7 Expand \(\left( z + \frac { 1 } { z } \right) ^ { 4 } \left( z - \frac { 1 } { z } \right) ^ { 2 }\) and, by substituting \(z = \cos \theta + \mathrm { i } \sin \theta\), find integers \(p , q , r , s\) such that
$$64 \sin ^ { 2 } \theta \cos ^ { 4 } \theta = p + q \cos 2 \theta + r \cos 4 \theta + s \cos 6 \theta$$
Using the substitution \(x = 2 \cos \theta\), show that
$$\int _ { 1 } ^ { 2 } x ^ { 4 } \sqrt { } \left( 4 - x ^ { 2 } \right) \mathrm { d } x = \frac { 4 } { 3 } \pi + \sqrt { } 3$$
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Question 7:
Answer Marks
Guidance
Answer/Working Marks
Guidance
\(\left(z+\frac{1}{z}\right)^4\left(z-\frac{1}{z}\right)^2 = \left(z^2+2+\frac{1}{z^2}\right)\left(z^4-2+\frac{1}{z^4}\right)\) M1A1
Complete strategy and getting halfway
\(= z^6+2z^4-z^2-4-\frac{1}{z^2}+\frac{2}{z^4}+\frac{1}{z^6}\) A1
Fully correct
\(= \left(z^6+\frac{1}{z^6}\right)+2\left(z^4+\frac{1}{z^4}\right)-\left(z^2+\frac{1}{z^2}\right)-4\) M1
Grouping
\(16\cos^4\theta\cdot(-4\sin^2\theta) = 2\cos6\theta+4\cos4\theta-2\cos2\theta-4\) A1(L)
\(64\cos^4\theta\sin^2\theta = 4+2\cos2\theta-4\cos4\theta-2\cos6\theta\) A1(R)
Part Mark: 6
\(x=2\cos\theta\), \(\frac{dx}{d\theta}=-2\sin\theta\); \(x=1\Rightarrow\theta=\frac{\pi}{3}\), \(x=2\Rightarrow\theta=0\) —
\(-\int_{\pi/3}^{0}16\cos^4\theta\cdot4\sin^2\theta\,d\theta = \int_0^{\pi/3}64\cos^4\theta\sin^2\theta\,d\theta\) M1
Sets up substitution
\(= \int_0^{\pi/3}(4+2\cos2\theta-4\cos4\theta-2\cos6\theta)\,d\theta\) M1
Uses result obtained above
\(= \left[4\theta+\sin2\theta-\sin4\theta-\frac{1}{3}\sin6\theta\right]_0^{\pi/3}\) A1
\(= \left[\frac{4\pi}{3}+\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right] = \frac{4\pi}{3}+\sqrt{3}\) (AG) A1
Part Mark: 4; Total: [10]
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## Question 7:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(z+\frac{1}{z}\right)^4\left(z-\frac{1}{z}\right)^2 = \left(z^2+2+\frac{1}{z^2}\right)\left(z^4-2+\frac{1}{z^4}\right)$ | M1A1 | Complete strategy and getting halfway |
| $= z^6+2z^4-z^2-4-\frac{1}{z^2}+\frac{2}{z^4}+\frac{1}{z^6}$ | A1 | Fully correct |
| $= \left(z^6+\frac{1}{z^6}\right)+2\left(z^4+\frac{1}{z^4}\right)-\left(z^2+\frac{1}{z^2}\right)-4$ | M1 | Grouping |
| $16\cos^4\theta\cdot(-4\sin^2\theta) = 2\cos6\theta+4\cos4\theta-2\cos2\theta-4$ | A1(L) | |
| $64\cos^4\theta\sin^2\theta = 4+2\cos2\theta-4\cos4\theta-2\cos6\theta$ | A1(R) | Part Mark: 6 |
| $x=2\cos\theta$, $\frac{dx}{d\theta}=-2\sin\theta$; $x=1\Rightarrow\theta=\frac{\pi}{3}$, $x=2\Rightarrow\theta=0$ | — | |
| $-\int_{\pi/3}^{0}16\cos^4\theta\cdot4\sin^2\theta\,d\theta = \int_0^{\pi/3}64\cos^4\theta\sin^2\theta\,d\theta$ | M1 | Sets up substitution |
| $= \int_0^{\pi/3}(4+2\cos2\theta-4\cos4\theta-2\cos6\theta)\,d\theta$ | M1 | Uses result obtained above |
| $= \left[4\theta+\sin2\theta-\sin4\theta-\frac{1}{3}\sin6\theta\right]_0^{\pi/3}$ | A1 | |
| $= \left[\frac{4\pi}{3}+\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right] = \frac{4\pi}{3}+\sqrt{3}$ (AG) | A1 | Part Mark: 4; Total: **[10]** |
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7 Expand $\left( z + \frac { 1 } { z } \right) ^ { 4 } \left( z - \frac { 1 } { z } \right) ^ { 2 }$ and, by substituting $z = \cos \theta + \mathrm { i } \sin \theta$, find integers $p , q , r , s$ such that
$$64 \sin ^ { 2 } \theta \cos ^ { 4 } \theta = p + q \cos 2 \theta + r \cos 4 \theta + s \cos 6 \theta$$
Using the substitution $x = 2 \cos \theta$, show that
$$\int _ { 1 } ^ { 2 } x ^ { 4 } \sqrt { } \left( 4 - x ^ { 2 } \right) \mathrm { d } x = \frac { 4 } { 3 } \pi + \sqrt { } 3$$
\hfill \mbox{\textit{CAIE FP1 2012 Q7 [10]}}