Difference between revisions of "1996 AHSME Problems/Problem 6"

Revision as of 19:44, 18 August 2011

If $f(x) = x^{(x+1)}(x+2)^{(x+3)}$ , then $f(0)+f(-1)+f(-2)+f(-3) =$

$\text{(A)}\ -\frac{8}{9}\qquad\text{(B)}\ 0\qquad\text{(C)}\ \frac{8}{9}\qquad\text{(D)}\ 1\qquad\text{(E)}\ \frac{10}{9}$

Plugging in $x=0$ into the function will give $0^1\cdot 2^3$ . Since $0^1 = 0$ , this gives $0$ .

Plugging in $x=-1$ into the function will give $(-1)^0 \cdot 1^2$ . Since $(-1)^0 = 1$ and $1^2 = 1$ , this gives $1$ .

Plugging in $x=-2$ will give a $0^1$ factor as the second term, giving an answer of $0$ .

Plugging in $x=-3$ will give $(-3)^{-2}\cdot (-1)^0$ . The last term is $1$ , while the first term is $\frac{1}{(-3)^2} = \frac{1}{9}$

Adding up all four values, the answer is $1 + \frac{1}{9} = \frac{10}{9}$ , and the right answer is $\boxed{E}$ .

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

@@ Line 13: / Line 13: @@
 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>.