IBDP Maths AHL 1.14 Conjugate roots of polynomial equations AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a) Using the binomial expansion:
\[ (\cos \theta + i \sin \theta)^5 = \sum_{k=0}^{5} \binom{5}{k} (\cos \theta)^{5-k} (i \sin \theta)^k \]
Expanding and simplifying:
\[ = \cos^5 \theta + 5i \cos^4 \theta \sin \theta – 10 \cos^3 \theta \sin^2 \theta – 10i \cos^2 \theta \sin^3 \theta + 5 \cos \theta \sin^4 \theta – i \sin^5 \theta \]
Grouping real and imaginary parts:
\[ \boxed{ \left( \cos^5 \theta – 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta \right) + i \left( 5 \cos^4 \theta \sin \theta – 10 \cos^2 \theta \sin^3 \theta – \sin^5 \theta \right) } \]
(b) By De Moivre’s theorem:
\[ (\cos \theta + i \sin \theta)^5 = \cos 5\theta + i \sin 5\theta \]
Equating imaginary parts:
\[ \sin 5\theta = 5 \cos^4 \theta \sin \theta – 10 \cos^2 \theta \sin^3 \theta – \sin^5 \theta \]
Substituting \( \cos^2 \theta = 1 – \sin^2 \theta \):
\[ \sin 5\theta = 5(1 – \sin^2 \theta)^2 \sin \theta – 10(1 – \sin^2 \theta) \sin^3 \theta – \sin^5 \theta \]
Expanding and simplifying:
\[ \boxed{ \sin 5\theta = 16\sin^5 \theta – 20\sin^3 \theta + 5\sin \theta } \]
(c) (i) For \( \theta = \frac{\pi}{5} \) and \( \theta = \frac{3\pi}{5} \), \( \sin 5\theta = 0 \):
\[ 16\sin^4 \theta – 20\sin^2 \theta + 5 = 0 \]
Thus, these angles satisfy the equation.
(c) (ii) Using the quadratic formula on \( 16x^2 – 20x + 5 = 0 \) where \( x = \sin^2 \theta \):
\[ x = \frac{5 \pm \sqrt{5}}{8} \]
Therefore:
\[ \sin \frac{\pi}{5} \sin \frac{3\pi}{5} = \sqrt{\frac{5 – \sqrt{5}}{8}} \times \sqrt{\frac{5 + \sqrt{5}}{8}} = \frac{\sqrt{5}}{4} \]
Thus:
\[ \boxed{ \sin \frac{\pi}{5} \sin \frac{3\pi}{5} = \frac{\sqrt{5}}{4} } \]
Markscheme
(a) \( (\cos \theta + i \sin \theta)^5 = \cos 5\theta + i \sin 5\theta \)
Using binomial expansion:
\( \cos 5\theta = \cos^5 \theta – 10\cos^3 \theta \sin^2 \theta + 5\cos \theta \sin^4 \theta \)
\( \sin 5\theta = 5\cos^4 \theta \sin \theta – 10\cos^2 \theta \sin^3 \theta + \sin^5 \theta \)
(b) Equating imaginary parts:
\( \sin 5\theta = 5\cos^4 \theta \sin \theta – 10\cos^2 \theta \sin^3 \theta + \sin^5 \theta \)
Substituting \( \cos^2 \theta = 1 – \sin^2 \theta \):
\( \sin 5\theta = 16\sin^5 \theta – 20\sin^3 \theta + 5\sin \theta \)
(c) (i) For \( \theta = \frac{\pi}{5} \) and \( \theta = \frac{3\pi}{5} \), \( \sin 5\theta = 0 \), so they satisfy \( 16\sin^4 \theta – 20\sin^2 \theta + 5 = 0 \).
(ii) Using the quadratic formula:
\( \sin^2 \theta = \frac{5 \pm \sqrt{5}}{8} \)
\( \sin \frac{\pi}{5} \sin \frac{3\pi}{5} = \sqrt{\frac{5 + \sqrt{5}}{8}} \times \sqrt{\frac{5 – \sqrt{5}}{8}} = \frac{\sqrt{5}}{4} \)
▶️ Answer/Explanation
Markscheme
(a) \(\boxed{\alpha + \beta + \gamma = \dfrac{7}{2}}\)
(b) \(p – 3i\) is also a root (seen anywhere)
Recognition of 5 roots and attempt to sum these roots
\(p + 3i + p – 3i + \frac{7}{2} = \frac{11}{2}\)
\(\boxed{p = 1}\)
(c) (i) Attempt to find product of 5 roots and equate to ±10
\((1 + 3i)(1 – 3i)\left(\frac{1}{2}\right)\alpha\beta = 10\)
\(\boxed{\alpha\beta = 2}\)
(ii) \(\boxed{\alpha = 1}\) and \(\boxed{\beta = 2}\)
Solution
(a) Using Vieta’s formulas for \(g(x) = 2x^3 – 7x^2 + dx + e\):
\(\alpha + \beta + \gamma = -\dfrac{-7}{2} = \dfrac{7}{2}\)
(b) For \(h(z) = 2z^5 – 11z^4 + \cdots – 20\):
Sum of roots: \(p + 3i + p – 3i + \alpha + \beta + \gamma = \dfrac{11}{2}\)
Substituting from part (a): \(2p + \dfrac{7}{2} = \dfrac{11}{2}\)
Solving gives \(p = 1\)
(c) (i) Product of roots for \(h(z)\):
\((1+3i)(1-3i)\left(\dfrac{1}{2}\right)\alpha\beta = \dfrac{20}{2}\)
\((1+9)\left(\dfrac{1}{2}\right)\alpha\beta = 10\)
\(\alpha\beta = 2\)
(ii) From \(\alpha + \beta = \dfrac{7}{2} – \dfrac{1}{2} = 3\) and \(\alpha\beta = 2\)
The positive integer solutions are \(\alpha = 1\) and \(\beta = 2\)