MATH100 December 2014
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Question 09 (b)
Consider the equation , where is some real number. let be any real number. Prove that the equation has at most two solutions.
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Try proving by contradiction.
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- Let and suppose that the equation has solutions . ie .
- Now since is a polynomial it is continuous and differentiable everywhere, so we can apply the Mean Value Theorem or Rolle’s theorem.
- By Rolle’s theorem there must be points with between and and between and so that with .
- But which only has a single real zero at .
- Hence the derivative cannot have distinct zeros, so we have reached a contradiction and so cannot have solutions.
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