A2 June 2022 Paper 1 Q9

EdexcelCurrent spec6 marksHyperbolic Functions Integration

(i)

(a) Explain why \(\displaystyle\int_0^{\infty}\cosh x\,\mathrm{d}x\) is an improper integral. (1)

(b) Show that \(\displaystyle\int_0^{\infty}\cosh x\,\mathrm{d}x\) is divergent. (3)

(ii) \[4\sinh x = p\cosh x \qquad \text{where } p \text{ is a real constant}\]Given that this equation has real solutions, determine the range of possible values for \(p\) (2)

Mark scheme (i)(a)

Scheme	Marks	AO
E. g. Because the interval being integrated over is unbounded. \(\cosh x\) is undefined at the limit of \(\infty\) the upper limit is infinite	B1	1.2
	(1)

Notes

B1: For a suitable explanation. Technically this should refer to the interval being unbounded, but this is unlikely to be seen. Accept “Because the upper limit is infinity”, but not “because it is infinity” without reference to what “it” is. Do not accept “the upper limit tends to infinity” or “the integral is unbounded”.

Mark scheme (i)(b)

Scheme	Marks	AO
\(\displaystyle\int_0^{\infty}\cosh x\,\mathrm{d}x = \lim_{t \to \infty}\int_0^{t}\cosh x\,\mathrm{d}x\) or \(\displaystyle\lim_{t \to \infty}\int_0^{t}\frac{1}{2}\left(\mathrm{e}^x + \mathrm{e}^{-x}\right)\mathrm{d}x\)	B1	2.5
\(\displaystyle\int_0^{t}\cosh x\,\mathrm{d}x = [\sinh x]_0^t = \sinh t\ (-0)\) or \(\displaystyle\frac{1}{2}\int_0^{t}\mathrm{e}^x + \mathrm{e}^{-x}\,\mathrm{d}x = \frac{1}{2}\left[\mathrm{e}^x - \mathrm{e}^{-x}\right]_0^t = \frac{1}{2}\left[\mathrm{e}^t - \mathrm{e}^{-t}\right]\left(-\frac{1}{2}\left[\mathrm{e}^0 - \mathrm{e}^0\right]\right)\)	M1	1.1b
When \(t \to \infty\) \(\mathrm{e}^t \to \infty\) and \(\mathrm{e}^{-t} \to 0\) therefore the integral is divergent	A1	2.4
	(3)

Notes

B1: Writes the integral in terms of a limit as \(t \to \infty\) (or other variable) with limits 0 and “\(t\)”, or implies the integral is a limit by subsequent working by correct language.

M1: Integrates \(\cosh x\) correctly either as \(\sinh x\) or in terms of exponentials and applies correctly the limits of 0 and “\(t\)”. The bottom limit zero may be implied. No need for the \(\lim\limits_{t \to \infty}\) for this mark but substitution of \(\infty\) is M0.

A1: cso States that (as \(t \to \infty\)) \(\sinh t \to \infty\) or \(\mathrm{e}^t \to \infty\) and \(\mathrm{e}^{-t} \to 0\) therefore divergent (or not convergent), or equivalent working. Accept \(\sinh t\) is undefined as \(t \to \infty\)

Mark scheme (ii)

Scheme	Marks	AO
\(4\sinh x = p\cosh x \Rightarrow \tanh x = \dfrac{p}{4}\) or \(4\tanh x = p\) Alternative \(\dfrac{4}{2}\left(\mathrm{e}^x - \mathrm{e}^{-x}\right) = \dfrac{p}{2}\left(\mathrm{e}^x + \mathrm{e}^{-x}\right) \Rightarrow 4\mathrm{e}^x - 4\mathrm{e}^{-x} = p\mathrm{e}^x + p\mathrm{e}^{-x}\) \(\mathrm{e}^{2x}(4 - p) = p + 4 \Rightarrow \mathrm{e}^{2x} = \dfrac{p + 4}{4 - p}\)	M1	3.1a
\(\left\{-1 \lt \dfrac{p}{4} \lt 1 \Rightarrow\right\} {-4} \lt p \lt 4\)	A1	2.2a
	(2)
	(6 marks)

Notes

M1: Divides through by \(\cosh x\) to find an expression involving \(\tanh x\)
Alternative: uses the correct exponential definitions and finds an expression for \(\mathrm{e}^{2x}\) or solves a quadratic in \(\mathrm{e}^{2x}\)

A1: Deduces the correct inequality for \(p\). Note \(|p| \lt 4\) is a correct inequality for \(p\).