| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Find range where function increasing/decreasing |
| Difficulty | Moderate -0.8 This is a straightforward C1 question testing standard differentiation applications: rearranging an inequality, solving a quadratic inequality by factorization, verifying a horizontal tangent, and finding an intersection point. All steps are routine procedures with no problem-solving insight required, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.03b Straight lines: parallel and perpendicular relationships1.07b Gradient as rate of change: dy/dx notation1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left(\text{Increasing} \Rightarrow\right) \frac{dy}{dx} > 0\); either \(20x - 6x^2 - 16 > 0\) | M1 | Correct interpretation of \(y\) increasing |
| \(\Rightarrow 6x^2 - 20x + 16 < 0\) *or* \((2)(10x - 3x^2 - 8) > 0 \Rightarrow 3x^2 - 10x + 8 < 0\) | A1 (2 marks) | CSO AG no errors in working; must see at least one of these steps before final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((3x-4)(x-2)\) | M1 | Correct factors or correct use of quadratic formula as far as \(\frac{10 \pm \sqrt{4}}{6}\) |
| CVs are \(\frac{4}{3}\) and \(2\) | A1 | Condone \(\frac{8}{6}\) and \(\frac{12}{6}\) here but not in final line |
| Sign diagram / sketch showing \(+\ -\ +\) with \(\frac{4}{3}\) and \(2\) | M1 | Sketch or sign diagram |
| \(\frac{4}{3} < x < 2\) | A1 (4 marks) | or \(2 > x > \frac{4}{3}\); accept \(x < 2\) AND \(x > \frac{4}{3}\); but not \(x < 2\) OR \(x > \frac{4}{3}\); nor \(x < 2,\ x > \frac{4}{3}\); Mark their final line as their answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = 2\); \(\left(\frac{dy}{dx} =\right) 40 - 24 - 16\) | M1 | Sub \(x = 2\) into \(\frac{dy}{dx}\) and simplify terms |
| \(\frac{dy}{dx} = 0 \Rightarrow\) tangent at \(P\) is parallel to the \(x\)-axis | A1 (2 marks) | Must be all correct working plus statement |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = 3\); \(\frac{dy}{dx} = 20 \times 3 - 6 \times 3^2 - 16\) | M1 | Must attempt to sub \(x = 3\) into \(\frac{dy}{dx}\) |
| \((= 60 - 54 - 16) = -10\) | A1 | |
| Gradient of normal \(= \frac{1}{10}\) | A1\(\checkmark\) | \(\frac{-1}{\text{"their } -10\text{"}}\) |
| Normal: \((y - -1) = \text{'their grad'}(x - 3)\) | m1 | Normal attempted with correct coordinates used and gradient obtained from their \(\frac{dy}{dx}\) value |
| \(y + 1 = \frac{1}{10}(x-3)\) | A1 | Any correct form, e.g. \(10y = x - 13\) but must simplify \(--\) to \(+\) |
| (Equation of tangent at \(P\) is) \(y = 3\) | B1 | |
| \(x = 43\) | A1 (7 marks) | CSO; \(\Rightarrow R(43, 3)\) |
## Question 7:
**Part 7(a)(i):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left(\text{Increasing} \Rightarrow\right) \frac{dy}{dx} > 0$; either $20x - 6x^2 - 16 > 0$ | M1 | Correct interpretation of $y$ increasing |
| $\Rightarrow 6x^2 - 20x + 16 < 0$ *or* $(2)(10x - 3x^2 - 8) > 0 \Rightarrow 3x^2 - 10x + 8 < 0$ | A1 (2 marks) | CSO **AG** no errors in working; must see at least one of these steps before final answer |
**Part 7(a)(ii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $(3x-4)(x-2)$ | M1 | Correct factors or correct use of quadratic formula as far as $\frac{10 \pm \sqrt{4}}{6}$ |
| CVs are $\frac{4}{3}$ and $2$ | A1 | Condone $\frac{8}{6}$ and $\frac{12}{6}$ here but not in final line |
| Sign diagram / sketch showing $+\ -\ +$ with $\frac{4}{3}$ and $2$ | M1 | Sketch or sign diagram |
| $\frac{4}{3} < x < 2$ | A1 (4 marks) | or $2 > x > \frac{4}{3}$; accept $x < 2$ **AND** $x > \frac{4}{3}$; but **not** $x < 2$ **OR** $x > \frac{4}{3}$; **nor** $x < 2,\ x > \frac{4}{3}$; Mark their final line as their answer |
**Part 7(b)(i):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = 2$; $\left(\frac{dy}{dx} =\right) 40 - 24 - 16$ | M1 | Sub $x = 2$ into $\frac{dy}{dx}$ and simplify terms |
| $\frac{dy}{dx} = 0 \Rightarrow$ tangent at $P$ is parallel to the $x$-axis | A1 (2 marks) | Must be all correct working plus statement |
**Part 7(b)(ii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = 3$; $\frac{dy}{dx} = 20 \times 3 - 6 \times 3^2 - 16$ | M1 | Must attempt to sub $x = 3$ into $\frac{dy}{dx}$ |
| $(= 60 - 54 - 16) = -10$ | A1 | |
| Gradient of normal $= \frac{1}{10}$ | A1$\checkmark$ | $\frac{-1}{\text{"their } -10\text{"}}$ |
| Normal: $(y - -1) = \text{'their grad'}(x - 3)$ | m1 | Normal attempted with correct coordinates used and gradient obtained from their $\frac{dy}{dx}$ value |
| $y + 1 = \frac{1}{10}(x-3)$ | A1 | Any correct form, e.g. $10y = x - 13$ but must simplify $--$ to $+$ |
| (Equation of tangent at $P$ is) $y = 3$ | B1 | |
| $x = 43$ | A1 (7 marks) | CSO; $\Rightarrow R(43, 3)$ |7 The gradient, $\frac { \mathrm { d } y } { \mathrm {~d} x }$, of a curve $C$ at the point $( x , y )$ is given by
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = 20 x - 6 x ^ { 2 } - 16$$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that $y$ is increasing when $3 x ^ { 2 } - 10 x + 8 < 0$.
\item Solve the inequality $3 x ^ { 2 } - 10 x + 8 < 0$.
\end{enumerate}\item The curve $C$ passes through the point $P ( 2,3 )$.
\begin{enumerate}[label=(\roman*)]
\item Verify that the tangent to the curve at $P$ is parallel to the $x$-axis.
\item The point $Q ( 3 , - 1 )$ also lies on the curve. The normal to the curve at $Q$ and the tangent to the curve at $P$ intersect at the point $R$. Find the coordinates of $R$.\\
(7 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2012 Q7 [15]}}