| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Determine nature of stationary points |
| Difficulty | Moderate -0.8 This is a straightforward C1 differentiation question requiring only routine application of the power rule for polynomials. All parts involve standard procedures: differentiate using power rule, substitute a value, verify stationary point by showing derivative equals zero, and use second derivative test. No problem-solving insight or novel techniques required—purely mechanical application of basic calculus rules. |
| Spec | 1.07b Gradient as rate of change: dy/dx notation1.07d Second derivatives: d^2y/dx^2 notation1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| \((x^2 - 10x) + (y^2 + 12y) + 41 = 0\) | M1 | Attempt to complete the square |
| \((x-5)^2 - 25 + (y+6)^2 - 36 + 41 = 0\) | A1 | At least one bracket correct |
| \((x-5)^2 + (y+6)^2 = 20\) | A1 | Fully correct |
| Answer | Marks |
|---|---|
| Centre \(C = (5, -6)\) | B1 |
| Answer | Marks |
|---|---|
| \(r^2 = 20 = 4 \times 5\), so \(r = 2\sqrt{5}\), hence \(n = 2\) | M1 |
| Clear demonstration that \(r = 2\sqrt{5}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient of \(CA\): \(\frac{-2-(-6)}{3-5} = \frac{4}{-2} = -2\) | M1 | Finding gradient of radius |
| Gradient of tangent \(= \frac{1}{2}\) | M1 | Using perpendicular gradient |
| \(y-(-2) = \frac{1}{2}(x-3)\) | M1 | Equation through \(A\) |
| \(2y + 4 = x - 3 \Rightarrow x - 2y = 7\) | A1 | Rearranging to required form |
| \(p = -2\), \(q = 7\) | A1 | Both values correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(BC^2 = CA^2 + AB^2\) or use of Pythagoras with tangent-radius relationship | M1 | |
| \(CA = 2\sqrt{5}\), \(BC = \sqrt{CA^2 + AB^2}\)... \(BC^2 = AB^2 + 20\) | M1 | |
| \(AB = \sqrt{BC^2 - r^2} = \sqrt{36+20} = \sqrt{56} = 2\sqrt{14}\) | A1 | Correct answer |
# Question 7:
## Part (a):
| $(x^2 - 10x) + (y^2 + 12y) + 41 = 0$ | M1 | Attempt to complete the square |
| $(x-5)^2 - 25 + (y+6)^2 - 36 + 41 = 0$ | A1 | At least one bracket correct |
| $(x-5)^2 + (y+6)^2 = 20$ | A1 | Fully correct |
## Part (b)(i):
| Centre $C = (5, -6)$ | B1 | |
## Part (b)(ii):
| $r^2 = 20 = 4 \times 5$, so $r = 2\sqrt{5}$, hence $n = 2$ | M1 | |
| Clear demonstration that $r = 2\sqrt{5}$ | A1 | |
## Part (c):
| Gradient of $CA$: $\frac{-2-(-6)}{3-5} = \frac{4}{-2} = -2$ | M1 | Finding gradient of radius |
| Gradient of tangent $= \frac{1}{2}$ | M1 | Using perpendicular gradient |
| $y-(-2) = \frac{1}{2}(x-3)$ | M1 | Equation through $A$ |
| $2y + 4 = x - 3 \Rightarrow x - 2y = 7$ | A1 | Rearranging to required form |
| $p = -2$, $q = 7$ | A1 | Both values correct |
## Part (d):
| $BC^2 = CA^2 + AB^2$ or use of Pythagoras with tangent-radius relationship | M1 | |
| $CA = 2\sqrt{5}$, $BC = \sqrt{CA^2 + AB^2}$... $BC^2 = AB^2 + 20$ | M1 | |
| $AB = \sqrt{BC^2 - r^2} = \sqrt{36+20} = \sqrt{56} = 2\sqrt{14}$ | A1 | Correct answer |
---7 The volume, $V \mathrm {~m} ^ { 3 }$, of water in a tank at time $t$ seconds is given by
$$V = \frac { 1 } { 3 } t ^ { 6 } - 2 t ^ { 4 } + 3 t ^ { 2 } , \quad \text { for } t \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Find:
\begin{enumerate}[label=(\roman*)]
\item $\frac { \mathrm { d } V } { \mathrm {~d} t }$;\\
(3 marks)
\item $\frac { \mathrm { d } ^ { 2 } V } { \mathrm {~d} t ^ { 2 } }$.\\
(2 marks)
\end{enumerate}\item Find the rate of change of the volume of water in the tank, in $\mathrm { m } ^ { 3 } \mathrm {~s} ^ { - 1 }$, when $t = 2$.
\item \begin{enumerate}[label=(\roman*)]
\item Verify that $V$ has a stationary value when $t = 1$.
\item Determine whether this is a maximum or minimum value.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 Q7 [14]}}