| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Solve |f(x)| > k using sketch |
| Difficulty | Challenging +1.2 This is a comprehensive curve sketching question requiring multiple techniques (asymptotes, stationary points, modulus transformations, and inequality solving), but each individual step follows standard A-level procedures. The algebraic manipulation is routine, and the modulus inequality solving, while requiring careful case analysis, is a standard Further Maths technique. More demanding than a typical pure maths question due to the multi-part nature and Further Maths content, but not requiring exceptional insight. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02l Modulus function: notation, relations, equations and inequalities1.02n Sketch curves: simple equations including polynomials1.02o Sketch reciprocal curves: y=a/x and y=a/x^21.02s Modulus graphs: sketch graph of |ax+b|1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = \frac{3}{2}\) | B1 | |
| \(-2x^2+x+10 = (2x-3)(-x-1)+7 \Rightarrow y = -x-1\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(2x-3)(1-4x)-2(10+x-2x^2)}{(2x-3)^2}\) | M1 | |
| \(4x^2-12x+23=0 \quad \left(\text{or } \frac{\mathrm{d}y}{\mathrm{d}x} = -1-\frac{14}{(2x-3)^2}\right)\) | A1 | |
| \(12^2-4(4)(23) = -224 < 0\) (or \(y'<0\)) \(\Rightarrow\) No turning points | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Sketch with correct axes and asymptotes | B1 | |
| Correct branches | B1 | |
| \((0,-\frac{10}{3}), (-2,0), (\frac{5}{2},0)\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Sketch showing correct shape from part (c) | B1 FT | Follow through from their sketch in (c) |
| Correct shape at infinity (two outer branches going to same direction) | B1 | Correct shape at infinity |
| \(\frac{10+x-2x^2}{2x-3} = 4\) or \(\frac{10+x-2x^2}{2x-3} = -4\) | M2 | Setting equal to \(\pm 4\) |
| \(2x^2+7x-22=0\) or \(2x^2-9x+2=0\) | ||
| \(x = -\frac{11}{2}, 2\) or \(x = \frac{1}{4}(9 \pm \sqrt{65})\) | A1 | |
| \(-\frac{11}{2} < x < \frac{1}{4}(9-\sqrt{65})\) and \(2 < x < \frac{1}{4}(9+\sqrt{65})\) | A1 FT |
## Question 6:
**Part 6(a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = \frac{3}{2}$ | B1 | |
| $-2x^2+x+10 = (2x-3)(-x-1)+7 \Rightarrow y = -x-1$ | M1 A1 | |
**Part 6(b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(2x-3)(1-4x)-2(10+x-2x^2)}{(2x-3)^2}$ | M1 | |
| $4x^2-12x+23=0 \quad \left(\text{or } \frac{\mathrm{d}y}{\mathrm{d}x} = -1-\frac{14}{(2x-3)^2}\right)$ | A1 | |
| $12^2-4(4)(23) = -224 < 0$ (or $y'<0$) $\Rightarrow$ No turning points | A1 | |
**Part 6(c):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Sketch with correct axes and asymptotes | B1 | |
| Correct branches | B1 | |
| $(0,-\frac{10}{3}), (-2,0), (\frac{5}{2},0)$ | B1 | |
## Question 6(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Sketch showing correct shape from part (c) | **B1 FT** | Follow through from their sketch in (c) |
| Correct shape at infinity (two outer branches going to same direction) | **B1** | Correct shape at infinity |
| $\frac{10+x-2x^2}{2x-3} = 4$ or $\frac{10+x-2x^2}{2x-3} = -4$ | **M2** | Setting equal to $\pm 4$ |
| $2x^2+7x-22=0$ or $2x^2-9x+2=0$ | | |
| $x = -\frac{11}{2}, 2$ or $x = \frac{1}{4}(9 \pm \sqrt{65})$ | **A1** | |
| $-\frac{11}{2} < x < \frac{1}{4}(9-\sqrt{65})$ and $2 < x < \frac{1}{4}(9+\sqrt{65})$ | **A1 FT** | |
---6 The curve $C$ has equation $\mathrm { y } = \frac { 10 + \mathrm { x } - 2 \mathrm { x } ^ { 2 } } { 2 \mathrm { x } - 3 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the equations of the asymptotes of $C$.
\item Show that $C$ has no turning points.
\item Sketch $C$, stating the coordinates of the intersections with the axes.
\item Sketch the curve with equation $y = \left| \frac { 10 + x - 2 x ^ { 2 } } { 2 x - 3 } \right|$ and find in exact form the set of values of $x$ for which $\left| \frac { 10 + x - 2 x ^ { 2 } } { 2 x - 3 } \right| < 4$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2020 Q6 [15]}}