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Inverse hyperbolic functionsEdexcel International A Level Further Maths: Subtopic test

10 questions, 27 marks

Edexcel International A Level Further Maths

Inverse hyperbolic functions

Total 27 marks

Name

Class

Date

  1. 1
    The inverse hyperbolic functions have the logarithmic forms arsinh⁡x=ln⁡(x+x2+1)\operatorname{arsinh}x=\ln\left(x+\sqrt{x^2+1}\right) for all real xx, arcosh⁡x=ln⁡(x+x2−1)\operatorname{arcosh}x=\ln\left(x+\sqrt{x^2-1}\right) for x≥1x\ge1, and artanh⁡x=12ln⁡1+x1−x\operatorname{artanh}x=\frac12\ln\frac{1+x}{1-x} for ∣x∣<1|x|<1.
    (a)
    Find the exact value of arsinh⁡34\operatorname{arsinh}\frac34.
    [1 mark]
    • Aln⁡54\ln\frac54
    • Bln⁡2\ln2
    • C−ln⁡2-\ln2
    • Dln⁡74\ln\frac74
    (b)
    Find the exact value of artanh⁡12\operatorname{artanh}\frac12.
    [1 mark]
    • Aln⁡3\ln3
    • B12ln⁡13\frac12\ln\frac13
    • Cln⁡32\ln\frac32
    • D12ln⁡3\frac12\ln3
    (c)
    Find the exact value of arcosh⁡5\operatorname{arcosh}5, giving your answer in the form ln⁡(a+b6)\ln\left(a+b\sqrt6\right).
    [2 marks]

    Total for question 1: 4 marks

  2. 2
    The function ff is defined by f(x)=arcosh⁡xf(x)=\operatorname{arcosh}x.
    (a)
    What is the largest possible domain of ff?
    [1 mark]
    • Ax≥1x\ge1
    • Ball real xx
    • C∣x∣<1|x|<1
    • Dx≥0x\ge0
    (b)
    Find the value of xx for which f(x)=ln⁡3f(x)=\ln3.
    [1 mark]
    • A43\frac43
    • B33
    • C53\frac53
    • D103\frac{10}{3}
    (c)
    Explain how the graph of y=arcosh⁡xy=\operatorname{arcosh}x is related to the graph of y=cosh⁡xy=\cosh x.
    [2 marks]

    Total for question 2: 4 marks

  3. 3
    For real xx, y=arsinh⁡xy=\operatorname{arsinh}x means that x=sinh⁡yx=\sinh y.
    (a)
    Prove that arsinh⁡x=ln⁡(x+x2+1)\operatorname{arsinh}x=\ln\left(x+\sqrt{x^2+1}\right).
    [3 marks]
    (b)
    Using the result in part (a), show that arsinh⁡(−x)=−arsinh⁡x\operatorname{arsinh}(-x)=-\operatorname{arsinh}x.
    [4 marks]

    Total for question 3: 7 marks

  4. 4
    For ∣x∣<1|x|<1, y=artanh⁡xy=\operatorname{artanh}x means that x=tanh⁡yx=\tanh y.
    (a)
    Prove that artanh⁡x=12ln⁡1+x1−x\operatorname{artanh}x=\frac12\ln\frac{1+x}{1-x}, and explain why the result needs ∣x∣<1|x|<1.
    [6 marks]
    (b)
    Using logarithmic forms, solve artanh⁡x=arsinh⁡43−ln⁡2\operatorname{artanh}x=\operatorname{arsinh}\frac43-\ln2, giving the exact value of xx.
    [6 marks]

    Total for question 4: 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).