| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Sketch absolute value of function |
| Difficulty | Standard +0.3 This is a structured multi-part question covering standard C3 techniques: sign change for roots, differentiation for turning points, polynomial division, and curve sketching including absolute values. Each part follows routine procedures with clear guidance, making it slightly easier than average despite the multiple steps. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02m Graphs of functions: difference between plotting and sketching1.02s Modulus graphs: sketch graph of |ax+b|1.07n Stationary points: find maxima, minima using derivatives1.09a Sign change methods: locate roots |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(f(-2) = 16 + 8 - 8(=16) > 0\); \(f(-1) = 1 + 4 - 8(= -3) < 0\); Change of sign (and continuity) \(\Rightarrow\) root in interval \((-2, -1)\); ft their calculation as long as there is a sign change | B1, B1, B1ft | (3 marks) |
| (b) \(\frac{dy}{dx} = 4x^3 - 4 = 0 \Rightarrow x = 1\); Turning point is \((1, -11)\) | M1, A1, A1 | (3 marks) |
| (c) \(a = 2, b = 4, c = 4\) | B1, B1, B1 | (3 marks) |
| (d) Sketch showing cubic curve passing through approximately \((−2, 0)\), with turning point in correct quadrant only, and passing through \(2\) and \(−8\) | B1, B1ft, B1 | ft their turning point in correct quadrant only (3 marks) |
| (e) Sketch showing shape with turning point at \((1,11)\) | B1 | (1 mark) |
(a) $f(-2) = 16 + 8 - 8(=16) > 0$; $f(-1) = 1 + 4 - 8(= -3) < 0$; Change of sign (and continuity) $\Rightarrow$ root in interval $(-2, -1)$; ft their calculation as long as there is a sign change | B1, B1, B1ft | (3 marks)
(b) $\frac{dy}{dx} = 4x^3 - 4 = 0 \Rightarrow x = 1$; Turning point is $(1, -11)$ | M1, A1, A1 | (3 marks)
(c) $a = 2, b = 4, c = 4$ | B1, B1, B1 | (3 marks)
(d) Sketch showing cubic curve passing through approximately $(−2, 0)$, with turning point in correct quadrant only, and passing through $2$ and $−8$ | B1, B1ft, B1 | ft their turning point in correct quadrant only (3 marks)
(e) Sketch showing shape with turning point at $(1,11)$ | B1 | (1 mark)
---7.
$$f ( x ) = x ^ { 4 } - 4 x - 8$$
\begin{enumerate}[label=(\alph*)]
\item Show that there is a root of $\mathrm { f } ( x ) = 0$ in the interval $[ - 2 , - 1 ]$.
\item Find the coordinates of the turning point on the graph of $y = \mathrm { f } ( x )$.
\item Given that $\mathrm { f } ( x ) = ( x - 2 ) \left( x ^ { 3 } + a x ^ { 2 } + b x + c \right)$, find the values of the constants, $a , b$ and $c$.
\item In the space provided on page 21, sketch the graph of $y = \mathrm { f } ( x )$.
\item Hence sketch the graph of $y = | \mathrm { f } ( x ) |$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2007 Q7 [13]}}