A2 June 2022 Paper 2 Q8

AQACurrent spec10 marksDifferentiation & Maclaurin

(a) The function \(\mathrm{f}\) is defined as \(\mathrm{f}(x) = \sec x\)

(i) Show that \(\mathrm{f}^{(4)}(0) = 5\) [4 marks]

(ii) Hence find the first three non-zero terms of the Maclaurin series for \(\mathrm{f}(x) = \sec x\) [2 marks]

(b) Prove that\[\lim_{x \to 0}\left(\frac{\sec x - \cosh x}{x^4}\right) = \frac{1}{6}\] [4 marks]

Mark scheme (a)

Scheme	Marks	AO
(i) Differentiates twice correctly	B1	1.1b
Differentiates three times	M1	1.1a
Obtains a correct expression for the fourth derivative of \(\mathrm{f}(x)\)	A1	1.1b
Obtains the required result from correct working	A1	1.1b
	(4)
(ii) Substitutes zero and uses in Maclaurin expansion. Condone no division by factorial	M1	1.1a
Obtains correct result (allow factorial notation)	A1	1.1b
	(2)

Typical solution

(i)

\[\begin{aligned}\mathrm{f}'(x) &= \sec x\tan x \\ \mathrm{f}''(x) &= \sec x(\sec^2 x) + \tan x(\sec x\tan x) \\ &= \sec x(\sec^2 x + \tan^2 x) \\ \mathrm{f}^{(3)}(x) &= \sec x\tan x(\sec^2 x + \tan^2 x) \\ &\quad + \sec x(2\sec x(\sec x\tan x) + 2\tan x\sec^2 x) \\ &= 5\sec^3 x\tan x + \sec x\tan^3 x \\ \mathrm{f}^{(4)}(x) &= \tan x\left(15\sec^2 x(\sec x\tan x)\right) \\ &\quad + 5\sec^5 x + \sec x(3\tan^2 x\sec^2 x) \\ &\quad + \tan^3 x(\sec x\tan x) \\ &= 18\sec^3 x\tan^2 x + 5\sec^5 x + \tan^4 x\sec x\end{aligned}\]\[\mathrm{f}^{(4)}(0) = 0 + 5 + 0 = 5\]

(ii)

\[\mathrm{f}(0) = 1\]\[\mathrm{f}'(0) = 0\]\[\mathrm{f}''(0) = 1\]\[\mathrm{f}^{(3)}(0) = 0\]\[\mathrm{f}(x) = 1 + \frac{x^2}{2} + \frac{5x^4}{24} + \cdots\]

Mark scheme (b)

Scheme	Marks	AO
Deduces correct expansion of \(\cosh x\) or justifies the use of l’Hôpital’s rule (must see 0/0 at least once)	B1	2.2a
Substitutes expansions into expression PI or applies l’Hôpital’s rule four times	M1	1.1a
Simplifies correctly	A1	1.1b
Completes a rigorous argument to prove the required result, including a clear demonstration of the limiting process. Must show evidence of higher powers until the limit is taken or must justify l’Hôpital’s rule at each stage	R1	2.1
	(4)
	(10 marks)

Typical solution

\[\begin{aligned}\cosh x &= \frac{1}{2}\left(1 + x + \frac{x^2}{2!} + \cdots\right) + \frac{1}{2}\left(1 - x + \frac{x^2}{2!} + \cdots\right) \\ &= 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots\end{aligned}\]\[\begin{aligned}&\lim_{x \to 0}\left(\frac{\sec x - \cosh x}{x^4}\right) \\ &= \lim_{x \to 0}\frac{\left(1 + \frac{x^2}{2!} + \frac{5x^4}{4!} + \cdots\right) - \left(1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots\right)}{x^4} \\ &= \lim_{x \to 0}\frac{\frac{x^4}{6} + \text{higher powers}}{x^4} \\ &= \lim_{x \to 0}\left(\frac{1}{6} + \text{higher powers}\right) = \frac{1}{6}\end{aligned}\]\[\therefore \lim_{x \to 0}\left(\frac{\sec x - \cosh x}{x^4}\right) = \frac{1}{6}\]