Difference between revisions of "1996 AHSME Problems/Problem 6"
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Plugging in <math>x=-2</math> will give a <math>0^1</math> factor as the second term, giving an answer of <math>0</math>. | Plugging in <math>x=-2</math> will give a <math>0^1</math> factor as the second term, giving an answer of <math>0</math>. | ||
− | Plugging in <math>x=-3</math> will give <math>(-3)^{-2}\cdot -1^0</math>. The last term is <math>1</math>, while the first term is <math>\frac{1}{(-3)^2} = \frac{1}{9}</math> | + | Plugging in <math>x=-3</math> will give <math>(-3)^{-2}\cdot (-1)^0</math>. The last term is <math>1</math>, while the first term is <math>\frac{1}{(-3)^2} = \frac{1}{9}</math> |
Adding up all four values, the answer is <math>1 + \frac{1}{9} = \frac{10}{9}</math>, and the right answer is <math>\boxed{E}</math>. | Adding up all four values, the answer is <math>1 + \frac{1}{9} = \frac{10}{9}</math>, and the right answer is <math>\boxed{E}</math>. |
Revision as of 20:44, 18 August 2011
Problem 6
If , then
Solution
Plugging in into the function will give
. Since
, this gives
.
Plugging in into the function will give
. Since
and
, this gives
.
Plugging in will give a
factor as the second term, giving an answer of
.
Plugging in will give
. The last term is
, while the first term is
Adding up all four values, the answer is , and the right answer is
.
See also
1996 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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 |