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2014/15 AHSMC Part 1 Problem 8

Problem 8 of 16

Let ff be a function such that f(x)+3f(1x)=x2f(x) + 3f\bigl(\dfrac{1}{x}\bigr) = x^2 for any nonzero real number xx. Then the value of f(−3)f(-3) is

Answer choices

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Solution

Replacing xx by −3-3 and then by −13-\dfrac{1}{3} in the given equation, we obtain f(−3)+3f(−13)=9f(-3) + 3f\bigl(-\dfrac{1}{3}\bigr) = 9 and respectively f(−13)+3f(−3)=19f\bigl(-\dfrac{1}{3}\bigr) + 3f(-3) = \dfrac{1}{9}. Solving for f(−3)f(-3) we get f(−3)=−1312f(-3) = -\dfrac{13}{12}. The answer is (D).

Alternative Solution: Replacing xx by 1x\dfrac{1}{x} in the given equation, we obtain f(1x)+3f(x)=1x2f\bigl(\dfrac{1}{x}\bigr) + 3f(x) = \dfrac{1}{x^2}. Solving for f(x)f(x) we get f(x)=3−x48x2f(x) = \dfrac{3-x^4}{8x^2}. Hence f(−3)=−3936=−1312f(-3) = -\dfrac{39}{36} = -\dfrac{13}{12}.

University of Alberta