A2 June 2022 Paper 2 Q9

EdexcelCurrent spec8 marksDifferentiation & Maclaurin Hyperbolic Functions

\[y = \cosh^n x \qquad n \geqslant 5\]

(a)

(i) Show that\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = n^2\cosh^n x - n(n - 1)\cosh^{n-2} x\] (4)

(ii) Determine an expression for \(\dfrac{\mathrm{d}^4y}{\mathrm{d}x^4}\) (2)

(b) Hence determine the first three non-zero terms of the Maclaurin series for \(y\), giving each coefficient in simplest form. (2)

Mark scheme (a)(i)

Scheme	Marks	AO
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \ldots\cosh^{n-1} x\sinh x\) \(\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \ldots\cosh^{n-2} x\sinh^2 x + \ldots\cosh^{n-1} x\cosh x\) Alternatively \(y = \left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^{n}\) leading to \(\frac{\mathrm{d}y}{\mathrm{d}x} = \ldots\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^{n-1}\left(\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right)\) \(\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \ldots\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^{n-2}\left(\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right)^{2} + \ldots\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^{n}\)	M1	1.1b
\(\frac{\mathrm{d}y}{\mathrm{d}x} = n\cosh^{n-1} x\sinh x\) \(\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = n(n - 1)\cosh^{n-2} x\sinh^2 x + n\cosh^n x\) Alternatively \(\frac{\mathrm{d}y}{\mathrm{d}x} = n\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^{n-1}\left(\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right)\) \(\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = n(n - 1)\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^{n-2}\left(\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right)^{2} + n\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^{n}\)	A1	2.1
\(\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = n(n - 1)\cosh^{n-2} x\left(\cosh^2 x - 1\right) + n\cosh^n x\)	M1	2.1
\(\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = n^2\cosh^n x - n(n - 1)\cosh^{n-2} x\) * cso	A1*	1.1b
	(4)

Notes

M1: Uses the chain rule and product rule to find the first and second derivatives which must be of the required form, condone sign slips
Alternatively uses the exponential definition and uses the chain rule and product rule to find the first and second derivatives which must be of the required form.

A1: Correct unsimplified first and second derivatives, may be in exponential form.

M1: Uses the identity \(\pm\cosh^2 x \pm \sinh^2 x = 1\)

A1*: Achieves the printed answer with no errors or omissions e.g. missing \(x\)’s

Mark scheme (a)(ii)

Scheme	Marks	AO
\(\frac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = \ldots\cosh^{n-1} x\sinh x - \ldots\cosh^{n-3} x\sinh x\) \(\frac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} = \ldots\cosh^{n-2} x\sinh^2 x + \ldots\cosh^n x - \ldots\cosh^{n-4} x\sinh^2 x - \ldots\cosh^{n-2} x\)	M1	1.1b
\(\frac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = n^3\cosh^{n-1} x\sinh x - n(n - 1)(n - 2)\cosh^{n-3} x\sinh x\) \(\frac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} = n^3(n - 1)\cosh^{n-2} x\sinh^2 x + n^3\cosh^n x\) \(\qquad - n(n - 1)(n - 2)(n - 3)\cosh^{n-4} x\sinh^2 x - n(n - 1)(n - 2)\cosh^{n-2} x\)	A1	1.1b
	(2)

Notes

(The printed M1 line is cut off at the edge of the page after “\(-\ldots\cosh\)”; its last term is completed here to match the A1 line.)

M1: Uses the chain rule and product rule to find the third and fourth derivatives which must be of the required form, condone sign slips

A1: Correct fourth derivative, does not need to be simplified ISW

Alternative 1

Scheme	Marks	AO
using \(\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = n^2y - n(n - 1)\cosh^{n-2} x\) leading to \(\frac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = n^2\frac{\mathrm{d}y}{\mathrm{d}x} - \ldots\cosh^{n-3} x\sinh x\) \(\frac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} = n^2\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} - \ldots\cosh^{n-4} x\sinh^2 x - \ldots\cosh^{n-2} x\)	M1	1.1b
\(\frac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = n^2\frac{\mathrm{d}y}{\mathrm{d}x} - n(n - 1)(n - 2)\cosh^{n-3} x\sinh x\) \(\frac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} = n^2\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} - n(n - 1)(n - 2)(n - 3)\cosh^{n-4} x\sinh^2 x - n(n - 1)(n - 2)\cosh^{n-2} x\)	A1	1.1b
	(2)

M1: Using \(\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = n^2y - n(n - 1)\cosh^{n-2} x\) to find the third and fourth derivatives which must be of the required form, condone sign slips

A1: Correct fourth derivative, does not need to be simplified ISW

Alternative 2

Scheme	Marks	AO
\(y = \cosh^n x \Rightarrow \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = n^2\cosh^n x - n(n - 1)\cosh^{n-2} x\) \(y = \cosh^{n-2} x \Rightarrow \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \ldots\cosh^{n-2} x - \ldots\cosh^{n-4} x\) \(\frac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} = n^2\left[n^2\cosh^n x - n(n - 1)\cosh^{n-2} x\right] - n(n - 1)\left[\ldots\cosh^{n-2} x - \ldots\cosh^{n-4} x\right]\)	M1	1.1b
\(y = \cosh^n x \Rightarrow \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = n^2\cosh^n x - n(n - 1)\cosh^{n-2} x\) \(y = \cosh^{n-2} x \Rightarrow \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = (n - 2)^2\cosh^{n-2} x - (n - 2)(n - 3)\cosh^{n-4} x\) \(\frac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} = n^2\left[n^2\cosh^n x - n(n - 1)\cosh^{n-2} x\right]\) \(\qquad - n(n - 1)\left[(n - 2)^2\cosh^{n-2} x - (n - 2)(n - 3)\cosh^{n-4} x\right]\)	A1	1.1b
	(2)

M1: Using \(y = \cosh^n x \Rightarrow \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = n^2\cosh^n x - n(n - 1)\cosh^{n-2} x\)
\(y = \cosh^{n-2} x \Rightarrow \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \ldots\cosh^{n-2} x - \ldots\cosh^{n-4} x\) leading to
\(\frac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} = n^2\left[n^2\cosh^n x - n(n - 1)\cosh^{n-2} x\right] - n(n - 1)\left[\text{their } \dfrac{\mathrm{d}\left(\cosh^{n-2} x\right)}{\mathrm{d}x}\right]\)

A1: Correct fourth derivative, does not need to be simplified ISW

Alternative 3

Scheme	Marks	AO
Using \(\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = n^2\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^{n} - n(n - 1)\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^{n-2}\) leading to \(\frac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = \ldots\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^{n-1}\left(\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right) - \ldots\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^{n-3}\left(\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right)\) \(\frac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} = \ldots\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^{n-2}\left(\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right)^{2} + \ldots\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^{n}\) \(\qquad - \ldots\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^{n-4}\left(\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right)^{2} - \ldots\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^{n-2}\)	M1	1.1b
\(\frac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = n^3\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^{n-1}\left(\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right) - n(n - 1)(n - 2)\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^{n-3}\left(\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right)\) \(\frac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} = n^3(n - 1)\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^{n-2}\left(\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right)^{2} + n^3\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^{n}\) \(\qquad - n(n - 1)(n - 2)(n - 3)\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^{n-4}\left(\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right)^{2} - n(n - 1)(n - 2)\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^{n-2}\)	A1	1.1b
	(2)

(Corrected from the printed mark scheme: in Alternative 3 the printed scheme gives the power of the second term of \(\frac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}}\) as \(n - 2\) in both lines; it should be \(n\), i.e. \(n^3\cosh^n x\).)

M1: Uses the exponential definition and uses the chain rule and product rule to find the third and fourth derivatives which must be of the required form.

A1: Correct fourth derivative, does not need to be simplified ISW

Mark scheme (b)

Scheme	Marks	AO
When \(x = 0\) \(y = 1,\quad y^\prime = 0,\quad y^{\prime\prime} = n^2 - n(n - 1),\quad y^{(3)} = 0,\quad y^{(4)} = n^3 - n(n - 1)(n - 2)\) Uses their values in the expansion \(y = y(0) + xy^\prime(0) + \dfrac{x^2}{2!}y^{\prime\prime}(0) + \dfrac{x^3}{3!}y^{(3)}(0) + \dfrac{x^4}{4!}y^{(4)}(0) + \ldots\)	M1	1.1b
\(y = 1 + \dfrac{nx^2}{2} + \dfrac{\left(3n^2 - 2n\right)x^4}{24} + \ldots\) cso	A1	2.5
	(2)
	(8 marks)

Notes

M1: Attempts the evaluation of all four of their derivatives at \(x = 0\) and applies the Maclaurin formula with their values. Note that \(y^{(1)}(0) = 0\) and \(y^{(3)}(0) = 0\) may be implied as they will have a multiple of \(\sinh 0\). If their \(y^{(3)}(0) \neq 0\) they allow this mark for their first 3 non-zero terms

A1: Correct simplified expansion from correct derivatives cso