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Date November 2019 Marks available 8 Reference code 19N.3.AHL.TZ0.Hca_3
Level Additional Higher Level Paper Paper 3 Time zone Time zone 0
Command term Find Question number Hca_3 Adapted from N/A

Question

The function f is defined by f(x)=arcsin(2x), where 12x12.

By finding a suitable number of derivatives of f, find the first two non-zero terms in the Maclaurin series for f.

[8]
a.

Hence or otherwise, find limx0arcsin(2x)2x(2x)3.

[3]
b.

Markscheme

f(x)=arcsin(2x)

f(x)=214x2       M1A1

Note: Award M1A0 for f(x)=114x2

f(x)=8x(14x2)32        A1

EITHER

f(x)=8(14x2)328x(32(8x)(14x2)12)(14x2)3(=8(14x2)32+96x2(14x2)12(14x2)3)        A1

OR

f(x)=8(14x2)32+8x(32(14x2)52)(8x)(=8(14x2)32+96x2(14x2)52)        A1

THEN

substitute x=0 into f or any of its derivatives         (M1)

f(0)=0f(0)=2 and f(0)=0        A1

f(0)=8

the Maclaurin series is

f(x)=2x+8x36+(=2x+4x33+)         (M1)A1

[8 marks]

a.

METHOD 1

limx0arcsin(2x)2x(2x)3=limx02x+4x33+2x8x3       M1

=limx043+ terms with x8         (M1)

=16        A1

Note: Condone the omission of +… in their working.

 

METHOD 2

limx0arcsin(2x)2x(2x)3=00  indeterminate form, using L’Hôpital’s rule

=limx0214x2224x2         M1

=00  indeterminate form, using L’Hôpital’s rule again

=limx08x(14x2)3248x(=limx016(14x2)32)         M1

Note: Award M1 only if their previous expression is in indeterminate form.

=16        A1

Note: Award FT for use of their derivatives from part (a).

 

[3 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 5 —Calculus » AHL 5.19—Maclaurin series
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