| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2009 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Classify nature of stationary points |
| Difficulty | Moderate -0.8 This is a straightforward C1 differentiation question requiring routine application of power rule, evaluation at given points, and second derivative test. All steps are standard textbook exercises with no problem-solving insight needed, making it easier than average but not trivial due to the multiple parts and polynomial degree. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = 5x^4 + 40x\) | M1 | One of these powers correct |
| A1 | One of these terms correct | |
| A1 | All correct (no \(+ c\) etc) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = -2\): \(\frac{dy}{dx} = 5\times(-2)^4 + (40\times -2)\) | M1 | Substitute \(x = -2\) into their \(\frac{dy}{dx}\) |
| \(\frac{dy}{dx} = 5\times16 + (40\times-2) = 0\) | ||
| \(\Rightarrow P\) is stationary point | A1 | CSO Shown \(= 0\) plus statement, e.g. "st pt", "as required", "grad \(= 0\)" etc |
| Or their \(\frac{dy}{dx} = 0 \Rightarrow x^n = k\) | (M1) | |
| \(x^3 = -8 \Rightarrow x = -2\) | (A1) | CSO \(x = 0\) need not be considered |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{d^2y}{dx^2} = 20x^3 + 40\) | B1\(\checkmark\) | Correct, ft their \(\frac{dy}{dx}\) |
| \(= 20\times(-2)^3 + 40\) | M1 | Subst \(x = -2\) into their second derivative |
| \((= -160 + 40) = -120\) | A1 | CSO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Maximum (value) | E1\(\checkmark\) | Accept minimum if their \(c(i)\) answer \(> 0\) and correctly interpreted; Parts (i) and (ii) may be combined but \(-120\) must be seen to award A1 in part (c)(i) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| When \(x = 1\), \(y = 13\) | B1 | |
| When \(x = 1\), \(\frac{dy}{dx} = 5 + 40\) | M1 | Sub \(x = 1\) into their \(\frac{dy}{dx}\) |
| \(y = (\text{their } 45)x + k\) OE | m1 | ft their \(\frac{dy}{dx}\) |
| Tangent has equation \(y - 13 = 45(x-1)\) | A1 | CSO OE \(y = 45x + c\), \(c = -32\) |
## Question 3:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 5x^4 + 40x$ | M1 | One of these powers correct |
| | A1 | One of these terms correct |
| | A1 | All correct (no $+ c$ etc) |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = -2$: $\frac{dy}{dx} = 5\times(-2)^4 + (40\times -2)$ | M1 | Substitute $x = -2$ into their $\frac{dy}{dx}$ |
| $\frac{dy}{dx} = 5\times16 + (40\times-2) = 0$ | | |
| $\Rightarrow P$ is stationary point | A1 | CSO Shown $= 0$ plus statement, e.g. "st pt", "as required", "grad $= 0$" etc |
| **Or** their $\frac{dy}{dx} = 0 \Rightarrow x^n = k$ | (M1) | |
| $x^3 = -8 \Rightarrow x = -2$ | (A1) | CSO $x = 0$ need not be considered |
### Part (c)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d^2y}{dx^2} = 20x^3 + 40$ | B1$\checkmark$ | Correct, ft their $\frac{dy}{dx}$ |
| $= 20\times(-2)^3 + 40$ | M1 | Subst $x = -2$ into their second derivative |
| $(= -160 + 40) = -120$ | A1 | CSO |
### Part (c)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Maximum (value) | E1$\checkmark$ | Accept minimum if their $c(i)$ answer $> 0$ and correctly interpreted; Parts (i) and (ii) may be combined but $-120$ must be seen to award A1 in part (c)(i) |
### Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| When $x = 1$, $y = 13$ | B1 | |
| When $x = 1$, $\frac{dy}{dx} = 5 + 40$ | M1 | Sub $x = 1$ into their $\frac{dy}{dx}$ |
| $y = (\text{their } 45)x + k$ OE | m1 | ft their $\frac{dy}{dx}$ |
| Tangent has equation $y - 13 = 45(x-1)$ | A1 | CSO OE $y = 45x + c$, $c = -32$ |
---3 The curve with equation $y = x ^ { 5 } + 20 x ^ { 2 } - 8$ passes through the point $P$, where $x = - 2$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Verify that the point $P$ is a stationary point of the curve.
\item \begin{enumerate}[label=(\roman*)]
\item Find the value of $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ at the point $P$.
\item Hence, or otherwise, determine whether $P$ is a maximum point or a minimum point.
\end{enumerate}\item Find an equation of the tangent to the curve at the point where $x = 1$.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2009 Q3 [13]}}