A2 June 2025 Paper 1 Q2

OCR ACurrent spec5 marksHyperbolic Functions

(a) Given that \(y = \cosh^{-1}\left(\tfrac{1}{3}x\right)\), find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(x\). [1]

(b) Determine an equation of the normal to the curve \(y = \cosh^{-1}\left(\tfrac{1}{3}x\right)\) at the point where \(x = 5\).
Give your answer in the form \(ax + by = c + \ln d\), where \(a\), \(b\), \(c\) and \(d\) are integers. [4]

Mark scheme (a)

Scheme	Marks	AO
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{\sqrt{x^2 - 9}}\)	B1	1.1
	[1]

Notes

B1: Accept any equivalent (un-simplified) forms e.g. \(\dfrac{1}{3} \times \dfrac{1}{\sqrt{\left(\frac{x}{3}\right)^2 - 1}}\) or \(\dfrac{1}{\sqrt{x^2 - 3^2}}\). Invisible brackets e.g. \(\dfrac{1}{3} \times \dfrac{1}{\sqrt{\frac{x}{3}^2 - 1}}\) is B0 unless correctly recovered. ISW if correct un-simplified form is simplified incorrectly

Mark scheme (b)

Scheme	Marks	AO
\(m_N = -4\)	B1	1.1
\(y = \ln\left(\dfrac{5}{3} + \sqrt{\left(\dfrac{5}{3}\right)^2 - 1}\right) \quad (= \ln 3)\)	B1	1.1
\(y - \ln 3 = -4(x - 5)\)	M1	1.1
\(4x + y = 20 + \ln 3\)	A1	1.1
	[4]

Notes

B1: Correct normal gradient (soi) – possibly BC

B1: For correct un-simplified (or simplified) \(y\)-coordinate in logarithmic form when \(x = 5\) – ISW if simplified incorrectly

M1: Any complete correct method for the equation of a straight line with \(x = 5\), their \(y\) coordinate (which must be of the form \(\ln(a)\) where \(a \gt 0\)) and any non-zero gradient. If using \(y = mx + c\) then must explicitly find \(c\) using \(x = 5\), any non-zero \(m\) and their \(y\) which must be of the form as stated above (but allow errors in the evaluation of \(c\))

A1: oe but must be integer values e.g. \(8x + 2y = 40 + \ln 9\) and must be of the form \(ax + by = c + \ln d\) e.g. allow \(-\ln 3 - 20 = -y - 4x\)