| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2014 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Solve trigonometric equation via iteration |
| Difficulty | Moderate -0.3 This is a standard C3 fixed point iteration question with routine steps: sign change to show root exists, algebraic rearrangement (straightforward), then mechanical application of iteration formula. All parts follow textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Radians: \(f(0.2) = -0.4\), \(f(0.4) = 0.3\) or considers smaller subset of \([0.2, 0.4]\) OR Degrees: \(f(0.2) = -0.4\), \(f(0.4) = 0.2\) or considers smaller subset of \([0.2, 0.4]\) | M1 | Gives two answers with at least one correct to 1sf. May work in degrees or radians. Maximum 6/7 for degrees. If larger interval chosen, this is M0 |
| Change of sign implies root | A1 | Both values correct to at least one decimal place, and reason given (e.g. change of sign, or \(f(0.2)<0\), \(f(0.4)>0\), or product \(f(0.2)f(0.4)<0\)) and conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sec x + 3x - 2 = 0 \Rightarrow 3x = 2 - \sec x\) and so \(x = \dfrac{2}{3} - \dfrac{1}{3\cos x}\) * | B1 | Starts with equation equal to zero, rearranges correctly with no errors and at least one intermediate step |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Radians: \(x_1 = 0.3177\), \(x_2 = 0.3158\), \(x_3 = 0.3160\) OR Degrees: \(x_1 = 0.3333\), \(x_2 = 0.3333\), \(x_3 = 0.3333\) | M1, A1, A1 | M1: Substitutes \(x_0 = 0.3\) into \(x = \dfrac{2}{3} - \dfrac{1}{3\cos x} \Rightarrow x_1 =\) (implied by \(x_1 = \dfrac{2}{3} - \dfrac{1}{3\cos 0.3}\), or answers rounding to 0.32 rads or 0.33 degrees). A1: \(x_1\) awrt 0.3177 4dp (rads) or awrt 0.3333 4dp (degrees). A1: \(x_2\) = awrt 0.3158, \(x_3\) = awrt 0.3160 (rads). NB \(x_2\) = awrt 0.3333, \(x_3\) = awrt 0.3333 (degrees) — this mark is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.316\) (radians) OR \(0.333\) (degrees) | B1 | 0.316 stated to 3dp (independent of part (c)) for radians or 0.333 for degrees. Whole answer must maintain consistent units. Degree answers have maximum M1A1B1M1A1A0B1 i.e. 6/7 |
# Question 1:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Radians: $f(0.2) = -0.4$, $f(0.4) = 0.3$ or considers smaller subset of $[0.2, 0.4]$ **OR** Degrees: $f(0.2) = -0.4$, $f(0.4) = 0.2$ or considers smaller subset of $[0.2, 0.4]$ | M1 | Gives two answers with at least one correct to 1sf. May work in degrees or radians. Maximum 6/7 for degrees. If larger interval chosen, this is M0 |
| Change of sign implies root | A1 | Both values correct to at least one decimal place, **and** reason given (e.g. change of sign, or $f(0.2)<0$, $f(0.4)>0$, or product $f(0.2)f(0.4)<0$) **and** conclusion |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sec x + 3x - 2 = 0 \Rightarrow 3x = 2 - \sec x$ and so $x = \dfrac{2}{3} - \dfrac{1}{3\cos x}$ * | B1 | Starts with equation equal to zero, rearranges correctly with **no errors** and at least one intermediate step |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Radians: $x_1 = 0.3177$, $x_2 = 0.3158$, $x_3 = 0.3160$ **OR** Degrees: $x_1 = 0.3333$, $x_2 = 0.3333$, $x_3 = 0.3333$ | M1, A1, A1 | M1: Substitutes $x_0 = 0.3$ into $x = \dfrac{2}{3} - \dfrac{1}{3\cos x} \Rightarrow x_1 =$ (implied by $x_1 = \dfrac{2}{3} - \dfrac{1}{3\cos 0.3}$, or answers rounding to 0.32 rads or 0.33 degrees). A1: $x_1$ awrt 0.3177 4dp (rads) or awrt 0.3333 4dp (degrees). A1: $x_2$ = awrt 0.3158, $x_3$ = awrt 0.3160 (rads). NB $x_2$ = awrt 0.3333, $x_3$ = awrt 0.3333 (degrees) — **this mark is A0** |
## Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.316$ (radians) **OR** $0.333$ (degrees) | B1 | 0.316 stated to 3dp (independent of part (c)) for radians or 0.333 for degrees. Whole answer must maintain consistent units. Degree answers have maximum M1A1B1M1A1A0B1 i.e. 6/7 |1.
$$f ( x ) = \sec x + 3 x - 2 , \quad - \frac { \pi } { 2 } < x < \frac { \pi } { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item Show that there is a root of $\mathrm { f } ( x ) = 0$ in the interval $[ 0.2,0.4 ]$
\item Show that the equation $\mathrm { f } ( x ) = 0$ can be written in the form
$$x = \frac { 2 } { 3 } - \frac { 1 } { 3 \cos x }$$
The solution of $\mathrm { f } ( x ) = 0$ is $\alpha$, where $\alpha = 0.3$ to 1 decimal place.
\item Starting with $x _ { 0 } = 0.3$, use the iterative formula
$$x _ { n + 1 } = \frac { 2 } { 3 } - \frac { 1 } { 3 \cos x _ { n } }$$
to calculate the values of $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$, giving your answers to 4 decimal places.
\item State the value of $\alpha$ correct to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2014 Q1 [7]}}