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

Revision as of 14:07, 10 August 2020

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

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

Let $x^2 = a$ and $y^2 = b.$

then

and

By solving we find--

$a=1$

$b=0$

However $a=x^2$ and $b=y^2$

Therefore $y=0$ , and $x=1,-1$

Thus, the only solutions are $(0,1)$ , and $(0,-1)$

So there are only 2 solutions

$\rightarrow$ $\fbox{C}$

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 1: / Line 1: @@
-== Problem ==
+==Problem 14==
-The number of distinct points common to the curves <math>x^2+4y^2=1</math> and <math>4x^2+y^2=4</math> is:
-<math>\text{(A) } 0 \quad \text{(B) } 1 \quad \text{(C) } 2 \quad \text{(D) } 3 \quad \text{(E) } 4</math>
+Let <math>f(t)=\frac{t}{1-t}</math>, <math>t \not= 1</math>.  If <math>y=f(x)</math>, then <math>x</math> can be expressed as
+<math>\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)</math>
+[[1967 AHSME Problems/Problem 14|Solution]]
 == Solution ==