A2 June 2025 Paper 1 Q14

OCR MEICurrent spec10 marksHyperbolic Functions Matrices

(a) By using the definition of \(\cosh x\) and \(\sinh x\) in terms of \(\mathrm{e}^x\) and \(\mathrm{e}^{-x}\), show that \(\cosh^2 x + \sinh^2 x \equiv \cosh 2x\). [2]

(b) The transformation T of the plane has associated matrix \(\mathbf{M}\), where \(\mathbf{M} = \begin{pmatrix} \cosh x & \sinh x \\ \sinh x & \cosh x \end{pmatrix}\) and \(x \gt 0\).
Show that T transforms the unit square with coordinates \((0, 0)\), \((1, 0)\), \((0, 1)\) and \((1, 1)\) to a rhombus of unit area. [6]

(c) You are given that the length of each side of the rhombus is 2 units.
Determine the exact value of \(x\). Give your answer in logarithmic form. [2]

Mark scheme (a)

Scheme	Marks	AO
\(\cosh^2 x + \sinh^2 x = \frac{1}{4}(\mathrm{e}^x + \mathrm{e}^{-x})^2 + \frac{1}{4}(\mathrm{e}^x - \mathrm{e}^{-x})^2\)	M1	2.1
\(= \frac{1}{4}(\mathrm{e}^{2x} + 2 + \mathrm{e}^{-2x} + \mathrm{e}^{2x} - 2 + \mathrm{e}^{-2x})\) \(= \frac{1}{2}(\mathrm{e}^{2x} + \mathrm{e}^{-2x}) = \cosh 2x\)	A1	2.2a
	[2]

Notes

M1: LHS correctly written in exponential form. Must be seen.

A1: AG expanded and simplified to \(\frac{1}{2}(\mathrm{e}^{2x} + \mathrm{e}^{-2x})\) must be seen before final answer. Complete argument required, www.

Mark scheme (b)

Scheme	Marks	AO
\(\begin{pmatrix} \cosh x & \sinh x \\ \sinh x & \cosh x \end{pmatrix}\begin{pmatrix} 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \end{pmatrix}\)	M1	2.1
\(= \begin{pmatrix} 0 & \cosh x & \sinh x & \cosh x + \sinh x \\ 0 & \sinh x & \cosh x & \cosh x + \sinh x \end{pmatrix}\)	A1	1.1
Suppose vertices are O, P, Q, R respectively \(\mathrm{OP}^2 = \mathrm{OQ}^2 = \cosh^2 x + \sinh^2 x\) \(\mathrm{PR}^2 = (\cosh x + \sinh x - \sinh x)^2 + (\cosh x + \sinh x - \cosh x)^2 = \cosh^2 x + \sinh^2 x\) \(\mathrm{QR}^2 = (\cosh x + \sinh x - \sinh x)^2 + (\cosh x + \sinh x - \cosh x)^2 = \cosh^2 x + \sinh^2 x\)	M1	3.1a
So \(\mathrm{OP} = \mathrm{OQ} = \mathrm{PR} = \mathrm{QR}\) so rhombus	A1	2.2a
\(\det\mathbf{M} = \cosh^2 x - \sinh^2 x\)	M1	2.1
\(= 1 \Rightarrow\) transformation preserves area, so area of rhombus is 1	A1	3.2a
	[6]

Notes

M1: finding images by matrix multiplication or at least two correct images

A1: or all correct images found

M1: explicitly giving two side lengths of image or showing that there are two pairs of parallel sides. Lengths or their squares could be used.

A1: Complete method to show image is rhombus and concluding, e.g. showing all four sides are equal (working must be seen for PR and QR). Could also show there are two pairs of parallel sides and two adjacent sides are equal length; that P and Q are reflections in \(y = x\), R lies on \(y = x\) and OP = PR; that two diagonals are perpendicular bisectors of each other. Results must be shown not just stated.

M1: must be seen; or for a complete method to find area of image

A1: \(= 1\), and some comment about the effect of the transformation if det M used. Accept “[area scale] factor = 1”, “area stays same”, “area = \(1 \times 1 = 1\)” etc., but not just “so area = 1”.

Mark scheme (c)

Scheme	Marks	AO
\(2 = \sqrt{\cosh^2 x + \sinh^2 x}\) or \(2 = \sqrt{\cosh 2x}\)	M1	3.1a
\(x = \dfrac{1}{2}\ln\left(4 + \sqrt{15}\right)\)	A1	1.1
	[2]

Notes

M1: or \(4 = \cosh^2 x + \sinh^2 x\) or \(4 = \cosh 2x\) oe

A1: oe e.g. \(x = \ln\left(\frac{\sqrt{5} + \sqrt{3}}{\sqrt{2}}\right)\). Do not condone missing brackets. Answer must be supported by some working. ISW if error manipulating logs only.