| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Multiple transformations including squared |
| Difficulty | Challenging +1.2 This is a comprehensive curve sketching question requiring multiple techniques: finding asymptotes (vertical and oblique via polynomial division), stationary points via quotient rule, and transformations involving modulus and y². While it has many parts and requires careful execution, each individual step uses standard A-level techniques without requiring novel insight. The transformations in part (d) are routine applications of reflection principles taught in Further Maths, making this moderately above average but not exceptionally challenging. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02m Graphs of functions: difference between plotting and sketching1.02n Sketch curves: simple equations including polynomials1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = 3\) | B1 | States vertical asymptote |
| \(y = \dfrac{(x-3)(x+5)+16}{x-3} = x + 5 + \dfrac{16}{x-3}\) | M1 | Finds oblique asymptote |
| \(y = x + 5\) | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{dy}{dx} = 1 - \dfrac{16}{(x-3)^2} = 0 \Rightarrow (x-3)^2 = 16\) | M1 | Differentiates and sets equal to zero |
| \(x = -1, 7\) | A1 | Finds \(x\)-coordinates |
| \((-1, 0),\ (7, 16)\) | A1 | States coordinates of turning points |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Sketch with axes and labelled asymptotes | B1FT | Axes and labelled asymptotes |
| Upper branch correct | B1 | Upper branch correct |
| Lower branch correct and no additional branches | B1 | Lower branch correct and no additional branches |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Clear labels, axes and vertical asymptote | B1FT | Clear labels, axes and their vertical asymptote |
| \(y = \left\lvert\dfrac{x^2+2x+1}{x-3}\right\rvert\) correct | B1FT | \(y = \left\lvert\dfrac{x^2+2x+1}{x-3}\right\rvert\) correct, FT from their sketch in (c) |
| Upper branch of \(y^2 = \dfrac{x^2+2x+1}{x-3}\) (positive square root) | B1 | Upper branch of \(y^2 = \dfrac{x^2+2x+1}{x-3}\) |
| Lower branch of \(y^2 = \dfrac{x^2+2x+1}{x-3}\) (negative square root), FT from previous mark | B1FT | Lower branch of \(y^2 = \dfrac{x^2+2x+1}{x-3}\), FT from previous mark |
| Total: 4 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = 3$ | B1 | States vertical asymptote |
| $y = \dfrac{(x-3)(x+5)+16}{x-3} = x + 5 + \dfrac{16}{x-3}$ | M1 | Finds oblique asymptote |
| $y = x + 5$ | A1 | |
| **Total: 3** | | |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{dy}{dx} = 1 - \dfrac{16}{(x-3)^2} = 0 \Rightarrow (x-3)^2 = 16$ | M1 | Differentiates and sets equal to zero |
| $x = -1, 7$ | A1 | Finds $x$-coordinates |
| $(-1, 0),\ (7, 16)$ | A1 | States coordinates of turning points |
| **Total: 3** | | |
## Question 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Sketch with axes and labelled asymptotes | B1FT | Axes and labelled asymptotes |
| Upper branch correct | B1 | Upper branch correct |
| Lower branch correct and no additional branches | B1 | Lower branch correct and no additional branches |
| **Total: 3** | | |
## Question 7(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Clear labels, axes and vertical asymptote | B1FT | Clear labels, axes and their vertical asymptote |
| $y = \left\lvert\dfrac{x^2+2x+1}{x-3}\right\rvert$ correct | B1FT | $y = \left\lvert\dfrac{x^2+2x+1}{x-3}\right\rvert$ correct, FT from their sketch in **(c)** |
| Upper branch of $y^2 = \dfrac{x^2+2x+1}{x-3}$ (positive square root) | B1 | Upper branch of $y^2 = \dfrac{x^2+2x+1}{x-3}$ |
| Lower branch of $y^2 = \dfrac{x^2+2x+1}{x-3}$ (negative square root), FT from previous mark | B1FT | Lower branch of $y^2 = \dfrac{x^2+2x+1}{x-3}$, FT from previous mark |
| **Total: 4** | | |7 The curve $C$ has equation $\mathrm { y } = \frac { \mathrm { x } ^ { 2 } + 2 \mathrm { x } + 1 } { \mathrm { x } - 3 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the equations of the asymptotes of $C$.
\item Find the coordinates of the turning points on $C$.
\item Sketch $C$.
\item Sketch the curves with equations $y = \left| \frac { x ^ { 2 } + 2 x + 1 } { x - 3 } \right|$ and $y ^ { 2 } = \frac { x ^ { 2 } + 2 x + 1 } { x - 3 }$ on a single diagram, clearly identifying each curve.
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2023 Q7 [13]}}