| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simultaneous equations |
| Type | Tangency condition for line and curve |
| Difficulty | Standard +0.3 This is a standard AS-level tangency problem requiring substitution of the line equation into the curve, rearranging to a quadratic, and applying the discriminant condition (b² - 4ac = 0) for tangency. The algebra is straightforward with no conceptual surprises, making it slightly easier than average. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.02o Sketch reciprocal curves: y=a/x and y=a/x^21.02q Use intersection points: of graphs to solve equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{x}\) shape in 1st quadrant | M1 | 1.1b – Must not cross either axis; acceptable curvature; negative gradient changing from \(-\infty\) to 0; condone "slips of the pencil" |
| Correct (both branches in quadrants 1, 2 and 3 with correct curvature including asymptotic behaviour) | A1 | 1.1b |
| Asymptote \(y = 1\) | B1 | 1.2 – May appear on diagram or in text; must be only horizontal asymptote offered |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Combines equations \(\Rightarrow \frac{k^2}{x} + 1 = -2x + 5\) | M1 | 1.1b – Attempts to combine \(y = \frac{k^2}{x} + 1\) with \(y = -2x+5\) to form equation in just \(x\) |
| \((\times x) \Rightarrow k^2 + 1x = -2x^2 + 5x \Rightarrow 2x^2 - 4x + k^2 = 0\) * | A1* | 2.1 – Multiplies by \(x\) (processed line must be seen); no slips; condone different ordering e.g. \(2x^2 + k^2 - 4x = 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Attempts to set \(b^2 - 4ac = 0\) | M1 | 3.1a – For the given equation; if \(a,b,c\) stated, only accept \(a=2, b=\pm4, c=k^2\), so \(4^2 - 4 \times 2 \times k^2 = 0\); alternatively completes square: \((x-1)^2 = 1 - \frac{1}{2}k^2 \Rightarrow\) "\(1 - \frac{1}{2}k^2\)"\(= 0\) |
| \(8k^2 = 16\) | A1 | 1.1b – Or exact simplified equivalent e.g. \(8k^2 - 16 = 0\) |
| \(k = \pm\sqrt{2}\) | A1 | 1.1b – Following correct \(a\), \(b\) and \(c\) if stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Sets gradient of curve \(= -2 \Rightarrow -\frac{k^2}{x^2} = -2 \Rightarrow x = (\pm)\frac{k}{\sqrt{2}}\) and attempts to substitute into \(2x^2 - 4x + k^2 = 0\) | M1 | |
| \(2k^2 = (\pm)2\sqrt{2}k\) | A1 | |
| \(k = \pm\sqrt{2}\) | A1 |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{x}$ shape in 1st quadrant | M1 | 1.1b – Must not cross either axis; acceptable curvature; negative gradient changing from $-\infty$ to 0; condone "slips of the pencil" |
| Correct (both branches in quadrants 1, 2 and 3 with correct curvature including asymptotic behaviour) | A1 | 1.1b |
| Asymptote $y = 1$ | B1 | 1.2 – May appear on diagram or in text; must be only horizontal asymptote offered |
**(3 marks)**
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Combines equations $\Rightarrow \frac{k^2}{x} + 1 = -2x + 5$ | M1 | 1.1b – Attempts to combine $y = \frac{k^2}{x} + 1$ with $y = -2x+5$ to form equation in just $x$ |
| $(\times x) \Rightarrow k^2 + 1x = -2x^2 + 5x \Rightarrow 2x^2 - 4x + k^2 = 0$ * | A1* | 2.1 – Multiplies by $x$ (processed line must be seen); no slips; condone different ordering e.g. $2x^2 + k^2 - 4x = 0$ |
**(2 marks)**
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempts to set $b^2 - 4ac = 0$ | M1 | 3.1a – For the given equation; if $a,b,c$ stated, only accept $a=2, b=\pm4, c=k^2$, so $4^2 - 4 \times 2 \times k^2 = 0$; alternatively completes square: $(x-1)^2 = 1 - \frac{1}{2}k^2 \Rightarrow$ "$1 - \frac{1}{2}k^2$"$= 0$ |
| $8k^2 = 16$ | A1 | 1.1b – Or exact simplified equivalent e.g. $8k^2 - 16 = 0$ |
| $k = \pm\sqrt{2}$ | A1 | 1.1b – Following correct $a$, $b$ and $c$ if stated |
**(3 marks)**
**Alternative via differentiation:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sets gradient of curve $= -2 \Rightarrow -\frac{k^2}{x^2} = -2 \Rightarrow x = (\pm)\frac{k}{\sqrt{2}}$ and attempts to substitute into $2x^2 - 4x + k^2 = 0$ | M1 | |
| $2k^2 = (\pm)2\sqrt{2}k$ | A1 | |
| $k = \pm\sqrt{2}$ | A1 | |\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$y = \frac { k ^ { 2 } } { x } + 1 \quad x \in \mathbb { R } , x \neq 0$$
where $k$ is a constant.\\
(a) Sketch $C$ stating the equation of the horizontal asymptote.
The line $l$ has equation $y = - 2 x + 5$\\
(b) Show that the $x$ coordinate of any point of intersection of $l$ with $C$ is given by a solution of the equation
$$2 x ^ { 2 } - 4 x + k ^ { 2 } = 0$$
(c) Hence find the exact values of $k$ for which $l$ is a tangent to $C$.
\hfill \mbox{\textit{Edexcel AS Paper 1 2019 Q7 [8]}}