| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2015 |
| Session | January |
| Marks | 15 |
| 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 structured multi-part question covering standard differentiation applications: finding a normal line (routine), solving simultaneous equations (algebraic manipulation), and using the discriminant condition for tangency (standard technique). While part (c) requires setting up a discriminant condition, all parts follow well-established procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02d Quadratic functions: graphs and discriminant conditions1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = 6x - 4\) | M1A1 | M1: evidence of differentiation, \(x^n \to x^{n-1}\) at least once; A1: both terms correct |
| At \((1,1)\) gradient of curve is \(2\), so gradient of normal is \(-\frac{1}{2}\) | M1 | Substitutes \(x=1\), uses perpendicular property |
| \(\therefore (y-1) = -\frac{1}{2}(x-1)\) and so \(x + 2y - 3 = 0\) | M1 A1* | Correct method for linear equation using \((1,1)\) and changed gradient; printed answer [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Eliminate \(x\) or \(y\): \(2(3x^2-4x+2)+x-3=0\) or \(y=3(3-2y)^2-4(3-2y)+2\) | M1 | May make sign slips |
| Solve three term quadratic e.g. \(6x^2-7x+1=0\) or \(12y^2-29y+17=0\) | M1 | Solve three term quadratic to give \(x=\) or \(y=\) |
| \(x=\frac{1}{6}\) or \(y=1\frac{5}{12}\) | A1 | One correct coordinate |
| Both \(x=\frac{1}{6}\) and \(y=1\frac{5}{12}\), i.e. \((\frac{1}{6}, 1\frac{5}{12})\) or \((0.17, 1.42)\) | A1 | Both correct; ignore \((1,1)\) listed as well [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2(3x^2-4x+2)+kx-3=0\) | M1 | Eliminate \(y\) (condone small copying errors) |
| \(6x^2+(k-8)x+1=0\) | dM1 | Collect into 3 term quadratic; identify \(a\), \(b\), \(c\) |
| Uses condition for equal roots: \(b^2=4ac\) | ddM1 | Dependent on both previous M marks |
| \((k-8)^2=24\), i.e. \(k^2-16k+40=0\) | A1* | Need \((k-8)^2=24\) before stating printed answer [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(k = 8\pm\sqrt{24}\) or \(8\pm 2\sqrt{6}\) or \(\frac{16\pm\sqrt{96}}{2}\) | M1A1 | M1: solve by formula or completing the square; A1: correct exact answer [2] |
# Question 13:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 6x - 4$ | M1A1 | M1: evidence of differentiation, $x^n \to x^{n-1}$ at least once; A1: both terms correct |
| At $(1,1)$ gradient of curve is $2$, so gradient of normal is $-\frac{1}{2}$ | M1 | Substitutes $x=1$, uses perpendicular property |
| $\therefore (y-1) = -\frac{1}{2}(x-1)$ and so $x + 2y - 3 = 0$ | M1 A1* | Correct method for linear equation using $(1,1)$ and changed gradient; printed answer **[5]** |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Eliminate $x$ or $y$: $2(3x^2-4x+2)+x-3=0$ or $y=3(3-2y)^2-4(3-2y)+2$ | M1 | May make sign slips |
| Solve three term quadratic e.g. $6x^2-7x+1=0$ or $12y^2-29y+17=0$ | M1 | Solve three term quadratic to give $x=$ or $y=$ |
| $x=\frac{1}{6}$ or $y=1\frac{5}{12}$ | A1 | One correct coordinate |
| Both $x=\frac{1}{6}$ and $y=1\frac{5}{12}$, i.e. $(\frac{1}{6}, 1\frac{5}{12})$ or $(0.17, 1.42)$ | A1 | Both correct; ignore $(1,1)$ listed as well **[4]** |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2(3x^2-4x+2)+kx-3=0$ | M1 | Eliminate $y$ (condone small copying errors) |
| $6x^2+(k-8)x+1=0$ | dM1 | Collect into 3 term quadratic; identify $a$, $b$, $c$ |
| Uses condition for equal roots: $b^2=4ac$ | ddM1 | Dependent on both previous M marks |
| $(k-8)^2=24$, i.e. $k^2-16k+40=0$ | A1* | Need $(k-8)^2=24$ before stating printed answer **[4]** |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $k = 8\pm\sqrt{24}$ or $8\pm 2\sqrt{6}$ or $\frac{16\pm\sqrt{96}}{2}$ | M1A1 | M1: solve by formula or completing the square; A1: correct exact answer **[2]** |
---13. The curve $C$ has equation
$$y = 3 x ^ { 2 } - 4 x + 2$$
The line $l _ { 1 }$ is the normal to the curve $C$ at the point $P ( 1,1 )$
\begin{enumerate}[label=(\alph*)]
\item Show that $l _ { 1 }$ has equation
$$x + 2 y - 3 = 0$$
The line $l _ { 1 }$ meets curve $C$ again at the point $Q$.
\item By solving simultaneous equations, determine the coordinates of the point $Q$.
Another line $l _ { 2 }$ has equation $k x + 2 y - 3 = 0$, where $k$ is a constant.
\item Show that the line $l _ { 2 }$ meets the curve $C$ once only when
$$k ^ { 2 } - 16 k + 40 = 0$$
\item Find the two exact values of $k$ for which $l _ { 2 }$ is a tangent to $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2015 Q13 [15]}}