A2 June 2025 Paper 1 Q18

AQACurrent spec15 marksHyperbolic Functions Integration

18 The diagram shows part of the graph of \(y = 15\operatorname{cosech} x\) and part of the graph of \(y = 4\sinh x + \dfrac{1}{2}\)

Graphs of y = 15 cosech x (two branches, in the first and third quadrants, asymptotic to both axes) and y = 4 sinh x + 1/2 (an increasing curve crossing the y-axis just above O); the curves intersect once in the first quadrant and once in the third quadrant

(a) Solve the inequality\[15\operatorname{cosech} x \lt 4\sinh x + \frac{1}{2}\]

Give your answer in logarithmic form. [4 marks]

(b) Given that\[\mathrm{f}(x) = \ln\left(\tanh\left(\frac{1}{2}x\right)\right) \qquad (x \gt 0)\]

Show that

\[\mathrm{f}^{\prime}(x) = \operatorname{cosech} x\] [4 marks]

(c) The shaded region \(R\) is enclosed by the positive \(x\)-axis, the positive \(y\)-axis, the graph of \(y = 4\sinh x + \dfrac{1}{2}\), the graph of \(y = 15\operatorname{cosech} x\) and the line \(x = \ln 9\)

The same two graphs, with the region R shaded in the first quadrant: bounded by the y-axis, the x-axis, the curve y = 4 sinh x + 1/2 up to the intersection point, then the curve y = 15 cosech x down to a vertical line

Find the area of \(R\)

Give your answer in the form \(\dfrac{p}{q} + \ln r + s\ln\left(\dfrac{t}{3}\right)\) where \(p\), \(q\), \(r\), \(s\) and \(t\) are integers. [7 marks]

Mark scheme (a)

Scheme	Marks	AO
Forms and solves a quadratic equation or inequality in \(\sinh x\) or Forms a quartic expression in \(\mathrm{e}^x\)	M1	3.1a
Obtains \(-2\) and \(\frac{15}{8}\) or Obtains \(4\mathrm{e}^{4x} + \mathrm{e}^{3x} - 68\mathrm{e}^{2x} - \mathrm{e}^x + 4\)	A1	1.1b
Obtains at least one of \(\ln 4\) or \(\ln(\sqrt{5} - 2)\) OE	M1	1.1a
Obtains \(\ln(\sqrt{5} - 2) \lt x \lt 0,\ x \gt \ln 4\) OE	A1	2.2a
	(4)

Typical solution

Let \(s = \sinh x\)

\[\frac{15}{s} = 4s + 0.5\]\[8s^2 + s - 30 = 0\]\[\sinh x = -2, \tfrac{15}{8}\]\[x = \ln 4, \ln(\sqrt{5} - 2)\]

Solution of inequality:

\[\ln(\sqrt{5} - 2) \lt x \lt 0,\ x \gt \ln 4\]

Mark scheme (b)

Scheme	Marks	AO
Uses the chain rule.	M1	3.1a
Obtains a correct expression for the derivative.	A1	1.1b
Uses hyperbolic identities to express their \(\mathrm{f}^{\prime}(x)\) in terms of only \(\sinh\left(\tfrac{1}{2}x\right)\) and \(\cosh\left(\tfrac{1}{2}x\right)\)	M1	2.2a
Completes a reasoned argument to show that \(\mathrm{f}^{\prime}(x) = \operatorname{cosech} x\) AG	R1	2.1
	(4)

Typical solution

\[\mathrm{f}(x) = \ln\left(\tanh\left(\tfrac{1}{2}x\right)\right)\]\[\begin{aligned}\mathrm{f}^{\prime}(x) &= \frac{\frac{1}{2}\operatorname{sech}^2\left(\frac{1}{2}x\right)}{\tanh\left(\frac{1}{2}x\right)} \\ &= \frac{\cosh\left(\frac{1}{2}x\right)}{2\sinh\left(\frac{1}{2}x\right)\cosh^2\left(\frac{1}{2}x\right)} \\ &= \frac{1}{2\sinh\left(\frac{1}{2}x\right)\cosh\left(\frac{1}{2}x\right)} \\ &= \frac{1}{\sinh x} \\ &= \operatorname{cosech} x\end{aligned}\]

Mark scheme (c)

Scheme	Marks	AO
Deduces that the area of \(R\) is the sum of two integrals (at least one correct).	B1	2.2a
Deduces the correct limits for \(A_1\) and \(A_2\)	B1	2.2a
Obtains \(\pm\left(4\cosh x + \frac{1}{2}x\right)\)	B1	1.1b
Obtains \(\pm 15\ln\left(\tanh\left(\frac{1}{2}x\right)\right)\)	B1	1.1b
Obtains \(\pm\left(\frac{9}{2} + \ln 2\right)\) or AWRT \(\pm 5.19\)	M1	1.1a
Obtains \(\pm 15\ln\dfrac{4}{3}\) or AWRT \(\pm 4.32\)	M1	1.1a
Obtains \(\frac{9}{2} + \ln 2 + 15\ln\frac{4}{3}\)	R1	2.1
	(7)
	(15 marks)

Typical solution

Area of \(R = A_1 + A_2\), where

\[\begin{aligned} A_1 &= \int_0^{\ln 4}\left(4\sinh x + \tfrac{1}{2}\right)\mathrm{d}x = \left[4\cosh x + \tfrac{1}{2}x\right]_0^{\ln 4} \\ &= \left(4\cosh(\ln 4) + \tfrac{1}{2}\ln 4\right) - 4 \\ &= \tfrac{9}{2} + \ln 2 \end{aligned}\]\[\begin{aligned} A_2 &= \int_{\ln 4}^{\ln 9} 15\operatorname{cosech} x\,\mathrm{d}x = 15\left[\ln\left(\tanh\left(\tfrac{1}{2}x\right)\right)\right]_{\ln 4}^{\ln 9} \\ &= 15\left\{\ln(\tanh(\ln 3)) - \ln(\tanh(\ln 2))\right\} \\ &= 15\left\{\ln\frac{4}{5} - \ln\frac{3}{5}\right\} = 15\ln\frac{4}{3} \end{aligned}\]

Area of \(R = \frac{9}{2} + \ln 2 + 15\ln\frac{4}{3}\)