Difference between revisions of "1982 AHSME Problems/Problem 12"

Revision as of 15:44, 31 January 2019

Problem

Let $f(x) = ax^7+bx^3+cx-5$ , where $a,b$ and $c$ are constants. If $f(-7) = 7$ , then $f(7)$ equals

$\text {(A)} -17 \qquad \text {(B)} -7 \qquad \text {(C)} 14 \qquad \text {(D)} 21\qquad \text {(E)} \text{not uniquely determined}$

Solution

$f(x)$ is an odd function shifted down 5 units. Thus, it can be written as $f(x)=g(x)-5$ where $g(x)=ax^6+bx^3+cx$ . Thus: $f(-7)=g(-7)-5=7$ and $g(-7)=12$ . Using this and the fact $g(x)$ is odd, we can evaluate $f(7)$ , which is:

$f(7) = g(7)-5 = -g(-7)-5 = -12-5 = -17$

Therefore, the answer is $\boxed{ \textbf{A}}$ .

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Difference between revisions of "1982 AHSME Problems/Problem 12"

Revision as of 15:44, 31 January 2019

Problem

Solution