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2016/17 AHSMC Part 1 Problem 12

Problem 12 of 16

Let f ⁣:R→Rf \colon \mathbb{R} \to \mathbb{R} be a function such that xf(x)+(1−x)f(−x)=x2+x+1x f(x) + (1-x)f(-x) = x^2 + x + 1 for any real number xx. The greatest real number MM for which f(x)≥Mf(x) \ge M for all real numbers xx, is

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Solution

If xx is replaced by −x-x in the given equation then −xf(−x)+(1+x)f(x)=x2−x+1. -x f(-x) + (1+x)f(x) = x^2 - x + 1. Using the given equation and the one that is obtained above, one obtains by subtraction that f(−x)=f(x)+2xf(-x) = f(x) + 2x, so xf(x)+(1−x)(f(x)+2x)=x2+x+1x f(x) + (1-x)(f(x)+2x) = x^2 + x + 1 and thus f(x)=x2+x+1−2x(1−x)=3x2−x+1=3(x−16)2+1112 f(x) = x^2 + x + 1 - 2x(1-x) = 3x^2 - x + 1 = 3\Bigl(x - \frac{1}{6}\Bigr)^2 + \frac{11}{12} and hence f(x)≥1112f(x) \ge \dfrac{11}{12}. If x=16x = \dfrac{1}{6} then we get f(16)=1112f\bigl(\dfrac{1}{6}\bigr) = \dfrac{11}{12}. The answer is (E).

University of Alberta