| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Topic | Fixed Point Iteration |
| Type | Show root in interval |
| Difficulty | Moderate -0.3 This is a standard numerical methods question covering routine A-level topics: change of sign verification (straightforward substitution), basic sketching, and iteration convergence using gradient conditions. While multi-part with 11 marks total, each component requires only standard techniques without novel insight—slightly easier than a typical A-level question due to the guided structure and elementary nature of the tasks. |
| Spec | 1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(f(0.7) = 0.0648 > 0\); \(f(0.8) = -0.103 < 0\); Sign change hence root | M1, A1 [2] | Evaluate at both 0.7 and 0.8. Conclude by referring to sign change oe or CWO. |
| (ii) Graph of \(y = x\) and \(y = \cos x\) | B1, B1 [2] | Sketch both graphs... ... in correct proportion to each other and intercepts correct. |
| (iii) \(\frac{dy}{dx} = -\sin x\); since \(0 < x < \pi/2\) the magnitude of \(-\sin x\) is less than 1, therefore the iteration converges | B1, M1, A1 [3] | State correct derivative. Consider magnitude of gradient, either in general terms or at specific value(s). Allow use of \(^4/_4\) as a specific value. Conclude using \( |
| \(x\) | \(0.7\) | \(0.8\) |
| \(dy/dx\) | \(-0.64...\) | \(-0.71...\) |
| Answer | Marks | Guidance |
|---|---|---|
| (iv) | M1, A1 [2] | First two segments. At least 5 segments. Allow (ii) and (iv) on the same graph. |
| (v) \(\cos(0.73905) - 0.73905 = +5.879...\times 10^{-5}\); \(\cos(0.73915) - 0.73915 = -1.085...\times 10^{-4}\); By the sign change rule \(\alpha\) lies in that interval and therefore rounds to \(0.7391\) to 4 dp. | M1, A1 [2] | Evaluate at both 0.73905 and 0.73915 (or values closer to the root). Conclude by referring to sign change oe or CWO. |
**(i)** $f(0.7) = 0.0648 > 0$; $f(0.8) = -0.103 < 0$; Sign change hence root | M1, A1 [2] | Evaluate at both 0.7 and 0.8. Conclude by referring to sign change oe or CWO.
**(ii)** Graph of $y = x$ and $y = \cos x$ | B1, B1 [2] | Sketch both graphs... ... in correct proportion to each other and intercepts correct.
**(iii)** $\frac{dy}{dx} = -\sin x$; since $0 < x < \pi/2$ the magnitude of $-\sin x$ is less than 1, therefore the iteration converges | B1, M1, A1 [3] | State correct derivative. Consider magnitude of gradient, either in general terms or at specific value(s). Allow use of $^4/_4$ as a specific value. Conclude using $|F'(x)| < 1$. Allow $-1 < F'(x) < 0$. A0 for $|F'(x)| < 1$. A0 for $|F'(x)| < 1$ from $0 < x < \pi/2$, unless end point clearly dealt with.
| $x$ | $0.7$ | $0.8$ |
| $dy/dx$ | $-0.64...$ | $-0.71...$ |
magnitude of gradient in the region is less than 1 therefore the iteration converges
**(iv)** | M1, A1 [2] | First two segments. At least 5 segments. Allow (ii) and (iv) on the same graph.
**(v)** $\cos(0.73905) - 0.73905 = +5.879...\times 10^{-5}$; $\cos(0.73915) - 0.73915 = -1.085...\times 10^{-4}$; By the sign change rule $\alpha$ lies in that interval and therefore rounds to $0.7391$ to 4 dp. | M1, A1 [2] | Evaluate at both 0.73905 and 0.73915 (or values closer to the root). Conclude by referring to sign change oe or CWO.\begin{enumerate}[label=(\roman*)]
\item Use a change of sign to verify that the equation $\cos x - x = 0$ has a root $\alpha$ between $x = 0.7$ and $x = 0.8$. [2]
\item Sketch, on a single diagram, the curve $y = \cos x$ and the line $y = x$ for $0 \leqslant x \leqslant \frac{1}{2}\pi$, giving the coordinates of all points of intersection with the coordinate axes. [2]
\end{enumerate}
An iteration of the form $x_{n+1} = \cos(x_n)$ is to be used to find $\alpha$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item By considering the gradient of $y = \cos x$, show that this iteration will converge. [3]
\item On a copy of your sketch from part (ii), illustrate how this iteration converges to $\alpha$. [2]
\item Use a change of sign to verify that $\alpha = 0.7391$ to 4 decimal places. [2]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2016 Q7 [11]}}