| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2018 |
| Session | January |
| 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 the inequality, expanding, factorizing, and interpreting the solution), these are all routine techniques covered in C1/C2 with no novel insight required. Slightly easier than average due to its predictable structure and straightforward algebra. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02g Inequalities: linear and quadratic in single variable |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(b^2 - 4ac = 8^2 - 4(p-2)(p+4)\) | M1 | Attempts \(b^2-4ac\) with at least two of \(a\), \(b\), \(c\) correct. No \(x\)'s present |
| \(8^2 - 4(p-2)(p+4) < 0\) | A1 | Correct un-simplified inequality, not the final printed answer |
| \(64 < 4p^2 + 8p - 32\) | ||
| \(p^2 + 2p - 24 > 0\) | A1* | Correct solution with intermediate working, inequality sign correct before final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(p^2+2p-24=0 \Rightarrow (p+1)^2-1-24=0\) or \((p=)\frac{-2\pm\sqrt{2^2-4\times1\times(-24)}}{2\times1}\) | M1 | Attempt to solve \(p^2+2p-24=0\) leading to two critical values |
| \(p = 4, -6\) | A1 | Correct critical values |
| \(p < \text{"-6"}\), \(p > \text{"4"}\) | M1 | Chooses outside region for two critical values |
| \(p < -6\) or \(p > 4\) | A1 | Correct inequalities. Note \(p<-6\) and \(p>4\) scores M1A0 |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $b^2 - 4ac = 8^2 - 4(p-2)(p+4)$ | M1 | Attempts $b^2-4ac$ with at least two of $a$, $b$, $c$ correct. No $x$'s present |
| $8^2 - 4(p-2)(p+4) < 0$ | A1 | Correct un-simplified inequality, not the final printed answer |
| $64 < 4p^2 + 8p - 32$ | | |
| $p^2 + 2p - 24 > 0$ | A1* | Correct solution with intermediate working, inequality sign correct before final answer |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $p^2+2p-24=0 \Rightarrow (p+1)^2-1-24=0$ or $(p=)\frac{-2\pm\sqrt{2^2-4\times1\times(-24)}}{2\times1}$ | M1 | Attempt to solve $p^2+2p-24=0$ leading to two critical values |
| $p = 4, -6$ | A1 | Correct critical values |
| $p < \text{"-6"}$, $p > \text{"4"}$ | M1 | Chooses outside region for two critical values |
| $p < -6$ or $p > 4$ | A1 | Correct inequalities. Note $p<-6$ **and** $p>4$ scores M1A0 |
---4. The equation\\
$( p - 2 ) x ^ { 2 } + 8 x + ( p + 4 ) = 0 , \quad$ where $p$ is a constant has no real roots.
\begin{enumerate}[label=(\alph*)]
\item Show that $p$ satisfies $p ^ { 2 } + 2 p - 24 > 0$
\item Hence find the set of possible values of $p$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2018 Q4 [7]}}