| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Solve exponential equation via iteration |
| Difficulty | Standard +0.3 This is a straightforward multi-part question covering standard C3 techniques: change of sign for roots, algebraic rearrangement using logarithms, applying a given iterative formula (just substitution), and Simpson's rule. All parts are routine applications with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(2) = -1\), \(f(3) = 22\); change of sign | M1 | Both values correct |
| \(\Rightarrow 2 < \alpha < 3\) | A1 | Must have both statement and interval in words or symbols; OR comparing 2 sides: at 2, \(2^2 < 5\); at 3, \(3^3 > 5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^x = 5 \Rightarrow \ln x^x = \ln 5\) | ||
| \(x \ln x = \ln 5\) | M1 | Taking logs and using rule of logs |
| \(\ln x = \frac{\ln 5}{x}\) | A1 | Must see this line |
| \(x = e^{\frac{\ln 5}{x}}\) | A1 | AG, all correct including middle line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x_2 = 2.236\) | B1 | |
| \(x_3 = 2.054\) | B1 | Ignore any further values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x\) values: \(0.5, 0.7, 0.9, 1.1, 1.3, 1.5, 1.7\) with correct \(y\) values | B1 | All 7 correct \(x\) values (and no extras used); PI by correct \(y\) values |
| At least 5 correct \(y\) values to 3dp or better | B1 | In exact form or decimal values, rounded or truncated (PI by correct answer) |
| \(\frac{1}{3} \times 0.2[4.2929 + 2.5353 + 4(4.2209 + 3.8895 + 3.1629) + 2(4.0905 + 3.5935)]\) | M1 | Correct use of Simpson's rule using \(\frac{1}{3}\) and \(0.2\) oe and their 7 \(y\) values (of which 5 are correct to 2dp), either listed or totalled |
| \(= 4.49\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(6 - \text{their (di)} = 1.51\) | M1 | PI by correct answer |
| \(= 1.51\) | A1F | SC1 for \(-1.51\) |
# Question 2:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(2) = -1$, $f(3) = 22$; change of sign | M1 | Both values correct |
| $\Rightarrow 2 < \alpha < 3$ | A1 | Must have both statement and interval in words or symbols; OR comparing 2 sides: at 2, $2^2 < 5$; at 3, $3^3 > 5$ |
**Note:** Condone $2 \leq x \leq 3$; allow '$x$', 'root' for $\alpha$, but not 'it'
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^x = 5 \Rightarrow \ln x^x = \ln 5$ | | |
| $x \ln x = \ln 5$ | M1 | Taking logs **and** using rule of logs |
| $\ln x = \frac{\ln 5}{x}$ | A1 | Must see this line |
| $x = e^{\frac{\ln 5}{x}}$ | A1 | AG, all correct including middle line |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_2 = 2.236$ | B1 | |
| $x_3 = 2.054$ | B1 | Ignore any further values |
## Part (di)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x$ values: $0.5, 0.7, 0.9, 1.1, 1.3, 1.5, 1.7$ with correct $y$ values | B1 | All 7 correct $x$ values (and no extras used); PI by correct $y$ values |
| At least 5 correct $y$ values to 3dp or better | B1 | In exact form or decimal values, rounded or truncated (PI by correct answer) |
| $\frac{1}{3} \times 0.2[4.2929 + 2.5353 + 4(4.2209 + 3.8895 + 3.1629) + 2(4.0905 + 3.5935)]$ | M1 | Correct use of Simpson's rule using $\frac{1}{3}$ and $0.2$ oe and their 7 $y$ values (of which 5 are correct to 2dp), either listed or totalled |
| $= 4.49$ | A1 | CAO |
**Note:** 4.49 with no working scores 0/4
## Part (dii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $6 - \text{their (di)} = 1.51$ | M1 | PI by correct answer |
| $= 1.51$ | A1F | **SC1** for $-1.51$ |
**Note:** 1.51 scores 2/2; $1.51\ldots$ with NMS scores 0/2
---2 The curve with equation $y = x ^ { x }$, where $x > 0$, intersects the line $y = 5$ at a single point, where $x = \alpha$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\alpha$ lies between 2 and 3 .
\item Show that the equation $x ^ { x } = 5$ can be rearranged into the form
$$x = \mathrm { e } ^ { \left( \frac { \ln 5 } { x } \right) }$$
\item Use the iterative formula
$$x _ { n + 1 } = \mathrm { e } ^ { \left( \frac { \ln 5 } { x _ { n } } \right) }$$
with $x _ { 1 } = 2$ to find the values of $x _ { 2 }$ and $x _ { 3 }$, giving your answers to three decimal places.
\item \begin{enumerate}[label=(\roman*)]
\item Use Simpson's rule with 7 ordinates ( 6 strips) to find an approximation to
$$\int _ { 0.5 } ^ { 1.7 } \left( 5 - x ^ { x } \right) \mathrm { d } x$$
giving your answer to three significant figures.
\item Hence find an approximation to $\int _ { 0.5 } ^ { 1.7 } x ^ { x } \mathrm {~d} x$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2016 Q2 [15]}}