| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2009 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Rational curve intersections |
| Difficulty | Moderate -0.3 This is a straightforward C1 question requiring substitution to find a coordinate, sketching two standard curves (a cubic and rectangular hyperbola), and counting intersections graphically. While it involves multiple parts and curve sketching, all techniques are routine for C1 with no problem-solving insight needed—slightly easier than average due to the guided structure. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02n 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 | Marks | Guidance |
| \((a =)\ (1+1)^2(2-1) = \underline{4}\); \((1, 4)\) or \(y = 4\) also acceptable | B1 | (1 mark total) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Shape \(\bigvee\) or \(\bigwedge\) anywhere | B1 | Must have one max and one min, no further turning points |
| Minimum at \((-1, 0)\); can be \(-1\) on \(x\)-axis; allow \((0,-1)\) if marked on \(x\)-axis | B1 | Marked in correct place; but 1 is B0 |
| \((2, 0)\) and \((0, 2)\) can be 2 on axes | B1 | |
| Top branch in 1st quadrant with 2 intersections | B1 | Branch fully within 1st quadrant having 2 intersections (not just touching) |
| Bottom branch in 3rd quadrant | B1 | Ignore any intersections for this branch (5 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| (2 intersections therefore) \(\underline{2}\) (roots) | B1ft | For statement about number of roots compatible with sketch; no sketch is B0; answer 2 incompatible with sketch is B0 (1 mark total) |
# Question 8:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(a =)\ (1+1)^2(2-1) = \underline{4}$; $(1, 4)$ or $y = 4$ also acceptable | B1 | (1 mark total) |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Shape $\bigvee$ or $\bigwedge$ anywhere | B1 | Must have one max and one min, no further turning points |
| Minimum at $(-1, 0)$; can be $-1$ on $x$-axis; allow $(0,-1)$ if marked on $x$-axis | B1 | Marked in correct place; but 1 is B0 |
| $(2, 0)$ and $(0, 2)$ can be 2 on axes | B1 | |
| Top branch in 1st quadrant with 2 intersections | B1 | Branch fully within 1st quadrant having 2 intersections (not just touching) |
| Bottom branch in 3rd quadrant | B1 | Ignore any intersections for this branch (5 marks total) |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| (2 intersections therefore) $\underline{2}$ (roots) | B1ft | For statement about number of roots compatible with sketch; no sketch is B0; answer 2 incompatible with sketch is B0 (1 mark total) |
---8. The point $P ( 1 , a )$ lies on the curve with equation $y = ( x + 1 ) ^ { 2 } ( 2 - x )$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$.
\item On the axes below sketch the curves with the following equations:
\begin{enumerate}[label=(\roman*)]
\item $y = ( x + 1 ) ^ { 2 } ( 2 - x )$,
\item $y = \frac { 2 } { x }$.
On your diagram show clearly the coordinates of any points at which the curves meet the axes.
\end{enumerate}\item With reference to your diagram in part (b) state the number of real solutions to the equation
$$( x + 1 ) ^ { 2 } ( 2 - x ) = \frac { 2 } { x } .$$
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{871f5957-180d-4379-88ce-186432f57bad-10_1347_1344_1245_297}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2009 Q8 [7]}}