| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Find stationary points coordinates |
| Difficulty | Standard +0.3 This is a straightforward optimization problem requiring basic differentiation to find a maximum, plus some contextual reasoning about modeling. Part (b) is routine calculus (differentiate, set to zero, solve linear equation). Parts (a), (c), and (d) test understanding of mathematical modeling in context but require only simple observations (speed limits exist; adding a constant shifts the curve vertically including at v=0 where C should be zero; adjusting coefficients is more appropriate). Slightly easier than average due to the simple quadratic function and inclusion of non-mathematical reasoning questions. |
| Spec | 1.02z Models in context: use functions in modelling1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Maximum speed of the car, or model will show consumption eventually becoming negative, or model may not apply for above 80 mph | B1 | or eg, doesn't drive faster than 80, or speed limit. Condone eg "Maximum number of miles car can drive" |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d}{dv}\!\left(\frac{12}{5}v - \frac{3}{125}v^2\right) = 0\) \(\left(\Rightarrow \frac{12}{5} - \frac{6v}{125} = 0\right)\) | M1 | Attempt differentiate \(C\) & equate to 0 |
| \(v = 50\) | A1 | |
| \(\frac{d^2}{dv^2}\!\left(\frac{12}{5}v - \frac{3}{125}v^2\right) = -\frac{6}{125}\) when \(v=50\), or any correct method showing SP is a maximum | M1 | Must be correct |
| Maximum speed is 50 mph | A1 | Units essential. Dep only on 1st M1 |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = -\frac{b}{2a}\left(= -\frac{\frac{12}{5}}{2\times(-3/125)}\right)\) Attempt complete square | M1 | |
| \(v = 50\) | A1 | |
| Coefficient of \(v^2\) negative, hence stationary point is a maximum | M1 | |
| Maximum speed is 50 mph | A1 | Units essential |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{12}{5}v - \frac{3}{125}v^2 = 0\) \((v=0\) or \(100)\) & correct sketch graph & \(v=50\) seen on graph as giving maximum | M1, B1, M1 | Working must be seen. NB: B1 mark can be gained without working to justify the graph. |
| Maximum speed is 50 mph | A1 | Units essential |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(v=0\) does not give \(C=0\) oe | B1 | They will not consume fuel at 0 mph oe |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| eg \(k\!\left(\frac{12}{5}v - \frac{3}{125}v^2\right)\) with any \(k>1\) | B1 | or "Increase both constants by the same factor" B1B1 |
| B1 | or with numerical value of \(k(>1)\) B1B1. SC: "Increase both constants" B1B0 | |
| Alternative: eg \((1+k)\!\left(\frac{12}{5}v - \frac{3}{125}v^2\right)\) where \(k>0\) | B1, B1 | |
| [2] |
# Question 5:
## Part (a):
| Maximum speed of the car, or model will show consumption eventually becoming negative, or model may not apply for above 80 mph | B1 | or eg, doesn't drive faster than 80, or speed limit. Condone eg "Maximum number of miles car can drive" |
|---|---|---|
| | [1] | |
## Part (b):
| $\frac{d}{dv}\!\left(\frac{12}{5}v - \frac{3}{125}v^2\right) = 0$ $\left(\Rightarrow \frac{12}{5} - \frac{6v}{125} = 0\right)$ | M1 | Attempt differentiate $C$ & equate to 0 |
|---|---|---|
| $v = 50$ | A1 | |
| $\frac{d^2}{dv^2}\!\left(\frac{12}{5}v - \frac{3}{125}v^2\right) = -\frac{6}{125}$ when $v=50$, or any correct method showing SP is a maximum | M1 | Must be correct |
| Maximum speed is 50 mph | A1 | Units essential. Dep only on 1st M1 |
| | [4] | |
**Alternative method 1:**
| $v = -\frac{b}{2a}\left(= -\frac{\frac{12}{5}}{2\times(-3/125)}\right)$ Attempt complete square | M1 | |
|---|---|---|
| $v = 50$ | A1 | |
| Coefficient of $v^2$ negative, hence stationary point is a maximum | M1 | |
| Maximum speed is 50 mph | A1 | Units essential |
**Alternative method 2:**
| $\frac{12}{5}v - \frac{3}{125}v^2 = 0$ $(v=0$ or $100)$ & correct sketch graph & $v=50$ seen on graph as giving maximum | M1, B1, M1 | Working must be seen. NB: B1 mark can be gained without working to justify the graph. |
|---|---|---|
| Maximum speed is 50 mph | A1 | Units essential |
| | [4] | |
## Part (c):
| $v=0$ does not give $C=0$ oe | B1 | They will not consume fuel at 0 mph oe |
|---|---|---|
| | [1] | |
## Part (d):
| eg $k\!\left(\frac{12}{5}v - \frac{3}{125}v^2\right)$ with any $k>1$ | B1 | or "Increase both constants by the same factor" B1B1 |
|---|---|---|
| | B1 | or with numerical value of $k(>1)$ B1B1. SC: "Increase both constants" B1B0 |
| **Alternative:** eg $(1+k)\!\left(\frac{12}{5}v - \frac{3}{125}v^2\right)$ where $k>0$ | B1, B1 | |
| | [2] | |
---5 The fuel consumption of a car, $C$ miles per gallon, varies with the speed, $v$ miles per hour. Jamal models the fuel consumption of his car by the formula\\
$C = \frac { 12 } { 5 } v - \frac { 3 } { 125 } v ^ { 2 }$, for $0 \leqslant v \leqslant 80$.
\begin{enumerate}[label=(\alph*)]
\item Suggest a reason why Jamal has included an upper limit in his model.
\item Determine the speed that gives the maximum fuel consumption.
Amaya's car does more miles per gallon than Jamal's car. She proposes to model the fuel consumption of her car using a formula of the form\\
$C = \frac { 12 } { 5 } v - \frac { 3 } { 125 } v ^ { 2 } + k$, for $0 \leqslant v \leqslant 80$, where $k$ is a positive constant.
\item Give a reason why this model is not suitable.
\item Suggest a different change to Jamal's formula which would give a more suitable model.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q5 [8]}}