| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2010 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Point position relative to circle |
| Difficulty | Moderate -0.3 This is a standard C1 circle question requiring completing the square to find centre/radius, basic geometric reasoning about position, and solving a quadratic inequality. All techniques are routine for this level, though part (c) requires careful algebraic manipulation across multiple steps. Slightly easier than average due to heavy scaffolding through 'show that' parts. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = xe^x + e^x\) | M1, A1 | Product rule OE |
| At stationary point(s) \(e^x(x+1) = 0\) | m1 | |
| \(e^x > 0\) | E1 | OE e.g. accept \(e^x \neq 0\) |
| Only one value of \(x\) for st. pt. Curve has exactly one st pt | A1 | CSO with conclusion |
| Stationary point is \((-1, -e^{-1})\) | A1 | Total: 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Stationary point is \((-1, k - e^{-1})\) | B1F | Or E1 for \(y = xe^x\) to \(y = xe^x + k\) is a vertical translation of \(k\) units |
| St. pt is on \(x\)-axis, so \(k = e^{-1}\) | B1 | Total: 2 |
## Question 7(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = xe^x + e^x$ | M1, A1 | Product rule OE |
| At stationary point(s) $e^x(x+1) = 0$ | m1 | |
| $e^x > 0$ | E1 | OE e.g. accept $e^x \neq 0$ |
| Only one value of $x$ for st. pt. Curve has exactly one st pt | A1 | CSO with conclusion |
| Stationary point is $(-1, -e^{-1})$ | A1 | **Total: 6** |
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## Question 7(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Stationary point is $(-1, k - e^{-1})$ | B1F | Or E1 for $y = xe^x$ to $y = xe^x + k$ is a vertical translation of $k$ units |
| St. pt is on $x$-axis, so $k = e^{-1}$ | B1 | **Total: 2** |
---7 A circle with centre $C$ has equation $x ^ { 2 } + y ^ { 2 } - 4 x + 12 y + 15 = 0$.
\begin{enumerate}[label=(\alph*)]
\item Find:
\begin{enumerate}[label=(\roman*)]
\item the coordinates of $C$;
\item the radius of the circle.
\end{enumerate}\item Explain why the circle lies entirely below the $x$-axis.
\item The point $P$ with coordinates $( 5 , k )$ lies outside the circle.
\begin{enumerate}[label=(\roman*)]
\item Show that $P C ^ { 2 } = k ^ { 2 } + 12 k + 45$.
\item Hence show that $k ^ { 2 } + 12 k + 20 > 0$.
\item Find the possible values of $k$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2010 Q7 [13]}}