A2 June 2023 Paper 2 Q9

OCR ACurrent spec9 marksDifferentiation & Maclaurin Induction

9 A function is defined by \(y = \mathrm{f}(t)\) where \(\mathrm{f}(t) = \ln(1 + at)\) and \(a\) is a constant.

(a) By considering \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\), \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}t^{2}}\), \(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}t^{3}}\) and \(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}t^{4}}\), make a conjecture for a general formula for \(\dfrac{\mathrm{d}^{n}y}{\mathrm{d}t^{n}}\) in terms of \(n\) and \(a\) for any integer \(n \geqslant 1\). [3]

(b) Use induction to prove the formula conjectured in part (a). [4]

(c) In the case where \(\mathrm{f}(t) = \ln(1 + 2t)\), find the rate at which the 6^th derivative of \(\mathrm{f}(t)\) is varying when \(t = \dfrac{3}{2}\). [2]

Mark scheme (a)

Scheme	Marks	AO
\(\dfrac{\mathrm{d}y}{\mathrm{d}t} = \dfrac{a}{1 + at} = a(1 + at)^{-1}\)	M1	1.1
\(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}t^{2}} = -a^2(1 + at)^{-2}\) and \(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}t^{3}} = 2a^3(1 + at)^{-3}\) and \(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}t^{4}} = 2(-3)a^4(1 + at)^{-4}\) oe	A1	1.1
\(\dfrac{\mathrm{d}^{n}y}{\mathrm{d}t^{n}} = (-1)^{n-1}a^n(n - 1)!(1 + at)^{-n}\)	A1	2.2b
	[3]

Notes

M1: Differentiating correctly first derivative, or \(\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{a}{1 + at} = \left(a^{-1} + t\right)^{-1}\) etc

A1: (1st) For all; constants need not be evaluated and may be unsimplified, e.g. could see \((-1)^2\) or \((-1)^3\)

A1: (2nd) Could be \((-1)^{n+1}\) oe, or \((-1)^{n-1}(n - 1)!\left(a^{-1} + t\right)^{-n}\). Omission of factorial term can score a possible B1M1 in part (b)

Mark scheme (b)

Scheme	Marks	AO
Basis case: \(n = 1\) \(\dfrac{\mathrm{d}^{1}y}{\mathrm{d}t^{1}} = (-1)^{1-1}a^1(1 - 1)!(1 + at)^{-1}\) \(= (-1)^0 a \times 0! \times (1 + at)^{-1} = a(1 + at)^{-1}\) But \(\dfrac{\mathrm{d}^{1}y}{\mathrm{d}t^{1}} = \dfrac{\mathrm{d}y}{\mathrm{d}t} = a(1 + at)^{-1}\) So true for \(n = 1\)	*B1ft	2.1
Assume true for \(n = k\) i.e. \(\dfrac{\mathrm{d}^{k}y}{\mathrm{d}t^{k}} = (-1)^{k-1}a^k(k - 1)!(1 + at)^{-k}\) \(\therefore \dfrac{\mathrm{d}^{k+1}y}{\mathrm{d}t^{k+1}} = \dfrac{\mathrm{d}}{\mathrm{d}t}\left(\dfrac{\mathrm{d}^{k}y}{\mathrm{d}t^{k}}\right) = \dfrac{\mathrm{d}}{\mathrm{d}t}\left((-1)^{k-1}a^k(k - 1)!(1 + at)^{-k}\right)\)	M1	3.1a
\(= (-1)^{k-1}a^k(k - 1)! \times (-k)a(1 + at)^{-k-1}\) \(= (-1)^{k-1}(-1) \times a \times a^k \times k(k - 1)! \times (1 + at)^{-(k+1)}\) \(= (-1)^k a^{k+1}k!(1 + at)^{-(k+1)}\) \(\left(= (-1)^{(k+1)-1}a^{k+1}((k + 1) - 1)!(1 + at)^{-(k+1)}\right)\) (which is the formula with \(n = k + 1\))	*A1	2.2a
So true for \(n = k\) implies true for \(n = k + 1\). But true for \(n = 1\). Therefore true for all integers \(n \geqslant 1\)	dep*A1	2.4
	[4]

Notes

*B1ft: Convincingly showing that their conjecture works for \(n = 1\) for their 1^st derivative in part (a); accept omission of working shown in 1^st line here

M1: Forming the inductive hypothesis and making it clear that the \((k + 1)\)^th derivative is the derivative of the \(k\)^th derivative

*A1: Differentiating and rewriting into correct form. Some intermediate working and/or justification must be seen. Could see substitution \(u = 1 + at\) etc

dep*A1: full correct argument

Mark scheme (c)

Scheme	Marks	AO
\(\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\dfrac{\mathrm{d}^{6}y}{\mathrm{d}t^{6}}\right) = \dfrac{\mathrm{d}^{7}y}{\mathrm{d}t^{7}} = (-1)^6 a^7(7 - 1)!(1 + at)^{-7}\)	M1	3.1a
\(a = 2,\ t = \dfrac{3}{2} \Rightarrow \dfrac{\mathrm{d}^{7}y}{\mathrm{d}t^{7}} = 2^7 \times 720 \times (1 + 3)^{-7} = \dfrac{720}{128} = \dfrac{45}{8}\)	A1	1.1
	[2]

Notes

M1: Considering the seventh derivative.

A1: 5.625. Ignore attempt at units.