| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Rational functions with parameters: analysis depending on parameter sign/range |
| Difficulty | Challenging +1.2 This is a Further Maths question on rational functions requiring asymptote identification (standard technique), proving a range exclusion (requires rearranging to quadratic discriminant analysis), and sketching. While multi-step, the techniques are systematic and well-practiced in Further Maths courses, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 1.02n Sketch curves: simple equations including polynomials |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = 1\) | B1 | States vertical asymptote |
| \(y = \frac{(x-1)(ax+a+1)+a}{x-1} = ax+a+1+\frac{a}{x-1}\) | M1 | Finds oblique asymptote |
| \(y = ax + a + 1\) | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(ax^2 + x - 1 = yx - y\) leading to \(ax^2 - (y-1)x + (y-1) = 0\) | M1 A1 | Forms a three term quadratic in \(x\) |
| \((y-1)^2 - 4a(y-1) < 0\) | M1 | Correct inequality using discriminant |
| \(1 < y < 1 + 4a\) | A1 | AG. Clear method to reach given answer e.g. \((y-1)(y-1-4a) < 0\) |
| Alternative: \((y-1)^2 - 4a(y-1) \geq 0\) | M1 | Correct inequality using discriminant. Must be clear looking for where values of \(y\) exist |
| \(1 < y < 1+4a\) | A1 | AG. Clear method e.g. \((y-1)(y-1-4a) \geq 0\) |
| Second Alternative: \(\frac{dy}{dx} = \frac{(x-1)(2ax+1)-(ax^2+x-1)}{(x-1)^2} = 0\) | M1 | |
| \(x = 0\) or \(x = 2\) | A1 | |
| Proves \((0,1)\) is local maximum and \((2, 4a+1)\) is local minimum | M1 | Either \(\frac{d^2y}{dx^2}\) or gradient either side |
| Reference to position of asymptote or two distinct branches to justify \(1 < y < 1+4a\) | A1 | AG |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Asymptotes identified, FT from part 3(a) | B1 FT | Asymptotes identified. FT from part 3a |
| Two branches in correct position relative to correct asymptotes and of correct shape. Good asymptotic behaviour. | B1 | Two branches in correct position relative to correct asymptotes |
| Lower branch in all four quadrants from correct asymptotes | B1 | Lower branch in all four quadrants from correct asymptotes |
| Total: 3 |
## Question 3:
**Part 3(a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = 1$ | B1 | States vertical asymptote |
| $y = \frac{(x-1)(ax+a+1)+a}{x-1} = ax+a+1+\frac{a}{x-1}$ | M1 | Finds oblique asymptote |
| $y = ax + a + 1$ | A1 | |
| **Total: 3** | | |
**Part 3(b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $ax^2 + x - 1 = yx - y$ leading to $ax^2 - (y-1)x + (y-1) = 0$ | M1 A1 | Forms a three term quadratic in $x$ |
| $(y-1)^2 - 4a(y-1) < 0$ | M1 | Correct inequality using discriminant |
| $1 < y < 1 + 4a$ | A1 | AG. Clear method to reach given answer e.g. $(y-1)(y-1-4a) < 0$ |
| **Alternative:** $(y-1)^2 - 4a(y-1) \geq 0$ | M1 | Correct inequality using discriminant. Must be clear looking for where values of $y$ exist |
| $1 < y < 1+4a$ | A1 | AG. Clear method e.g. $(y-1)(y-1-4a) \geq 0$ |
| **Second Alternative:** $\frac{dy}{dx} = \frac{(x-1)(2ax+1)-(ax^2+x-1)}{(x-1)^2} = 0$ | M1 | |
| $x = 0$ or $x = 2$ | A1 | |
| Proves $(0,1)$ is local maximum and $(2, 4a+1)$ is local minimum | M1 | Either $\frac{d^2y}{dx^2}$ or gradient either side |
| Reference to position of asymptote or two distinct branches to justify $1 < y < 1+4a$ | A1 | AG |
| **Total: 4** | | |
**Part 3(c):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Asymptotes identified, FT from part 3(a) | B1 FT | Asymptotes identified. FT from part **3a** |
| Two branches in correct position relative to correct asymptotes and of correct shape. Good asymptotic behaviour. | B1 | Two branches in correct position relative to correct asymptotes |
| Lower branch in all four quadrants from correct asymptotes | B1 | Lower branch in all four quadrants from correct asymptotes |
| **Total: 3** | | |3 A curve $C$ has equation $\mathrm { y } = \frac { \mathrm { a } \mathrm { x } ^ { 2 } + \mathrm { x } - 1 } { \mathrm { x } - 1 }$, where $a$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Find the equations of the asymptotes of $C$.
\item Show that there is no point on $C$ for which $1 < \mathrm { y } < 1 + 4 \mathrm { a }$.
\item Sketch C. You do not need to find the coordinates of the intersections with the axes.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2022 Q3 [10]}}