| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2009 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Normal meets curve/axis — further geometry |
| Difficulty | Moderate -0.3 This is a standard C1 differentiation question requiring routine application of differentiation rules (including x^{-1}), finding tangent/normal equations, and basic coordinate geometry. While it has multiple parts and requires careful arithmetic, all techniques are straightforward textbook exercises with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(\frac{dy}{dx} = -4 + 8x^{-2}\) | M1A1 | M1 for 4 or \(8x^{-2}\) (ignore signs); A1 for both terms correct including signs |
| \(x = 2 \Rightarrow m = -4 + 2 = -2\) | M1 | for substituting \(x = 2\) into their \(\frac{dy}{dx}\) (must be different from their \(y\)) |
| \(y = 9 - 8 - \frac{8}{2} = -3\) | B1 | for \(y_P = -3\), but not if clearly found from given equation of tangent. First 4 marks could be earned in part (b) |
| Equation of tangent: \(y + 3 = -2(x-2) \rightarrow y = 1 - 2x\) (*) | M1 A1cso | M1 for attempt to find equation of tangent at \(P\); A1cso for correct work leading to printed answer, allow equivalents with \(2x\), \(y\) and 1 terms such as \(2x + y - 1 = 0\) |
| Total | (6) | NO DIFFERENTIATION ATTEMPTED: Just assuming \(m = -2\) is M0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| Gradient of normal \(= \frac{1}{2}\) | B1ft | for correct use of perpendicular gradient rule. Follow through their \(m\), but if \(m \neq -2\) there must be clear evidence that \(m\) is thought to be gradient of tangent |
| \(\frac{y+3}{x-2} = \frac{1}{2}\) or better equivalent, e.g. \(y = \frac{1}{2}x - 4\) | M1A1 | M1 for attempt to find normal at \(P\) using changed gradient and their \(y_P\); A1 for any correct form as specified (correct answer only) |
| Total | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \((A:)\ \frac{1}{2}\), \((B:)\ 8\) | B1, B1 | 1st B1 for \(\frac{1}{2}\) and 2nd B1 for \(8\) |
| Area of triangle \(= \frac{1}{2}(x_B \pm x_A) \times y_P\) with values for all of \(x_B\), \(x_A\) and \(y_P\) | M1 | for full method for area of triangle \(ABP\). Follow through their \(x_A\), \(x_B\) and \(y_P\); mark awarded 'generously' condoning sign errors. Final answer must be positive for A1, with negatives in working condoned. Determinant: Area \(= \frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = \frac{1}{2}\begin{vmatrix} 2 & -3 & 1 \\ 0.5 & 0 & 1 \\ 8 & 0 & 1 \end{vmatrix} = \ldots\) (attempt to multiply out required for M1). Alternative: \(AP = \sqrt{(2-0.5)^2+(-3)^2}\), \(BP = \sqrt{(2-8)^2+(-3)^2}\), Area \(= \frac{1}{2}AP \times BP = \ldots\) M1. Intersections with \(y\)-axis instead of \(x\)-axis: Only the M mark is available B0 B0 M1 A0 |
| \(\frac{1}{2}\left(8 - \frac{1}{2}\right) \times 3 = \frac{45}{4}\) or \(11.25\) | A1 | |
| Total | (4) [13] |
# Question 11:
## Part (a):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $\frac{dy}{dx} = -4 + 8x^{-2}$ | M1A1 | M1 for 4 or $8x^{-2}$ (ignore signs); A1 for both terms correct including signs |
| $x = 2 \Rightarrow m = -4 + 2 = -2$ | M1 | for substituting $x = 2$ into their $\frac{dy}{dx}$ (must be different from their $y$) |
| $y = 9 - 8 - \frac{8}{2} = -3$ | B1 | for $y_P = -3$, but not if clearly found from given equation of tangent. First 4 marks could be earned in part (b) |
| Equation of tangent: $y + 3 = -2(x-2) \rightarrow y = 1 - 2x$ (*) | M1 A1cso | M1 for attempt to find equation of tangent at $P$; A1cso for correct work leading to printed answer, allow equivalents with $2x$, $y$ and 1 terms such as $2x + y - 1 = 0$ |
| **Total** | **(6)** | NO DIFFERENTIATION ATTEMPTED: Just assuming $m = -2$ is M0 |
## Part (b):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Gradient of normal $= \frac{1}{2}$ | B1ft | for correct use of perpendicular gradient rule. Follow through their $m$, but if $m \neq -2$ there must be clear evidence that $m$ is thought to be gradient of tangent |
| $\frac{y+3}{x-2} = \frac{1}{2}$ or better equivalent, e.g. $y = \frac{1}{2}x - 4$ | M1A1 | M1 for attempt to find normal at $P$ using changed gradient and their $y_P$; A1 for any correct form as specified (correct answer only) |
| **Total** | **(3)** | |
## Part (c):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $(A:)\ \frac{1}{2}$, $(B:)\ 8$ | B1, B1 | 1st B1 for $\frac{1}{2}$ and 2nd B1 for $8$ |
| Area of triangle $= \frac{1}{2}(x_B \pm x_A) \times y_P$ with values for all of $x_B$, $x_A$ and $y_P$ | M1 | for full method for area of triangle $ABP$. Follow through their $x_A$, $x_B$ and $y_P$; mark awarded 'generously' condoning sign errors. Final answer must be positive for A1, with negatives in working condoned. Determinant: Area $= \frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = \frac{1}{2}\begin{vmatrix} 2 & -3 & 1 \\ 0.5 & 0 & 1 \\ 8 & 0 & 1 \end{vmatrix} = \ldots$ (attempt to multiply out required for M1). Alternative: $AP = \sqrt{(2-0.5)^2+(-3)^2}$, $BP = \sqrt{(2-8)^2+(-3)^2}$, Area $= \frac{1}{2}AP \times BP = \ldots$ M1. Intersections with $y$-axis instead of $x$-axis: Only the M mark is available B0 B0 M1 A0 |
| $\frac{1}{2}\left(8 - \frac{1}{2}\right) \times 3 = \frac{45}{4}$ or $11.25$ | A1 | |
| **Total** | **(4) [13]** | |\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$y = 9 - 4 x - \frac { 8 } { x } , \quad x > 0$$
The point $P$ on $C$ has $x$-coordinate equal to 2 .\\
(a) Show that the equation of the tangent to $C$ at the point $P$ is $y = 1 - 2 x$.\\
(b) Find an equation of the normal to $C$ at the point $P$.
The tangent at $P$ meets the $x$-axis at $A$ and the normal at $P$ meets the $x$-axis at $B$.\\
(c) Find the area of triangle $A P B$.\\
\hfill \mbox{\textit{Edexcel C1 2009 Q11 [13]}}