| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2008 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Polynomial with line intersection |
| Difficulty | Moderate -0.8 This is a straightforward C1 question requiring basic sketching of a factored quadratic and line, then solving a simple quadratic equation. All steps are routine: expanding brackets, equating expressions, and factoring x²-x-2=(x-2)(x+1). No problem-solving insight needed, just methodical application of standard techniques. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.02q Use intersection points: of graphs to solve equations1.03a Straight lines: equation forms y=mx+c, ax+by+c=0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(L\): straight line with positive gradient and negative intercept on \(y\)-axis cutting at \((\frac{1}{4}, 0)\) and \((0, -1)\) | B1 | Line must cross both axes but need not reach the curve |
| \(L\): Condone 0.33 or better for \(\frac{1}{4}\) | B1 | |
| \(C\): attempt at parabola \(\cup\) or \(\cap\) through \((-3,0)\) and \((1,0)\) values \(-3\) and \(1\) stated as intercepts on \(x\)-axis | B1 | |
| \(\cup\) shaped graph – vertex below \(x\)-axis and cutting \(x\)-axis twice through \((0,-3)\) and minimum point to left of \(y\)-axis | M1 | |
| A1 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| \((x + 3)(x - 1) = 3x - 1\) \(x^2 + 3x - x - 3 - 3x + 1 = 0\) \(\Rightarrow x^2 - x - 2 = 0\) | M1 | |
| A1 | 2 | AG; must have "\(= 0\)" and no errors |
| Answer | Marks | Guidance |
|---|---|---|
| \((x-2)(x+1) = 0\) \(\Rightarrow x = 2, -1\) | M1, A1 | \((x \pm 1)(x \pm 2)\) or use of formula (one slip) correct values imply M1A1 |
| Substitute one value of \(x\) to find \(y\) | m1 | |
| Points of intersection \((2, 5)\) and \((-1, -4)\) | A1 | 4 |
**1(a)**
| $L$: straight line with positive gradient and negative intercept on $y$-axis cutting at $(\frac{1}{4}, 0)$ and $(0, -1)$ | B1 | Line must cross both axes but need not reach the curve |
| $L$: Condone 0.33 or better for $\frac{1}{4}$ | B1 | |
| $C$: attempt at parabola $\cup$ or $\cap$ through $(-3,0)$ and $(1,0)$ values $-3$ and $1$ stated as intercepts on $x$-axis | B1 | |
| $\cup$ shaped graph – vertex below $x$-axis and cutting $x$-axis twice through $(0,-3)$ and minimum point to left of $y$-axis | M1 | |
| | A1 | 5 |
**1(b)**
| $(x + 3)(x - 1) = 3x - 1$ $x^2 + 3x - x - 3 - 3x + 1 = 0$ $\Rightarrow x^2 - x - 2 = 0$ | M1 | |
| | A1 | 2 | AG; must have "$= 0$" and no errors |
**1(c)**
| $(x-2)(x+1) = 0$ $\Rightarrow x = 2, -1$ | M1, A1 | $(x \pm 1)(x \pm 2)$ or use of formula (one slip) correct values imply M1A1 |
| Substitute one value of $x$ to find $y$ | m1 | |
| Points of intersection $(2, 5)$ and $(-1, -4)$ | A1 | 4 | May say $x = 2, y = 5$ etc SC: $(2, 5) \Rightarrow$ B2 $(-1,-4) \Rightarrow$ B2 without working |
---1 The straight line $L$ has equation $y = 3 x - 1$ and the curve $C$ has equation
$$y = ( x + 3 ) ( x - 1 )$$
\begin{enumerate}[label=(\alph*)]
\item Sketch on the same axes the line $L$ and the curve $C$, showing the values of the intercepts on the $x$-axis and the $y$-axis.
\item Show that the $x$-coordinates of the points of intersection of $L$ and $C$ satisfy the equation $x ^ { 2 } - x - 2 = 0$.
\item Hence find the coordinates of the points of intersection of $L$ and $C$.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2008 Q1 [11]}}