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Difference between revisions of "1977 AHSME Problems/Problem 22"

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== Problem 22 ==
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If <math>f(x)</math> is a real valued function of the real variable <math>x</math>, and <math>f(x)</math> is not identically zero,
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and for all <math>a</math> and <math>b</math> <math>f(a+b)+f(a-b)=2f(a)+2f(b)</math>, then for all <math>x</math> and <math>y</math>
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<math>\textbf{(A) }f(0)=1\qquad
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\textbf{(B) }f(-x)=-f(x)\qquad
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\textbf{(C) }f(-x)=f(x)\qquad \\
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\textbf{(D) }f(x+y)=f(x)+f(y) \qquad \\
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\textbf{(E) }\text{there is a positive real number }T\text{ such that }f(x+T)=f(x)    </math>
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== Solution ==
 
We can start by finding the value of <math>f(0)</math>.  
 
We can start by finding the value of <math>f(0)</math>.  
 
Let <math>a = b = 0</math>
 
Let <math>a = b = 0</math>
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~~JustinLee2017
 
~~JustinLee2017
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==See Also==
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{{AHSME box|year=1977|num-b=21|num-a=23}}
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{{MAA Notice}}

Latest revision as of 23:29, 12 February 2021

Problem 22

If $f(x)$ is a real valued function of the real variable $x$, and $f(x)$ is not identically zero, and for all $a$ and $b$ $f(a+b)+f(a-b)=2f(a)+2f(b)$, then for all $x$ and $y$

$\textbf{(A) }f(0)=1\qquad \textbf{(B) }f(-x)=-f(x)\qquad \textbf{(C) }f(-x)=f(x)\qquad \\ \textbf{(D) }f(x+y)=f(x)+f(y) \qquad \\ \textbf{(E) }\text{there is a positive real number }T\text{ such that }f(x+T)=f(x)$

Solution

We can start by finding the value of $f(0)$. Let $a = b = 0$ \[f(0) + f(0) = 2f(0) + 2f(0) \Rrightarrow 2f(0) = 4f(0) \Rrightarrow f(0) = 0\] Thus, $A$ is not true. To check $B$, we let $a = 0$. We have \[f(0+b) + f(0-b) = 2f(0) + 2f(b)\] \[f(b) + f(-b) = 0 + 2f(b)\] \[f(-b) = f(b)\] Thus, $B$ is not true, but $C$ is. Thus, the answer is $\boxed{C}$

~~JustinLee2017

See Also

1977 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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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