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Difference between revisions of "1969 AHSME Problems/Problem 14"

(Created page with "== Problem == The complete set of <math>x</math>-values satisfying the inequality <math>\frac{x^2-4}{x^2-1}>0</math> is the set of all <math>x</math> such that: <math>\text{(A)...")
 
(Solution to Problem 14)
 
(One intermediate revision by one other user not shown)
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== Solution ==
 
== Solution ==
−
<math>\fbox{A}</math>
+
Factor the [[difference of squares]].
 +
<cmath>\frac{(x+2)(x-2)}{(x+1)(x-1)}>0</cmath>
 +
Note that the graph intersects the x-axis at when <math>x = \pm2</math> or <math>x \pm 1</math>, so check the sign of the result to see if it is positive.  After testing, <math>x<-2</math> or <math>-1<x<1</math> or <math>x>2</math>, so the answer is <math>\boxed{\textbf{(A)}}</math>.
  
 
== See also ==
 
== See also ==
−
{{AHSME box|year=1969|num-b=13|num-a=15}}   
+
{{AHSME 35p box|year=1969|num-b=13|num-a=15}}   
  
 
[[Category: Introductory Algebra Problems]]
 
[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 03:35, 7 June 2018

Problem

The complete set of $x$-values satisfying the inequality $\frac{x^2-4}{x^2-1}>0$ is the set of all $x$ such that:

$\text{(A) } x>2 \text{ or } x<-2 \text{ or} -1<x<1\quad \text{(B) } x>2 \text{ or } x<-2\quad \\ \text{(C) } x>1 \text{ or } x<-2\qquad\qquad\qquad\quad \text{(D) } x>1 \text{ or } x<-1\quad \\ \text{(E) } x \text{ is any real number except 1 or -1}$

Solution

Factor the difference of squares. \[\frac{(x+2)(x-2)}{(x+1)(x-1)}>0\] Note that the graph intersects the x-axis at when $x = \pm2$ or $x \pm 1$, so check the sign of the result to see if it is positive. After testing, $x<-2$ or $-1<x<1$ or $x>2$, so the answer is $\boxed{\textbf{(A)}}$.

See also

1969 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
All AHSME Problems and Solutions

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