AQA Paper 2 2024 June — Question 11

Exam BoardAQA
ModulePaper 2 (Paper 2)
Year2024
SessionJune
TopicProof

11
  1. A student states that 3 is the smallest value of \(k\) in the interval \(3 < k < 4\) Explain the error in the student's statement.
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  2. The student's teacher says there is no smallest value of \(k\) in the interval \(3 < k < 4\) The teacher gives the following correct proof: Step 1: Assume there is a smallest number in the interval \(3 < k < 4\) and let this smallest number be \(x\) Step 2: let \(y = \frac { 3 + x } { 2 }\)
    Step 3: \(3 < y < x\) which is a contradiction.
    Step 4: Therefore, there is no smallest number in interval \(3 < k < 4\) 11
    1. Explain the contradiction stated in Step 3
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  3. (ii) Prove that there is no largest value of \(k\) in the interval \(3 < k < 4\)
    \section*{END OF SECTION A TURN OVER FOR SECTION B}