Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1996 AHSME Problems/Problem 6"
 
(4 intermediate revisions by one other user not shown)
Line 1: Line 1:
−
==Problem 6==
+
==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>
Line 9: Line 9:
 
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>.
−
 
  
 
==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 13:07, 5 July 2013

Problem

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 (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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1996_AHSME_Problems/Problem_6&oldid=56473"