IMC 2020 Q20
20. For any fixed value of \(x\), which of the following four expressions has the largest value?
\((x + 1)(x - 1)\) \((x + \tfrac{1}{2})(x - \tfrac{1}{2})\) \((x + \tfrac{1}{3})(x - \tfrac{1}{3})\) \((x + \tfrac{1}{4})(x - \tfrac{1}{4})\)
- A \((x + 1)(x - 1)\)
- B \((x + \tfrac{1}{2})(x - \tfrac{1}{2})\)
- C \((x + \tfrac{1}{3})(x - \tfrac{1}{3})\)
- D \((x + \tfrac{1}{4})(x - \tfrac{1}{4})\)
- E it depends on the value of \(x\)
Intermediate Mathematical Challenge 2020, Question 20. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: D \((x + \tfrac{1}{4})(x - \tfrac{1}{4})\)
The first four expressions expand to \(x^2 - 1\); \(x^2 - \tfrac{1}{4}\); \(x^2 - \tfrac{1}{9}\); \(x^2 - \tfrac{1}{16}\) respectively.
Note that \(\tfrac{1}{16} < \tfrac{1}{9} < \tfrac{1}{4} < 1\). Therefore the least value is \(x^2 - \tfrac{1}{16}\), that is \((x + \tfrac{1}{4})(x - \tfrac{1}{4})\).
This result is irrespective of the value of \(x\).
Solution © UK Mathematics Trust, from the IMC 2020 solutions. Longer versions with investigations: Solutions and investigations (PDF).