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 −12⩽x⩽12.
By finding a suitable number of derivatives of f, find the first two non-zero terms in the Maclaurin series for f.
Hence or otherwise, find limx→0arcsin(2x)−2x(2x)3.
Markscheme
f(x)=arcsin(2x)
f′(x)=2√1−4x2 M1A1
Note: Award M1A0 for f′(x)=1√1−4x2
f″(x)=8x(1−4x2)32 A1
EITHER
f‴(x)=8(1−4x2)32−8x(32(−8x)(1−4x2)12)(1−4x2)3(=8(1−4x2)32+96x2(1−4x2)12(1−4x2)3) A1
OR
f‴(x)=8(1−4x2)−32+8x(−32(1−4x2)−52)(−8x)(=8(1−4x2)−32+96x2(1−4x2)−52) A1
THEN
substitute x=0 into f or any of its derivatives (M1)
f(0)=0, f′(0)=2 and f″(0)=0 A1
f‴(0)=8
the Maclaurin series is
f(x)=2x+8x36+…(=2x+4x33+…) (M1)A1
[8 marks]
METHOD 1
limx→0arcsin(2x)−2x(2x)3=limx→02x+4x33+…−2x8x3 M1
=limx→043+… terms with x8 (M1)
=16 A1
Note: Condone the omission of +… in their working.
METHOD 2
limx→0arcsin(2x)−2x(2x)3=00 indeterminate form, using L’Hôpital’s rule
=limx→02√1−4x2−224x2 M1
=00 indeterminate form, using L’Hôpital’s rule again
=limx→08x(1−4x2)3248x(=limx→016(1−4x2)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]