Difference between revisions of "1982 AHSME Problems/Problem 12"
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==Solution== | ==Solution== | ||
| − | <math>f(x)</math> is an odd function shifted down 5 units. Thus, it can be written as <math>f(x)=g(x)-5</math> where <math>g(x)=ax^ | + | <math>f(x)</math> is an odd function shifted down 5 units. Thus, it can be written as <math>f(x)=g(x)-5</math> where <math>g(x)=ax^7+bx^3+cx</math>. Thus: <math>f(-7)=g(-7)-5=7</math> and <math>g(-7)=12</math>. Using this and the fact <math>g(x)</math> is odd, we can evaluate <math>f(7)</math>, which is: |
<cmath>f(7) = g(7)-5 = -g(-7)-5 = -12-5 = -17</cmath> | <cmath>f(7) = g(7)-5 = -g(-7)-5 = -12-5 = -17</cmath> | ||
Therefore, the answer is <math>\boxed{ \textbf{A}}</math>. | Therefore, the answer is <math>\boxed{ \textbf{A}}</math>. | ||
Latest revision as of 15:44, 31 January 2019
Problem
Let 
, where 
and 
are constants. If 
, then 
equals
Solution
is an odd function shifted down 5 units. Thus, it can be written as 
where 
. Thus: 
and 
. Using this and the fact 
is odd, we can evaluate , which is:
![\[f(7) = g(7)-5 = -g(-7)-5 = -12-5 = -17\]](http://latex.artofproblemsolving.com/2/4/b/24bc0b0eaaefcba009989a2f6396241c89e3eb4a.png)
Therefore, the answer is 
.
