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2018/19 AHSMC Part 1 Problem 9

Problem 9 of 16

Let kk be a fixed positive real number. Suppose kx2≤10kx^2 \le 10 has exactly five positive integer solutions. The number of positive integer solutions to kx≤1kx \le 1 must be

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Solution

The given inequality may be written as follows: kx2≤10  ⟺  x2≤10k  ⟺  x∈[−10k, 10k]. kx^2 \le 10 \iff x^2 \le \frac{10}{k} \iff x \in \left[ -\sqrt{\frac{10}{k}},\ \sqrt{\frac{10}{k}} \right]. The inequality kx2≤10kx^2 \le 10 has exactly five positive integer solutions if and only if 10k∈[5,6)  ⟺  10k∈[25,36)  ⟺  1k∈[2.5,3.6). \sqrt{\frac{10}{k}} \in [5,6) \iff \frac{10}{k} \in [25,36) \iff \frac{1}{k} \in [2.5, 3.6). Since the inequality kx≤1kx \le 1 is equivalent to x≤1kx \le \dfrac{1}{k}, the number of positive integer solutions is two if 1k∈[2.5,3)\dfrac{1}{k} \in [2.5, 3) and three if 1k∈[3,3.6)\dfrac{1}{k} \in [3, 3.6). The answer is (B).