| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polynomial Division & Manipulation |
| Type | Stationary Points of Rational Functions |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths question requiring quotient rule differentiation, discriminant analysis for stationary points, asymptote identification, and curve sketching of rational functions. While it involves several techniques and is from FP1 (inherently harder), each part follows standard procedures without requiring novel insight—the discriminant inequality and asymptote work are routine for Further Maths students. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02n Sketch curves: simple equations including polynomials1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| \((\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta\) | B1 | |
| \(\cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta\) | M1A1 | If line 1 missing award B0M1A1M1M1A0 i.e. 4/6 |
| \(\sin 3\theta = 3\cos^2\theta\sin\theta - \sin^3\theta\), \(\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta\) | M1 | |
| \(\tan 3\theta = \frac{3\cos^2\theta\sin\theta - \sin^3\theta}{\cos^3\theta - 3\cos\theta\sin^2\theta}\) | M1 | |
| \(\therefore \tan 3\theta = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}\) (*) (AG) | A1 | Part mark: 6 |
| \(\frac{\pi}{12}, \frac{5\pi}{12}, \frac{9\pi}{12} = \frac{3\pi}{4}\) | B1 (one), B1 (all) | Part mark: 2 |
| Put \(\tan 3\theta = 1\) in (*) \(\Rightarrow t^3 - 3t^2 - 3t + 1 = 0\) | M1 | |
| Roots are \(\tan\frac{\pi}{12}, \tan\frac{5\pi}{12}, \tan\frac{3\pi}{4}\) | A1 (one), A1 (all) | Part mark: 3 |
| \((t+1)(t^2 - 4t + 1) = 0 \Rightarrow t = -1, 2\pm\sqrt{3}\) | M1 | |
| \(\tan\frac{3\pi}{4} = -1\), \(\tan\frac{\pi}{12} = 2 - \sqrt{3}\) | A1 (one) | |
| \(\tan\frac{5\pi}{12} = 2 + \sqrt{3}\) | A1 (all) | Part mark: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{dy}{dx} = \frac{(x+3)(2x+\lambda) - (x^2+\lambda x - 6\lambda^2)}{(x+3)^2}\) | M1 | |
| \(= \frac{x^2 + 6x + 3\lambda + 6\lambda^2}{(x+3)^2}\) | A1 | |
| \(\frac{dy}{dx} = 0\) has distinct roots if \(36 - 12\lambda - 24\lambda^2 > 0 \Rightarrow 3 - \lambda - 2\lambda^2 > 0\) | M1A1 | |
| \((3 + 2\lambda)(1-\lambda) > 0\) | A1 | |
| \(\Rightarrow -\frac{3}{2} < \lambda < 1\) (AG) | A1 | No equality; Part mark: 5 |
| (ii) \(\frac{x^2 + \lambda x - 6\lambda^2}{x+3} \equiv x + \lambda - 3 + \frac{9 - 3\lambda - 6\lambda^2}{x+3}\) | M1 | Tolerate error on remainder term |
| Asymptotes: \(x = -3\) and \(y = x + \lambda - 3\) | B1, A1 | Part mark: 3 |
| (iii) For \(0 < \lambda < 1\): Axes and asymptotes; upper branch with minimum below \(x\)-axis; lower branch with maximum | B1, B1, B1 | Part mark: 3 |
| (iv) For \(\lambda > 3\): Axes and asymptotes (oblique asymptote has positive \(y\) intercept); left-hand branch; right-hand branch | B1, B1, B1 | Part mark: 3 |
## Question 11 (EITHER):
**De Moivre's theorem route:**
$(\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta$ | B1 |
$\cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta$ | M1A1 | If line 1 missing award B0M1A1M1M1A0 i.e. 4/6
$\sin 3\theta = 3\cos^2\theta\sin\theta - \sin^3\theta$, $\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta$ | M1 |
$\tan 3\theta = \frac{3\cos^2\theta\sin\theta - \sin^3\theta}{\cos^3\theta - 3\cos\theta\sin^2\theta}$ | M1 |
$\therefore \tan 3\theta = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}$ (*) (AG) | A1 | Part mark: 6
$\frac{\pi}{12}, \frac{5\pi}{12}, \frac{9\pi}{12} = \frac{3\pi}{4}$ | B1 (one), B1 (all) | Part mark: 2
Put $\tan 3\theta = 1$ in (*) $\Rightarrow t^3 - 3t^2 - 3t + 1 = 0$ | M1 |
Roots are $\tan\frac{\pi}{12}, \tan\frac{5\pi}{12}, \tan\frac{3\pi}{4}$ | A1 (one), A1 (all) | Part mark: 3
$(t+1)(t^2 - 4t + 1) = 0 \Rightarrow t = -1, 2\pm\sqrt{3}$ | M1 |
$\tan\frac{3\pi}{4} = -1$, $\tan\frac{\pi}{12} = 2 - \sqrt{3}$ | A1 (one) |
$\tan\frac{5\pi}{12} = 2 + \sqrt{3}$ | A1 (all) | Part mark: 3
**Total: [14]**
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## Question 11 (OR):
**(i)** $\frac{dy}{dx} = \frac{(x+3)(2x+\lambda) - (x^2+\lambda x - 6\lambda^2)}{(x+3)^2}$ | M1 |
$= \frac{x^2 + 6x + 3\lambda + 6\lambda^2}{(x+3)^2}$ | A1 |
$\frac{dy}{dx} = 0$ has distinct roots if $36 - 12\lambda - 24\lambda^2 > 0 \Rightarrow 3 - \lambda - 2\lambda^2 > 0$ | M1A1 |
$(3 + 2\lambda)(1-\lambda) > 0$ | A1 |
$\Rightarrow -\frac{3}{2} < \lambda < 1$ (AG) | A1 | No equality; Part mark: 5
**(ii)** $\frac{x^2 + \lambda x - 6\lambda^2}{x+3} \equiv x + \lambda - 3 + \frac{9 - 3\lambda - 6\lambda^2}{x+3}$ | M1 | Tolerate error on remainder term
Asymptotes: $x = -3$ and $y = x + \lambda - 3$ | B1, A1 | Part mark: 3
**(iii)** For $0 < \lambda < 1$: Axes and asymptotes; upper branch with minimum below $x$-axis; lower branch with maximum | B1, B1, B1 | Part mark: 3
**(iv)** For $\lambda > 3$: Axes and asymptotes (oblique asymptote has positive $y$ intercept); left-hand branch; right-hand branch | B1, B1, B1 | Part mark: 3
Deduct 1 mark overall for wrong forms at infinity.
**Total: [14]**The curve $C$ has equation
$$y = \frac { x ^ { 2 } + \lambda x - 6 \lambda ^ { 2 } } { x + 3 }$$
where $\lambda$ is a constant such that $\lambda \neq 1$ and $\lambda \neq - \frac { 3 } { 2 }$.\\
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and deduce that if $C$ has two stationary points then $- \frac { 3 } { 2 } < \lambda < 1$.\\
(ii) Find the equations of the asymptotes of $C$.\\
(iii) Draw a sketch of $C$ for the case $0 < \lambda < 1$.\\
(iv) Draw a sketch of $C$ for the case $\lambda > 3$.
\hfill \mbox{\textit{CAIE FP1 2011 Q11 OR}}