| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Increasing/decreasing intervals |
| Difficulty | Moderate -0.8 This is a straightforward C1 question testing routine differentiation and application of second derivative test. Part (a)(i) requires simple differentiation of a polynomial, (a)(ii) applies standard minimum point verification, and part (b) involves basic manipulation and solving a quadratic inequality—all standard textbook exercises with no problem-solving insight required. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.07d Second derivatives: d^2y/dx^2 notation1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\dfrac{d^2y}{dx^2} = 27 - 12x\) | M1 A1 | One term correct; all correct (no \(+c\) etc) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\dfrac{dy}{dx} = 54 + 27\times\left(-\dfrac{3}{2}\right) - 6\times\left(-\dfrac{3}{2}\right)^2\) | M1 | |
| \(\dfrac{dy}{dx} = 54 - \dfrac{81}{2} - \dfrac{54}{4} = 0\) | A1 | Convincingly showing \(\dfrac{dy}{dx}=0\) and \(\dfrac{dy}{dx}=\ldots\) must appear on at least one line |
| \(\dfrac{d^2y}{dx^2} = 27 - 12\times\left(-\dfrac{3}{2}\right)\) | M1 | Correct substitution into "their" \(\dfrac{d^2y}{dx^2}\) |
| \(\dfrac{d^2y}{dx^2} = 27+18\ (=45) > 0\), \(\Rightarrow P\) is minimum point | A1cso | Correct working and \(\dfrac{d^2y}{dx^2}\) used and value shown to be \(>0\) with correct statement(s); must earn 3 previous marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| (Decreasing so) \(54+27x-6x^2 < 0\); \(6x^2-27x-54>0\); \(2x^2-9x-18>0\) | M1 A1 | AG be convinced |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((2x+3)(x-6)\) | M1 | Correct factors or correct use of formula as far as \(\dfrac{9\pm\sqrt{225}}{4}\) |
| CVs are \(x=-\dfrac{3}{2}\), \(x=6\) | A1 | Condone equivalent fractions |
| Sign diagram: \(+\) then \(-\) then \(+\) about \(-\dfrac{3}{2}\) and \(6\) | M1 | Use of sign diagram or graph |
| \(x < -\dfrac{3}{2},\quad x>6\) | A1 | Fractions must be simplified for final mark; no ISW |
# Question 8:
## Part (a)(i)
| Working | Mark | Guidance |
|---------|------|----------|
| $\dfrac{d^2y}{dx^2} = 27 - 12x$ | M1 A1 | One term correct; all correct (no $+c$ etc) |
## Part (a)(ii)
| Working | Mark | Guidance |
|---------|------|----------|
| $\dfrac{dy}{dx} = 54 + 27\times\left(-\dfrac{3}{2}\right) - 6\times\left(-\dfrac{3}{2}\right)^2$ | M1 | |
| $\dfrac{dy}{dx} = 54 - \dfrac{81}{2} - \dfrac{54}{4} = 0$ | A1 | Convincingly showing $\dfrac{dy}{dx}=0$ **and** $\dfrac{dy}{dx}=\ldots$ must appear on at least one line |
| $\dfrac{d^2y}{dx^2} = 27 - 12\times\left(-\dfrac{3}{2}\right)$ | M1 | Correct substitution into "their" $\dfrac{d^2y}{dx^2}$ |
| $\dfrac{d^2y}{dx^2} = 27+18\ (=45) > 0$, $\Rightarrow P$ is minimum point | A1cso | Correct working and $\dfrac{d^2y}{dx^2}$ used and value shown to be $>0$ with correct statement(s); must earn 3 previous marks |
## Part (b)(i)
| Working | Mark | Guidance |
|---------|------|----------|
| (Decreasing so) $54+27x-6x^2 < 0$; $6x^2-27x-54>0$; $2x^2-9x-18>0$ | M1 A1 | AG be convinced |
## Part (b)(ii)
| Working | Mark | Guidance |
|---------|------|----------|
| $(2x+3)(x-6)$ | M1 | Correct factors or correct use of formula as far as $\dfrac{9\pm\sqrt{225}}{4}$ |
| CVs are $x=-\dfrac{3}{2}$, $x=6$ | A1 | Condone equivalent fractions |
| Sign diagram: $+$ then $-$ then $+$ about $-\dfrac{3}{2}$ and $6$ | M1 | Use of sign diagram or graph |
| $x < -\dfrac{3}{2},\quad x>6$ | A1 | Fractions must be simplified for final mark; no ISW |8 The gradient, $\frac { \mathrm { d } y } { \mathrm {~d} x }$, at the point $( x , y )$ on a curve is given by
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = 54 + 27 x - 6 x ^ { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.
\item The curve passes through the point $P \left( - 1 \frac { 1 } { 2 } , 4 \right)$.
Verify that the curve has a minimum point at $P$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Show that at the points on the curve where $y$ is decreasing
$$2 x ^ { 2 } - 9 x - 18 > 0$$
\item Solve the inequality $2 x ^ { 2 } - 9 x - 18 > 0$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2016 Q8 [12]}}