Difference between revisions of "1996 AHSME Problems/Problem 6"
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− | ==Problem | + | ==Problem== |
If <math> f(x) = x^{(x+1)}(x+2)^{(x+3)} </math>, then <math> f(0)+f(-1)+f(-2)+f(-3) = </math> | If <math> f(x) = x^{(x+1)}(x+2)^{(x+3)} </math>, then <math> f(0)+f(-1)+f(-2)+f(-3) = </math> | ||
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Plugging in <math>x=0</math> into the function will give <math>0^1\cdot 2^3</math>. Since <math>0^1 = 0</math>, this gives <math>0</math>. | Plugging in <math>x=0</math> into the function will give <math>0^1\cdot 2^3</math>. Since <math>0^1 = 0</math>, this gives <math>0</math>. | ||
− | Plugging in <math>x=-1</math> into the function will give <math>-1^0 \cdot 1^2</math>. Since <math>-1^0 = 1</math> and <math>1^2 = 1</math>, this gives <math>1</math>. | + | Plugging in <math>x=-1</math> into the function will give <math>(-1)^0 \cdot 1^2</math>. Since <math>(-1)^0 = 1</math> and <math>1^2 = 1</math>, this gives <math>1</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=-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>. | ||
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==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=5|num-a=7}} | {{AHSME box|year=1996|num-b=5|num-a=7}} | ||
+ | {{MAA Notice}} |
Latest revision as of 14:07, 5 July 2013
Problem
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.