| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Solving quadratics and applications |
| Type | Completing the square, form and properties |
| Difficulty | Moderate -0.3 This is a standard two-part question on completing the square and coordinate geometry. Part (a) is routine algebraic manipulation, while part (b) requires finding intercepts and calculating a triangle area using a formula. Both are textbook exercises requiring no novel insight, making it slightly easier than average for A-level. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02f Solve quadratic equations: including in a function of unknown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(b = 2\) | B1 | Implied from expression, e.g. look for \(-2(x \pm ...)^2\). Beware \(21 + 2(x-3)^2\) which would be B0 |
| \(\ldots \pm \ldots(x \pm 3)^2\) | M1 | Sight of \((x \pm 3)^2\) |
| \(f(x) = 21 - 2(x-3)^2\) | A1 | Condone \(21 - 2(x + -3)^2\). May state values of \(a\), \(b\), \(c\). If contradiction between expression and stated values treat as isw |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(R\) is \((0, -4)\) or "\(h\)" \(= 4\) | B1 | Coordinate may be seen on diagram. Allow just \(-4\) rather than full coordinate on \(y\)-axis. May be implied by working |
| \(f(x) - 7 = 14 - 2(x-3)^2 \Rightarrow x = \ldots\) or \(f(x) - 7 = -4 + 12x - 2x^2 \Rightarrow x = \ldots\) (NB \(x = 3 \pm \sqrt{7}\)) | M1 | Attempts to solve \(f(x) - 7 = 0\). Score for attempt to solve a 3TQ of form \(-2x^2 + 12x + C = 0\) where \(C \neq 3\), or \(A - 2(x-3)^2 = 0\) where \(A \neq 21\) |
| \(\text{Area} = \frac{1}{2} \times \left("3+\sqrt{7}" - "3-\sqrt{7}"\right) \times "4"\) | dM1 | Fully correct method for area of triangle. Dependent on previous M1. Score for \(\frac{1}{2} \times (\beta - \alpha) \times\) "\(y\)-intercept" where \(\alpha, \beta\) are roots. Should be \(\frac{1}{2} \times (2\sqrt{\ldots}) \times\) "\(y\)-intercept" |
| \(= 4\sqrt{7}\) | A1 | Cao. Allow e.g. \(= 2\sqrt{28}\), \(\sqrt{112}\). May work in decimals but must state correct exact answer \(4\sqrt{7}\) if it follows from correct method |
# Question 2(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $b = 2$ | B1 | Implied from expression, e.g. look for $-2(x \pm ...)^2$. Beware $21 + 2(x-3)^2$ which would be B0 |
| $\ldots \pm \ldots(x \pm 3)^2$ | M1 | Sight of $(x \pm 3)^2$ |
| $f(x) = 21 - 2(x-3)^2$ | A1 | Condone $21 - 2(x + -3)^2$. May state values of $a$, $b$, $c$. If contradiction between expression and stated values treat as isw |
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# Question 2(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $R$ is $(0, -4)$ or "$h$" $= 4$ | B1 | Coordinate may be seen on diagram. Allow just $-4$ rather than full coordinate on $y$-axis. May be implied by working |
| $f(x) - 7 = 14 - 2(x-3)^2 \Rightarrow x = \ldots$ or $f(x) - 7 = -4 + 12x - 2x^2 \Rightarrow x = \ldots$ (NB $x = 3 \pm \sqrt{7}$) | M1 | Attempts to solve $f(x) - 7 = 0$. Score for attempt to solve a 3TQ of form $-2x^2 + 12x + C = 0$ where $C \neq 3$, or $A - 2(x-3)^2 = 0$ where $A \neq 21$ |
| $\text{Area} = \frac{1}{2} \times \left("3+\sqrt{7}" - "3-\sqrt{7}"\right) \times "4"$ | dM1 | Fully correct method for area of triangle. Dependent on previous M1. Score for $\frac{1}{2} \times (\beta - \alpha) \times$ "$y$-intercept" where $\alpha, \beta$ are roots. Should be $\frac{1}{2} \times (2\sqrt{\ldots}) \times$ "$y$-intercept" |
| $= 4\sqrt{7}$ | A1 | Cao. Allow e.g. $= 2\sqrt{28}$, $\sqrt{112}$. May work in decimals but must state correct exact answer $4\sqrt{7}$ if it follows from correct method |
---2.
$$f ( x ) = 3 + 12 x - 2 x ^ { 2 }$$
(a) Express $\mathrm { f } ( x )$ in the form\\
2. $\mathrm { f } ( x ) = 3 + 12 x - 2 x ^ { 2 }$\\
(a) Express $\mathrm { f } ( x )$ in the form
$$\begin{aligned}
& \qquad a - b ( x + c ) ^ { 2 } \\
& \text { where } a , b \text { and } c \text { are integers to be found. } \\
& \text { he curve with equation } y = \mathrm { f } ( x ) - 7 \text { crosses the } x \text {-axis at the points } P \text { and } Q \text { and crosses } \\
& \text { te } y \text {-axis at the point } R \text {. } \\
& \text { F) Find the area of the triangle } P Q R \text {, giving your answer in the form } m \sqrt { n } \text { where } m \text { and } \\
& n \text { are integers to be found. }
\end{aligned}$$
$\_\_\_\_$ "
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\hfill \mbox{\textit{Edexcel P1 2020 Q2 [7]}}