FP3 June 2014 Q4

EdexcelOld spec7 marksHyperbolic Functions

4. Using the definitions of hyperbolic functions in terms of exponentials,

(a) show that \[\mathrm{sech}^2 x = 1 - \tanh^2 x\] (3)

(b) solve the equation \[4\sinh x - 3\cosh x = 3\] (4)

Mark scheme (a)

Scheme	Marks
\(\tanh x = \dfrac{\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}}{\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}}\) or \(\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{\mathrm{e}^x + \mathrm{e}^{-x}}\) or \(\dfrac{\mathrm{e}^{2x} - 1}{\mathrm{e}^{2x} + 1}\) Use of the correct exponential form of \(\tanh x\)	M1
\(1 - \tanh^2 x = 1 - \left(\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{\mathrm{e}^x + \mathrm{e}^{-x}}\right)^2 = \dfrac{(\mathrm{e}^{2x} + \mathrm{e}^{-2x} + 2) - (\mathrm{e}^{2x} + \mathrm{e}^{-2x} - 2)}{(\mathrm{e}^x + \mathrm{e}^{-x})^2}\) Attempts \(1 - \tanh^2 x\) with their \(\tanh x\), obtains a common denominator and expands the numerator correctly – three terms from \((a + b)^2\) at least once	dM1
\(= \dfrac{2\mathrm{e}^x \cdot 2\mathrm{e}^{-x}}{(\mathrm{e}^x + \mathrm{e}^{-x})^2}\)
\(= \dfrac{4}{(\mathrm{e}^x + \mathrm{e}^{-x})^2} = \mathrm{sech}^2 x\ ^\) Correct completion with no errors*	A1*
	(3)

Notes

Allow candidates to process both sides and ‘meet in the middle’
Note that it is possible to start from \(\mathrm{sech}^2 x\) and obtain \(1 - \tanh^2 x\) by reversing the above work

Mark scheme (b)

Scheme	Marks
Ignore any imaginary solutions in (b)
\(4\left(\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right) - 3\left(\dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right) = 3\) Substitutes the correct* exponential forms for \(\sinh x\) and \(\cosh x\)*	M1
\(\mathrm{e}^x - 7\mathrm{e}^{-x} = 6\)
\(\mathrm{e}^{2x} - 6\mathrm{e}^x - 7 = 0\) Obtains a quadratic in \(\mathrm{e}^x\)	M1
\((\mathrm{e}^x + 1)(\mathrm{e}^x - 7) = 0 \Rightarrow \mathrm{e}^x = \ldots\) Attempt to solve their 3TQ in \(\mathrm{e}^x\) as far as \(\mathrm{e}^x = \ldots\)	M1
\(x = \ln 7\) or awrt 1.95	A1
	(4)
	(7 marks)

Notes

Alternatives for (b)

Scheme	Marks
\(4\sinh x - 3\cosh x = 3 \Rightarrow 4\sinh x = 3 + 3\cosh x\) \(\Rightarrow 7\cosh^2 x - 18\cosh x - 25 = 0\) M1: Attempt to square correctly and obtains a quadratic in \(\cosh x\)	M1
\(7\cosh^2 x - 18\cosh x - 25 = 0 \Rightarrow (7\cosh x - 25)(\cosh x + 1) = 0\)
\(\cosh x = \tfrac{25}{7} \Rightarrow \tfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2} = \tfrac{25}{7} \Rightarrow 7\mathrm{e}^{2x} - 50\mathrm{e}^x + 7 = 0\) Uses the correct form of \(\cosh x\) in terms of exponentials to obtain a 3TQ in \(\mathrm{e}^x\)	M1
\(7\mathrm{e}^{2x} - 50\mathrm{e}^x + 7 = 0 \Rightarrow (7\mathrm{e}^x - 1)(\mathrm{e}^x - 7) = 0 \Rightarrow \mathrm{e}^x = \ldots\) Attempt to solve their 3TQ as far as \(\mathrm{e}^x = \ldots\)	M1
\(x = \ln 7\) or awrt 1.95 No other values	A1

Scheme	Marks
\(4\sinh x - 3\cosh x = 3 \Rightarrow 4\sinh x - 3 = 3\cosh x\) \(\Rightarrow 7\sinh^2 x - 24\sinh x = 0\) M1: Attempt to square correctly and obtains a quadratic in \(\sinh x\)	M1
\(7\sinh^2 x - 24\sinh x = 0 \Rightarrow \sinh x(7\sinh x - 24) = 0\)
\(\sinh x = \tfrac{24}{7} \Rightarrow \tfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2} = \tfrac{24}{7} \Rightarrow 7\mathrm{e}^{2x} - 48\mathrm{e}^x - 7 = 0\) Uses the correct form of \(\sinh x\) in terms of exponentials to obtain a 3TQ in \(\mathrm{e}^x\)	M1
\(7\mathrm{e}^{2x} - 48\mathrm{e}^x - 7 = 0 \Rightarrow (7\mathrm{e}^x + 1)(\mathrm{e}^x - 7) = 0 \Rightarrow \mathrm{e}^x = \ldots\) Attempt to solve their 3TQ as far as \(\mathrm{e}^x = \ldots\)	M1
\(x = \ln 7\) or awrt 1.95 No other values	A1

Scheme	Marks
\(4\sinh x - 3\cosh x = 3 \Rightarrow 4\tanh x - 3 = 3\,\mathrm{sech}\,x\) \(\Rightarrow 25\tanh^2 x - 24\tanh x = 0\) M1: Attempt to square correctly and obtains a quadratic in \(\tanh x\)	M1
\(25\tanh^2 x - 24\tanh x = 0 \Rightarrow \tanh x(25\tanh x - 24) = 0\)
\(\tanh x = \tfrac{24}{25} \Rightarrow \tfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{\mathrm{e}^x + \mathrm{e}^{-x}} = \tfrac{24}{25} \Rightarrow \mathrm{e}^{2x} = 49\) Uses the correct form of \(\tanh x\) in terms of exponentials to obtain a 2TQ in \(\mathrm{e}^x\)	M1
\(\mathrm{e}^{2x} = 49 \Rightarrow \mathrm{e}^x = \ldots\) Attempt to solve their 2TQ as far as \(\mathrm{e}^x = \ldots\)	M1
\(x = \ln 7\) or awrt 1.95 No other values	A1

(corrected from the printed mark scheme: in the first two alternatives the guidance is printed the other way round, “a quadratic in \(\sinh x\)” for the \(\cosh x\) quadratic and “a quadratic in \(\cosh x\)” for the \(\sinh x\) quadratic)