AQA C1 2008 January — Question 2 11 marks

Exam BoardAQA
ModuleC1 (Core Mathematics 1)
Year2008
SessionJanuary
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStationary points and optimisation
TypeClassify nature of stationary points
DifficultyModerate -0.8 This is a straightforward C1 stationary points question requiring only standard differentiation techniques (power rule), solving a simple equation (x³ = 8), and applying the second derivative test. All parts are routine textbook exercises with no problem-solving insight needed, making it easier than average but not trivial due to the x⁴ term and multiple parts.
Spec1.07d Second derivatives: d^2y/dx^2 notation1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.07p Points of inflection: using second derivative

2 The curve with equation \(y = x ^ { 4 } - 32 x + 5\) has a single stationary point, \(M\).
  1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
  2. Hence find the \(x\)-coordinate of \(M\).
    1. Find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\).
    2. Hence, or otherwise, determine whether \(M\) is a maximum or a minimum point.
  3. Determine whether the curve is increasing or decreasing at the point on the curve where \(x = 0\).

Part (a)
AnswerMarks Guidance
\(\frac{dy}{dx} = 4x^3 - 32\)M1, A1, A1 Reduce one power by 1; One term correct; All correct (no \(+ c\) etc)
3
Part (b)
AnswerMarks Guidance
Stationary point \(\Rightarrow \frac{dy}{dx} = 0\) \(\Rightarrow x^3 = 8\) \(\Rightarrow x = 2\)M1, A1\(\uparrow\), A1 \(x^n = k\) following from their \(\frac{dy}{dx}\); CSO
3
Part (c)(i)
AnswerMarks Guidance
\(\frac{d^2y}{dx^2} = 12x^2\)B1\(\uparrow\) FT their \(\frac{dy}{dx}\)
1
Part (c)(ii)
AnswerMarks Guidance
When \(x = 2\), \(\frac{d^2y}{dx^2}\) considered \(\Rightarrow\) minimum pointM1, E1\(\uparrow\) Or complete test with \(2 \pm \epsilon\) using \(\frac{dy}{dx}\)
2
Part (d)
AnswerMarks Guidance
Putting \(x = 0\) into their \(\frac{dy}{dx}\) (\(= -32\)) \(\frac{dy}{dx} < 0 \Rightarrow\) decreasingM1, A1\(\uparrow\) Allow "increasing" if their \(\frac{dy}{dx} > 0\)
2
11
## Part (a)
$\frac{dy}{dx} = 4x^3 - 32$ | M1, A1, A1 | Reduce one power by 1; One term correct; All correct (no $+ c$ etc)

| | 3 |

## Part (b)
Stationary point $\Rightarrow \frac{dy}{dx} = 0$ $\Rightarrow x^3 = 8$ $\Rightarrow x = 2$ | M1, A1$\uparrow$, A1 | $x^n = k$ following from their $\frac{dy}{dx}$; CSO

| | 3 |

## Part (c)(i)
$\frac{d^2y}{dx^2} = 12x^2$ | B1$\uparrow$ | FT their $\frac{dy}{dx}$

| | 1 |

## Part (c)(ii)
When $x = 2$, $\frac{d^2y}{dx^2}$ considered $\Rightarrow$ minimum point | M1, E1$\uparrow$ | Or complete test with $2 \pm \epsilon$ using $\frac{dy}{dx}$

| | 2 |

## Part (d)
Putting $x = 0$ into their $\frac{dy}{dx}$ ($= -32$) $\frac{dy}{dx} < 0 \Rightarrow$ decreasing | M1, A1$\uparrow$ | Allow "increasing" if their $\frac{dy}{dx} > 0$

| | 2 |

| | 11 |

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2 The curve with equation $y = x ^ { 4 } - 32 x + 5$ has a single stationary point, $M$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Hence find the $x$-coordinate of $M$.
\item \begin{enumerate}[label=(\roman*)]
\item Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.
\item Hence, or otherwise, determine whether $M$ is a maximum or a minimum point.
\end{enumerate}\item Determine whether the curve is increasing or decreasing at the point on the curve where $x = 0$.
\end{enumerate}

\hfill \mbox{\textit{AQA C1 2008 Q2 [11]}}