| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Line tangent to curve, find k for tangency |
| Difficulty | Moderate -0.3 This is a standard C1 tangent problem requiring students to set the line equal to the curve, form a quadratic equation, and use the discriminant condition (b²-4ac=0) for tangency. While it involves multiple steps, the technique is routine and commonly practiced at this level, making it slightly easier than average. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02n Sketch curves: simple equations including polynomials |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| U-shaped parabola symmetric about \(y\)-axis | B1 | |
| Graph passes through \((0, 8)\) | B1 | |
| Shape and position for \(L\) | M1 | |
| Both \(\left(-\frac{k}{3}, 0\right)\) and \((0, k)\) | A1 | Allow marks even if on the same diagram |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equate \(\frac{1}{3}x^2 + 8 = 3x + k\) and collect terms on one side | M1 | |
| \(\frac{1}{3}x^2 - 3x + (8-k)\) | A1 | |
| Method 1a: Uses \(b^2=4ac\): \(9 = 4\times\frac{1}{3}\times(8-k) \Rightarrow k=...\) | dM1, dM1 | |
| Method 1b: Attempt \(\frac{1}{3}(x-\frac{9}{2})^2 - \lambda + 8 - k\); deduce \(k = 8 - \lambda\) | dM1, dM1 | |
| \(k = \frac{5}{4}\) oe | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts to set \(\frac{dy}{dx} = 3\) | M1 | |
| \(\frac{2}{3}x = 3 \Rightarrow x = 4.5\) | A1 | |
| Method 2a: Substitutes \(x=4.5\) into \(y=\frac{1}{3}x^2+8 \Rightarrow y=14.75\); substitutes both \(x\) and \(y\) into \(y=3x+k\) to find \(k\) | dM1, dM1 | |
| Method 2b: Substitutes \(x=4.5\) into \(\frac{1}{3}x^2+8=3x+k\); finds \(k=\) | dM1, dM1 | |
| \(k = 1.25\) oe | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Shape for C: \(\cup\) approximately symmetrical about \(y\)-axis | B1 | |
| Coordinates of \((0, 8)\) | B1 | There must be a graph; accept graph crossing positive \(y\)-axis with only 8 marked; accept \((8,0)\) if given on \(y\)-axis |
| Shape for \(L\): straight line with positive gradient and positive intercept | M1 | |
| Coordinates of \((0, k)\) and \((-k/3, 0)\), or \(k\) marked on \(y\)-axis and \(-k/3\) marked on \(x\)-axis | A1 | Accept \((k, 0)\) on \(y\)-axis and \((0, -k/3)\) on \(x\)-axis |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equate curves \(\frac{1}{3}x^2 + 8 = 3x + k\) and collect \(x\) terms and \((8-k)\) terms to one side | M1 | |
| Achieves e.g. \(\frac{1}{3}x^2 - 3x + (8-k)\) | A1 | |
| Uses \(b^2 = 4ac\) or \(b^2 - 4ac = 0\) | dM1 | Depends on previous M mark |
| Solves their \(b^2 = 4ac\) leading to \(k =\) ... | dM1 | Depends on previous M mark |
| \(k = \dfrac{5}{4}\) | A1 | cso; accept equivalents like 1.25 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses completion of the square | dM1 | Depends on previous M mark |
| Uses \(k = 8 - \lambda\) | dM1 | Depends on previous M mark |
| \(k = \dfrac{5}{4}\) | A1 | cso; accept equivalents like 1.25 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equate \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3\) | M1 | Not given just for derivative |
| Solves to get \(x = 4.5\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitutes \(x = 4.5\) into equation for \(C\) to give \(y\)-coordinate | dM1 | |
| Substitutes both \(x\) and \(y\) into \(y = 3x + k\) to find \(k\) | dM1 | |
| \(k = \dfrac{5}{4}\) | A1 | cso; accept equivalents like 1.25 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitutes \(x = 4.5\) into \(\frac{1}{3}x^2 + 8 = 3x + k\) | dM1 | |
| Finds \(k\) | dM1 | |
| \(k = \dfrac{5}{4}\) | A1 | cso; accept equivalents like 1.25 |
## Question 9:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| U-shaped parabola symmetric about $y$-axis | B1 | |
| Graph passes through $(0, 8)$ | B1 | |
| Shape and position for $L$ | M1 | |
| Both $\left(-\frac{k}{3}, 0\right)$ and $(0, k)$ | A1 | Allow marks even if on the same diagram |
**Part (b) – Method 1:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equate $\frac{1}{3}x^2 + 8 = 3x + k$ and collect terms on one side | M1 | |
| $\frac{1}{3}x^2 - 3x + (8-k)$ | A1 | |
| Method 1a: Uses $b^2=4ac$: $9 = 4\times\frac{1}{3}\times(8-k) \Rightarrow k=...$ | dM1, dM1 | |
| Method 1b: Attempt $\frac{1}{3}(x-\frac{9}{2})^2 - \lambda + 8 - k$; deduce $k = 8 - \lambda$ | dM1, dM1 | |
| $k = \frac{5}{4}$ oe | A1 | |
**Part (b) – Method 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to set $\frac{dy}{dx} = 3$ | M1 | |
| $\frac{2}{3}x = 3 \Rightarrow x = 4.5$ | A1 | |
| Method 2a: Substitutes $x=4.5$ into $y=\frac{1}{3}x^2+8 \Rightarrow y=14.75$; substitutes both $x$ and $y$ into $y=3x+k$ to find $k$ | dM1, dM1 | |
| Method 2b: Substitutes $x=4.5$ into $\frac{1}{3}x^2+8=3x+k$; finds $k=$ | dM1, dM1 | |
| $k = 1.25$ oe | A1 | |
# Question (a) [Curve C and Line L]:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Shape for C: $\cup$ approximately symmetrical about $y$-axis | B1 | |
| Coordinates of $(0, 8)$ | B1 | There must be a graph; accept graph crossing positive $y$-axis with only 8 marked; accept $(8,0)$ if given on $y$-axis |
| Shape for $L$: straight line with positive gradient and positive intercept | M1 | |
| Coordinates of $(0, k)$ and $(-k/3, 0)$, or $k$ marked on $y$-axis and $-k/3$ marked on $x$-axis | A1 | Accept $(k, 0)$ on $y$-axis and $(0, -k/3)$ on $x$-axis |
---
# Question (b) [Finding k]:
**Method 1a:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equate curves $\frac{1}{3}x^2 + 8 = 3x + k$ and collect $x$ terms and $(8-k)$ terms to one side | M1 | |
| Achieves e.g. $\frac{1}{3}x^2 - 3x + (8-k)$ | A1 | |
| Uses $b^2 = 4ac$ or $b^2 - 4ac = 0$ | dM1 | Depends on previous M mark |
| Solves their $b^2 = 4ac$ leading to $k =$ ... | dM1 | Depends on previous M mark |
| $k = \dfrac{5}{4}$ | A1 | cso; accept equivalents like 1.25 |
**Method 1b:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses completion of the square | dM1 | Depends on previous M mark |
| Uses $k = 8 - \lambda$ | dM1 | Depends on previous M mark |
| $k = \dfrac{5}{4}$ | A1 | cso; accept equivalents like 1.25 |
**Method 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equate $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3$ | M1 | Not given just for derivative |
| Solves to get $x = 4.5$ | A1 | |
**Method 2a:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $x = 4.5$ into equation for $C$ to give $y$-coordinate | dM1 | |
| Substitutes both $x$ and $y$ into $y = 3x + k$ to find $k$ | dM1 | |
| $k = \dfrac{5}{4}$ | A1 | cso; accept equivalents like 1.25 |
**Method 2b:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $x = 4.5$ into $\frac{1}{3}x^2 + 8 = 3x + k$ | dM1 | |
| Finds $k$ | dM1 | |
| $k = \dfrac{5}{4}$ | A1 | cso; accept equivalents like 1.25 |
---9. The curve $C$ has equation $y = \frac { 1 } { 3 } x ^ { 2 } + 8$
The line $L$ has equation $y = 3 x + k$, where $k$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Sketch $C$ and $L$ on separate diagrams, showing the coordinates of the points at which $C$ and $L$ cut the axes.
Given that line $L$ is a tangent to $C$,
\item find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2014 Q9 [9]}}