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 | Find | Question number | 12 | Adapted from | N/A |
Question
The function f is defined by f(x)=ex sin x, where x∈ℝ.
The function g is defined by g(x)=ex cos x, where x∈ℝ.
Find the Maclaurin series for f(x) up to and including the x3 term.
Hence, find an approximate value for ∫10ex2 sin(x2)dx.
Show that g(x) satisfies the equation g''(x)=2(g'(x)-g(x)).
Hence, deduce that g(4)(x)=2(g'''(x)-g''(x)).
Using the result from part (c), find the Maclaurin series for g(x) up to and including the x4 term.
Hence, or otherwise, determine the value of limx→0ex cos x-1-xx3.
Markscheme
METHOD 1
recognition of both known series (M1)
ex=1+x1!+x22!+… and sin x=x-x33!+x55!+…
attempt to multiply the two series up to and including x3 term (M1)
ex sin x=(1+x1!+x22!+…)(x-x33!+x55!+…)
=x-x33!+x2+x32!+… (A1)
ex sin x=x+x2+13x3+… A1
METHOD 2
f(x)=ex sin x
f'(x)=ex cos x+ex sin x A1
f''(x)=ex cos x-ex sin x+ex sin x+ex cos x (=2ex cos x)
f'''(x)=2ex cos x-2ex sin x
f''(x)=2ex cos x and f'''(x)=2ex(cos x-sin x) A1
substitute x=0 into f or its derivatives to obtain Maclaurin series (M1)
ex sin x=0+x1!×1+x22!×2+x33!×2+…
ex sin x=x+x2+13x3+… A1
[4 marks]
ex2 sin (x2)=x2+x4+13x6+… (A1)
substituting their expression and attempt to integrate M1
∫10ex2 sin(x2)dx≈∫10(x2+x4+13x6)dx
Note: Condone absence of limits up to this stage.
=[x33+x55+x721]10 A1
=61105 A1
[4 marks]
attempt to use product rule at least once M1
g'(x)=ex cos x-ex sin x A1
g''(x)=ex cos x-ex sin x-ex sin x-ex cos x(=-2ex sin x) A1
EITHER
2(g'(x)-g(x))=2(ex cos x-ex sin x-ex cos x)=-2ex sin x A1
OR
g''(x)=2(ex cos x-ex sin x-ex cos x) A1
THEN
g''(x)=2(g'(x)-g(x)) AG
Note: Accept working with each side separately to obtain -2ex sin x.
[4 marks]
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 -4ex cos x.
[1 mark]
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-23!x3-44!x4+…(=1+x-13x3-16x4+…) A1
Note: Do not award any marks for approaches that do not use the part (c) result.
[5 marks]
METHOD 1
limx→0ex cos x-1-xx3=limx→0(1+x-13x3-16x4+…)-1-xx3 M1
=limx→0(-13-16x+…) (A1)
=-13 A1
Note: Condone the omission of +… in their working.
METHOD 2
limx→0ex cos x-1-xx3=00 indeterminate form, attempt to apply l'Hôpital's rule M1
=limx→0ex cos x-ex sin x-13x2(=limx→0g'(x)-13x2)
=00, using l'Hôpital's rule again
=limx→0-2ex sin x6x(=limx→0g''(x)6x)
=00, using l'Hôpital's rule again
=limx→0-2ex sin x-2ex cos x6(=limx→0g'''(x)6) A1
=-13 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).