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

Revision as of 14:09, 10 August 2020

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

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

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

\[y=\frac{1}{1-x}\RightArrow y(1-x)=1 \RightArrow y-yx=1\RightArrow yx=y-1\RightArrow x=\frac{y-1}{y}\] (Error compiling LaTeX. Unknown error_msg)

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

@@ 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
+<cmath>y=\frac{1}{1-x}\RightArrow y(1-x)=1 \RightArrow y-yx=1\RightArrow yx=y-1\RightArrow x=\frac{y-1}{y}</cmath>
-         <math>a+4b=1</math>
-and
-         <math>4a+b=4</math>
-By solving we find--
-<math>a=1</math>
-<math>b=0</math>
-However <cmath>a=x^2</cmath>
-and <cmath>b=y^2</cmath>
-Therefore
-<math>y=0</math>, and <math>x=1,-1</math>
-Thus, the only solutions are <math>(0,1)</math>, and <math>(0,-1)</math>
-So there are only 2 solutions
-<math>\rightarrow</math> <math>\fbox{C}</math>
 == See also ==