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03 Oct 2019, 06:24
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C
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[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
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
03 Oct 2019, 08:35
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MathRevolution
f1(x) = x/(1+x)
f2(x) = f1(f1(x)) = f1(x/(1+x)) = [x/(1+x)]/[1 + x/(1+x)] = x/(1+x + x) = x/(1+2x)
Similarly, f3(x) = f1(f2(x)) = f1(x/(1+2x)) = [x/(1+2x)]/[1 + x/(1+2x)] = x/(1+3x)
So, fn(x) = x/(1+nx)
Given, fn(2/3) = 2/255
—> [2/3]/[1+2/3n] = 2/255
—> 2/(3+2n) = 2/255
—> 3+2n = 255
—> 2n = 252
—> n = 126
IMO Option C
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General Discussion
06 Oct 2019, 19:21
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=>
f1(x) = \(\frac{x}{(1+x)}.\)
f2(x) = f1(f1(x)) = \([\frac{x}{(1+x)}]/[1+\frac{x}{(1+x)}] = [\frac{x}{(1+x)}]/[\frac{(1+2x)}{(1+x)}] = \frac{x}{(1+2x)}.\)
f3(x) = f1(f2(x)) =\([\frac{x}{(1+2x)}]/[\frac{1+x}{(1+2x)}] = [\frac{x}{(1+2x)}]/[\frac{(1+3x)}{(1+2x)}] = \frac{x}{(1+3x)}.\)
…
fn(x) = f1(fn-1(x)) = \([\frac{x}{(1+(n-1)x})]/[\frac{1+x}{(1+(n-1)x)}]\)
\(= [\frac{x}{(1+(n-1)x)}]/[\frac{(1+nx)}{(1+(n-1)x)}]\)
\(= \frac{x}{(1+nx)}.\)
We have fn\((\frac{2}{3}) = (\frac{2}{3})/[1+n(\frac{2}{3})] = \frac{2}{(3+2n)} = \frac{2}{255}\) or \(2n + 3 = 255.\)
Then we have \(n = 126\).
Therefore, C is the answer.
Answer: C
f1(x) = \(\frac{x}{(1+x)}.\)
f2(x) = f1(f1(x)) = \([\frac{x}{(1+x)}]/[1+\frac{x}{(1+x)}] = [\frac{x}{(1+x)}]/[\frac{(1+2x)}{(1+x)}] = \frac{x}{(1+2x)}.\)
f3(x) = f1(f2(x)) =\([\frac{x}{(1+2x)}]/[\frac{1+x}{(1+2x)}] = [\frac{x}{(1+2x)}]/[\frac{(1+3x)}{(1+2x)}] = \frac{x}{(1+3x)}.\)
…
fn(x) = f1(fn-1(x)) = \([\frac{x}{(1+(n-1)x})]/[\frac{1+x}{(1+(n-1)x)}]\)
\(= [\frac{x}{(1+(n-1)x)}]/[\frac{(1+nx)}{(1+(n-1)x)}]\)
\(= \frac{x}{(1+nx)}.\)
We have fn\((\frac{2}{3}) = (\frac{2}{3})/[1+n(\frac{2}{3})] = \frac{2}{(3+2n)} = \frac{2}{255}\) or \(2n + 3 = 255.\)
Then we have \(n = 126\).
Therefore, C is the answer.
Answer: C
