| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2008 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Quadratic equation real roots |
| Difficulty | Moderate -0.3 This is a straightforward discriminant problem requiring students to apply the condition b²-4ac < 0 for no real roots, then solve a quadratic inequality. It's slightly easier than average because it's a standard textbook exercise with clear steps and the 'show that' in part (a) provides the key inequality, requiring only routine algebraic manipulation and factorization. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02g Inequalities: linear and quadratic in single variable |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(x^2 + kx + (8-k) = 0\) (= 0 can be implied) | M1 | \(8 - k\) need not be bracketed |
| \(b^2 - 4ac = k^2 - 4(8-k)\) | M1 | |
| \(b^2 - 4ac < 0 \Rightarrow k^2 + 4k - 32 < 0\) | A1also |
| Answer | Marks |
|---|---|
| (b) \((k+8)(k-4) = 0\) | M1 |
| \(k = ...\) | M1 |
| \(k = -8\) | A1 |
| \(k = 4\) | A1 |
| Choosing 'inside' region (between the two \(k\) values) | M1 |
| \(-8 < k < 4\) or \(4 > k > -8\) | A1 |
(a) $x^2 + kx + (8-k) = 0$ (= 0 can be implied) | M1 | $8 - k$ need not be bracketed
$b^2 - 4ac = k^2 - 4(8-k)$ | M1 |
$b^2 - 4ac < 0 \Rightarrow k^2 + 4k - 32 < 0$ | A1also |
**Total for (a): 3 marks**
(b) $(k+8)(k-4) = 0$ | M1 |
$k = ...$ | M1 |
$k = -8$ | A1 |
$k = 4$ | A1 |
Choosing 'inside' region (between the two $k$ values) | M1 |
$-8 < k < 4$ or $4 > k > -8$ | A1 |
**Total for (b): 4 marks**
**Total: 7 marks**
---8. The equation
$$x ^ { 2 } + k x + 8 = k$$
has no real solutions for $x$.
\begin{enumerate}[label=(\alph*)]
\item Show that $k$ satisfies $k ^ { 2 } + 4 k - 32 < 0$.
\item Hence find the set of possible values of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2008 Q8 [7]}}