| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Show discriminant inequality, then solve |
| Difficulty | Moderate -0.3 This is a standard C1 question on quadratic intersections requiring equating expressions, rearranging to standard form, and using the discriminant condition for two distinct roots. All steps are routine applications of well-practiced techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions |
| Answer | Marks | Guidance |
|---|---|---|
| \(2x^2 - x - 1 = 2kx - 3k\) | equated and multiplied out and all 5 terms on one side and \(= 0\) | |
| \(2x^2 - x - 1 - 2kx + 3k = 0 \Rightarrow 2x^2 - (2k + 1)x + 3k - 1 = 0\) | B1 | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| \((2k + 1)^2 - 4 \times 2(3k - 1)\) | M1 | clear attempt at \(b^2 - 4ac\) |
| \((2k + 1)^2 - 4 \times 2(3k - 1) > 0\) | B1 | discriminant \(> 0\) which must appear before the printed answer |
| \(4k^2 + 4k + 1 - 24k + 8 > 0 \Rightarrow 4k^2 - 20k + 9 > 0\) | A1 cso | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(4k^2 - 20k + 9 = (2k - 9)(2k - 1)\) | M1 | correct factors or correct use of formula as far as \(\frac{20 \pm \sqrt{256}}{8}\); condone \(\frac{4}{8}, \frac{36}{8}\) etc here but must combine sums of fractions |
| critical values are \(\frac{1}{2}\) and \(\frac{9}{2}\) | A1 | |
| M1 | sketch or sign diagram including values; \(\quad + \quad - \quad +\) \(\quad\quad\quad 0.5 \quad 4.5\) | |
| \(k < \frac{1}{2}, k > \frac{9}{2}\) Take their final line as their answer | A1 | 4 marks |
**8(a)**
$2x^2 - x - 1 = 2kx - 3k$ | | equated and multiplied out and all 5 terms on one side and $= 0$
$2x^2 - x - 1 - 2kx + 3k = 0 \Rightarrow 2x^2 - (2k + 1)x + 3k - 1 = 0$ | B1 | 1 mark | **AG** (correct with no trailing = signs etc)
**8(b)(i)**
$(2k + 1)^2 - 4 \times 2(3k - 1)$ | M1 | clear attempt at $b^2 - 4ac$
$(2k + 1)^2 - 4 \times 2(3k - 1) > 0$ | B1 | discriminant $> 0$ which must appear before the printed answer
$4k^2 + 4k + 1 - 24k + 8 > 0 \Rightarrow 4k^2 - 20k + 9 > 0$ | A1 cso | 3 marks | **AG** (all working correct with no missing brackets etc)
**8(b)(ii)**
$4k^2 - 20k + 9 = (2k - 9)(2k - 1)$ | M1 | correct factors or correct use of formula as far as $\frac{20 \pm \sqrt{256}}{8}$; condone $\frac{4}{8}, \frac{36}{8}$ etc here but must combine sums of fractions
critical values are $\frac{1}{2}$ and $\frac{9}{2}$ | A1 |
| M1 | sketch or sign diagram including values; $\quad + \quad - \quad +$ $\quad\quad\quad 0.5 \quad 4.5$
$k < \frac{1}{2}, k > \frac{9}{2}$ **Take their final line as their answer** | A1 | 4 marks | fractions must be simplified; condone use of OR but not AND
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**TOTAL: 75 marks**8 A curve has equation $y = 2 x ^ { 2 } - x - 1$ and a line has equation $y = k ( 2 x - 3 )$, where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that the $x$-coordinate of any point of intersection of the curve and the line satisfies the equation
$$2 x ^ { 2 } - ( 2 k + 1 ) x + 3 k - 1 = 0$$
\item The curve and the line intersect at two distinct points.
\begin{enumerate}[label=(\roman*)]
\item Show that $4 k ^ { 2 } - 20 k + 9 > 0$.
\item Find the possible values of $k$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2013 Q8 [8]}}