FP3 June 2014 (R) Q3

3. The curve \(C\) has equation \[y = \frac{1}{2}\ln(\coth x), \quad x > 0\]

(a) Show that \[\frac{\mathrm{d}y}{\mathrm{d}x} = -\mathrm{cosech}\,2x\] (3)

The points \(A\) and \(B\) lie on \(C\).

The \(x\) coordinates of \(A\) and \(B\) are \(\ln 2\) and \(\ln 3\) respectively.

(b) Find the length of the arc \(AB\), giving your answer in the form \(p\ln q\), where \(p\) and \(q\) are rational numbers. (6)

Mark scheme (a)

Scheme	Marks
\(y = \dfrac{1}{2}\ln(\coth x)\)
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2} \times \dfrac{1}{\coth x} \times -\mathrm{cosech}^2 x\) M1: Correct use of the chain rule. Allow an expression of the form \(\dfrac{k}{\coth x} \times \mathrm{f}(x)\) where \(\mathrm{f}(x)\) is a hyperbolic function. A1: Correct differentiation	M1A1
\(= \dfrac{-1}{2\sinh x\cosh x} = \dfrac{-1}{\sinh 2x} = -\mathrm{cosech}\,2x\ ^\) Completes to printed answer with at least one line of working (e.g as shown) and no errors*	A1*
	(3)

Scheme	Marks
\(\mathrm{e}^{2y} = \coth x \Rightarrow 2\mathrm{e}^{2y}\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\mathrm{cosech}^2 x\) M1: Makes \(\mathrm{e}^y\) the subject and attempt to differentiate with respect to \(x\) A1: Correct differentiation	M1A1
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-\mathrm{cosech}^2 x}{2\coth x} = \dfrac{-1}{\sinh 2x} = -\mathrm{cosech}\,2x\ ^\) Completes to printed answer with no errors*	A1*
	(3)

Mark scheme (b)

Scheme	Marks
\(S = \displaystyle\int_{(\ln 2)}^{(\ln 3)} (1 + \mathrm{cosech}^2 2x)^{\frac{1}{2}}\,\mathrm{d}x\) Substitutes \(\mathrm{cosech}\,2x\) into a correct formula (limits not needed)	M1
\(S = \displaystyle\int_{(\ln 2)}^{(\ln 3)} \coth 2x\,\mathrm{d}x\) Use of \(1 + \mathrm{cosech}^2 2x = \coth^2 2x\)	M1
\(S = \left[\dfrac{1}{2}\ln(\sinh 2x)\right]_{\ln 2}^{\ln 3}\) Correct integration	A1
\(S = \dfrac{1}{2}\ln(\sinh(2\ln 3)) - \dfrac{1}{2}\ln(\sinh(2\ln 2))\) Uses the limits ln2 and ln3 and subtracts either way round. Dependent on first M.	dM1
\(S = \dfrac{1}{2}\ln\left(\dfrac{9 - \frac{1}{9}}{2}\right)\left(\dfrac{2}{4 - \frac{1}{4}}\right)\) Uses the exponential form of \(\sinh x\) and combines ln’s to give an expression in terms of ln only. Dependent on the first and 3^rd M.	ddM1
\(S = \tfrac{1}{2}\ln\tfrac{64}{27}\) or \(\tfrac{3}{2}\ln\tfrac{4}{3}\)	A1
	(6)
	(9 marks)