N2007 P1 Q9

N2007 P1 Q9

11 marks
Tutorial

The diagram shows the graph of \({y = e^x - 3x}\). The two roots of the equation \({{\mathrm{e}}^{x}}-3x=0\) are denoted by \({\alpha}\) and \({\beta}\), where \({\alpha < \beta}\).

  1. Find the values of \({\alpha}\) and \({\beta}\), each correct to \(3\) decimal places.[2]

A sequence of real numbers \({{x}_{1}}\), \({{x}_{2}}\), \({{x}_{3}}\), … satisfies the recurrence relation

\({{x}_{n+1}}=\frac{1}{3}{{\mathrm{e}}^{{{x}_{n}}}}\) for \(n\ge 1\).

  1. Prove algebraically that, if the sequence converges, then it converges to either \({\alpha}\) or \({\beta}\).[2]
  2. Use a calculator to determine the behaviour of the sequence for each of the cases \({x_1 = 0}\), \({x_1 = 1}\), \({x_1 = 2}\).[3]
  3. By considering \(x_{n+1}-x_n\), prove that
    \(x_{n+1}<x_n\) if \(\alpha < x_{n} < \beta\),
    \(x_{n+1}>x_n\) if \( x_n < \alpha \) or \(x_n>\beta\).[2]
  4. State briefly how the results in part (iv) relate to the behaviours determined in part (iii). [2]

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Answer:(i) \(\alpha=0.619\), \(\beta=1.512\) (ii) Any finite limit is \(\alpha\) or \(\beta\) (iii) \(x_1=0\): increasing to \(\alpha\), \(x_1=1\): decreasing to \(\alpha\), \(x_1=2\): increasing and divergent (iv)(a) \(\alpha<x_n<\beta\) (b) \(x_n<\alpha\) or \(x_n>\beta\) (v) These intervals explain the three behaviours in (iii).

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