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2020/21 AHSMC Part 1 Problem 6

Problem 6 of 16

Let ff be the function such that f(x)=ax9+bx5+cx3−7f(x) = ax^9 + bx^5 + cx^3 - 7, where aa, bb, and cc are real numbers. If f(−9)=9f(-9) = 9 then f(9)f(9) is equal to

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Solution

f(9)=a⋅99+b⋅95+c⋅93−7f(−9)=9=a⋅(−9)9+b⋅(−9)5+c⋅(−9)3−7\begin{align*} f(9) &= a \cdot 9^9 + b \cdot 9^5 + c \cdot 9^3 - 7 \tag{1} \\ f(-9) &= 9 = a \cdot (-9)^9 + b \cdot (-9)^5 + c \cdot (-9)^3 - 7 \tag{2} \end{align*} Adding equations (1) and (2), we obtain f(9)+f(−9)=f(9)+9=−14. f(9) + f(-9) = f(9) + 9 = -14. From this, f(9)=−14−9=−23. f(9) = -14 - 9 = -23. The answer is (A).

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