Difference between revisions of "1990 AHSME Problems/Problem 8"
(Created page with "== Problem == The number of real solutions of the equation <cmath>|x-2|+|x-3|=1</cmath> is <math>\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 3\quad \te...") |
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== Solution == | == Solution == | ||
− | <math>\fbox{E}</math> | + | For <math>2\le x\le 3</math>, the left-hand side is <math>(x-2)-(x-3)</math> which is <math>1</math> for all <math>x</math> in the interval. <math>\fbox{E}</math> |
== See also == | == See also == |
Latest revision as of 03:55, 4 February 2016
Problem
The number of real solutions of the equation is
Solution
For , the left-hand side is which is for all in the interval.
See also
1990 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.