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Difference between revisions of "1984 AHSME Problems/Problem 16"

(Created page with "==Problem== The function <math> f(x) </math> satisfies <math> f(2+x)=f(2-x) </math> for all real numbers <math> x </math>. If the equation <math> f(x)=0 </math> has exactly f...")
 
 
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==Solution==
 
==Solution==
 
Let one of the roots be <math> r_1 </math>. Also, define <math> x </math> such that <math> 2+x=r_1 </math>. Thus, we have <math> f(2+x)=f(r_1)=0 </math> and <math> f(2+x)=f(2-x) </math>. Therefore, we have <math> f(2-x)=0 </math>, and <math> 2-x </math> is also a root. Let this root be <math> r_2 </math>. The sum <math> r_1+r_2=2+x+2-x=4 </math>. Similarly, we can let <math> r_3 </math> be a root and define <math> y </math> such that <math> 2+y=r_3 </math>, and we will find <math> 2-y </math> is also a root, say, <math> r_4 </math>, so <math> r_3+r_4=2+y+2-y=4 </math>. Therefore, <math> r_1+r_2+r_3+r_4=4+4=8, \boxed{\text{E}} </math>.
 
Let one of the roots be <math> r_1 </math>. Also, define <math> x </math> such that <math> 2+x=r_1 </math>. Thus, we have <math> f(2+x)=f(r_1)=0 </math> and <math> f(2+x)=f(2-x) </math>. Therefore, we have <math> f(2-x)=0 </math>, and <math> 2-x </math> is also a root. Let this root be <math> r_2 </math>. The sum <math> r_1+r_2=2+x+2-x=4 </math>. Similarly, we can let <math> r_3 </math> be a root and define <math> y </math> such that <math> 2+y=r_3 </math>, and we will find <math> 2-y </math> is also a root, say, <math> r_4 </math>, so <math> r_3+r_4=2+y+2-y=4 </math>. Therefore, <math> r_1+r_2+r_3+r_4=4+4=8, \boxed{\text{E}} </math>.
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==Solution 2==
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The graph of this function is symmetric around 2. Therefore, two roots will be greater than <math>2</math> and the other two roots will be less than <math>2</math>. These four roots are symmetric around <math>2</math>, so the average of the four roots is <math>2</math>. Then, the sum is <math>2\times4=8</math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1984|num-b=15|num-a=17}}
 
{{AHSME box|year=1984|num-b=15|num-a=17}}
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{{MAA Notice}}

Latest revision as of 12:53, 24 January 2024

Problem

The function $f(x)$ satisfies $f(2+x)=f(2-x)$ for all real numbers $x$. If the equation $f(x)=0$ has exactly four distinct real roots, then the sum of these roots is

$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 8$

Solution

Let one of the roots be $r_1$. Also, define $x$ such that $2+x=r_1$. Thus, we have $f(2+x)=f(r_1)=0$ and $f(2+x)=f(2-x)$. Therefore, we have $f(2-x)=0$, and $2-x$ is also a root. Let this root be $r_2$. The sum $r_1+r_2=2+x+2-x=4$. Similarly, we can let $r_3$ be a root and define $y$ such that $2+y=r_3$, and we will find $2-y$ is also a root, say, $r_4$, so $r_3+r_4=2+y+2-y=4$. Therefore, $r_1+r_2+r_3+r_4=4+4=8, \boxed{\text{E}}$.

Solution 2

The graph of this function is symmetric around 2. Therefore, two roots will be greater than $2$ and the other two roots will be less than $2$. These four roots are symmetric around $2$, so the average of the four roots is $2$. Then, the sum is $2\times4=8$.

See Also

1984 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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

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