| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Sketch rational with quadratic numerator |
| Difficulty | Standard +0.3 This is a standard Further Maths curve sketching question requiring asymptote identification (vertical at x=-1/2, oblique via polynomial division), stationary points via quotient rule, and a sketch. While it involves multiple techniques, each step follows routine procedures with no novel insight required, 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^21.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = -\frac{1}{2}\) | B1 | Vertical asymptote |
| \(y = \frac{(2x+1)(\frac{1}{2}x - \frac{1}{4}) + \frac{1}{4}}{2x+1} = \frac{1}{2}x - \frac{1}{4} + \frac{1}{4(2x+1)}\) | M1 | Polynomial division method |
| \(y = \frac{1}{2}x - \frac{1}{4}\) | A1 | Oblique asymptote |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{1}{2} - \frac{1}{2(2x+1)^2} = 0 \Rightarrow (2x+1)^2 = 1\) | M1 | Differentiating and setting to zero |
| \(x = 0, -1\) | A1 | |
| \((0, 0)\), \((-1, -1)\) | A1 | Correct coordinates of both stationary points |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Axes and asymptotes correct | B1 | |
| Upper branch correct | B1 | |
| Lower branch correct | B1 |
## Question 3:
### Part 3(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = -\frac{1}{2}$ | B1 | Vertical asymptote |
| $y = \frac{(2x+1)(\frac{1}{2}x - \frac{1}{4}) + \frac{1}{4}}{2x+1} = \frac{1}{2}x - \frac{1}{4} + \frac{1}{4(2x+1)}$ | M1 | Polynomial division method |
| $y = \frac{1}{2}x - \frac{1}{4}$ | A1 | Oblique asymptote |
### Part 3(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = \frac{1}{2} - \frac{1}{2(2x+1)^2} = 0 \Rightarrow (2x+1)^2 = 1$ | M1 | Differentiating and setting to zero |
| $x = 0, -1$ | A1 | |
| $(0, 0)$, $(-1, -1)$ | A1 | Correct coordinates of both stationary points |
### Part 3(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Axes and asymptotes correct | B1 | |
| Upper branch correct | B1 | |
| Lower branch correct | B1 | |
---3 The curve $C$ has equation $\mathrm { y } = \frac { \mathrm { x } ^ { 2 } } { 2 \mathrm { x } + 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the equations of the asymptotes of $C$.
\item Find the coordinates of the stationary points on $C$.
\item Sketch $C$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2020 Q3 [9]}}