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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 Find Question number 12 Adapted from N/A

Question

The function f is defined by f(x)=exsinx, where x.

The function g is defined by g(x)=excosx, where x.

Find the Maclaurin series for f(x) up to and including the x3 term.

[4]
a.

Hence, find an approximate value for 01ex2sinx2dx.

[4]
b.

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

[4]
c.i.

Hence, deduce that g4x=2g'''x-g''x.

[1]
c.ii.

Using the result from part (c), find the Maclaurin series for g(x) up to and including the x4 term.

[5]
d.

Hence, or otherwise, determine the value of limx0excosx-1-xx3.

[3]
e.

Markscheme

METHOD 1

recognition of both known series          (M1)

ex=1+x1!+x22!+  and sinx=x-x33!+x55!+

attempt to multiply the two series up to and including x3 term           (M1)

exsinx=1+x1!+x22!+x-x33!+x55!+

=x-x33!+x2+x32!+           (A1)

exsinx=x+x2+13x3+          A1

 

METHOD 2

fx=exsinx

f'x=excosx+exsinx          A1

f''x=excosx-exsinx+exsinx+excosx =2excosx

f'''x=2excosx-2exsinx

f''x=2excosx  and  f'''x=2excosx-sinx          A1

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

exsinx=0+x1!×1+x22!×2+x33!×2+

exsinx=x+x2+13x3+          A1

 

[4 marks]

a.

ex2sinx2=x2+x4+13x6+           (A1)

substituting their expression and attempt to integrate              M1

01ex2sinx2dx01x2+x4+13x6dx

 

Note: Condone absence of limits up to this stage.

 

=x33+x55+x72101          A1

=61105          A1

 

[4 marks]

b.

attempt to use product rule at least once             M1

g'(x)=excosx-exsinx          A1

g''(x)=excosx-exsinx-exsinx-excosx=-2exsinx          A1


EITHER

2g'(x)-gx=2excosx-exsinx-excosx=-2exsinx          A1


OR

g''(x)=2excosx-exsinx-excosx          A1


THEN

g''(x)=2g'(x)-gx          AG

 

Note: Accept working with each side separately to obtain -2exsinx.

 

[4 marks]

c.i.

g'''(x)=2g''(x)-g'x          A1

g4x=2g'''x-g''x          AG

 

Note: Accept working with each side separately to obtain -4excosx.

 

[1 mark]

c.ii.

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

g0=1, g'0=1, g''0=0          A1

g'''0=-2, g40=-4           (A1)

attempt to substitute into Maclaurin formula          (M1)

gx=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]

d.

METHOD 1

limx0excosx-1-xx3=limx01+x-13x3-16x4+-1-xx3         M1

=limx0-13-16x+           (A1)

=-13          A1

 

Note: Condone the omission of + in their working.

 

METHOD 2

limx0excosx-1-xx3=00 indeterminate form, attempt to apply l'Hôpital's rule         M1

=limx0excosx-exsinx-13x2=limx0g'x-13x2

=00, using l'Hôpital's rule again

=limx0-2exsinx6x=limx0g''x6x

=00, using l'Hôpital's rule again

=limx0-2exsinx-2excosx6=limx0g'''x6          A1

=-13          A1

 

[3 marks]

e.

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).

a.
[N/A]
b.
[N/A]
c.i.
[N/A]
c.ii.
[N/A]
d.
[N/A]
e.

Syllabus sections

Topic 5 —Calculus » AHL 5.19—Maclaurin series
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