| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Sketch |f(x)| only |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring polynomial division for asymptotes, algebraic manipulation to prove a range restriction, and sketching both the rational function and its modulus. While individual parts use standard techniques, the combination of proving the range gap and accurately sketching |f(x)| with discontinuities requires careful analysis beyond routine A-level work. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02l Modulus function: notation, relations, equations and inequalities1.02n Sketch curves: simple equations including polynomials1.02s Modulus graphs: sketch graph of |ax+b| |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = 1\) | B1 | States vertical asymptote |
| \(x^2 + x - 1 = (x-1)(x+2) + 1 \Rightarrow y = x + 2\) | M1 A1 | Finds oblique asymptote |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(yx - y = x^2 + x - 1 \Rightarrow x^2 + (1-y)x + y - 1 = 0\) | M1 A1 | Forms quadratic in \(x\) |
| \((1-y)^2 - 4(y-1) < 0 \Rightarrow y^2 - 6y + 5 < 0\) | M1 | Uses that discriminant is negative if there are no values of \(x\) |
| \(1 < y < 5\) | A1 | AG |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| [Graph with axes and asymptotes shown] | B1 | Axes and asymptotes |
| [Branches correct] | B1 | Branches correct |
| \((0,1)\), \(\left(-\frac{1}{2}+\frac{1}{2}\sqrt{5}, 0\right)\), \(\left(-\frac{1}{2}-\frac{1}{2}\sqrt{5}, 0\right)\) | B1 | States coordinates of intersections with axes; accept \((0.618, 0)\) and \((-1.62, 0)\) |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| [Graph showing \(\ | y\ | \) transformation with correct shape] |
| B1 | Correct shape at extremities | |
| Total: 2 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = 1$ | B1 | States vertical asymptote |
| $x^2 + x - 1 = (x-1)(x+2) + 1 \Rightarrow y = x + 2$ | M1 A1 | Finds oblique asymptote |
| **Total: 3** | | |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $yx - y = x^2 + x - 1 \Rightarrow x^2 + (1-y)x + y - 1 = 0$ | M1 A1 | Forms quadratic in $x$ |
| $(1-y)^2 - 4(y-1) < 0 \Rightarrow y^2 - 6y + 5 < 0$ | M1 | Uses that discriminant is negative if there are no values of $x$ |
| $1 < y < 5$ | A1 | AG |
| **Total: 4** | | |
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| [Graph with axes and asymptotes shown] | B1 | Axes and asymptotes |
| [Branches correct] | B1 | Branches correct |
| $(0,1)$, $\left(-\frac{1}{2}+\frac{1}{2}\sqrt{5}, 0\right)$, $\left(-\frac{1}{2}-\frac{1}{2}\sqrt{5}, 0\right)$ | B1 | States coordinates of intersections with axes; accept $(0.618, 0)$ and $(-1.62, 0)$ |
| **Total: 3** | | |
## Question 6(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| [Graph showing $\|y\|$ transformation with correct shape] | B1 FT | FT from sketch in (c) with both branches |
| | B1 | Correct shape at extremities |
| **Total: 2** | | |6 The curve $C$ has equation $\mathrm { y } = \frac { \mathrm { x } ^ { 2 } + \mathrm { x } - 1 } { \mathrm { x } - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the equations of the asymptotes of $C$.
\item Show that there is no point on $C$ for which $1 < y < 5$.
\item Find the coordinates of the intersections of $C$ with the axes, and sketch $C$.
\item Sketch the curve with equation $\mathrm { y } = \left| \frac { \mathrm { x } ^ { 2 } + \mathrm { x } - 1 } { \mathrm { x } - 1 } \right|$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2020 Q6 [12]}}