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...")
 
 
Line 12: Line 12:
  
 
== 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 \[|x-2|+|x-3|=1\] is

$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 3\quad \text{(E) more than } 3$

Solution

For $2\le x\le 3$, the left-hand side is $(x-2)-(x-3)$ which is $1$ for all $x$ in the interval. $\fbox{E}$

See also

1990 AHSME (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS