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A function fn(x) is defined as f1(x) = x/(1+x) and fn(x)=f1(

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A function fn(x) is defined as f1(x) = x/(1+x) and fn(x)=f1(

by Max@Math Revolution » Thu Oct 03, 2019 4:59 am
[GMAT math practice question]

A function fn(x) is defined as f1(x) = x/(1+x) and fn(x)=f1(fn-1(x)). If fn(2/3)=2/255, what is n?

A. 122
B.124
C.126
D. 128
E. 130
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by deloitte247 » Sat Oct 05, 2019 10:14 am
$$f_n\left(x\right)=f_1\left(f_{n-1}\left(x\right)\right)==>\ This\ is\ a\ composite\ function$$
$$f_1\left(x\right)=\frac{x}{\left(1+x\right)}$$
If n=2
$$f_2\left(x\right)=f_1\left(f_{2-1}\left(x\right)\right)=f_1\left(f_1\left(x\right)\right)$$
$$f_2\left(x\right)=f_1\left[\frac{x}{\left(1+x\right)}\right]=>\frac{\frac{x}{\left(1+x\right)}}{1+\frac{x}{1+x}}$$
$$f_2\left(x\right)=\frac{x}{\left(1+x\right)}\cdot\frac{\left(1+x\right)}{\left(1+x\right)+x}\ =\frac{x}{1+2x}$$
If n=3
$$f_3\left(x\right)=f_1\left(f_{3-1}\left(x\right)\right)=f_1\left(f_2\left(x\right)\right)$$
$$f_3\left(x\right)=f_1\left[\frac{x}{\left(1+2x\right)}\right]=>\frac{\frac{x}{\left(1+2x\right)}}{1+\frac{x}{1+2x}}$$
$$f_3\left(x\right)=\frac{x}{\left(1+2x\right)}\cdot\frac{\left(1+2x\right)}{\left(1+2x\right)+x}\ =\frac{x}{1+3x}$$
If n=4
$$f_4\left(x\right)=f_1\left(f_{4-1}\left(x\right)\right)=f_1\left(f_3\left(x\right)\right)$$
$$f_4\left(x\right)=f_1\left[\frac{x}{\left(1+3x\right)}\right]=>\frac{\frac{x}{\left(1+3x\right)}}{1+\frac{x}{1+3x}}$$
$$f_4\left(x\right)=\frac{x}{\left(1+3x\right)}\cdot\frac{\left(1+3x\right)}{\left(1+3x\right)+x}\ =\frac{x}{1+4x}$$
$$Therefore,\ we\ can\ assert\ that\ f_n\left(x\right)=\frac{x}{1+n\left(x\right)}$$
$$Given\ that\ f_n\left(\frac{2}{3}\right)=\frac{2}{255}$$
$$x=\frac{2}{3}$$
$$Therefore,\ \frac{x}{1+n\left(x\right)}=\frac{2}{255}=>\frac{\frac{2}{3}}{1+n\left(\frac{2}{3}\right)}=\frac{2}{255}$$
$$\frac{2}{3}\cdot\frac{1}{1+\frac{2n}{3}}=\frac{2}{255}$$
$$\frac{2}{3\left(1+\frac{2n}{3}\right)}=\frac{2}{255}$$
$$\frac{2}{3+2n}=\frac{2}{255}$$
$$2\cdot255=2\left(3+2n\right)$$
$$510=6+4n$$
$$n=\frac{504}{4}=126$$
$$Answer\ =\ option\ C$$
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by Max@Math Revolution » Sun Oct 06, 2019 5:58 pm
=>

f1(x) = x/(1+x).
f2(x) = f1(f1(x)) = [x/(1+x)]/[1+x/(1+x)] = [x/(1+x)]/[(1+2x)/(1+x)] = x/(1+2x).
f3(x) = f1(f2(x)) = [x/(1+2x)]/[1+x/(1+2x)] = [x/(1+2x)]/[(1+3x)/(1+2x)] = x/(1+3x).
...
fn(x) = f1(fn-1(x)) = [x/(1+(n-1)x)]/[1+x/(1+(n-1)x)]
= [x/(1+(n-1)x)]/[(1+nx)/(1+(n-1)x)]
= x/(1+nx).

We have fn(2/3) = (2/3)/[1+n(2/3)] = 2/(3+2n) = 2/255 or 2n + 3 = 255.
Then we have n = 126.

Therefore, C is the answer.
Answer: C
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