| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Quadratic equation real roots |
| Difficulty | Moderate -0.3 This is a straightforward application of the discriminant condition for no real roots. Part (a) requires rearranging to standard form and setting b²-4ac < 0, which is routine algebraic manipulation. Part (b) involves solving a quadratic inequality by factorization, a standard C1 skill. The question is slightly easier than average as it's a direct textbook-style exercise with clear steps and no conceptual surprises. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02g Inequalities: linear and quadratic in single variable |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(b^2 - 4ac = (4k)^2 - 4(-2)(20+13k)\) | M1 | Attempts \(b^2 - 4ac\) with \(a = \pm(20 \pm 13k)\), \(b = \pm 4k\), \(c = \pm 2\) |
| \((4k)^2 - 4(-2)(20+13k)\) | A1 | Correct unsimplified expression |
| \(b^2 - 4ac < 0 \Rightarrow (4k)^2 - 4(-2)(20+13k) < 0\) | M1 | Uses \(b^2 - 4ac < 0\); "\(<0\)" must appear before final answer |
| \(16k^2 + 160 + 104k < 0 \Rightarrow 2k^2 + 13k + 20 < 0^*\) | A1* | Reaches printed answer with no errors, including correct bracketing |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2k^2 + 13k + 20 = 0 \Rightarrow (2k+5)(k+4) = 0 \Rightarrow k = \ldots\) | M1 | Attempt to solve the given quadratic to find 2 values for \(k\) |
| \(k = -\frac{5}{2}, -4\) | A1 | Both correct. Allow \(\frac{-13 \pm 3}{4}\) for this mark but not \(\sqrt{9}\) for 3 |
| \(-4 < k < -\frac{5}{2}\) | M1A1 | M1: chooses 'inside' region for their critical values. A1: Allow \(k \in (-4, -\frac{5}{2})\) or just \((-4, -\frac{5}{2})\). Note: \(-\frac{5}{2} < k < -4\) is M0A0 |
## Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $b^2 - 4ac = (4k)^2 - 4(-2)(20+13k)$ | M1 | Attempts $b^2 - 4ac$ with $a = \pm(20 \pm 13k)$, $b = \pm 4k$, $c = \pm 2$ |
| $(4k)^2 - 4(-2)(20+13k)$ | A1 | Correct unsimplified expression |
| $b^2 - 4ac < 0 \Rightarrow (4k)^2 - 4(-2)(20+13k) < 0$ | M1 | Uses $b^2 - 4ac < 0$; "$<0$" must appear before final answer |
| $16k^2 + 160 + 104k < 0 \Rightarrow 2k^2 + 13k + 20 < 0^*$ | A1* | Reaches printed answer with no errors, including correct bracketing |
## Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2k^2 + 13k + 20 = 0 \Rightarrow (2k+5)(k+4) = 0 \Rightarrow k = \ldots$ | M1 | Attempt to solve the given quadratic to find 2 values for $k$ |
| $k = -\frac{5}{2}, -4$ | A1 | Both correct. Allow $\frac{-13 \pm 3}{4}$ for this mark but not $\sqrt{9}$ for 3 |
| $-4 < k < -\frac{5}{2}$ | M1A1 | M1: chooses 'inside' region for their critical values. A1: Allow $k \in (-4, -\frac{5}{2})$ or just $(-4, -\frac{5}{2})$. Note: $-\frac{5}{2} < k < -4$ is M0A0 |\begin{enumerate}
\item The equation $20 x ^ { 2 } = 4 k x - 13 k x ^ { 2 } + 2$, where $k$ is a constant, has no real roots.\\
(a) Show that $k$ satisfies the inequality
\end{enumerate}
$$2 k ^ { 2 } + 13 k + 20 < 0$$
(b) Find the set of possible values for $k$.\\
\hfill \mbox{\textit{Edexcel C1 2018 Q7 [8]}}