| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Rational curve intersections |
| Difficulty | Standard +0.3 This is a standard P1 question combining rational function sketching with intersection conditions. Part (a) requires routine asymptote identification and axis intercepts. Part (b) involves substituting the line into the curve equation and applying the discriminant condition (b²-4ac>0), which is a well-practiced technique. Part (c) requires solving a quadratic inequality, also standard. While multi-part, each step follows predictable A-level methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02g Inequalities: linear and quadratic in single variable1.02n Sketch curves: simple equations including polynomials1.02o Sketch reciprocal curves: y=a/x and y=a/x^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Negative reciprocal shape (top left/bottom right) | B1 | No clear vertical or horizontal overlaps. Condone pen slips if no clear turning point |
| Intercept at \(\left(0, \frac{1}{2}\right)\) | B1 | Allow \(\frac{1}{2}\) marked on \(y\)-axis. Condone \(\left(\frac{1}{2}, 0\right)\) if in correct position. Cannot award without sketch |
| \(x=2\), \(y=0\) asymptotes | B1 | Must state both. Do not allow "\(x\)-axis" for \(y=0\). Cannot award without sketch |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4x+k = \frac{1}{2-x} \Rightarrow (4x+k)(2-x)=1 \Rightarrow 8x+2k-4x^2-kx-1(=0)\) | M1 | Sets line = curve, multiplies by \((2-x)\), expands and collects terms. Condone absence of \(=0\) |
| \(4x^2+(k-8)x+1-2k=0\); \(a=4, b=k-8, c=1-2k\) or \(a=-4, b=8-k, c=2k-1\) | A1 | Correct equation unsimplified with all terms on one side. May be implied by correct \(a,b,c\) substituted into discriminant |
| \((k-8)^2 - 4\times4(1-2k) > 0\) | M1 | Attempts discriminant using their \(a,b,c\). Condone invisible brackets and ignore inequality sign for this mark |
| \(k^2-16k+64-16+32k > 0 \Rightarrow k^2+16k+48>0\) | A1* | Correctly expands, sets \(>0\), reaches given answer with no errors. Must have used correct inequality or stated \(b^2-4ac>0\) somewhere |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(k^2+16k+48=0 \Rightarrow (k+12)(k+4)=0 \Rightarrow k=\ldots\) | M1 | Solves given quadratic to obtain 2 values of \(k\) |
| \(k=-12,\ -4\) | A1 | Both values correct |
| \(k < -12\) or \(k > -4\) | M1 | Attempts outside regions. \(\leq\) or \(\geq\) condoned for this mark. Cannot be scored from diagram alone |
| \(k < -12\) or \(k > -4\) | A1 | Must be in terms of \(k\). Accept set notation equivalents. Do NOT allow '\(k<-12\) and \(k>-4\)' or '\(-12>k>-4\)' |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Negative reciprocal shape (top left/bottom right) | B1 | No clear vertical or horizontal overlaps. Condone pen slips if no clear turning point |
| Intercept at $\left(0, \frac{1}{2}\right)$ | B1 | Allow $\frac{1}{2}$ marked on $y$-axis. Condone $\left(\frac{1}{2}, 0\right)$ if in correct position. Cannot award without sketch |
| $x=2$, $y=0$ asymptotes | B1 | Must state both. Do not allow "$x$-axis" for $y=0$. Cannot award without sketch |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4x+k = \frac{1}{2-x} \Rightarrow (4x+k)(2-x)=1 \Rightarrow 8x+2k-4x^2-kx-1(=0)$ | M1 | Sets line = curve, multiplies by $(2-x)$, expands and collects terms. Condone absence of $=0$ |
| $4x^2+(k-8)x+1-2k=0$; $a=4, b=k-8, c=1-2k$ or $a=-4, b=8-k, c=2k-1$ | A1 | Correct equation unsimplified with all terms on one side. May be implied by correct $a,b,c$ substituted into discriminant |
| $(k-8)^2 - 4\times4(1-2k) > 0$ | M1 | Attempts discriminant using their $a,b,c$. Condone invisible brackets and ignore inequality sign for this mark |
| $k^2-16k+64-16+32k > 0 \Rightarrow k^2+16k+48>0$ | A1* | Correctly expands, sets $>0$, reaches given answer with no errors. Must have used correct inequality or stated $b^2-4ac>0$ somewhere |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k^2+16k+48=0 \Rightarrow (k+12)(k+4)=0 \Rightarrow k=\ldots$ | M1 | Solves given quadratic to obtain 2 values of $k$ |
| $k=-12,\ -4$ | A1 | Both values correct |
| $k < -12$ or $k > -4$ | M1 | Attempts outside regions. $\leq$ or $\geq$ condoned for this mark. Cannot be scored from diagram alone |
| $k < -12$ or $k > -4$ | A1 | Must be in terms of $k$. Accept set notation equivalents. Do NOT allow '$k<-12$ and $k>-4$' or '$-12>k>-4$' |
---7. The curve $C$ has equation
$$y = \frac { 1 } { 2 - x }$$
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of $C$. On your sketch you should show the coordinates of any points of intersection with the coordinate axes and state clearly the equations of any asymptotes.
The line $l$ has equation $y = 4 x + k$, where $k$ is a constant.
Given that $l$ meets $C$ at two distinct points,
\item show that
$$k ^ { 2 } + 16 k + 48 > 0$$
\item Hence find the range of possible values for $k$.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2020 Q7 [11]}}