| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Solving quadratics and applications |
| Type | Quadratic trajectory/projectile model |
| Difficulty | Moderate -0.8 This is a straightforward applied quadratic question requiring routine techniques: solving a quadratic equation, interpreting a constant in context, completing the square, and reading maximum values from vertex form. All parts are standard textbook exercises with no novel problem-solving required, making it easier than average A-level questions. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02e Complete the square: quadratic polynomials and turning points1.02z Models in context: use functions in modelling |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sets \(H = 0 \Rightarrow 1.8 + 0.4d - 0.002d^2 = 0\) | M1 | |
| \(d = \frac{-0.4 \pm \sqrt{(0.4)^2 - 4(-0.002)(1.8)}}{2 \times -0.002}\) | dM1 | Solves using appropriate method; if completing the square stated must be correct: \((d-100)^2 = 10900 \Rightarrow d = \ldots\) |
| Distance \(=\) awrt \(204\) m only | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States the initial height of the arrow above the ground | B1 | Do not allow "it is the height of the archer" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1.8 + 0.4d - 0.002d^2 = -0.002(d^2 - 200d) + 1.8\) | M1 | Takes out common factor of \(-0.002\) from at least \(d^2\) and \(d\) terms |
| \(= -0.002\!\left((d-100)^2 - 10000\right) + 1.8\) | M1 | Completes the square for \((d^2 - 200d)\) term |
| \(= 21.8 - 0.002(d-100)^2\) | A1 | Or exact equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) \(22.1\) metres | B1ft | For their \(21.8 + 0.3 = 22.1\) m |
| (ii) \(100\) metres | B1ft | For their \(100\) m |
# Question 11:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $H = 0 \Rightarrow 1.8 + 0.4d - 0.002d^2 = 0$ | M1 | |
| $d = \frac{-0.4 \pm \sqrt{(0.4)^2 - 4(-0.002)(1.8)}}{2 \times -0.002}$ | dM1 | Solves using appropriate method; if completing the square stated must be correct: $(d-100)^2 = 10900 \Rightarrow d = \ldots$ |
| Distance $=$ awrt $204$ m only | A1 | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States the initial height of the arrow above the ground | B1 | Do not allow "it is the height of the archer" |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1.8 + 0.4d - 0.002d^2 = -0.002(d^2 - 200d) + 1.8$ | M1 | Takes out common factor of $-0.002$ from at least $d^2$ and $d$ terms |
| $= -0.002\!\left((d-100)^2 - 10000\right) + 1.8$ | M1 | Completes the square for $(d^2 - 200d)$ term |
| $= 21.8 - 0.002(d-100)^2$ | A1 | Or exact equivalent |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| (i) $22.1$ metres | B1ft | For their $21.8 + 0.3 = 22.1$ m |
| (ii) $100$ metres | B1ft | For their $100$ m |11. An archer shoots an arrow.
The height, $H$ metres, of the arrow above the ground is modelled by the formula
$$H = 1.8 + 0.4 d - 0.002 d ^ { 2 } , \quad d \geqslant 0$$
where $d$ is the horizontal distance of the arrow from the archer, measured in metres.\\
Given that the arrow travels in a vertical plane until it hits the ground,
\begin{enumerate}[label=(\alph*)]
\item find the horizontal distance travelled by the arrow, as given by this model.
\item With reference to the model, interpret the significance of the constant 1.8 in the formula.
\item Write $1.8 + 0.4 d - 0.002 d ^ { 2 }$ in the form
$$A - B ( d - C ) ^ { 2 }$$
where $A , B$ and $C$ are constants to be found.
It is decided that the model should be adapted for a different archer.\\
The adapted formula for this archer is
$$H = 2.1 + 0.4 d - 0.002 d ^ { 2 } , \quad d \geqslant 0$$
Hence or otherwise, find, for the adapted model
\item \begin{enumerate}[label=(\roman*)]
\item the maximum height of the arrow above the ground.
\item the horizontal distance, from the archer, of the arrow when it is at its maximum height.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 Q11 [9]}}