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Date	May 2022	Marks available	4	Reference code	22M.1.AHL.TZ1.12
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 1
Command term	Hence and Find	Question number	12	Adapted from	N/A

Question

The function $f$ is defined by $f (x) = e^{x} \sin x$ , where $x \in ℝ$ .

The function $g$ is defined by $g (x) = e^{x} \cos x$ , where $x \in ℝ$ .

Find the Maclaurin series for $f (x)$ up to and including the $x^{3}$ term.

[4]

Hence, find an approximate value for $\int_{0}^{1} e^{x^{2}} \sin (x^{2}) d x$ .

[4]

Show that $g (x)$ satisfies the equation $g'' (x) = 2 (g' (x) - g (x))$ .

[4]

c.i.

Hence, deduce that $g^{(4)} (x) = 2 (g''' (x) - g'' (x))$ .

[1]

c.ii.

Using the result from part (c), find the Maclaurin series for $g (x)$ up to and including the $x^{4}$ term.

[5]

Hence, or otherwise, determine the value of $\lim_{x \to 0} \frac{e^{x} \cos x - 1 - x}{x^{3}}$ .

[3]

Markscheme

METHOD 1

recognition of both known series (M1)

$e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \dots$ and $\sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + \dots$

attempt to multiply the two series up to and including $x^{3}$ term (M1)

$e^{x} \sin x = (1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \dots) (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + \dots)$

$= x - \frac{x^{3}}{3!} + x^{2} + \frac{x^{3}}{2!} + \dots$ (A1)

$e^{x} \sin x = x + x^{2} + \frac{1}{3} x^{3} + \dots$ A1

METHOD 2

$f (x) = e^{x} \sin x$

$f' (x) = e^{x} \cos x + e^{x} \sin x$ A1

$f'' (x) = e^{x} \cos x - e^{x} \sin x + e^{x} \sin x + e^{x} \cos x (= 2 e^{x} \cos x)$

$f''' (x) = 2 e^{x} \cos x - 2 e^{x} \sin x$

$f'' (x) = 2 e^{x} \cos x$ and $f''' (x) = 2 e^{x} (\cos x - \sin x)$ A1

substitute $x = 0$ into $f$ or its derivatives to obtain Maclaurin series (M1)

$e^{x} \sin x = 0 + \frac{x}{1!} \times 1 + \frac{x^{2}}{2!} \times 2 + \frac{x^{3}}{3!} \times 2 + \dots$

$e^{x} \sin x = x + x^{2} + \frac{1}{3} x^{3} + \dots$ A1

[4 marks]

$e^{x^{2}} \sin (x^{2}) = x^{2} + x^{4} + \frac{1}{3} x^{6} + \dots$ (A1)

substituting their expression and attempt to integrate M1

$\int_{0}^{1} e^{x^{2}} \sin (x^{2}) d x \approx \int_{0}^{1} (x^{2} + x^{4} + \frac{1}{3} x^{6}) d x$

Note: Condone absence of limits up to this stage.

$= {[\frac{x^{3}}{3} + \frac{x^{5}}{5} + \frac{x^{7}}{21}]}_{0}^{1}$ A1

$= \frac{61}{105}$ A1

[4 marks]

attempt to use product rule at least once M1

$g' (x) = e^{x} \cos x - e^{x} \sin x$ A1

$g'' (x) = e^{x} \cos x - e^{x} \sin x - e^{x} \sin x - e^{x} \cos x (= - 2 e^{x} \sin x)$ A1

EITHER

$2 (g' (x) - g (x)) = 2 (e^{x} \cos x - e^{x} \sin x - e^{x} \cos x) = - 2 e^{x} \sin x$ A1

$g'' (x) = 2 (e^{x} \cos x - e^{x} \sin x - e^{x} \cos x)$ A1

THEN

$g'' (x) = 2 (g' (x) - g (x))$ AG

Note: Accept working with each side separately to obtain $- 2 e^{x} \sin x$ .

[4 marks]

c.i.

$g''' (x) = 2 (g'' (x) - g' (x))$ A1

$g^{(4)} (x) = 2 (g''' (x) - g'' (x))$ AG

Note: Accept working with each side separately to obtain $- 4 e^{x} \cos x$ .

[1 mark]

c.ii.

attempt to substitute $x = 0$ into a derivative (M1)

$g (0) = 1, g' (0) = 1, g'' (0) = 0$ A1

$g''' (0) = - 2, g^{(4)} (0) = - 4$ (A1)

attempt to substitute into Maclaurin formula (M1)

$g (x) = 1 + x - \frac{2}{3!} x^{3} - \frac{4}{4!} x^{4} + \dots (= 1 + x - \frac{1}{3} x^{3} - \frac{1}{6} x^{4} + \dots)$ A1

Note: Do not award any marks for approaches that do not use the part (c) result.

[5 marks]

METHOD 1

$\lim_{x \to 0} \frac{e^{x} \cos x - 1 - x}{x^{3}} = \lim_{x \to 0} \frac{(1 + x - \frac{1}{3} x^{3} - \frac{1}{6} x^{4} + \dots) - 1 - x}{x^{3}}$ M1

$= \lim_{x \to 0} (- \frac{1}{3} - \frac{1}{6} x + \dots)$ (A1)

$= - \frac{1}{3}$ A1

Note: Condone the omission of $+ \dots$ in their working.

METHOD 2

$\lim_{x \to 0} \frac{e^{x} \cos x - 1 - x}{x^{3}} = \frac{0}{0}$ indeterminate form, attempt to apply l'Hôpital's rule M1

$= \lim_{x \to 0} \frac{e^{x} \cos x - e^{x} \sin x - 1}{3 x^{2}} (= \lim_{x \to 0} \frac{g' (x) - 1}{3 x^{2}})$

$= \frac{0}{0}$ , using l'Hôpital's rule again

$= \lim_{x \to 0} \frac{- 2 e^{x} \sin x}{6 x} (= \lim_{x \to 0} \frac{g'' (x)}{6 x})$

$= \frac{0}{0}$ , using l'Hôpital's rule again

$= \lim_{x \to 0} \frac{- 2 e^{x} \sin x - 2 e^{x} \cos x}{6} (= \lim_{x \to 0} \frac{g''' (x)}{6})$ A1

$= - \frac{1}{3}$ A1

[3 marks]

Examiners report

Part (a) was well answered using both methods. A number failed to see the connection between parts (a) and (b). Far too often, a candidate could not add three fractions together in part (b). There were many good responses to part (c) with candidates showing results on both sides are equal. A number of candidates failed to use the result from (c) in part (d). There were some good responses to part (e), with candidates working successfully with the series from (d) or applying l'Hôpital's rule. In particular, some responses were missing the appropriate limit notation and candidates following method 2 did not always show that the initial expression was of an indeterminate form before applying l'Hôpital's rule. Many candidates did not attempt parts (d) and (e).

[N/A]

c.i.

[N/A]

c.ii.

[N/A]