Difference between revisions of "1967 AHSME Problems/Problem 14"

Latest revision as of 19:42, 22 August 2023

Problem

Let $f(t)=\frac{t}{1-t}$ , $t \not= 1$ . If $y=f(x)$ , then $x$ can be expressed as

$\textbf{(A)}\ f\left(\frac{1}{y}\right)\qquad \textbf{(B)}\ -f(y)\qquad \textbf{(C)}\ -f(-y)\qquad \textbf{(D)}\ f(-y)\qquad \textbf{(E)}\ f(y)$

Solution

Since we know that $y=f(x)$ , we can solve for $y$ in terms of $x$ . This gives us

$y=\frac{x}{1-x}$

$\Rightarrow y(1-x)=x$

$\Rightarrow y-yx=x$

$\Rightarrow y=yx+x$

$\Rightarrow y=x(y+1)$

$\Rightarrow x=\frac{y}{y+1}$

Therefore, we want to find the function with $y$ that outputs $\frac{y}{y+1}$ Listing out the possible outputs from each of the given functions we get $f\left(\frac{1}{y}\right)=\frac{1}{y-1}$

$f(y)=\frac{1}{1-y}$

$f(-y)=\frac{-y}{y+1}$

$-f(y)=\frac{y}{y-1}$

$-f(-y)=\frac{y}{y+1}$

Since $-f(-y)=\frac{y}{y+1}=x$ the answer must be $\boxed{C}$ .

1967 AHSC (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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 10: / Line 10: @@
 == Solution ==
-Let <math>x^2 = a</math> and <math>y^2 = b.</math>
+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
-then
+<math>y=\frac{x}{1-x}</math>
-         <math>a+4b=1</math>
+<math>\Rightarrow y(1-x)=x</math>
-and
+<math>\Rightarrow y-yx=x</math>
-         <math>4a+b=4</math>
+<math>\Rightarrow y=yx+x</math>
-By solving we find--
+<math>\Rightarrow y=x(y+1)</math>
-<math>a=1</math>
+<math>\Rightarrow x=\frac{y}{y+1}</math>
-<math>b=0</math>
+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>
-However <cmath>a=x^2</cmath>
+<math>f(y)=\frac{1}{1-y}</math>
-and <cmath>b=y^2</cmath>
-Therefore
+<math>f(-y)=\frac{-y}{y+1}</math>
-<math>y=0</math>, and <math>x=1,-1</math>
-Thus, the only solutions are <math>(0,1)</math>, and <math>(0,-1)</math>
+<math>-f(y)=\frac{y}{y-1}</math>
+<math>-f(-y)=\frac{y}{y+1}</math>
-So there are only 2 solutions
+Since <math>-f(-y)=\frac{y}{y+1}=x</math> the answer must be <math>\boxed{C}</math>.
-<math>\rightarrow</math> <math>\fbox{C}</math>
 == See also ==
-{{AHSME box|year=1967|num-b=13|num-a=15}}
+{{AHSME 40p box|year=1967|num-b=13|num-a=15}}
 [[Category:Introductory Algebra Problems]]
 {{MAA Notice}}

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Difference between revisions of "1967 AHSME Problems/Problem 14"

Latest revision as of 19:42, 22 August 2023

Problem

Solution

See also