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AQA A Level Maths MFP2 Further Pure 2 Unofficial Mark Scheme 22 June 2018

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Reply 1
Q1

(a) Show that f(rβˆ’1)βˆ’f(r)=k(2r+1)(2r+3)(2r+5)\mathrm{f}(r-1)-\mathrm{f}(r) = \dfrac{k}{(2r+1)(2r+3)(2r+5)}, where f(r)=1(2r+3)(2r+5)\mathrm{f}(r) = \dfrac{1}{(2r+3)(2r+5)}. [2 marks]

1(2r+1)(2r+3)βˆ’1(2r+3)(2r+5)\dfrac{1}{(2r+1)(2r+3)} - \dfrac{1}{(2r+3)(2r+5)}

(2r+5)βˆ’(2r+1)(2r+1)(2r+3)(2r+5)=4(2r+1)(2r+3)(2r+5)\dfrac{(2r+5) - (2r+1)}{(2r+1)(2r+3)(2r+5)} = \dfrac{4}{(2r+1)(2r+3)(2r+5)}

(b) Hence use the method of differences to find βˆ‘r=1N1((2r+1)(2r+3)(2r+5)\displaystyle \sum_{r=1}^{N} \frac{1}{((2r+1)(2r+3)(2r+5)}. [3 marks]

14βˆ‘r=1Nf(rβˆ’1)βˆ’f(r)\displaystyle \frac{1}{4}\sum_{r=1}^{N}\mathrm{f}(r-1) - \mathrm{f}(r)

14(f(0)βˆ’f(1)+f(1)βˆ’f(2)+β‹―+f(Nβˆ’2)βˆ’f(Nβˆ’1)+f(Nβˆ’1)βˆ’f(N))\frac{1}{4}(\mathrm{f}(0) - \mathrm{f}(1) + \mathrm{f}(1) - \mathrm{f}(2) + \dotsb + \mathrm{f}(N - 2) - \mathrm{f}(N - 1) + \mathrm{f}(N - 1) - \mathrm{f}(N))

By method of differences, 14(f(0)βˆ’f(N))\frac{1}{4}(\mathrm{f}(0) - \mathrm{f}(N))

So 160βˆ’14(2N+3)(2N+5)\dfrac{1}{60} - \dfrac{1}{4(2N+3)(2N+5)}.

Q2

z=βˆ’22+26iz=-2\sqrt{2} + 2\sqrt{6}\mathrm{i} can be expressed as reiΞΈr\mathrm{e}^{\mathrm{i}\theta} with r > 0 and
Unparseable latex formula:

-\pi < \theta \leqslant \pi

.

(a) Show that r=(2)nr = (\sqrt{2})^n. [2 marks]

r=∣z∣=(βˆ’22)2+(26)2=32=25=25/2=(21/2)5=(2)5r = |z| = \sqrt{(-2\sqrt{2})^2 + (2\sqrt{6})^2} = \sqrt{32} = \sqrt{2^5} = 2^{5/2} = (2^{1/2})^5 = (\sqrt{2})^5

(a) (ii) Find ΞΈ\theta. [1 mark]

ΞΈ=Ο€βˆ’tanβ‘βˆ’12622=2Ο€3\theta = \pi - \tan^{-1} \dfrac{2\sqrt{6}}{2\sqrt{2}} = \dfrac{2\pi}{3}

Q3 Prove by induction that un=2n+1βˆ’52n+1βˆ’3u_n = \dfrac{2^{n+1} - 5}{2^{n+1} - 3} if u1=βˆ’1u_1 = -1 and un+1=unβˆ’53unβˆ’7u_{n+1} = \dfrac{u_n - 5}{3u_n - 7} [6 marks]

Q4

(a) Express (1+e2x)(1+eβˆ’2x)(1+\mathrm{e}^{2x})(1+\mathrm{e}^{-2x}) in terms of cosh⁑x\cosh x. [3 marks]

1+e2x+eβˆ’2x+1=2+e2x+eβˆ’2x=2+2cosh⁑2x1 + \mathrm{e}^{2x} + \mathrm{e}^{-2x} + 1 = 2 + \mathrm{e}^{2x} + \mathrm{e}^{-2x} = 2 + 2\cosh 2x

2+2(cosh⁑2xβˆ’1)=4cosh⁑2x2+2(\cosh^2 x - 1) = 4\cosh^2 x

(b)Find ∫011(1+e2x)(1+eβˆ’2x)β€…β€Šdx\displaystyle \int^1_0 \frac{1}{(1+ \mathrm{e}^{2x})(1+ \mathrm{e}^{-2x})}\;\mathrm{d}x in terms of e\mathrm{e}. [4 marks]

14∫01sech2xβ€…β€Šdx=14[tanh⁑x]01=14(tanh⁑1βˆ’tanh⁑0)=14tanh⁑1\displaystyle \frac{1}{4} \int^{1}_0 \text{sech}^2 x\;\mathrm{d}x = \frac{1}{4}\left[\tanh x\right]^1_0 = \frac{1}{4}(\tanh 1 - \tanh 0) = \frac{1}{4}\tanh 1

14((e1βˆ’eβˆ’1)/2(e1+eβˆ’1)/2)=14(e2βˆ’1e2+1)\dfrac{1}{4}\left( \dfrac{( \mathrm{e}^{1}-\mathrm{e}^{-1} )/2}{(\mathrm{e}^{1}+\mathrm{e}^{-1})/2} \right) = \dfrac{1}{4}\left(\dfrac{\mathrm{e}^2 - 1}{\mathrm{e}^2 + 1}\right)

Q5

(a) On the Argand diagram, sketch the locus of points satisfying ∣zβˆ’2∣=∣z+4i∣|z - 2| = |z + 4\mathrm{i}|. [3 marks]

On the Argand diagram, a perpendicular bisector of the line connecting the points on the Argand diagram of (2, 0) and (0, -4)

(b) The complex number z1z_1 is such that ∣zβˆ’2∣=∣z+4i∣|z - 2| = |z + 4\mathrm{i}| and ∣z∣|z| is at its maximum value. [4 marks]

Using the Argand diagram, a more geometrical or algebraic approach can be used

z1=βˆ’35βˆ’65iz_1 = -\dfrac{3}{5} - \dfrac{6}{5}\mathrm{i}

Q6

(a) Sketch the graph of y=coshβ‘βˆ’1xy=\cosh^{-1} x. [2 marks]

This can be obtained off a graphical calculator

(b) Show that dydx=1x2βˆ’1\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{\sqrt{x^2-1}}. [3 marks]

cosh⁑y=xβ‡’dydxsinh⁑y=1β‡’dydx=1sinh⁑y\cosh y = x \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x}\sinh y = 1 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{\sinh y}

sinh⁑2y=cosh⁑2yβˆ’1=x2βˆ’1β‡’sinh⁑y=x2βˆ’1\sinh^2 y = \cosh^2 y-1 = x^2-1 \Rightarrow \sinh y = \sqrt{x^2-1}, since yβ©Ύ0y \geqslant 0 by the sketch so sinh⁑yβ©Ύ0\sinh y \geqslant 0

(c) Show that y=53βˆ’4x+coshβ‘βˆ’1(3x)y=\dfrac{5}{3} -4x + \cosh^{-1}(3x) has one stationary point and find its y-coordinate. [5 marks]

dydx=βˆ’4+39x2βˆ’1\dfrac{\mathrm{d}y}{\mathrm{d}x} = -4 + \dfrac{3}{\sqrt{9x^2-1}}

At stationary points, dy/dx = 0, 9x2βˆ’1=169β‡’x=Β±5129x^2-1 = \dfrac{16}{9} \Rightarrow x = \pm \dfrac{5}{12}, but 3xβ©Ύ13x \geqslant 1 and so x=512x = \dfrac{5}{12}

y=53βˆ’53+coshβ‘βˆ’154=coshβ‘βˆ’154=ln⁑(54+(54)2βˆ’1)=ln⁑2y=\dfrac{5}{3} - \dfrac{5}{3} + \cosh^{-1}\dfrac{5}{4} = \cosh^{-1} \dfrac{5}{4} = \ln \left(\dfrac{5}{4} + \sqrt{\left(\dfrac{5}{4}\right)^2 - 1}\right) = \ln 2

Q7

x=3βˆ’cos⁑2tx=3-\cos 2t, y=4sin⁑ty=4\sin t

(a) Show that the surface area is kΟ€βˆ«0Ο€2sin⁑tcos⁑t1+sin⁑2tβ€…β€Šdt\displaystyle k\pi \int ^{\frac{\pi}{2}}_0 \sin t \cos t \sqrt{1+\sin^2 t}\; \mathrm{d}t [4 marks]

dx/dt = 2 sin 2t, dy/dt = 4 cos t

2Ο€βˆ«0Ο€24sin⁑t4sin⁑22t+16cos⁑2tβ€…β€Šdt\displaystyle 2\pi \int^{\frac{\pi}{2}}_0 4\sin t \sqrt{4\sin^2 2t + 16\cos^2t}\;\mathrm{d}t

2Ο€βˆ«0Ο€24sin⁑t16sin⁑2tcos⁑2t+16cos⁑2tβ€…β€Šdt\displaystyle 2\pi \int^{\frac{\pi}{2}}_0 4\sin t \sqrt{16\sin^2 t\cos^2 t + 16\cos^2t}\;\mathrm{d}t

32Ο€βˆ«0Ο€2sin⁑tcos⁑t1+sin⁑2tβ€…β€Šdt\displaystyle 32\pi \int ^{\frac{\pi}{2}}_0 \sin t \cos t \sqrt{1+\sin^2 t}\; \mathrm{d}t

(b) Show that the surface area is Ο€3(n2+m)\dfrac{\pi}{3}(n\sqrt{2}+m) [5 marks]

u=1+sin⁑2tβ‡’dt=du2sin⁑tcos⁑tu=1+\sin^2 t \Rightarrow \mathrm{d}t = \dfrac{\mathrm{d}u}{2\sin t \cos t}

s=16Ο€βˆ«12uβ€…β€Šdu=16Ο€[23u32]12=Ο€3(642βˆ’32)\displaystyle s=16\pi \int^2_1 \sqrt{u}\; \mathrm{d}u = 16\pi \left[\dfrac{2}{3}u^{\frac{3}{2}} \right]^2_1 = \dfrac{\pi}{3}(64\sqrt{2} - 32)

Q8

2z3+5z+3=02z^3 + 5z + 3 = 0

(a) (i) Ξ±Ξ²+Ξ²Ξ³+Ξ³Ξ±\alpha\beta + \beta\gamma + \gamma\alpha [1 mark]

5/2

(a) (ii) Ξ±Ξ²Ξ³\alpha\beta\gamma [1 mark]

-3/2

(a) (iii) 1Ξ±+1Ξ²+1Ξ³\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma} [2 marks]

Ξ±Ξ²+Ξ²Ξ³+Ξ³Ξ±Ξ±Ξ²Ξ³=5/2βˆ’3/2=βˆ’53\dfrac{\alpha\beta+\beta\gamma+ \gamma\alpha}{\alpha\beta\gamma} = \dfrac{5/2}{-3/2} = -\dfrac{5}{3}

(b) (i) z2=1xz^2 = \dfrac{1}{x}, show that 9x3βˆ’25x2+mx+n=09x^3 - 25x^2 + mx + n = 0 [4 marks]

z(2z2+5)=βˆ’3β‡’z2(2z2+5)2=9β‡’1x(2x+5)2=9z(2z^2 + 5) = -3 \Rightarrow z^2(2z^2 + 5)^2 = 9 \Rightarrow \dfrac{1}{x}\left(\dfrac{2}{x} + 5\right)^2 = 9

9x3βˆ’25x2βˆ’20xβˆ’4=09x^3 - 25x^2 - 20x - 4 = 0

(b) (ii) 1Ξ±4+1Ξ²4+1Ξ³4\dfrac{1}{\alpha^4} + \dfrac{1}{\beta^4} + \dfrac{1}{\gamma^4} [4 marks]

x=1z2x = \dfrac{1}{z^2} so 1Ξ±2+1Ξ²2+1Ξ³2=259\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} + \dfrac{1}{\gamma^2} = \dfrac{25}{9} and 1Ξ±21Ξ²2+1Ξ²21Ξ³2+1Ξ³21Ξ±2=βˆ’209\dfrac{1}{\alpha^2} \dfrac{1}{\beta^2} + \dfrac{1}{\beta^2} \dfrac{1}{\gamma^2} + \dfrac{1}{\gamma^2} \dfrac{1}{ \alpha^2} = -\dfrac{20}{9}

So 1Ξ±4+1Ξ²4+1Ξ³4=(259)2βˆ’2(βˆ’209)=98581\dfrac{1}{\alpha^4} + \dfrac{1}{\beta^4} + \dfrac{1}{\gamma^4} = \left(\dfrac{25}{9}\right)^2 - 2\left(-\dfrac{20}{9}\right) = \dfrac{985}{81}

Q9

(a) Use de Moivre's theorem to show that cos⁑5θ=16cos⁑5θ+Acos⁑3θ+Bcos⁑θ\cos 5\theta = 16 \cos^5 \theta + A \cos^3 \theta + B \cos \theta. [5 marks]

de Moire's theorem for n=5n=5: cos⁑5θ+isin⁑5θ=(cos⁑θ+isin⁑θ)5\cos 5\theta + \mathrm{i}\sin 5\theta = (\cos \theta + \mathrm{i}\sin \theta)^5

Applying the binomial expansion to de Moivre's theorem, and using the identity sin⁑2ΞΈ=1βˆ’cos⁑2ΞΈ\sin^2\theta = 1-\cos^2 \theta gives
cos⁑5ΞΈ=16cos⁑5ΞΈβˆ’20cos⁑3ΞΈ+5cos⁑θ\cos 5\theta = 16 \cos^5 \theta - 20 \cos^3 \theta + 5\cos \theta

(b) (i) cos⁑5ΞΈ=0\cos 5\theta = 0, and cos⁑θ≠0\cos \theta \neq 0, find the possible values of cos⁑2ΞΈ\cos^2 \theta. [2 marks]

Using the de Moivre's theorem result, 16c5βˆ’20c3+5c=0β‡’16c4βˆ’20c2+5=016c^5 - 20c^3 + 5c = 0 \Rightarrow 16c^4 - 20c^2 + 5 = 0, where c=cos⁑θc = \cos \theta

16u2βˆ’20u+5=016u^2 - 20u + 5 = 0 where u=cos⁑2ΞΈu = \cos^2 \theta, u=5Β±58u = \dfrac{5\pm \sqrt{5}}{8}

(b) (ii) Deduce the value of cos⁑23Ο€10\cos^2 \dfrac{3\pi}{10} and show that cos⁑3Ο€5=1βˆ’58\cos \dfrac{3\pi}{5} = \dfrac{1-\sqrt{5}}{8}. [4 marks]

cos⁑5ΞΈ=0β‡’ΞΈ=nΟ€+Ο€2=(2n+1)Ο€2\cos 5\theta = 0 \Rightarrow \theta = n\pi + \dfrac{\pi}{2} = \dfrac{(2n+1)\pi}{2}, i.e. ΞΈ=Ο€10,β€…β€Š3Ο€10,β€…β€Š5Ο€10,β€…β€Š7Ο€10,β€…β€Š9Ο€10,β€…β€Šβ€¦\theta = \dfrac{\pi}{10},\;\dfrac{3\pi}{10},\;\dfrac{5\pi}{10},\;\dfrac{7\pi}{10},\;\dfrac{9\pi}{10}, \; \dots

cos⁑2Ο€10=cos⁑29Ο€10\cos^2 \dfrac{\pi}{10} = \cos^2 \dfrac{9\pi}{10}, cos⁑23Ο€10=cos⁑27Ο€10\cos^2 \dfrac{3\pi}{10} = \cos^2 \dfrac{7\pi}{10}, and cos⁑25Ο€10=0\cos^2 \dfrac{5\pi}{10} = 0, so we can ignore this value

cos⁑2θ\cos^2 \theta is decreasing for
Unparseable latex formula:

0 < \theta < \pi/2

and so
Unparseable latex formula:

\cos^2 \dfrac{\pi}{10} > \cos^2 \dfrac{3\pi}{10}

, and so cos⁑23Ο€10=5βˆ’58\cos^2 \dfrac{3\pi}{10} = \dfrac{5-\sqrt{5}}{8}

cos⁑3Ο€5=2cos⁑23Ο€10βˆ’1=5βˆ’54βˆ’1=1βˆ’54\cos \dfrac{3\pi}{5} = 2\cos^2 \dfrac{3\pi}{10} - 1 = \dfrac{5-\sqrt{5}}{4} - 1 = \dfrac{1-\sqrt{5}}{4}
(edited 5 years ago)
Reply 2
Original post by Integer123
Q1

(a) Show that f(rβˆ’1)βˆ’f(r)=k(2r+1)(2r+3)(2r+5)\mathrm{f}(r-1)-\mathrm{f}(r) = \dfrac{k}{(2r+1)(2r+3)(2r+5)}, where f(r)=1(2r+3)(2r+5)\mathrm{f}(r) = \dfrac{1}{(2r+3)(2r+5)}. [2 marks]

1(2r+1)(2r+3)βˆ’1(2r+3)(2r+5)\dfrac{1}{(2r+1)(2r+3)} - \dfrac{1}{(2r+3)(2r+5)}

(2r+5)βˆ’(2r+1)(2r+1)(2r+3)(2r+5)=4(2r+1)(2r+3)(2r+5)\dfrac{(2r+5) - (2r+1)}{(2r+1)(2r+3)(2r+5)} = \dfrac{4}{(2r+1)(2r+3)(2r+5)}

(b) Hence use the method of differences to find βˆ‘r=1N1((2r+1)(2r+3)(2r+5)\displaystyle \sum_{r=1}^{N} \frac{1}{((2r+1)(2r+3)(2r+5)}. [3 marks]

14βˆ‘r=1Nf(rβˆ’1)βˆ’f(r)\displaystyle \frac{1}{4}\sum_{r=1}^{N}\mathrm{f}(r-1) - \mathrm{f}(r)

14(f(0)βˆ’f(1)+f(1)βˆ’f(2)+β‹―+f(Nβˆ’2)βˆ’f(Nβˆ’1)+f(Nβˆ’1)βˆ’f(N))\frac{1}{4}(\mathrm{f}(0) - \mathrm{f}(1) + \mathrm{f}(1) - \mathrm{f}(2) + \dotsb + \mathrm{f}(N - 2) - \mathrm{f}(N - 1) + \mathrm{f}(N - 1) - \mathrm{f}(N))

By method of differences, 14(f(0)βˆ’f(N))\frac{1}{4}(\mathrm{f}(0) - \mathrm{f}(N))

So 160βˆ’14(2N+3)(2N+5)\dfrac{1}{60} - \dfrac{1}{4(2N+3)(2N+5)}.

Q8

2z3+5z+3=02z^3 + 5z + 3 = 0

(a) (i) Ξ±Ξ²+Ξ²Ξ³+Ξ³Ξ±\alpha\beta + \beta\gamma + \gamma\alpha [1 mark]

5/2

(a) (ii) Ξ±Ξ²Ξ³\alpha\beta\gamma [1 mark]

-3/2

(a) (iii) 1Ξ±+1Ξ²+1Ξ³\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma} [2 marks]

Ξ±Ξ²+Ξ²Ξ³+Ξ³Ξ±Ξ±Ξ²Ξ³=5/2βˆ’3/2=βˆ’53\dfrac{\alpha\beta+\beta\gamma+ \gamma\alpha}{\alpha\beta\gamma} = \dfrac{5/2}{-3/2} = -\dfrac{5}{3}

(b) (i) z2=1xz^2 = \dfrac{1}{x}, show that 9x3βˆ’25x2+mx+n=09x^3 - 25x^2 + mx + n = 0 [4 marks]

z(2z2+5)=βˆ’3β‡’z2(2z2+5)2=9β‡’1x(2x+5)2=9z(2z^2 + 5) = -3 \Rightarrow z^2(2z^2 + 5)^2 = 9 \Rightarrow \dfrac{1}{x}\left(\dfrac{2}{x} + 5\right)^2 = 9

9x3βˆ’25x2βˆ’20xβˆ’4=09x^3 - 25x^2 - 20x - 4 = 0

(b) (ii) 1Ξ±4+1Ξ²4+1Ξ³4\dfrac{1}{\alpha^4} + \dfrac{1}{\beta^4} + \dfrac{1}{\gamma^4} [4 marks]

x=1z2x = \dfrac{1}{z^2} so 1Ξ±2+1Ξ²2+1Ξ³2=259\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} + \dfrac{1}{\gamma^2} = \dfrac{25}{9} and 1Ξ±21Ξ²2+1Ξ²21Ξ³2+1Ξ³21Ξ±2=βˆ’209\dfrac{1}{\alpha^2} \dfrac{1}{\beta^2} + \dfrac{1}{\beta^2} \dfrac{1}{\gamma^2} + \dfrac{1}{\gamma^2} \dfrac{1}{ \alpha^2} = -\dfrac{20}{9}

So 1Ξ±4+1Ξ²4+1Ξ³4=(259)2βˆ’2(βˆ’209)=98581\dfrac{1}{\alpha^4} + \dfrac{1}{\beta^4} + \dfrac{1}{\gamma^4} = \left(\dfrac{25}{9}\right)^2 - 2\left(-\dfrac{20}{9}\right) = \dfrac{985}{81}


Fam where you applied? And got offers ?
Reply 3
I flopped so bad ffs
Ahahah question 8bii is so easy when you spot it but of course I couldn't work it out in the exam
You are honestly a lifesaver with these thank you !!
i thought that was quite a damn good exam. Pretty sure I got everything apart from the last part of the cubic question. I was doing it correctly but at the end for some reason my brain went oh the roots are 1/sqr root alphabeta and gamma ffs haha. Anyway hopefully it can still be 100 ums depending on boundaries.
Reply 7
Can I have your brain pls
Original post by Anonymouspsych
i thought that was quite a damn good exam. Pretty sure I got everything apart from the last part of the cubic question. I was doing it correctly but at the end for some reason my brain went oh the roots are 1/sqr root alphabeta and gamma ffs haha. Anyway hopefully it can still be 100 ums depending on boundaries.
Original post by Integer123
Q1

(a) Show that f(rβˆ’1)βˆ’f(r)=k(2r+1)(2r+3)(2r+5)\mathrm{f}(r-1)-\mathrm{f}(r) = \dfrac{k}{(2r+1)(2r+3)(2r+5)}, where f(r)=1(2r+3)(2r+5)\mathrm{f}(r) = \dfrac{1}{(2r+3)(2r+5)}. [2 marks]

1(2r+1)(2r+3)βˆ’1(2r+3)(2r+5)\dfrac{1}{(2r+1)(2r+3)} - \dfrac{1}{(2r+3)(2r+5)}

(2r+5)βˆ’(2r+1)(2r+1)(2r+3)(2r+5)=4(2r+1)(2r+3)(2r+5)\dfrac{(2r+5) - (2r+1)}{(2r+1)(2r+3)(2r+5)} = \dfrac{4}{(2r+1)(2r+3)(2r+5)}

(b) Hence use the method of differences to find βˆ‘r=1N1((2r+1)(2r+3)(2r+5)\displaystyle \sum_{r=1}^{N} \frac{1}{((2r+1)(2r+3)(2r+5)}. [3 marks]

14βˆ‘r=1Nf(rβˆ’1)βˆ’f(r)\displaystyle \frac{1}{4}\sum_{r=1}^{N}\mathrm{f}(r-1) - \mathrm{f}(r)

14(f(0)βˆ’f(1)+f(1)βˆ’f(2)+β‹―+f(Nβˆ’2)βˆ’f(Nβˆ’1)+f(Nβˆ’1)βˆ’f(N))\frac{1}{4}(\mathrm{f}(0) - \mathrm{f}(1) + \mathrm{f}(1) - \mathrm{f}(2) + \dotsb + \mathrm{f}(N - 2) - \mathrm{f}(N - 1) + \mathrm{f}(N - 1) - \mathrm{f}(N))

By method of differences, 14(f(0)βˆ’f(N))\frac{1}{4}(\mathrm{f}(0) - \mathrm{f}(N))

So 160βˆ’14(2N+3)(2N+5)\dfrac{1}{60} - \dfrac{1}{4(2N+3)(2N+5)}.

Q2

z=βˆ’22+26iz=-2\sqrt{2} + 2\sqrt{6}\mathrm{i} can be expressed as reiΞΈr\mathrm{e}^{\mathrm{i}\theta} with r > 0 and βˆ’Ο€<ΞΈβ©½Ο€-\pi < \theta \leqslant \pi.

(a) Show that r=(2)nr = (\sqrt{2})^n. [2 marks]

r=∣z∣=(βˆ’22)2+(26)2=32=25=25/2=(21/2)5=(2)5r = |z| = \sqrt{(-2\sqrt{2})^2 + (2\sqrt{6})^2} = \sqrt{32} = \sqrt{2^5} = 2^{5/2} = (2^{1/2})^5 = (\sqrt{2})^5

(a) (ii) Find ΞΈ\theta. [1 mark]

ΞΈ=Ο€βˆ’tanβ‘βˆ’12622=2Ο€3\theta = \pi - \tan^{-1} \dfrac{2\sqrt{6}}{2\sqrt{2}} = \dfrac{2\pi}{3}

Q3 Prove by induction that un=2n+1βˆ’52n+1βˆ’3u_n = \dfrac{2^{n+1} - 5}{2^{n+1} - 3} if u1=βˆ’1u_1 = -1 and un+1=unβˆ’53unβˆ’7u_{n+1} = \dfrac{u_n - 5}{3u_n - 7} [6 marks]

Q4

(a) Express (1+e2x)(1+eβˆ’2x)(1+\mathrm{e}^{2x})(1+\mathrm{e}^{-2x}) in terms of cosh⁑x\cosh x. [3 marks]

1+e2x+eβˆ’2x+1=2+e2x+eβˆ’2x=2+2cosh⁑2x1 + \mathrm{e}^{2x} + \mathrm{e}^{-2x} + 1 = 2 + \mathrm{e}^{2x} + \mathrm{e}^{-2x} = 2 + 2\cosh 2x

2+2(cosh⁑2xβˆ’1)=4cosh⁑2x2+2(\cosh^2 x - 1) = 4\cosh^2 x

(b) Find ∫011(1+e2x)(1+eβˆ’2x)β€…β€Šdx\displaystyle \int^1_0 \frac{1}{(1+ \mathrm{e}^{2x})(1+ \mathrm{e}^{-2x})}\;\mathrm{d}x in terms of e\mathrm{e}. [4 marks]

14∫01sech2xβ€…β€Šdx=14[tanh⁑x]01=14(tanh⁑1βˆ’tanh⁑0)=14tanh⁑1\displaystyle \frac{1}{4} \int^{1}_0 \text{sech}^2 x\;\mathrm{d}x = \frac{1}{4}\left[\tanh x\right]^1_0 = \frac{1}{4}(\tanh 1 - \tanh 0) = \frac{1}{4}\tanh 1

14((e1βˆ’eβˆ’1)/2(e1+eβˆ’1)/2)=14(e2βˆ’1e2+1)\dfrac{1}{4}\left( \dfrac{( \mathrm{e}^{1}-\mathrm{e}^{-1} )/2}{(\mathrm{e}^{1}+\mathrm{e}^{-1})/2} \right) = \dfrac{1}{4}\left(\dfrac{\mathrm{e}^2 - 1}{\mathrm{e}^2 + 1}\right)

Q5

(a) On the Argand diagram, sketch the locus of points satisfying ∣zβˆ’2∣=∣z+4i∣|z - 2| = |z + 4\mathrm{i}|. [3 marks]

On the Argand diagram, a perpendicular bisector of the line connecting the points on the Argand diagram of (2, 0) and (0, -4)

(b) The complex number z1z_1 is such that ∣zβˆ’2∣=∣z+4i∣|z - 2| = |z + 4\mathrm{i}| and ∣z∣|z| is at its maximum value. [4 marks]

Using the Argand diagram, a more geometrical or algebraic approach can be used

z1=βˆ’35βˆ’65iz_1 = -\dfrac{3}{5} - \dfrac{6}{5}\mathrm{i}

Q8

2z3+5z+3=02z^3 + 5z + 3 = 0

(a) (i) Ξ±Ξ²+Ξ²Ξ³+Ξ³Ξ±\alpha\beta + \beta\gamma + \gamma\alpha [1 mark]

5/2

(a) (ii) Ξ±Ξ²Ξ³\alpha\beta\gamma [1 mark]

-3/2

(a) (iii) 1Ξ±+1Ξ²+1Ξ³\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma} [2 marks]

Ξ±Ξ²+Ξ²Ξ³+Ξ³Ξ±Ξ±Ξ²Ξ³=5/2βˆ’3/2=βˆ’53\dfrac{\alpha\beta+\beta\gamma+ \gamma\alpha}{\alpha\beta\gamma} = \dfrac{5/2}{-3/2} = -\dfrac{5}{3}

(b) (i) z2=1xz^2 = \dfrac{1}{x}, show that 9x3βˆ’25x2+mx+n=09x^3 - 25x^2 + mx + n = 0 [4 marks]

z(2z2+5)=βˆ’3β‡’z2(2z2+5)2=9β‡’1x(2x+5)2=9z(2z^2 + 5) = -3 \Rightarrow z^2(2z^2 + 5)^2 = 9 \Rightarrow \dfrac{1}{x}\left(\dfrac{2}{x} + 5\right)^2 = 9

9x3βˆ’25x2βˆ’20xβˆ’4=09x^3 - 25x^2 - 20x - 4 = 0

(b) (ii) 1Ξ±4+1Ξ²4+1Ξ³4\dfrac{1}{\alpha^4} + \dfrac{1}{\beta^4} + \dfrac{1}{\gamma^4} [4 marks]

x=1z2x = \dfrac{1}{z^2} so 1Ξ±2+1Ξ²2+1Ξ³2=259\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} + \dfrac{1}{\gamma^2} = \dfrac{25}{9} and 1Ξ±21Ξ²2+1Ξ²21Ξ³2+1Ξ³21Ξ±2=βˆ’209\dfrac{1}{\alpha^2} \dfrac{1}{\beta^2} + \dfrac{1}{\beta^2} \dfrac{1}{\gamma^2} + \dfrac{1}{\gamma^2} \dfrac{1}{ \alpha^2} = -\dfrac{20}{9}

So 1Ξ±4+1Ξ²4+1Ξ³4=(259)2βˆ’2(βˆ’209)=98581\dfrac{1}{\alpha^4} + \dfrac{1}{\beta^4} + \dfrac{1}{\gamma^4} = \left(\dfrac{25}{9}\right)^2 - 2\left(-\dfrac{20}{9}\right) = \dfrac{985}{81}

Q9

(a) Use de Moivre's theorem to show that cos⁑5θ=16cos⁑5θ+Acos⁑3θ+Bcos⁑θ\cos 5\theta = 16 \cos^5 \theta + A \cos^3 \theta + B \cos \theta. [5 marks]

de Moire's theorem for n=5n=5: cos⁑5θ+isin⁑5θ=(cos⁑θ+isin⁑θ)5\cos 5\theta + \mathrm{i}\sin 5\theta = (\cos \theta + \mathrm{i}\sin \theta)^5

Applying the binomial expansion to de Moivre's theorem, and using the identity sin⁑2ΞΈ=1βˆ’cos⁑2ΞΈ\sin^2\theta = 1-\cos^2 \theta gives
cos⁑5ΞΈ=16cos⁑5ΞΈβˆ’20cos⁑3ΞΈ+5cos⁑θ\cos 5\theta = 16 \cos^5 \theta - 20 \cos^3 \theta + 5\cos \theta

(b) (i) cos⁑5ΞΈ=0\cos 5\theta = 0, and cos⁑θ≠0\cos \theta \neq 0, find the possible values of cos⁑2ΞΈ\cos^2 \theta. [2 marks]

Using the de Moivre's theorem result, 16c5βˆ’20c3+5c=0β‡’16c4βˆ’20c2+5=016c^5 - 20c^3 + 5c = 0 \Rightarrow 16c^4 - 20c^2 + 5 = 0, where c=cos⁑θc = \cos \theta

16u2βˆ’20u+5=016u^2 - 20u + 5 = 0 where u=cos⁑2ΞΈu = \cos^2 \theta, u=5Β±58u = \dfrac{5\pm \sqrt{5}}{8}

(b) (ii) Deduce the value of cos⁑23Ο€10\cos^2 \dfrac{3\pi}{10} and show that cos⁑3Ο€5=1βˆ’58\cos \dfrac{3\pi}{5} = \dfrac{1-\sqrt{5}}{8}. [4 marks]

cos⁑5ΞΈ=0β‡’ΞΈ=nΟ€+Ο€2=(2n+1)Ο€2\cos 5\theta = 0 \Rightarrow \theta = n\pi + \dfrac{\pi}{2} = \dfrac{(2n+1)\pi}{2}, i.e. ΞΈ=Ο€10,β€…β€Š3Ο€10,β€…β€Š5Ο€10,β€…β€Š7Ο€10,β€…β€Š9Ο€10,β€…β€Šβ€¦\theta = \dfrac{\pi}{10},\;\dfrac{3\pi}{10},\;\dfrac{5\pi}{10},\;\dfrac{7\pi}{10},\;\dfrac{9\pi}{10}, \; \dots

cos⁑2Ο€10=cos⁑29Ο€10\cos^2 \dfrac{\pi}{10} = \cos^2 \dfrac{9\pi}{10}, cos⁑23Ο€10=cos⁑27Ο€10\cos^2 \dfrac{3\pi}{10} = \cos^2 \dfrac{7\pi}{10}, and cos⁑25Ο€10=0\cos^2 \dfrac{5\pi}{10} = 0, so we can ignore this value

cos⁑2ΞΈ\cos^2 \theta is decreasing for 0<ΞΈ<Ο€/20 < \theta < \pi/2 and so cos⁑2Ο€10>cos⁑23Ο€10\cos^2 \dfrac{\pi}{10} > \cos^2 \dfrac{3\pi}{10}, and so cos⁑23Ο€10=5βˆ’58\cos^2 \dfrac{3\pi}{10} = \dfrac{5-\sqrt{5}}{8}

Unparseable latex formula:

\cos \dfrac{3\pi}{5} = 2\cos^2 \dfrac{3\pi}{10} - 1 = \dfrac{5-\qrt{5}}{4} - 1 = \dfrac{1-\sqrt{5}}{4}



for 4a I didn't notice it saying give it in terms of cosh x so i left it as 2+2cosh2x but in part b i converted this into 4cosh^2x during the integration. Does that mean I'll lose a mark :frown:
Reply 9
i ****ed this lads yw for lowering the grade boundaries.
Original post by dhchgch
i ****ed this lads yw for lowering the grade boundaries.


I'm sure you did better than you thought! A lot of people skipped some of the 4 markers (including me), so already everyone's lowered the grade boundaries for this yearπŸ˜‚
Original post by jess2157
I'm sure you did better than you thought! A lot of people skipped some of the 4 markers (including me), so already everyone's lowered the grade boundaries for this yearπŸ˜‚


do u remember what answer(s) you put for question 6 and 7? Like the last parts of them
Reply 12
Anyone got any rough ideas on grade boundaries?
Original post by Anonymouspsych
do u remember what answer(s) you put for question 6 and 7? Like the last parts of them


For question 6 my answer was weird. I tried over and over but couldn't fix it. I got that coshy = +-5/12, but then I couldn't do the inverse of either of those answers soooooooo pffftt I don't know about that.
For question 7, my final answer was Ο€/3[64sqrt(2)-32].
Original post by fos99e
Anyone got any rough ideas on grade boundaries?


I'm guessing they'll be pretty similar to last year, if not a couple of marks lower just because some of the 4 markers were quite hard.
Last year's grade boundaries were this by the way:

58/52/45/38 for A*/A/B/C...
Original post by jess2157
For question 6 my answer was weird. I tried over and over but couldn't fix it. I got that coshy = +-5/12, but then I couldn't do the inverse of either of those answers soooooooo pffftt I don't know about that.
For question 7, my final answer was Ο€/3[64sqrt(2)-32].


yess i think i put down the same as you for q 7 and yea for q6 you get coshy = +-5/12 but you need to reject the negative solution as coshy>=1
(edited 5 years ago)
Original post by Anonymouspsych
yess i think i put down the same as you for q 7 and yea for q6 you get coshy = +-5/12 but you need to reject the negative solution as coshy>=1 so u just use y=arccosh(5/12) = ln2 i believe.


But I don't understand ughhhhhhhh
For question 6, I got that coshy = +-5/12 and rejected the negative (because of the graph). But when I tried to do cosh-1(5/12), I couldn't get it on my calculator or by using the formula... I don't know where I went wrong haha!
Original post by jess2157
But I don't understand ughhhhhhhh
For question 6, I got that coshy = +-5/12 and rejected the negative (because of the graph). But when I tried to do cosh-1(5/12), I couldn't get it on my calculator or by using the formula... I don't know where I went wrong haha!


Oh my bad it wouldn't be coshy=5/12 as obviously coshy can't be less than 1. I did get an answer with ln something but i can't remember lol do you remember the question xD?
Original post by Anonymouspsych
Oh my bad it wouldn't be coshy=5/12 as obviously coshy can't be less than 1. I did get an answer with ln something but i can't remember lol do you remember the question xD?


Ugh dammit XD I just tried your answer (ln2) it came out with cosh(ln2) = 5/4, if that looks familiar?
The first part said you had to prove the differential of cosh-1x, which I got. In the end, you had to find the y-coordinate of the stationary point for y = (a constant) + (a constant)x - cosh-1(3x) or something, but that's all I remember
Original post by jess2157
Ugh dammit XD I just tried your answer (ln2) it came out with cosh(ln2) = 5/4, if that looks familiar?
The first part said you had to prove the differential of cosh-1x, which I got. In the end, you had to find the y-coordinate of the stationary point for y = (a constant) + (a constant)x - cosh-1(3x) or something, but that's all I remember


yesss ah i remember oh you're gonna be annoyed when you see this. so you worked out x correctly to be 5/12 but it was cosh-1(3x) not cosh-1(x)
so you are actually evaluating cosh-1(3*5/12) =cosh-1(5/4) and since 5/4>1 this will work with the cosh-1(x) formula with ln.

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