Multiple roots and starting value selection

A question is this type if and only if it involves an equation with multiple roots and requires analysis of which root will be found based on the choice of starting value, often with graphical interpretation.

2 questions

OCR FP2 2009 January Q5
5
\includegraphics[max width=\textwidth, alt={}, center]{b9f29713-bc86-4869-9e54-195208e5e81d-3_661_734_267_703} The diagram shows the curve with equation \(y = \mathrm { f } ( x )\), where $$f ( x ) = 2 x ^ { 3 } - 9 x ^ { 2 } + 12 x - 4.36$$ The curve has turning points at \(x = 1\) and \(x = 2\) and crosses the \(x\)-axis at \(x = \alpha , x = \beta\) and \(x = \gamma\), where \(0 < \alpha < \beta < \gamma\).
  1. The Newton-Raphson method is to be used to find the roots of the equation \(\mathrm { f } ( x ) = 0\), with \(x _ { 1 } = k\).
    (a) To which root, if any, would successive approximations converge in each of the cases \(k < 0\) and \(k = 1\) ?
    (b) What happens if \(1 < k < 2\) ?
  2. Sketch the curve with equation \(y ^ { 2 } = \mathrm { f } ( x )\). State the coordinates of the points where the curve crosses the \(x\)-axis and the coordinates of any turning points.
  3. Using the definitions of \(\cosh x\) and \(\sinh x\) in terms of \(\mathrm { e } ^ { x }\) and \(\mathrm { e } ^ { - x }\), show that $$1 + 2 \sinh ^ { 2 } x \equiv \cosh 2 x .$$
  4. Solve the equation $$\cosh 2 x - 5 \sinh x = 4$$ giving your answers in logarithmic form.
OCR MEI Paper 3 2024 June Q11
11 Fig. 11.1 shows the curve with equation \(\mathrm { y } = \mathrm { g } ( \mathrm { x } )\) where \(\mathrm { g } ( x ) = x \sin x + \cos x\) and the curve of the gradient function \(\mathrm { y } = \mathrm { g } ^ { \prime } ( \mathrm { x } )\) for \(- 2 \pi \leqslant x \leqslant 2 \pi\). \begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Fig. 11.1} \includegraphics[alt={},max width=\textwidth]{60e1e785-c34b-48ef-a63f-13a25fee186e-08_1136_1196_459_246}
\end{figure}
  1. Show that the \(x\)-coordinates of the points on the curve \(y = g ( x )\) where the gradient is 1 satisfy the equation \(\frac { 1 } { x } - \cos x = 0\). Fig. 11.2 shows part of the curve with equation \(y = \frac { 1 } { x } - \cos x\). \begin{figure}[h]
    \captionsetup{labelformat=empty} \caption{Fig. 11.2} \includegraphics[alt={},max width=\textwidth]{60e1e785-c34b-48ef-a63f-13a25fee186e-09_678_1363_424_239}
    \end{figure}
  2. Use the Newton-Raphson method with a suitable starting value to find the smallest positive \(x\)-coordinate of a point on the curve \(y = x \sin x + \cos x\) where the gradient is 1 . You should write down at least the following.
    • The iteration you use
    • The starting value
    • The solution correct to \(\mathbf { 4 }\) decimal places
    • Explain why \(x _ { 1 } = 3\) is not a suitable starting value for the Newton-Raphson method in part (b).