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

(Solution)
(Solution)
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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>.

Revision as of 19:43, 18 August 2011

Problem 6

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

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

Solution

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 $\boxed{E}$.

See also

1996 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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