| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Sketch graphs to show root existence |
| Difficulty | Standard +0.3 This is a straightforward fixed point iteration question requiring a simple graph sketch to show uniqueness, then mechanical application of a given iterative formula. The iteration converges quickly and requires only calculator work with no rearrangement needed. Slightly easier than average as it's purely procedural with no problem-solving or insight required. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02n Sketch curves: simple equations including polynomials1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sketch reasonable attempt at \(y = x^5\) | *B1 | Accept non-zero gradient at \(O\) but curvature to be correct in first and third quadrants |
| Sketch straight line with negative gradient | *B1 | Existing at least in (part of) first quadrant |
| Indicate in some way single point of intersection | B1 | 3 dep *B1 *B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Obtain correct first iterate | B1 | Allow if not part of subsequent iteration |
| Carry out process to find at least 3 iterates in all | M1 | |
| Obtain at least 1 correct iterate after the first | A1 | Allow for recovery after error; showing at least 3 d.p. in iterates |
| Conclude 2.175 | A1 | 4 answer required to precisely 3 d.p. |
| \([0 \to 2.21236 \to 2.17412 \to 2.17480 \to 2.17479;\) \(1 \to 2.19540 \to 2.17442 \to 2.17480 \to 2.17479;\) \(2 \to 2.17791 \to 2.17473 \to 2.17479 \to 2.17479;\) \(3 \to 2.15983 \to 2.17506 \to 2.17479 \to 2.17479]\) |
# Question 3:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sketch reasonable attempt at $y = x^5$ | *B1 | Accept non-zero gradient at $O$ but curvature to be correct in first and third quadrants |
| Sketch straight line with negative gradient | *B1 | Existing at least in (part of) first quadrant |
| Indicate in some way single point of intersection | B1 | **3** dep *B1 *B1 |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Obtain correct first iterate | B1 | Allow if not part of subsequent iteration |
| Carry out process to find at least 3 iterates in all | M1 | |
| Obtain at least 1 correct iterate after the first | A1 | Allow for recovery after error; showing at least 3 d.p. in iterates |
| Conclude 2.175 | A1 | **4** answer required to precisely 3 d.p. |
| $[0 \to 2.21236 \to 2.17412 \to 2.17480 \to 2.17479;$ $1 \to 2.19540 \to 2.17442 \to 2.17480 \to 2.17479;$ $2 \to 2.17791 \to 2.17473 \to 2.17479 \to 2.17479;$ $3 \to 2.15983 \to 2.17506 \to 2.17479 \to 2.17479]$ | | |
---3
\begin{enumerate}[label=(\alph*)]
\item It is given that $a$ and $b$ are positive constants. By sketching graphs of
$$y = x ^ { 5 } \quad \text { and } \quad y = a - b x$$
on the same diagram, show that the equation
$$x ^ { 5 } + b x - a = 0$$
has exactly one real root.
\item Use the iterative formula $x _ { n + 1 } = \sqrt [ 5 ] { 53 - 2 x _ { n } }$, with a suitable starting value, to find the real root of the equation $x ^ { 5 } + 2 x - 53 = 0$. Show the result of each iteration, and give the root correct to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{OCR C3 2007 Q3 [7]}}