Difference between revisions of "1967 AHSME Problems/Problem 14"
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Since we know that <math>y=f(x)</math>, we can solve for <math>y</math> in terms of <math>x</math>. This gives us | Since we know that <math>y=f(x)</math>, we can solve for <math>y</math> in terms of <math>x</math>. This gives us | ||
− | <math>y=\frac{ | + | <math>y=\frac{x}{1-x}</math> |
− | <math>\Rightarrow y(1-x)= | + | <math>\Rightarrow y(1-x)=x</math> |
− | <math>\Rightarrow y-yx= | + | <math>\Rightarrow y-yx=x</math> |
− | <math>\Rightarrow yx | + | <math>\Rightarrow y=yx+x</math> |
− | <math>\Rightarrow x | + | <math>\Rightarrow y=x(y+1)</math> |
− | Therefore, we want to find the function with <math>y</math> that outputs <math>\frac{y | + | <math>\Rightarrow x=\frac{y}{y+1}</math> |
− | Listing out the possible outputs from each of the given functions we get | + | |
+ | Therefore, we want to find the function with <math>y</math> that outputs <math>\frac{y}{y+1}</math> Listing out the possible outputs from each of the given functions we get | ||
<math>f\left(\frac{1}{y}\right)=\frac{1}{y-1}</math> | <math>f\left(\frac{1}{y}\right)=\frac{1}{y-1}</math> | ||
Revision as of 21:02, 29 May 2022
Problem
Let ,
. If
, then
can be expressed as
Solution
Since we know that , we can solve for
in terms of
. This gives us
Therefore, we want to find the function with that outputs
Listing out the possible outputs from each of the given functions we get
Since the answer must be
.
See also
1967 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.