| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2012 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive stationary point equation |
| Difficulty | Standard +0.3 This is a standard C3 numerical methods question requiring routine application of sign change, differentiation to find stationary points, and iterative formula application. Part (b) involves straightforward rearrangement of f'(x)=0, and all parts follow predictable textbook patterns with no novel problem-solving required. Slightly easier than average due to the guided structure. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07n Stationary points: find maxima, minima using derivatives1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(0.8) = 0.082\), \(f(0.9) = -0.089\) | M1 | Calculates both values; evidence includes both \(x\) substitutions, or one value correct to 1 sig fig |
| Change of sign \(\Rightarrow\) root in \((0.8, 0.9)\) | A1 | Requires both correct values, a reason (change of sign, \(<0>0\), \(+ve\ -ve\), \(f(0.8)f(0.9)<0\)) and a conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f'(x) = 2x - 3 - \sin\!\left(\frac{1}{2}x\right)\) | M1 A1 | M1 for attempt to differentiate; seeing \(2x\), \(3\) or \(\pm A\sin(\frac{1}{2}x)\) sufficient |
| Sets \(f'(x)=0 \Rightarrow x = \dfrac{3+\sin\!\left(\frac{1}{2}x\right)}{2}\) | M1 A1* | Must confirm they are setting \(f'(x)=0\) first; given answer so must be fully correct (cso) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Sub \(x_0=2\): \(x_{n+1} = \dfrac{3+\sin\!\left(\frac{1}{2}x_n\right)}{2}\) | M1 | Evidence could be awrt 1.9 or 1.5 (from degrees) |
| \(x_1 =\) awrt \(1.921\) | A1 | |
| \(x_2 =\) awrt \(1.91(0)\), \(x_3 =\) awrt \(1.908\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Choose interval \([1.90775, 1.90785]\) | M1 | Or tighter interval containing root \(= 1.907845522\) |
| \(f'(1.90775) = -0.00016\), \(f'(1.90785) = 0.0000076\) | M1 | At least one correct, rounded or truncated; \(f'(1.90775)\approx -0.0001\), \(f'(1.90785)\approx 0.000007\) |
| Change of sign \(\Rightarrow x = 1.9078\) | A1 | Both values correct with valid reason and minimal conclusion |
## Question 6:
**Part (a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(0.8) = 0.082$, $f(0.9) = -0.089$ | M1 | Calculates both values; evidence includes both $x$ substitutions, or one value correct to 1 sig fig |
| Change of sign $\Rightarrow$ root in $(0.8, 0.9)$ | A1 | Requires both correct values, a reason (change of sign, $<0>0$, $+ve\ -ve$, $f(0.8)f(0.9)<0$) and a conclusion |
**Part (b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f'(x) = 2x - 3 - \sin\!\left(\frac{1}{2}x\right)$ | M1 A1 | M1 for attempt to differentiate; seeing $2x$, $3$ or $\pm A\sin(\frac{1}{2}x)$ sufficient |
| Sets $f'(x)=0 \Rightarrow x = \dfrac{3+\sin\!\left(\frac{1}{2}x\right)}{2}$ | M1 A1* | Must confirm they are setting $f'(x)=0$ first; given answer so must be fully correct (cso) |
**Part (c):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sub $x_0=2$: $x_{n+1} = \dfrac{3+\sin\!\left(\frac{1}{2}x_n\right)}{2}$ | M1 | Evidence could be awrt 1.9 or 1.5 (from degrees) |
| $x_1 =$ awrt $1.921$ | A1 | |
| $x_2 =$ awrt $1.91(0)$, $x_3 =$ awrt $1.908$ | A1 | |
**Part (d):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Choose interval $[1.90775, 1.90785]$ | M1 | Or tighter interval containing root $= 1.907845522$ |
| $f'(1.90775) = -0.00016$, $f'(1.90785) = 0.0000076$ | M1 | At least one correct, rounded or truncated; $f'(1.90775)\approx -0.0001$, $f'(1.90785)\approx 0.000007$ |
| Change of sign $\Rightarrow x = 1.9078$ | A1 | Both values correct with valid reason and minimal conclusion |
---6.
$$f ( x ) = x ^ { 2 } - 3 x + 2 \cos \left( \frac { 1 } { 2 } x \right) , \quad 0 \leqslant x \leqslant \pi$$
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $\mathrm { f } ( x ) = 0$ has a solution in the interval $0.8 < x < 0.9$
The curve with equation $y = \mathrm { f } ( x )$ has a minimum point $P$.
\item Show that the $x$-coordinate of $P$ is the solution of the equation
$$x = \frac { 3 + \sin \left( \frac { 1 } { 2 } x \right) } { 2 }$$
\item Using the iteration formula
$$x _ { n + 1 } = \frac { 3 + \sin \left( \frac { 1 } { 2 } x _ { n } \right) } { 2 } , \quad x _ { 0 } = 2$$
find the values of $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$, giving your answers to 3 decimal places.
\item By choosing a suitable interval, show that the $x$-coordinate of $P$ is 1.9078 correct to 4 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2012 Q6 [12]}}