| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Applied differentiation |
| Type | Stationary points and nature classification |
| Difficulty | Standard +0.3 This is a multi-part question covering standard C3 techniques: showing a root lies in an interval (substitution), algebraic rearrangement (routine manipulation), iteration (calculator work), and finding stationary points via differentiation. Part (b)(ii) requires understanding transformations but is still mechanical. All parts are textbook-standard with no novel insight required, making it slightly easier than average. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)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 |
| Let \(g(x) = 6\ln x - 8x + x^2 + 3\) | M1 | Correct function formed |
| \(g(5) = 6\ln 5 - 40 + 25 + 3 = 6\ln 5 - 12 \approx -2.35 < 0\) | ||
| \(g(6) = 6\ln 6 - 48 + 36 + 3 = 6\ln 6 - 9 \approx 1.75 > 0\) | A1 | Sign change and conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = 4 + \sqrt{13-6\ln x}\) | M1 | |
| \((x-4)^2 = 13-6\ln x\) | M1 | Squaring both sides |
| \(x^2 - 8x + 16 = 13 - 6\ln x\) | A1 | Rearranging to given form \(6\ln x + x^2 - 8x + 3 = 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x_1 = 5\) | ||
| \(x_2 = 4 + \sqrt{13 - 6\ln 5} \approx 5.258\) | B1 | Correct \(x_2\) |
| \(x_3 = 4 + \sqrt{13 - 6\ln(5.258)} \approx 5.205\) | B1 | Correct \(x_3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f'(x) = \frac{6}{x} + 2x - 8 = 0\) | M1 | Correct differentiation |
| \(2x^2 - 8x + 6 = 0 \Rightarrow x^2 - 4x + 3 = 0\) | M1 | Forming quadratic |
| \((x-1)(x-3) = 0 \Rightarrow x = 1, x = 3\) | A1 | Both x-values |
| \(f(1) = 0 + 1 - 8 + 3 = -4\); point \((1, -4)\) | A1 | |
| \(f(3) = 6\ln 3 + 9 - 24 + 3 = 6\ln 3 - 12\); point \((3, 6\ln 3 - 12)\) | A1 | Both exact coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = 2f(x-4)\): replace \(x\) with \(x-4\), multiply \(y\) by 2 | M1 | Correct transformation logic |
| Stationary points at \((5, -8)\) and \((7, 12\ln 3 - 24)\) | A1 | Both correct |
# Question 3:
## Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $g(x) = 6\ln x - 8x + x^2 + 3$ | M1 | Correct function formed |
| $g(5) = 6\ln 5 - 40 + 25 + 3 = 6\ln 5 - 12 \approx -2.35 < 0$ | | |
| $g(6) = 6\ln 6 - 48 + 36 + 3 = 6\ln 6 - 9 \approx 1.75 > 0$ | A1 | Sign change and conclusion |
## Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = 4 + \sqrt{13-6\ln x}$ | M1 | |
| $(x-4)^2 = 13-6\ln x$ | M1 | Squaring both sides |
| $x^2 - 8x + 16 = 13 - 6\ln x$ | A1 | Rearranging to given form $6\ln x + x^2 - 8x + 3 = 0$ |
## Part (a)(iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x_1 = 5$ | | |
| $x_2 = 4 + \sqrt{13 - 6\ln 5} \approx 5.258$ | B1 | Correct $x_2$ |
| $x_3 = 4 + \sqrt{13 - 6\ln(5.258)} \approx 5.205$ | B1 | Correct $x_3$ |
## Part (b)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f'(x) = \frac{6}{x} + 2x - 8 = 0$ | M1 | Correct differentiation |
| $2x^2 - 8x + 6 = 0 \Rightarrow x^2 - 4x + 3 = 0$ | M1 | Forming quadratic |
| $(x-1)(x-3) = 0 \Rightarrow x = 1, x = 3$ | A1 | Both x-values |
| $f(1) = 0 + 1 - 8 + 3 = -4$; point $(1, -4)$ | A1 | |
| $f(3) = 6\ln 3 + 9 - 24 + 3 = 6\ln 3 - 12$; point $(3, 6\ln 3 - 12)$ | A1 | Both exact coordinates |
## Part (b)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = 2f(x-4)$: replace $x$ with $x-4$, multiply $y$ by 2 | M1 | Correct transformation logic |
| Stationary points at $(5, -8)$ and $(7, 12\ln 3 - 24)$ | A1 | Both correct |
I can see these are answer space pages (blank lined pages for students to write on) from an AQA A-level Mathematics paper (P/Jun15/MPC3), along with Question 4. However, the **mark scheme is not included** in these images — these pages show only:
- Blank answer spaces for Question 3 (pages 7–9)
- The question text for Question 4 (page 10)
- Blank answer space for Question 4 (page 11)
I can provide the **worked solutions** for Question 4 based on the question text visible:
---3
\begin{enumerate}[label=(\alph*)]
\item It is given that the curves with equations $y = 6 \ln x$ and $y = 8 x - x ^ { 2 } - 3$ intersect at a single point where $x = \alpha$.
\begin{enumerate}[label=(\roman*)]
\item Show that $\alpha$ lies between 5 and 6 .
\item Show that the equation $x = 4 + \sqrt { 13 - 6 \ln x }$ can be rearranged into the form
$$6 \ln x + x ^ { 2 } - 8 x + 3 = 0$$
\item Use the iterative formula
$$x _ { n + 1 } = 4 + \sqrt { 13 - 6 \ln x _ { n } }$$
with $x _ { 1 } = 5$ to find the values of $x _ { 2 }$ and $x _ { 3 }$, giving your answers to three decimal places.
\end{enumerate}\item A curve has equation $y = \mathrm { f } ( x )$ where $\mathrm { f } ( x ) = 6 \ln x + x ^ { 2 } - 8 x + 3$.
\begin{enumerate}[label=(\roman*)]
\item Find the exact values of the coordinates of the stationary points of the curve.
\item Hence, or otherwise, find the exact values of the coordinates of the stationary points of the curve with equation
$$y = 2 \mathrm { f } ( x - 4 )$$
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2015 Q3 [14]}}