Challenging +1.3 This is a multi-part Further Maths question on roots of unity requiring systematic application of standard techniques: finding nth roots, multiplying conjugate pairs, and factorizing polynomials using roots. While it involves several steps and Further Maths content (making it harder than typical A-level), the methods are direct applications of FP1 syllabus material without requiring novel insight or particularly challenging problem-solving.
11 Answer only one of the following two alternatives.
EITHER
State the fifth roots of unity in the form \(\cos \theta + \mathrm { i } \sin \theta\), where \(- \pi < \theta \leqslant \pi\).
Simplify
$$\left( x - \left[ \cos \frac { 2 } { 5 } \pi + i \sin \frac { 2 } { 5 } \pi \right] \right) \left( x - \left[ \cos \frac { 2 } { 5 } \pi - i \sin \frac { 2 } { 5 } \pi \right] \right) .$$
Hence find the real factors of
$$x ^ { 5 } - 1$$
Express the six roots of the equation
$$x ^ { 6 } - x ^ { 3 } + 1 = 0$$
as three conjugate pairs, in the form \(\cos \theta \pm \mathrm { i } \sin \theta\).
Hence find the real factors of
$$x ^ { 6 } - x ^ { 3 } + 1$$
OR
Given that
$$y ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 6 y ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } + 3 y ^ { 3 } = 25 \mathrm { e } ^ { - 2 x }$$
and that \(v = y ^ { 3 }\), show that
$$\frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } - 6 \frac { \mathrm {~d} v } { \mathrm {~d} x } + 9 v = 75 \mathrm { e } ^ { - 2 x }$$
Find the particular solution for \(y\) in terms of \(x\), given that when \(x = 0 , y = 2\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1\).
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11 Answer only one of the following two alternatives.
EITHER
State the fifth roots of unity in the form $\cos \theta + \mathrm { i } \sin \theta$, where $- \pi < \theta \leqslant \pi$.
Simplify
$$\left( x - \left[ \cos \frac { 2 } { 5 } \pi + i \sin \frac { 2 } { 5 } \pi \right] \right) \left( x - \left[ \cos \frac { 2 } { 5 } \pi - i \sin \frac { 2 } { 5 } \pi \right] \right) .$$
Hence find the real factors of
$$x ^ { 5 } - 1$$
Express the six roots of the equation
$$x ^ { 6 } - x ^ { 3 } + 1 = 0$$
as three conjugate pairs, in the form $\cos \theta \pm \mathrm { i } \sin \theta$.
Hence find the real factors of
$$x ^ { 6 } - x ^ { 3 } + 1$$
OR
Given that
$$y ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 6 y ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } + 3 y ^ { 3 } = 25 \mathrm { e } ^ { - 2 x }$$
and that $v = y ^ { 3 }$, show that
$$\frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } - 6 \frac { \mathrm {~d} v } { \mathrm {~d} x } + 9 v = 75 \mathrm { e } ^ { - 2 x }$$
Find the particular solution for $y$ in terms of $x$, given that when $x = 0 , y = 2$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 1$.
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\hfill \mbox{\textit{CAIE FP1 2013 Q11}}