A2 June 2022 Paper 1 Q6

AQACurrent spec8 marksHyperbolic Functions

(a) Given that \(|x| \lt 1\), prove that\[\tanh^{-1} x = \frac{1}{2}\ln\left(\frac{1 + x}{1 - x}\right)\] [4 marks]

(b) Solve the equation\[20\operatorname{sech}^2 x - 11\tanh x = 16\]

Give your answer in logarithmic form. [4 marks]

Mark scheme (a)

Scheme	Marks	AO
Recalls exponential definition of tanh	B1	1.2
Multiplies by their denominator and multiplies by \(\mathrm{e}^y\) (or \(\mathrm{e}^x\))	M1	3.1a
Obtains \(\mathrm{e}^{2y} = \dfrac{1 + x}{1 - x}\) or Obtains \(\mathrm{e}^{2x} = \dfrac{1 + y}{1 - y}\) OE	A1	1.1b
Completes a rigorous argument to show the required result	R1	2.1
	(4)

Typical solution

Let \(y = \tanh^{-1} x\)

Then \(x = \tanh y\)

\[x = \frac{\mathrm{e}^y - \mathrm{e}^{-y}}{\mathrm{e}^y + \mathrm{e}^{-y}}\]\[x(\mathrm{e}^y + \mathrm{e}^{-y}) = \mathrm{e}^y - \mathrm{e}^{-y}\]\[\mathrm{e}^y(x - 1) + \mathrm{e}^{-y}(x + 1) = 0\]\[\mathrm{e}^{2y}(x - 1) + (x + 1) = 0\]\[\mathrm{e}^{2y} = \frac{1 + x}{1 - x}\]\[2y = \ln\left(\frac{1 + x}{1 - x}\right)\]\[\tanh^{-1} x = \frac{1}{2}\ln\left(\frac{1 + x}{1 - x}\right)\]

as required.

Mark scheme (b)

Scheme	Marks	AO
Uses appropriate hyperbolic identity (condone sign errors) to obtain an equation in one hyperbolic function or substitutes exponential form, condone error in sum/difference of exponential terms	M1	1.1a
Solves their quadratic equation to obtain two solutions in \(\tanh x\) or quadratic in \(\mathrm{e}^{2x}\) to obtain two solutions or quartic in \(\mathrm{e}^x\) to obtain at least two solutions	M1	1.1a
Obtains \(\tanh x = -\dfrac{4}{5}\) and \(\dfrac{1}{4}\) OE or Obtains \(\mathrm{e}^{2x} = \dfrac{5}{3}\) and \(\dfrac{1}{9}\) OE or Obtains \(\mathrm{e}^x = \dfrac{\sqrt{15}}{3}\) and \(\dfrac{1}{3}\) OE	A1	1.1b
Obtains correct solutions (any correct exact log form) ISW	A1	1.1b
	(4)
	(8 marks)

Typical solution

\[20(1 - \tanh^2 x) - 11\tanh x = 16\]\[0 = 20\tanh^2 x + 11\tanh x - 4\]\[(5\tanh x + 4)(4\tanh x - 1) = 0\]\[\tanh x = -\frac{4}{5} \text{ or } \frac{1}{4}\]\[x = \tanh^{-1}\left(-\frac{4}{5}\right) = \frac{1}{2}\ln\left(\frac{1}{9}\right)\]\[\text{or } x = \tanh^{-1}\left(\frac{1}{4}\right) = \frac{1}{2}\ln\left(\frac{5}{3}\right)\]