| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2018 |
| 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 standard discriminant question requiring students to apply b²-4ac < 0 for no real roots, then solve a quadratic inequality. While it involves multiple steps (setting up discriminant, algebraic manipulation, solving inequality), these are routine techniques covered extensively in C1/C2 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 |
| \((k-4)x^2 - 4x + k - 2 = 0\); uses \(b^2-4ac\) with \(a=k-4\), \(b=-4\), \(c=k-2\) | M1 | Condone one slip; e.g. \(a=k+4\) |
| Uses \(b^2-4ac<0\) or \(b^2<4ac\); e.g. \(16-4(k-4)(k-2)<0\), \(16<4k^2-24k+32\) | dM1 | Uses correct inequality with correct \(a\), \(b\), \(c\); seen at least once before given answer |
| Proceeds correctly to \(k^2 - 6k + 4 > 0\) | A1* | CSO; inequality cannot just appear on last line; condone missing bracket on \(-4^2=16\) |
| Total | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts to solve \(k^2 - 6k + 4 = 0\) to give \(k=\) | M1 | Use formula, completion of square, or calculator (implied by awrt 5.24 and 0.76) |
| Critical values \(k = 3 \pm \sqrt{5}\) | A1 | May be unsimplified \(\frac{6\pm\sqrt{20}}{2}\); may be within inequality |
| \(k^2-6k+4>0\) gives \(k > 3+\sqrt{5}\) or \(k < 3-\sqrt{5}\) | M1, A1cao | M1: chooses outside region; do not award for diagram/table alone; allow boundary inclusion; \(3+\sqrt{5} < k < 3-\sqrt{5}\) scores M1 A0 |
| Total | [4] |
# Question 8:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(k-4)x^2 - 4x + k - 2 = 0$; uses $b^2-4ac$ with $a=k-4$, $b=-4$, $c=k-2$ | M1 | Condone one slip; e.g. $a=k+4$ |
| Uses $b^2-4ac<0$ or $b^2<4ac$; e.g. $16-4(k-4)(k-2)<0$, $16<4k^2-24k+32$ | dM1 | Uses correct inequality with correct $a$, $b$, $c$; seen at least once before given answer |
| Proceeds correctly to $k^2 - 6k + 4 > 0$ | A1* | CSO; inequality cannot just appear on last line; condone missing bracket on $-4^2=16$ |
| **Total** | **[3]** | |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to solve $k^2 - 6k + 4 = 0$ to give $k=$ | M1 | Use formula, completion of square, or calculator (implied by awrt 5.24 and 0.76) |
| Critical values $k = 3 \pm \sqrt{5}$ | A1 | May be unsimplified $\frac{6\pm\sqrt{20}}{2}$; may be within inequality |
| $k^2-6k+4>0$ gives $k > 3+\sqrt{5}$ or $k < 3-\sqrt{5}$ | M1, A1cao | M1: chooses outside region; do not award for diagram/table alone; allow boundary inclusion; $3+\sqrt{5} < k < 3-\sqrt{5}$ scores M1 A0 |
| **Total** | **[4]** | |
---8. The equation $( k - 4 ) x ^ { 2 } - 4 x + k - 2 = 0$, where $k$ is a constant, has no real roots.
\begin{enumerate}[label=(\alph*)]
\item Show that $k$ satisfies the inequality
$$k ^ { 2 } - 6 k + 4 > 0$$
\item Find the exact range of possible values for $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2018 Q8 [7]}}