CAIE FP1 2011 June — Question 11 EITHER

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionJune
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeDe Moivre to derive tan/cot identities
DifficultyChallenging +1.3 This is a standard Further Maths question using de Moivre's theorem to derive a triple angle formula, followed by routine algebraic manipulation. While it requires multiple steps and connects the cubic equation to tangent values, the techniques are well-practiced in FP1 and follow a predictable pattern. The final exact value calculations are straightforward once the angles are identified.
Spec1.05l Double angle formulae: and compound angle formulae4.02q De Moivre's theorem: multiple angle formulae4.05b Transform equations: substitution for new roots
Use de Moivre's theorem to prove that $$\tan 3 \theta = \frac { 3 \tan \theta - \tan ^ { 3 } \theta } { 1 - 3 \tan ^ { 2 } \theta }$$ State the exact values of \(\theta\), between 0 and \(\pi\), that satisfy \(\tan 3 \theta = 1\). Express each root of the equation \(t ^ { 3 } - 3 t ^ { 2 } - 3 t + 1 = 0\) in the form \(\tan ( k \pi )\), where \(k\) is a positive rational number. For each of these values of \(k\), find the exact value of \(\tan ( k \pi )\).
Question 11 (EITHER):
De Moivre's theorem route:
AnswerMarks 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
Total: [14]
Question 11 (OR):
AnswerMarks 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 maximumB1, B1, B1 Part mark: 3
(iv) For \(\lambda > 3\): Axes and asymptotes (oblique asymptote has positive \(y\) intercept); left-hand branch; right-hand branchB1, B1, B1 Part mark: 3
Deduct 1 mark overall for wrong forms at infinity.
Total: [14]
## 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]**

---

## 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]**
Use de Moivre's theorem to prove that

$$\tan 3 \theta = \frac { 3 \tan \theta - \tan ^ { 3 } \theta } { 1 - 3 \tan ^ { 2 } \theta }$$

State the exact values of $\theta$, between 0 and $\pi$, that satisfy $\tan 3 \theta = 1$.

Express each root of the equation $t ^ { 3 } - 3 t ^ { 2 } - 3 t + 1 = 0$ in the form $\tan ( k \pi )$, where $k$ is a positive rational number.

For each of these values of $k$, find the exact value of $\tan ( k \pi )$.

\hfill \mbox{\textit{CAIE FP1 2011 Q11 EITHER}}
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Q1 5 Q2 6 Q3 6 Q4 6 Q5 8 Q6 9 Q7 11 Q8 11 Q9 12 Q10 12 Q11 EITHER Q11 OR