2018 DHS P2 Q4

2018 DHS P2 Q4

8 marks
Prelims

The diagram below shows the curve \(y=\mathrm{{f}'}\left( x \right).\) It has turning points at \(\left( 0,6 \right),\left( 5,0 \right)\) and \(\left( 7,7 \right)\) and intersects the \(x\)-axis at \(\left( -3,0 \right)\) and \(\left( 8,0 \right).\)

  1. State the \(x\)-coordinates of the stationary points for the curve \(y=\mathrm{f}\left( x \right).\) Hence determine the nature of these stationary points.[3]
  2. State the range of values of \(x\) such that the curve \(y=\mathrm{f}\left( x \right)\) is concave downwards.[2]
  3. A student makes a claim on each of the following statements:
    1. If the curve \(y=\mathrm{f}\left( x \right)\) passes through the point \(\left( -3,7 \right),\) then \(\mathrm{f}\left( x \right)>7\) for \(-3 < x < 5.\)
    2. By using integration by parts, \(\displaystyle\displaystyle \displaystyle\int_{-3}^{8}{x\mathrm{f}\left( x \right)}\,\mathrm{d}x>32\mathrm{f}\left( 8 \right)-\frac{9}{2}\mathrm{f}\left( -3 \right).\)
      For each of the above statements, explain briefly whether the student is right to make such a claim.[3]
    For each of the above statements, explain briefly whether the student is right to make such a claim.[3]

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Answer:(i) \(x=-3\) minimum, \(x=5\) stationary point of inflexion, \(x=8\) maximum (ii) \(0<x<5\) or \(x>7\) (iii)(a) Right (iii)(b) Wrong because \(\int_{-3}^{8}xf(x)\,\mathrm dx<32f(8)-\frac92f(-3)\)

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