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

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    I'll start it below
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    Q1

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

    \dfrac{1}{(2r+1)(2r+3)} - \dfrac{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 \displaystyle \sum_{r=1}^{N} \frac{1}{((2r+1)(2r+3)(2r+5)}. [3 marks]

    \displaystyle \frac{1}{4}\sum_{r=1}^{N}\mathrm  {f}(r-1) - \mathrm{f}(r)

    \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, \frac{1}{4}(\mathrm{f}(0) - \mathrm{f}(N))

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

    Q2

    z=-2\sqrt{2} + 2\sqrt{6}\mathrm{i} can be expressed as r\mathrm{e}^{\mathrm{i}\theta} with r > 0 and -\pi < \theta \leqslant \pi.

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

    r = |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 \theta. [1 mark]

    \theta = \pi - \tan^{-1} \dfrac{2\sqrt{6}}{2\sqrt{2}} = \dfrac{2\pi}{3}

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

    Q4

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

    1 + \mathrm{e}^{2x} + \mathrm{e}^{-2x} + 1 = 2 + \mathrm{e}^{2x} + \mathrm{e}^{-2x} = 2 + 2\cosh 2x

    2+2(\cosh^2 x - 1) = 4\cosh^2 x

    (b)Find \displaystyle \int^1_0 \frac{1}{(1+ \mathrm{e}^{2x})(1+ \mathrm{e}^{-2x})}\;\mathrm{d}x in terms of \mathrm{e}. [4 marks]

    \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

    \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 |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 z_1 is such that |z - 2| = |z + 4\mathrm{i}| and |z| is at its maximum value. [4 marks]

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

    z_1 = -\dfrac{3}{5} - \dfrac{6}{5}\mathrm{i}

    Q6

    (a) Sketch the graph of y=\cosh^{-1} x. [2 marks]

    This can be obtained off a graphical calculator

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

    \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^2 y = \cosh^2 y-1 = x^2-1 \Rightarrow \sinh y = \sqrt{x^2-1}, since y \geqslant 0 by the sketch so \sinh y \geqslant 0

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

    \dfrac{\mathrm{d}y}{\mathrm{d}x} = -4 + \dfrac{3}{\sqrt{9x^2-1}}

    At stationary points, dy/dx = 0, 9x^2-1 = \dfrac{16}{9} \Rightarrow x = \pm \dfrac{5}{12}, but 3x \geqslant 1 and so x = \dfrac{5}{12}

    y=\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=3-\cos 2t, y=4\sin t

    (a) Show that the surface area is \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

    \displaystyle 2\pi \int^{\frac{\pi}{2}}_0 4\sin t \sqrt{4\sin^2 2t + 16\cos^2t}\;\mathrm{d}t

    \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

    \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 \dfrac{\pi}{3}(n\sqrt{2}+m) [5 marks]

    u=1+\sin^2 t \Rightarrow \mathrm{d}t = \dfrac{\mathrm{d}u}{2\sin t \cos t}

    \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

    2z^3 + 5z + 3 = 0

    (a) (i) \alpha\beta + \beta\gamma + \gamma\alpha [1 mark]

    5/2

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

    -3/2

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

    \dfrac{\alpha\beta+\beta\gamma+ \gamma\alpha}{\alpha\beta\gamma} = \dfrac{5/2}{-3/2} = -\dfrac{5}{3}

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

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

    9x^3 - 25x^2 - 20x - 4 = 0

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

    x = \dfrac{1}{z^2} so \dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} + \dfrac{1}{\gamma^2} = \dfrac{25}{9} and \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 \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 5\theta = 16 \cos^5 \theta + A \cos^3 \theta + B \cos \theta. [5 marks]

    de Moire's theorem for n=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^2\theta = 1-\cos^2 \theta gives
    \cos 5\theta = 16 \cos^5 \theta - 20 \cos^3 \theta + 5\cos \theta

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

    Using the de Moivre's theorem result, 16c^5 - 20c^3 + 5c = 0 \Rightarrow 16c^4 - 20c^2 + 5 = 0, where c = \cos \theta

    16u^2 - 20u + 5 = 0 where u = \cos^2 \theta, u = \dfrac{5\pm \sqrt{5}}{8}

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

    \cos 5\theta = 0 \Rightarrow \theta = n\pi + \dfrac{\pi}{2} = \dfrac{(2n+1)\pi}{2}, i.e. \theta = \dfrac{\pi}{10},\;\dfrac{3\pi}{1  0},\;\dfrac{5\pi}{10},\;\dfrac{7  \pi}{10},\;\dfrac{9\pi}{10}, \; \dots

    \cos^2 \dfrac{\pi}{10} = \cos^2 \dfrac{9\pi}{10}, \cos^2 \dfrac{3\pi}{10} = \cos^2 \dfrac{7\pi}{10}, and \cos^2 \dfrac{5\pi}{10} = 0, so we can ignore this value

    \cos^2 \theta is decreasing for 0 < \theta < \pi/2 and so \cos^2 \dfrac{\pi}{10} > \cos^2 \dfrac{3\pi}{10}, and so \cos^2 \dfrac{3\pi}{10} = \dfrac{5-\sqrt{5}}{8}

    \cos \dfrac{3\pi}{5} = 2\cos^2 \dfrac{3\pi}{10} - 1 = \dfrac{5-\sqrt{5}}{4} - 1 = \dfrac{1-\sqrt{5}}{4}
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    (Original post by Integer123)
    Q1

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

    \dfrac{1}{(2r+1)(2r+3)} - \dfrac{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 \displaystyle \sum_{r=1}^{N} \frac{1}{((2r+1)(2r+3)(2r+5)}. [3 marks]

    \displaystyle \frac{1}{4}\sum_{r=1}^{N}\mathrm  {f}(r-1) - \mathrm{f}(r)

    \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, \frac{1}{4}(\mathrm{f}(0) - \mathrm{f}(N))

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

    Q8

    2z^3 + 5z + 3 = 0

    (a) (i) \alpha\beta + \beta\gamma + \gamma\alpha [1 mark]

    5/2

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

    -3/2

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

    \dfrac{\alpha\beta+\beta\gamma+ \gamma\alpha}{\alpha\beta\gamma} = \dfrac{5/2}{-3/2} = -\dfrac{5}{3}

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

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

    9x^3 - 25x^2 - 20x - 4 = 0

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

    x = \dfrac{1}{z^2} so \dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} + \dfrac{1}{\gamma^2} = \dfrac{25}{9} and \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 \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 ?
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    I flopped so bad ffs
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    Ahahah question 8bii is so easy when you spot it but of course I couldn't work it out in the exam
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    You are honestly a lifesaver with these thank you !!
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    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.
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    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.
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    (Original post by Integer123)
    Q1

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

    \dfrac{1}{(2r+1)(2r+3)} - \dfrac{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 \displaystyle \sum_{r=1}^{N} \frac{1}{((2r+1)(2r+3)(2r+5)}. [3 marks]

    \displaystyle \frac{1}{4}\sum_{r=1}^{N}\mathrm  {f}(r-1) - \mathrm{f}(r)

    \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, \frac{1}{4}(\mathrm{f}(0) - \mathrm{f}(N))

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

    Q2

    z=-2\sqrt{2} + 2\sqrt{6}\mathrm{i} can be expressed as r\mathrm{e}^{\mathrm{i}\theta} with r > 0 and -\pi < \theta \leqslant \pi.

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

    r = |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 \theta. [1 mark]

    \theta = \pi - \tan^{-1} \dfrac{2\sqrt{6}}{2\sqrt{2}} = \dfrac{2\pi}{3}

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

    Q4

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

    1 + \mathrm{e}^{2x} + \mathrm{e}^{-2x} + 1 = 2 + \mathrm{e}^{2x} + \mathrm{e}^{-2x} = 2 + 2\cosh 2x

    2+2(\cosh^2 x - 1) = 4\cosh^2 x

    (b) Find \displaystyle \int^1_0 \frac{1}{(1+ \mathrm{e}^{2x})(1+ \mathrm{e}^{-2x})}\;\mathrm{d}x in terms of \mathrm{e}. [4 marks]

    \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

    \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 |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 z_1 is such that |z - 2| = |z + 4\mathrm{i}| and |z| is at its maximum value. [4 marks]

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

    z_1 = -\dfrac{3}{5} - \dfrac{6}{5}\mathrm{i}

    Q8

    2z^3 + 5z + 3 = 0

    (a) (i) \alpha\beta + \beta\gamma + \gamma\alpha [1 mark]

    5/2

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

    -3/2

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

    \dfrac{\alpha\beta+\beta\gamma+ \gamma\alpha}{\alpha\beta\gamma} = \dfrac{5/2}{-3/2} = -\dfrac{5}{3}

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

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

    9x^3 - 25x^2 - 20x - 4 = 0

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

    x = \dfrac{1}{z^2} so \dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} + \dfrac{1}{\gamma^2} = \dfrac{25}{9} and \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 \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 5\theta = 16 \cos^5 \theta + A \cos^3 \theta + B \cos \theta. [5 marks]

    de Moire's theorem for n=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^2\theta = 1-\cos^2 \theta gives
    \cos 5\theta = 16 \cos^5 \theta - 20 \cos^3 \theta + 5\cos \theta

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

    Using the de Moivre's theorem result, 16c^5 - 20c^3 + 5c = 0 \Rightarrow 16c^4 - 20c^2 + 5 = 0, where c = \cos \theta

    16u^2 - 20u + 5 = 0 where u = \cos^2 \theta, u = \dfrac{5\pm \sqrt{5}}{8}

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

    \cos 5\theta = 0 \Rightarrow \theta = n\pi + \dfrac{\pi}{2} = \dfrac{(2n+1)\pi}{2}, i.e. \theta = \dfrac{\pi}{10},\;\dfrac{3\pi}{1  0},\;\dfrac{5\pi}{10},\;\dfrac{7  \pi}{10},\;\dfrac{9\pi}{10}, \; \dots

    \cos^2 \dfrac{\pi}{10} = \cos^2 \dfrac{9\pi}{10}, \cos^2 \dfrac{3\pi}{10} = \cos^2 \dfrac{7\pi}{10}, and \cos^2 \dfrac{5\pi}{10} = 0, so we can ignore this value

    \cos^2 \theta is decreasing for 0 < \theta < \pi/2 and so \cos^2 \dfrac{\pi}{10} > \cos^2 \dfrac{3\pi}{10}, and so \cos^2 \dfrac{3\pi}{10} = \dfrac{5-\sqrt{5}}{8}

    \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
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    i ****ed this lads yw for lowering the grade boundaries.
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    (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 year😂
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    (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 year😂
    do u remember what answer(s) you put for question 6 and 7? Like the last parts of them
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    Anyone got any rough ideas on grade boundaries?
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    (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 π/3[64sqrt(2)-32].
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    (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...
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    (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 π/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
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    (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!
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    (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?
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    (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
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    (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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