| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Find range for no real roots |
| Difficulty | Standard +0.3 This is a standard discriminant problem requiring rearrangement to quadratic form, then applying b²-4ac < 0 for no real roots. It's slightly above average difficulty due to the algebraic manipulation needed (clearing fractions, rearranging) before applying the discriminant condition, but remains a routine P1 exercise with no novel insight required. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02g Inequalities: linear and quadratic in single variable |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{3}{x} + 5 = -2x + c\); multiplies through by \(x\): \(3 + 5x = -2x^2 + cx \Rightarrow \pm 2x^2 \ldots (=0)\) | M1 | Attempts to multiply through by \(x\) and moves all terms to one side; condone one term not multiplied by \(x\); all four terms must be on one side |
| \(2x^2 + (5-c)x + 3 (= 0)\) | A1 | Correct quadratic with terms in \(x\) factorised, or correct values of \(a\), \(b\), \(c\); "=0" not needed; ignore inequalities |
| Attempts \(b^2 - 4ac = (5-c)^2 - 24\) | M1 | Attempts \(b^2 - 4ac\) using their values; sufficient to see values substituted correctly; condone invisible brackets; must have a quadratic in \(x\) |
| Attempts \(b^2 - 4ac = 0 \Rightarrow (5-c)^2 - 24 = 0 \Rightarrow c = \ldots\) | dM1 | Dependent on previous M1; attempts to solve \(b^2 - 4ac = 0\) for at least one critical value of \(c\) |
| \((c =)\ 5 \pm 2\sqrt{6}\) | A1 | Correct critical values; accept \(5 \pm \sqrt{24}\) or exact equivalent of form \(\frac{a \pm \sqrt{b}}{c}\) |
| Attempt at inside region | M1 | Finds inside region for their critical values; allow \(\leq\) for one or both inequalities |
| \(5 - 2\sqrt{6} < c < 5 + 2\sqrt{6}\) | A1 | Must be in terms of \(c\) and must be exact; accept e.g. \(5-\sqrt{24} < c < 5+\sqrt{24}\); do NOT accept \(c > 5-2\sqrt{6}\) OR \(c < 5+2\sqrt{6}\) |
## Question 9:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{3}{x} + 5 = -2x + c$; multiplies through by $x$: $3 + 5x = -2x^2 + cx \Rightarrow \pm 2x^2 \ldots (=0)$ | M1 | Attempts to multiply through by $x$ and moves all terms to one side; condone one term not multiplied by $x$; all four terms must be on one side |
| $2x^2 + (5-c)x + 3 (= 0)$ | A1 | Correct quadratic with terms in $x$ factorised, or correct values of $a$, $b$, $c$; "=0" not needed; ignore inequalities |
| Attempts $b^2 - 4ac = (5-c)^2 - 24$ | M1 | Attempts $b^2 - 4ac$ using their values; sufficient to see values substituted correctly; condone invisible brackets; must have a quadratic in $x$ |
| Attempts $b^2 - 4ac = 0 \Rightarrow (5-c)^2 - 24 = 0 \Rightarrow c = \ldots$ | dM1 | Dependent on previous M1; attempts to solve $b^2 - 4ac = 0$ for at least one critical value of $c$ |
| $(c =)\ 5 \pm 2\sqrt{6}$ | A1 | Correct critical values; accept $5 \pm \sqrt{24}$ or exact equivalent of form $\frac{a \pm \sqrt{b}}{c}$ |
| Attempt at inside region | M1 | Finds inside region for their critical values; allow $\leq$ for one or both inequalities |
| $5 - 2\sqrt{6} < c < 5 + 2\sqrt{6}$ | A1 | Must be in terms of $c$ and must be exact; accept e.g. $5-\sqrt{24} < c < 5+\sqrt{24}$; do NOT accept $c > 5-2\sqrt{6}$ OR $c < 5+2\sqrt{6}$ |
---\begin{enumerate}
\item The equation
\end{enumerate}
$$\frac { 3 } { x } + 5 = - 2 x + c$$
where $c$ is a constant, has no real roots.\\
Find the range of possible values of $c$.\\
\hfill \mbox{\textit{Edexcel P1 2019 Q9 [7]}}