Edexcel A-level Further Maths coverage

Hyperbolic functions

Section CP-8
5 spec leafs

Notes and three levels of exam-style practice for each registered specification leaf in this section.

Open the printable pack
CP-8.1

Understand the definitions of hyperbolic functions sinh x, cosh x and tanh x, including their domains and ranges, and be able to sketch their graphs.

  • sinhx=exex2\sinh x=\dfrac{e^x-e^{-x}}2, coshx=ex+ex2\cosh x=\dfrac{e^x+e^{-x}}2 and tanhx=sinhxcoshx\tanh x=\dfrac{\sinh x}{\cosh x}.
  • All three functions have domain R\mathbb R; their ranges are R\mathbb R, [1,)[1,\infty) and (1,1)(-1,1) respectively.
  • sinhx\sinh x and tanhx\tanh x are odd and increasing, while coshx\cosh x is even with minimum point (0,1)(0,1); y=±1y=\pm1 are asymptotes of y=tanhxy=\tanh x.
  • Do not give tanhx\tanh x the closed range [1,1][-1,1]: its graph approaches but never reaches either horizontal asymptote.

Tier 1 · Easy

3 marks
ORIGINAL

State the domain, range and yy-intercept of y=coshxy=\cosh x.

Tier 2 · Standard

4 marks
ORIGINAL

Using the exponential definitions, find the exact values of sinh(ln3)\sinh(\ln3), cosh(ln3)\cosh(\ln3) and tanh(ln3)\tanh(\ln3).

Tier 3 · Hard

5 marks
ORIGINAL

Sketch y=23tanhxy=2-3\tanh x. Label its intercept and both horizontal asymptotes, and state its range.

CP-8.2

Differentiate and integrate hyperbolic functions.

  • The basic derivatives are (sinhx)=coshx(\sinh x)'=\cosh x, (coshx)=sinhx(\cosh x)'=\sinh x and (tanhx)=sech2x(\tanh x)'=\operatorname{sech}^2x.
  • Apply the chain, product and quotient rules exactly as for other differentiable functions; the derivative of coshx\cosh x has no minus sign.
  • Reverse these results when integrating, including the inner derivative: for example sinh(ax+b)dx=1acosh(ax+b)+C\int\sinh(ax+b)\,dx=\dfrac1a\cosh(ax+b)+C.
  • A common error is to copy circular-trigonometric signs, writing (coshx)=sinhx(\cosh x)'=-\sinh x instead of +sinhx+\sinh x.

Tier 1 · Easy

2 marks
ORIGINAL

Differentiate y=5cosh(4x)y=5\cosh(4x) with respect to xx.

Tier 2 · Standard

3 marks
ORIGINAL

Find (5cosh(3x)2sinhx)dx\displaystyle\int(5\cosh(3x)-2\sinh x)\,dx.

Tier 3 · Hard

6 marks
ORIGINAL

The function is f(x)=excosh(2x)f(x)=e^{-x}\cosh(2x). Find the exact coordinate of its stationary point and determine its nature.

CP-8.3

Understand and be able to use the definitions of the inverse hyperbolic functions and their domains and ranges.

  • y=arsinhxy=\operatorname{arsinh}x means x=sinhyx=\sinh y; it has domain R\mathbb R and range R\mathbb R.
  • y=arcoshxy=\operatorname{arcosh}x is the inverse of the restriction of cosh\cosh to x0x\geq0, so its domain is [1,)[1,\infty) and its range is [0,)[0,\infty).
  • y=artanhxy=\operatorname{artanh}x means x=tanhyx=\tanh y; it has domain (1,1)(-1,1) and range R\mathbb R.
  • Do not treat cosh\cosh as one-to-one on all real numbers: its principal inverse returns the non-negative value only.

Tier 1 · Easy

2 marks
ORIGINAL

State the domain and range of y=arcoshxy=\operatorname{arcosh}x.

Tier 2 · Standard

3 marks
ORIGINAL

Solve arsinh(2x1)=ln2\operatorname{arsinh}(2x-1)=\ln2 exactly.

Tier 3 · Hard

5 marks
ORIGINAL

Let u=artanh(1/3)u=\operatorname{artanh}(1/3). Find the exact value of xx satisfying arcoshx=2u\operatorname{arcosh}x=2u, without using decimal approximations.

CP-8.4

Derive and use the logarithmic forms of the inverse hyperbolic functions.

  • arsinhx=ln(x+x2+1)\operatorname{arsinh}x=\ln(x+\sqrt{x^2+1}) for every real xx.
  • arcoshx=ln(x+x21)\operatorname{arcosh}x=\ln(x+\sqrt{x^2-1}) for x1x\geq1, using the non-negative principal branch.
  • artanhx=12ln(1+x1x)\operatorname{artanh}x=\dfrac12\ln\left(\dfrac{1+x}{1-x}\right) for x<1|x|<1.
  • When deriving a logarithmic form, reject any algebraic root that would make ey0e^y\leq0, and retain the inverse function's domain restriction.

Tier 1 · Easy

2 marks
ORIGINAL

Use a logarithmic form to evaluate arsinh(3/4)\operatorname{arsinh}(3/4) exactly.

Tier 2 · Standard

4 marks
ORIGINAL

Solve arsinhx=ln5ln2\operatorname{arsinh}x=\ln5-\ln2, giving xx exactly.

Tier 3 · Hard

6 marks
ORIGINAL

Using logarithmic forms, prove that artanh(x1+x2)=arsinhx\operatorname{artanh}\left(\dfrac{x}{\sqrt{1+x^2}}\right)=\operatorname{arsinh}x for every real xx.

CP-8.5

Integrate functions of the form (x^2 + a^2)^(-1/2) and (x^2 - a^2)^(-1/2) and be able to choose substitutions to integrate associated functions.

  • For x2+a2\sqrt{x^2+a^2}, use x=asinhux=a\sinh u so that x2+a2=acoshu\sqrt{x^2+a^2}=a\cosh u.
  • For x2a2\sqrt{x^2-a^2} on xa>0x\geq a>0, use x=acoshux=a\cosh u so that x2a2=asinhu\sqrt{x^2-a^2}=a\sinh u.
  • The standard antiderivatives are arsinh(x/a)+C\operatorname{arsinh}(x/a)+C for (x2+a2)1/2(x^2+a^2)^{-1/2} and arcosh(x/a)+C\operatorname{arcosh}(x/a)+C for (x2a2)1/2(x^2-a^2)^{-1/2} on the stated branch.
  • Transform dxdx and the limits as well as the radical; omitting one of these is the usual source of an incorrect scale factor.

Tier 1 · Easy

2 marks
ORIGINAL

Find 1x2+16dx\displaystyle\int\dfrac{1}{\sqrt{x^2+16}}\,dx.

Tier 2 · Standard

6 marks
ORIGINAL

Using x=3sinhux=3\sinh u, evaluate 04x2+9dx\displaystyle\int_0^4\sqrt{x^2+9}\,dx exactly.

Tier 3 · Hard

6 marks
ORIGINAL

Use a hyperbolic substitution to evaluate 352x29x29dx\displaystyle\int_3^5\dfrac{2x^2-9}{\sqrt{x^2-9}}\,dx exactly.