| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| 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 straightforward discriminant question requiring students to apply b²-4ac < 0 for no real roots, then solve a quadratic inequality. Part (a) is routine algebraic manipulation, and part (b) involves completing the square or using the quadratic formula—both standard C1/C2 techniques with no novel problem-solving required. Slightly easier than average due to its predictable structure. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02g Inequalities: linear and quadratic in single variable |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses \(b^2 - 4ac = (6k+4)^2 - 4 \times 1 \times 3\) | B1 | Or equivalent |
| Multiplies out and uses \(b^2 - 4ac < 0\) | M1 | Alternatively find \(b^2 - 4ac\) then state \(b^2 - 4ac < 0\) or \(b^2 < 4ac\) |
| \(36k^2 + 48k + 16 - 12 < 0\) so \(9k^2 + 12k + 1 < 0\) | A1* | Given answer; must see intermediate line(s) including \(<0\) before final answer. Note: \(b^2-4ac = 4(9k^2+12k+1)\), so \(b^2-4ac<0 \Rightarrow 9k^2+12k+1<0\) is fine for B1 M1 A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Solves \(9k^2 + 12k + 1 = 0\) by formula or completing the square | M1 | Factorisation not suitable; answers appearing from graphical calculator score M0. Sight of either root \(\frac{-12-\sqrt{108}}{18}\) or \(\frac{-12+\sqrt{108}}{18}\) is evidence formula used. Sight of \(-\frac{6}{9} + \sqrt{\frac{27}{81}}\) could evidence completing the square |
| \(k = \frac{-12 \pm \sqrt{108}}{18}\) or \(k =\) awrt \(-1.24, -0.09\) | A1 | Accept both values; decimal equivalents awrt 2dp |
| Chooses region between two values and deduces \(\frac{-2-\sqrt{3}}{3} < k < \frac{-2+\sqrt{3}}{3}\) | M1, A1 cao | M1 for choosing inside region from their two roots. Accept \(-\frac{2}{3} - \frac{1}{\sqrt{3}} < k < -\frac{2}{3} + \frac{1}{\sqrt{3}}\). Accept equivalents such as \(\left(\frac{-2-\sqrt{3}}{3}, \frac{-2+\sqrt{3}}{3}\right)\), \(k > \frac{-2-\sqrt{3}}{3}\) and \(k < \frac{-2+\sqrt{3}}{3}\). Do NOT accept \(\left[\frac{-2-\sqrt{3}}{3}, \frac{-2+\sqrt{3}}{3}\right]\) or "\(k > \frac{-2-\sqrt{3}}{3}\) or \(k < \frac{-2+\sqrt{3}}{3}\)". If candidate writes \(\frac{-2-\sqrt{3}}{3} < x < \frac{-2+\sqrt{3}}{3}\) award M1 A0 |
# Question 9:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $b^2 - 4ac = (6k+4)^2 - 4 \times 1 \times 3$ | B1 | Or equivalent |
| Multiplies out and uses $b^2 - 4ac < 0$ | M1 | Alternatively find $b^2 - 4ac$ then state $b^2 - 4ac < 0$ or $b^2 < 4ac$ |
| $36k^2 + 48k + 16 - 12 < 0$ so $9k^2 + 12k + 1 < 0$ | A1* | Given answer; must see intermediate line(s) including $<0$ before final answer. Note: $b^2-4ac = 4(9k^2+12k+1)$, so $b^2-4ac<0 \Rightarrow 9k^2+12k+1<0$ is fine for B1 M1 A1* |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Solves $9k^2 + 12k + 1 = 0$ by formula or completing the square | M1 | Factorisation not suitable; answers appearing from graphical calculator score M0. Sight of either root $\frac{-12-\sqrt{108}}{18}$ or $\frac{-12+\sqrt{108}}{18}$ is evidence formula used. Sight of $-\frac{6}{9} + \sqrt{\frac{27}{81}}$ could evidence completing the square |
| $k = \frac{-12 \pm \sqrt{108}}{18}$ or $k =$ awrt $-1.24, -0.09$ | A1 | Accept both values; decimal equivalents awrt 2dp |
| Chooses region between two values and deduces $\frac{-2-\sqrt{3}}{3} < k < \frac{-2+\sqrt{3}}{3}$ | M1, A1 cao | M1 for choosing inside region from their two roots. Accept $-\frac{2}{3} - \frac{1}{\sqrt{3}} < k < -\frac{2}{3} + \frac{1}{\sqrt{3}}$. Accept equivalents such as $\left(\frac{-2-\sqrt{3}}{3}, \frac{-2+\sqrt{3}}{3}\right)$, $k > \frac{-2-\sqrt{3}}{3}$ and $k < \frac{-2+\sqrt{3}}{3}$. Do NOT accept $\left[\frac{-2-\sqrt{3}}{3}, \frac{-2+\sqrt{3}}{3}\right]$ or "$k > \frac{-2-\sqrt{3}}{3}$ or $k < \frac{-2+\sqrt{3}}{3}$". If candidate writes $\frac{-2-\sqrt{3}}{3} < x < \frac{-2+\sqrt{3}}{3}$ award M1 A0 |
---9. The equation $x ^ { 2 } + ( 6 k + 4 ) x + 3 = 0$, where $k$ is a constant, has no real roots.
\begin{enumerate}[label=(\alph*)]
\item Show that $k$ satisfies the inequality
$$9 k ^ { 2 } + 12 k + 1 < 0$$
\item Find the range of possible values for $k$, giving your boundaries as fully simplified surds.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2015 Q9 [7]}}