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FP3: Hyperbolic functionsEdexcel International A Level Further Maths: Topic test

20 questions, 54 marks

Edexcel International A Level Further Maths

FP3: Hyperbolic functions topic test

Total 54 marks

Name

Class

Date

  1. 1
    The real number xx satisfies sinh⁡x=34\sinh x=\frac34.
    (a)
    Find cosh⁡x\cosh x.
    [1 mark]
    • A74\frac{\sqrt7}{4}
    • B54\frac54
    • C2516\frac{25}{16}
    • D74\frac74
    (b)
    Find tanh⁡x\tanh x.
    [1 mark]
    • A43\frac43
    • B1516\frac{15}{16}
    • C34\frac34
    • D35\frac35
    (c)
    Find the exact value of xx.
    [2 marks]

    Total for question 1: 4 marks

  2. 2
    The real numbers aa and bb are defined by a=arsinh⁡512a=\operatorname{arsinh}\frac{5}{12} and b=arcosh⁡53b=\operatorname{arcosh}\frac53.
    (a)
    Find aa in exact logarithmic form.
    [1 mark]
    • Aln⁡32\ln\frac32
    • Bln⁡23\ln\frac23
    • Cln⁡1312\ln\frac{13}{12}
    • Dln⁡1712\ln\frac{17}{12}
    (b)
    Find bb in exact logarithmic form.
    [1 mark]
    • Aln⁡53\ln\frac53
    • Bln⁡43\ln\frac43
    • Cln⁡3\ln3
    • Dln⁡13\ln\frac13
    (c)
    Find the exact value of sinh⁡(a+b)\sinh(a+b).
    [2 marks]

    Total for question 2: 4 marks

  3. 3
    Consider the equation cosh⁡2x−3sinh⁡x=1\cosh2x-3\sinh x=1.
    (a)
    Show that the equation can be written as 2sinh⁡2x−3sinh⁡x=02\sinh^2x-3\sinh x=0.
    [3 marks]
    (b)
    Hence solve the equation, giving your non-zero answer in exact logarithmic form.
    [4 marks]

    Total for question 3: 7 marks

  4. 4
    The curve CC has equation y=3cosh⁡x−sinh⁡xy=3\cosh x-\sinh x and has exactly one stationary point, which is a minimum.
    (a)
    Find the exact coordinates of the minimum point of CC.
    [6 marks]
    (b)
    Solve the equation 3cosh⁡x−sinh⁡x=43\cosh x-\sinh x=4, giving your answers as exact logarithms.
    [6 marks]

    Total for question 4: 12 marks

  5. 5
    The real number xx satisfies cosh⁡x=135\cosh x=\frac{13}{5} and x>0x>0.
    (a)
    Find sinh⁡x\sinh x.
    [1 mark]
    • A1945\frac{\sqrt{194}}{5}
    • B512\frac{5}{12}
    • C125\frac{12}{5}
    • D16925\frac{169}{25}
    (b)
    Find cosh⁡2x\cosh2x.
    [1 mark]
    • A31325\frac{313}{25}
    • B33825\frac{338}{25}
    • C28825\frac{288}{25}
    • D31225\frac{312}{25}
    (c)
    Find the exact value of tanh⁡2x\tanh2x.
    [2 marks]

    Total for question 5: 4 marks

  6. 6
    The real number pp is defined by p=arsinh⁡2p=\operatorname{arsinh}2.
    (a)
    Find pp in exact logarithmic form.
    [1 mark]
    • Aln⁡(2+3)\ln\left(2+\sqrt3\right)
    • Bln⁡(5−2)\ln\left(\sqrt5-2\right)
    • Cln⁡(1+5)\ln\left(1+\sqrt5\right)
    • Dln⁡(2+5)\ln\left(2+\sqrt5\right)
    (b)
    Find cosh⁡p\cosh p.
    [1 mark]
    • A55
    • B5\sqrt5
    • C3\sqrt3
    • D15\frac{1}{\sqrt5}
    (c)
    Show that ep−e−p=4\mathrm{e}^p-\mathrm{e}^{-p}=4, using the logarithmic form of pp.
    [2 marks]

    Total for question 6: 4 marks

  7. 7
    This question is about the double-angle result sinh⁡2x≡2sinh⁡xcosh⁡x\sinh2x\equiv2\sinh x\cosh x.
    (a)
    Prove that sinh⁡2x≡2sinh⁡xcosh⁡x\sinh2x\equiv2\sinh x\cosh x, using the definitions of sinh⁡\sinh and cosh⁡\cosh in terms of exponentials.
    [3 marks]
    (b)
    Hence solve the equation sinh⁡2x=5sinh⁡x\sinh2x=5\sinh x, giving your answers in exact form.
    [4 marks]

    Total for question 7: 7 marks

  8. 8
    For x≥1x\ge1, let y=arcosh⁡xy=\operatorname{arcosh}x, where y≥0y\ge0.
    (a)
    Prove that arcosh⁡x=ln⁡(x+x2−1)\operatorname{arcosh}x=\ln\left(x+\sqrt{x^2-1}\right).
    [6 marks]
    (b)
    Use the result of part (a), and the logarithmic form of arsinh⁡\operatorname{arsinh}, to solve the equation arcosh⁡x=2arsinh⁡815\operatorname{arcosh}x=2\operatorname{arsinh}\frac{8}{15}.
    [6 marks]

    Total for question 8: 12 marks

End of questions

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).