Further Mathematics · OCR H245 · Topic 4.07

Hyperbolic functions

Definitions, differentiation and integration, inverse hyperbolic functions.

  • 3 sub-topics
  • Taught in Year 13
  • 37 practice questions
  • 3 animated lessons
Practice questions (37) Lessons Worksheets coming soon All OCR Further Mathematics topics

What the specification says

Every sub-topic OCR can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 4.07a-cDefinitionsY13
  2. 4.07dDifferentiation and integrationY13
  3. 4.07e-fInverse hyperbolic functionsY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 37 questions in this topic, free, no account needed.

Hyperbolic functions

Not needed for Year 12Easy

Hyperbolic definitions · Year 13 · 2 marks

  1. Find the exact value of cosh(ln 3) and sinh(ln 2).
    (2)
Show worked solutionHide worked solution
  1. cosh x = ½(eˣ + e⁻ˣ).cosh⁡(ln⁡3)\cosh(\ln 3)=12(3+13){}= \tfrac{1}{2}(3 + \tfrac{1}{3})=53{}= \tfrac{5}{3}
  2. sinh⁡(ln⁡2)\sinh(\ln 2)=12(2−12){}= \tfrac{1}{2}(2 - \tfrac{1}{2})=34{}= \tfrac{3}{4}

Total2

  • M
  • A
  • B
How did it go?

Report a problem with this question

Hyperbolic functions

Not needed for Year 12Medium

Hyperbolic identity · Year 13 · 2 marks

  1. Prove that cosh2⁡x−sinh2⁡x≡1\cosh^{2}x - \sinh^{2}x \equiv 1.
    (2)
Show worked solutionHide worked solution
  1. Square the exponential definitions.14(e2x+2+e−2x)\tfrac{1}{4}(\mathrm{e}^{2x} + 2 + \mathrm{e}^{-2x})−14(e2x−2+e−2x){}- \tfrac{1}{4}(\mathrm{e}^{2x} - 2 + \mathrm{e}^{-2x})
  2. =44=1= \tfrac{4}{4} = 1

Total2

  • M
  • A
  • B
How did it go?

Report a problem with this question

Hyperbolic functions

Not needed for Year 12Hard

Inverse hyperbolic functions · Year 13 · 6 marks

  1. (a)
    Prove that arsinh⁡x\operatorname{arsinh} x=ln⁡(x+x2+1){}= \ln\left(x + \sqrt{x^2 + 1}\right).
    (4)
  2. (b)
    Hence find the exact value of arsinh⁡512\operatorname{arsinh} \frac{5}{12}.
    (2)
Show worked solutionHide worked solution

Part (a)4 marks

  1. Start from x=sinh⁡yx = \sinh y.yy=arsinh⁡x{}= \operatorname{arsinh} x⇒x{}\Rightarrow x=sinh⁡y{}= \sinh y=ey−e−y2{}= \frac{\mathrm{e}^y - \mathrm{e}^{-y}}{2}
  2. A quadratic in ey\mathrm{e}^y.e2y\mathrm{e}^{2y}−2xey{}- 2x\mathrm{e}^y−1{}- 1=0{}= 0
  3. ey\mathrm{e}^y=x±x2+1{}= x \pm \sqrt{x^2 + 1}
  4. x−x2+1x - \sqrt{x^{2} + 1} is negative, so it is rejected.ey>0, so take \mathrm{e}^y > 0 \text{, so take }+: y{}+ : \ y=ln⁡(x+x2+1){}= \ln\left(x + \sqrt{x^2 + 1}\right)

Part (b)2 marks

  1. ln⁡(512+169144)\ln\left(\frac{5}{12} + \sqrt{\frac{169}{144}}\right)=ln⁡(512+1312){}= \ln\left(\frac{5}{12} + \frac{13}{12}\right)
  2. =ln⁡32= \ln \frac{3}{2}

Total6

  • M
  • A
  • B
How did it go?

Report a problem with this question