Difference between revisions of "1952 AHSME Problems/Problem 36"
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== Solution == | == Solution == | ||
| − | Factoring the numerator and denominator gives <cmath>\dfrac{(x+1)(x^{2}-x+1)}{(x+1)(x-1)} | + | Factoring the numerator using the sum of cubes identity and denominator using the difference of squares identity gives <cmath>\dfrac{(x+1)(x^{2}-x+1)}{(x+1)(x-1)}</cmath> |
| − | <math>\fbox{E}</math> | + | Cancelling out a factor of x+1 from the numerator and denominator gives <cmath>\dfrac{(x^{2}-x+1)}{(x-1)}</cmath> |
| + | Plugging in x= -1 gives <math>\dfrac{3}{-2}</math> or <math>\fbox{E}</math>. | ||
== See also == | == See also == | ||
Revision as of 21:47, 12 April 2020
Problem
To be continuous at 
, the value of 
is taken to be:
Solution
Factoring the numerator using the sum of cubes identity and denominator using the difference of squares identity gives ![\[\dfrac{(x+1)(x^{2}-x+1)}{(x+1)(x-1)}\]](http://latex.artofproblemsolving.com/c/1/7/c17d28fe7291e6df0322b381077c903f61df7939.png)
Cancelling out a factor of x+1 from the numerator and denominator gives ![\[\dfrac{(x^{2}-x+1)}{(x-1)}\]](http://latex.artofproblemsolving.com/c/1/9/c19cf166b39e0fb5071ab867ab4c689e99fdcd64.png)
Plugging in x= -1 gives 
or 
.
See also
| 1952 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 35 |
Followed by Problem 37 | |
| 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 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
