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Date May Specimen paper Marks available 4 Reference code SPM.1.AHL.TZ0.12
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term Show that Question number 12 Adapted from N/A

Question

The function f is defined by f(x)=esinx.

Find the first two derivatives of f(x) and hence find the Maclaurin series for f(x) up to and including the x2 term.

[8]
a.

Show that the coefficient of x3 in the Maclaurin series for f(x) is zero.

[4]
b.

Using the Maclaurin series for arctanx and e3x1, find the Maclaurin series for arctan(e3x1) up to and including the x3 term.

[6]
c.

Hence, or otherwise, find limx0f(x)1arctan(e3x1).

[3]
d.

Markscheme

attempting to use the chain rule to find the first derivative     M1

f(x)=(cosx)esinx       A1

attempting to use the product rule to find the second derivative      M1

f(x)=esinx(cos2xsinx) (or equivalent)        A1

attempting to find f(0)f(0) and f(0)       M1

f(0)=1; f(0)=(cos0)esin0=1f(0)=esin0(cos20sin0)=1        A1

substitution into the Maclaurin formula f(x)=f(0)+xf(0)+x22!f(0)+       M1

so the Maclaurin series for f(x) up to and including the x2 term is 1+x+x22      A1

[8 marks]

a.

METHOD 1

attempting to differentiate f(x)       M1

f(x)=(cosx)esinx(cos2xsinx)(cosx)esinx(2sinx+1) (or equivalent)        A2

substituting x=0 into their f(x)       M1

f(0)=1(10)1(0+1)=0

so the coefficient of x3 in the Maclaurin series for f(x) is zero    AG

 

METHOD 2

substituting sinx into the Maclaurin series for ex       (M1)

esinx=1+sinx+sin2x2!+sin3x3!+

substituting Maclaurin series for sinx       M1

esinx=1+(xx33!+)+(xx33!+)22!+(xx33!+)33!+     A1

coefficient of x3 is 13!+13!=0     A1

so the coefficient of x3 in the Maclaurin series for f(x) is zero    AG

 

[4 marks]

b.

substituting 3x into the Maclaurin series for ex       M1

e3x=1+3x+(3x)22!+(3x)33!+      A1

substituting (e3x1) into the Maclaurin series for arctanx    M1

arctan(e3x1)=(e3x1)(e3x1)33+(e3x1)55

=(3x+(3x)22!+(3x)33!+)(3x+(3x)22!+(3x)33!+)33+    A1

selecting correct terms from above      M1

=(3x+(3x)22!+(3x)33!)(3x)33

=3x+9x229x32     A1

[6 marks]

c.

METHOD 1

substitution of their series       M1

limx0x+x22+3x+9x22+       A1

=limx01+x2+3+9x2+

=13     A1

 

METHOD 2

use of l’Hôpital’s rule      M1

limx0(cosx)esinx3e3x1+(e3x1)2  (or equivalent)    A1

=13     A1

 

[3 marks]

d.

Examiners report

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Syllabus sections

Topic 5 —Calculus » AHL 5.19—Maclaurin series
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Topic 5 —Calculus

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