FP3 June 2014 (R) Q1

EdexcelOld spec5 marksHyperbolic Functions

1. Solve the equation \[5\tanh x + 7 = 5\,\mathrm{sech}\,x\]

Give each answer in the form \(\ln k\) where \(k\) is a rational number. (5)

Mark scheme

Scheme	Marks
\(5\tanh x + 7 = 5\,\mathrm{sech}\,x\)
\(5\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{\mathrm{e}^x + \mathrm{e}^{-x}} + 7 = \dfrac{10}{\mathrm{e}^x + \mathrm{e}^{-x}}\) The given equation correctly expressed in terms of exponentials in any form.	B1
\(5\tanh x + 7 = 5\,\mathrm{sech}\,x \Rightarrow 5\sinh x + 7\cosh x = 5\) \(\Rightarrow 5\dfrac{(\mathrm{e}^x - \mathrm{e}^{-x})}{2} + 7\dfrac{(\mathrm{e}^x + \mathrm{e}^{-x})}{2} = 5\) could also score B1
\(5(\mathrm{e}^x - \mathrm{e}^{-x}) + 7(\mathrm{e}^x + \mathrm{e}^{-x}) = 10\)
\(5(\mathrm{e}^{2x} - 1) + 7(\mathrm{e}^{2x} + 1) = 10\mathrm{e}^x\) Attempt quadratic in \(\mathrm{e}^x\)	M1
\(12\mathrm{e}^{2x} - 10\mathrm{e}^x + 2 = 0\) Correct quadratic	A1
\(6\mathrm{e}^{2x} - 5\mathrm{e}^x + 1 = 0 \Rightarrow (3\mathrm{e}^x - 1)(2\mathrm{e}^x - 1) = 0\) Solves their 3TQ in \(\mathrm{e}^x\)	M1
\(x = \ln\left(\tfrac{1}{3}\right),\ \ln\left(\tfrac{1}{2}\right)\) Both correct (Allow \(-\ln 3\) and/or \(-\ln 2\))	A1
	(5)
	(5 marks)

Notes

Alternative 1

Scheme	Marks
\(5\tanh x + 7 = 5\,\mathrm{sech}\,x \Rightarrow 25\tanh^2 x + 70\tanh x + 49 = 25\,\mathrm{sech}^2 x\)
\(50\tanh^2 x + 70\tanh x + 24 = 0\) Correct quadratic in \(\tanh x\)	B1
\(\tanh x = -\dfrac{4}{5},\ \tanh x = -\dfrac{3}{5}\) M1: Solves their 3TQ in \(\tanh x\) A1: Correct values	M1A1
\(\dfrac{\mathrm{e}^{2x} - 1}{\mathrm{e}^{2x} + 1} = -\dfrac{4}{5} \Rightarrow \mathrm{e}^{2x} = \dfrac{1}{9} \Rightarrow x = \ln\dfrac{1}{3}\) \(\dfrac{\mathrm{e}^{2x} - 1}{\mathrm{e}^{2x} + 1} = -\dfrac{3}{5} \Rightarrow \mathrm{e}^{2x} = \dfrac{1}{4} \Rightarrow x = \ln\dfrac{1}{2}\) M1: Uses the correct exponential form of \(\tanh x\) to obtain a value for \(x\) at least once A1 both answers correct	M1A1

Alternative 2

Scheme	Marks
\(5\sinh x + 7\cosh x = 5 \Rightarrow 49\cosh^2 x = 25 - 50\sinh x + 25\sinh^2 x\)
\(24\sinh^2 x + 50\sinh x + 24 = 0\) Correct quadratic in \(\sinh x\)	B1
\(\sinh x = -\dfrac{4}{3},\ \sinh x = -\dfrac{3}{4}\) M1: Solves their 3TQ in \(\sinh x\) A1: Correct values	M1A1
\(\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2} = -\dfrac{4}{3} \Rightarrow \mathrm{e}^x = \dfrac{1}{3} \Rightarrow x = \ln\dfrac{1}{3}\) \(\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2} = -\dfrac{3}{4} \Rightarrow \mathrm{e}^x = \dfrac{1}{2} \Rightarrow x = \ln\dfrac{1}{2}\) M1: Uses the correct exponential form of \(\sinh x\) to obtain a value for \(x\) at least once A1 both answers correct	M1A1

(corrected from the printed mark scheme: in Alternative 2 the exponential form of \(\sinh x\) is printed as \(\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{\mathrm{e}^x + \mathrm{e}^{-x}}\), which is \(\tanh x\); \(\sinh x = \dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\) gives the values shown)