| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Normal meets curve/axis — further geometry |
| Difficulty | Standard +0.3 This is a straightforward C1 differentiation question requiring finding a derivative using power rule (rewriting 18/x as 18x^{-1}), finding gradient at a point, then finding the normal, and solving a simultaneous equation. All steps are routine textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitutes \(x=2\) into \(y = 20 - 4x^2 - \frac{18}{x}\) and gets 3 | B1 | Must use curve equation, not line equation; substitution must be seen |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -4 + \dfrac{18}{x^2}\) | M1 A1 | M1 for differentiating negative power \(x^{-1} \to x^{-2}\); A1 correct expression |
| Substitute \(x=2 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2}\), then finds negative reciprocal \((-2)\) | dM1 | Dependent on first M1; states \(-2 \times \frac{1}{2} = -1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States/uses \(y - 3 = -2(x-2)\) or \(y = -2x + c\) with point \((2,3)\) | dM1 | Not their \(\frac{1}{2}\), \(-\frac{1}{2}\), 2, or \(-2\) from wrong working |
| \(y = -2x + 7\) | A1* | (6 marks); CSO — given answer, equation must be stated, no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Check that \((2,3)\) lies on line \(y = -2x+7\) | dM1 | |
| Deduce equation of normal as it has same gradient and passes through common point | A1* | CSO — both conditions must be stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Put \(20 - 4x - \dfrac{18}{x} = -2x + 7\) and simplify to give \(2x^2 - 13x + 18 = 0\) | M1 A1 | M1: equate two given expressions, collect terms to 3TQ (sign errors allowed); A1: correct 3TQ \(= 0\) |
| Or put \(y = 20 - 4\!\left(\frac{7-y}{2}\right) - \dfrac{18}{\left(\frac{7-y}{2}\right)}\) to give \(y^2 - y - 6 = 0\) | ||
| \((2x-9)(x-2)=0\) so \(x = \ldots\) or \((y-3)(y+2)=0\) so \(y = \ldots\) | dM1 | Attempt to solve quadratic by factorisation, formula, or completing the square |
| \(x = \dfrac{9}{2}\) or \(y = -2\) | A1 | Allow second answers; ignore \(x=2\) or \(y=3\) for this mark |
| \(x = \dfrac{9}{2},\ y = -2\) (i.e. point \(B = \left(\dfrac{9}{2}, -2\right)\)) | A1 | (5 marks); both correct; if \(x=2, y=3\) included and \(B\) not identified, last mark is A0 |
# Question 11(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $x=2$ into $y = 20 - 4x^2 - \frac{18}{x}$ and gets 3 | B1 | Must use curve equation, not line equation; substitution must be seen |
| $\dfrac{\mathrm{d}y}{\mathrm{d}x} = -4 + \dfrac{18}{x^2}$ | M1 A1 | M1 for differentiating negative power $x^{-1} \to x^{-2}$; A1 correct expression |
| Substitute $x=2 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2}$, then finds negative reciprocal $(-2)$ | dM1 | Dependent on first M1; states $-2 \times \frac{1}{2} = -1$ |
**Method 1:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| States/uses $y - 3 = -2(x-2)$ or $y = -2x + c$ with point $(2,3)$ | dM1 | Not their $\frac{1}{2}$, $-\frac{1}{2}$, 2, or $-2$ from wrong working |
| $y = -2x + 7$ | A1* | (6 marks); CSO — given answer, equation must be stated, no errors seen |
**Method 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Check that $(2,3)$ lies on line $y = -2x+7$ | dM1 | |
| Deduce equation of normal as it has same gradient and passes through common point | A1* | CSO — both conditions must be stated |
---
# Question 11(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Put $20 - 4x - \dfrac{18}{x} = -2x + 7$ and simplify to give $2x^2 - 13x + 18 = 0$ | M1 A1 | M1: equate two given expressions, collect terms to 3TQ (sign errors allowed); A1: correct 3TQ $= 0$ |
| Or put $y = 20 - 4\!\left(\frac{7-y}{2}\right) - \dfrac{18}{\left(\frac{7-y}{2}\right)}$ to give $y^2 - y - 6 = 0$ | | |
| $(2x-9)(x-2)=0$ so $x = \ldots$ or $(y-3)(y+2)=0$ so $y = \ldots$ | dM1 | Attempt to solve quadratic by factorisation, formula, or completing the square |
| $x = \dfrac{9}{2}$ or $y = -2$ | A1 | Allow second answers; ignore $x=2$ or $y=3$ for this mark |
| $x = \dfrac{9}{2},\ y = -2$ (i.e. point $B = \left(\dfrac{9}{2}, -2\right)$) | A1 | (5 marks); both correct; if $x=2, y=3$ included and $B$ not identified, last mark is A0 |
**Total: 11 marks**
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\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6db8acbd-7f61-46ff-8fdc-f0f4a8363aa6-17_700_1556_276_201}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A sketch of part of the curve $C$ with equation
$$y = 20 - 4 x - \frac { 18 } { x } , \quad x > 0$$
is shown in Figure 3.
Point $A$ lies on $C$ and has an $x$ coordinate equal to 2
\begin{enumerate}[label=(\alph*)]
\item Show that the equation of the normal to $C$ at $A$ is $y = - 2 x + 7$
The normal to $C$ at $A$ meets $C$ again at the point $B$, as shown in Figure 3 .
\item Use algebra to find the coordinates of $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2014 Q11 [11]}}