2-1/(-2-1/(x+3)+3)$ ,

$=-2-1/(1-1/(x+3))=-2-(x+3)/(x+3-1)$ ,

$=-2-(x+3)/(x+2)=-2-(x+2+1)/(x+2)$ ,

$=-2-((x+2)/(x+2)+1/(x+2))$

$rArr f_2(x)=-3-1/(x+2)$ .

$:. f_3(x)=f_1(f_2(x))$ ,

$=f_1(t)," say, where, "t=f_2(x)=-3-1/(x+2)$ ,

$=-2-1/(t+3)$ ,

$=-2-1/(-3-1/(x+2)+3)=-2-1/(-1/(x+2))=-2+x+2$ .

$rArr f_3(x)=x$ .

Clearly, then, $f_6(x)=f_9(x)=...=f_(3m)(x)=x, AA m in NN$ .

Since, $2001=3(667)$ , we have,

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

$color(magenta)("Enjoy Maths.!")$

If f_1(x)=-((2x+7)/(x+3)), f_(n+1)(x)=f_1(f_n(x))f1(x)=−(2x+7x+3),fn+1(x)=f1(fn(x))f_1(x)=-((2x+7)/(x+3)), f_(n+1)(x)=f_1(f_n(x)) then f_2001(2002)f2001(2002)f_2001(2002) is?