| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Line-curve intersection conditions |
| Difficulty | Standard +0.3 This is a standard discriminant problem requiring students to set the line equal to the curve, rearrange to a quadratic, and apply the condition that discriminant < 0 for no intersection. Part (b) involves solving a quadratic inequality. While it requires multiple steps, this is a routine C1 technique with no novel insight needed, making it slightly easier than average. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02g Inequalities: linear and quadratic in single variable |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2px^2 - 6px + 4p = \text{"} 3x - 7\text{"}\) or \(y = 2p\left(\frac{y+7}{3}\right)^2 - 6p\left(\frac{y+7}{3}\right) + 4p\) | M1 | Either: Compares the given quadratic expression with the given linear expression using \(<, >, =, \neq\) (may be implied). Or Rearranges \(y = 3x - 7\) to make \(x\) the subject and substitutes into the given quadratic |
| e.g. \(2px^2 - 6px + 4p - 3x + 7(= 0)\) or \(y = 2px^2 - 6px + 4p - 3x + 7\) | dM1 | Moves all terms to one side allowing sign errors only. Ignore \(> 0, < 0, = 0\) etc. Dependent on the first method mark. Terms do not need to be collected |
| \(b^2 - 4ac = (-6p-3)^2 - 4(2p)(4p+7)\) or \(b^2 - 4ac = (10p-9)^2 - 4(2p)(8p)\) | ddM1 | Attempts to use \(b^2 - 4ac\) with their \(a\), \(b\) and \(c\) where \(a = \pm 2p\), \(b = \pm(-6p \pm 3)\) and \(c = \pm(4p \pm 7)\) or for the quadratic in \(y\): \(a = \pm 2p\), \(b = \pm(10p \pm 9)\) and \(c = \pm 8p\). Dependent on both method marks. |
| \(4p^2 - 20p + 9 < 0\) | A1* | Obtains printed answer with no errors seen (Allow \(0 > 4p^2 - 20p + 9\)) but this \(< 0\) must have been seen at some stage before the last line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((2p-9)(2p-1) = 0 \Rightarrow p = \ldots\) to obtain \(p =\) | M1 | Attempt to solve the given quadratic to find 2 values for \(p\) |
| \(p = \frac{9}{2}, \frac{1}{2}\) | A1 | Both correct. May be implied by e.g. \(p < \frac{9}{2}\), \(p < \frac{1}{2}\). Allow equivalent values e.g. \(4.5\), \(\frac{36}{8}\), \(0.5\) etc. If they use the quadratic formula allow \(\frac{20 \pm 16}{8}\) for this mark but not \(\sqrt{256}\) for \(16\) |
| \(\frac{1}{2} < p < 4\frac{1}{2}\) | M1 A1 | M1: Chooses 'inside' region i.e. Lower Limit \(< p <\) Upper Limit or e.g. Lower Limit \(\leq p \leq\) Upper Limit. A1: Allow \(p \in (\frac{1}{2}, 4\frac{1}{2})\) or just \((\frac{1}{2}, 4\frac{1}{2})\) and allow \(p > \frac{1}{2}\) and \(p < 4\frac{1}{2}\) and \(4\frac{1}{2} > p > \frac{1}{2}\) but \(p > \frac{1}{2}\), \(p < 4\frac{1}{2}\) scores M1A0. \(\frac{1}{2} > p > 4\frac{1}{2}\) scores M0A0 |
# Question 8:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2px^2 - 6px + 4p = \text{"} 3x - 7\text{"}$ **or** $y = 2p\left(\frac{y+7}{3}\right)^2 - 6p\left(\frac{y+7}{3}\right) + 4p$ | M1 | Either: Compares the given quadratic expression with the given linear expression using $<, >, =, \neq$ (may be implied). **Or** Rearranges $y = 3x - 7$ to make $x$ the subject and substitutes into the given quadratic |
| e.g. $2px^2 - 6px + 4p - 3x + 7(= 0)$ or $y = 2px^2 - 6px + 4p - 3x + 7$ | dM1 | Moves all terms to one side allowing sign errors only. Ignore $> 0, < 0, = 0$ etc. **Dependent on the first method mark.** Terms do not need to be collected |
| $b^2 - 4ac = (-6p-3)^2 - 4(2p)(4p+7)$ or $b^2 - 4ac = (10p-9)^2 - 4(2p)(8p)$ | ddM1 | Attempts to use $b^2 - 4ac$ with their $a$, $b$ and $c$ where $a = \pm 2p$, $b = \pm(-6p \pm 3)$ and $c = \pm(4p \pm 7)$ or for the quadratic in $y$: $a = \pm 2p$, $b = \pm(10p \pm 9)$ and $c = \pm 8p$. **Dependent on both method marks.** |
| $4p^2 - 20p + 9 < 0$ | A1* | Obtains printed answer with **no errors** seen (Allow $0 > 4p^2 - 20p + 9$) **but** this $< 0$ must have been seen at some stage before the last line |
**[4]**
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(2p-9)(2p-1) = 0 \Rightarrow p = \ldots$ to obtain $p =$ | M1 | Attempt to solve the **given** quadratic to find 2 values for $p$ |
| $p = \frac{9}{2}, \frac{1}{2}$ | A1 | Both correct. May be implied by e.g. $p < \frac{9}{2}$, $p < \frac{1}{2}$. Allow equivalent values e.g. $4.5$, $\frac{36}{8}$, $0.5$ etc. If they use the quadratic formula allow $\frac{20 \pm 16}{8}$ for this mark but not $\sqrt{256}$ for $16$ |
| $\frac{1}{2} < p < 4\frac{1}{2}$ | M1 A1 | M1: Chooses 'inside' region i.e. Lower Limit $< p <$ Upper Limit or e.g. Lower Limit $\leq p \leq$ Upper Limit. A1: Allow $p \in (\frac{1}{2}, 4\frac{1}{2})$ or just $(\frac{1}{2}, 4\frac{1}{2})$ and allow $p > \frac{1}{2}$ **and** $p < 4\frac{1}{2}$ and $4\frac{1}{2} > p > \frac{1}{2}$ but $p > \frac{1}{2}$, $p < 4\frac{1}{2}$ scores M1A0. $\frac{1}{2} > p > 4\frac{1}{2}$ scores M0A0 |
**Allow working in terms of $x$ in (b) but the answer must be in terms of $p$ for the final A mark.**
**[4]**
**[8 marks total]**
---8. The straight line with equation $y = 3 x - 7$ does not cross or touch the curve with equation $y = 2 p x ^ { 2 } - 6 p x + 4 p$, where $p$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $4 p ^ { 2 } - 20 p + 9 < 0$
\item Hence find the set of possible values of $p$.\\
VILM SIHI NITIIIUMI ON OC\\
VILV SIHI NI III HM ION OC\\
VALV SIHI NI JIIIM ION OO
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2016 Q8 [8]}}