Date | November 2020 | Marks available | 4 | Reference code | 20N.2.AHL.TZ0.F_5 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Show that | Question number | F_5 | Adapted from | N/A |
Question
Assuming the Maclaurin series for cos x and ln(1+x), show that the Maclaurin series for cos(ln(1+x)) is
1-12x2+12x3-512x4+…
By differentiating the series in part (a), show that the Maclaurin series for sin(ln(1+x)) is x-12x2+16x3+… .
Hence determine the Maclaurin series for tan(ln(1+x)) as far as the term in x3.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
METHOD 1
attempts to substitute ln(x+1)=x-12x2+13x3-14x4+… into
cos x=1-12x2+124x4-… M1
cos x=(ln(1+x))=1-12(x-12x2+13x3+…)2+124(x+…)4+… A1
attempts to expand the RHS up to and including the x4 term M1
=1-12(x2-x3+14x4+23x4…)+124x4+… A1
=1-12x2+12x3-512x4+… AG
METHOD 2
attempts to substitute ln(x+1) into cos x=1-12x2+124x4-… M1
cos (ln(1+x))=1-12(ln(1+x))2+124(ln(1+x))4-…
attempts to find the Maclaurin series for (ln(1+x))2 up to and including the x4 term M1
(ln(1+x))2=x2-x3+1112x4-… A1
(ln(1+x))2=x4-…
=1-12(x2-x3+1112x4+…)+124x4+… A1
=1-12x2+12x3-512x4+… AG
[4 marks]
-sin(ln(1+x))×11+x=-x+32x2-53x3+… A1A1
sin(ln(1+x))=-(1+x)(-x+32x2-53x3+…)
attempts to expand the RHS up to and including the x3 term M1
=x-32x2+53x3+x2-32x3+… A1
=x-12x2+16x3+… AG
[4 marks]
METHOD 1
let tan(ln(1+x))=a0+a1x+a2x2+a3x3+…
uses sin(ln(1+x))=cos(ln(1+x))×tan(ln(1+x)) to form M1
x-12x2+16x3+…=(1-12x2+12x3+…)(a0+a1x+a2x2+a3x3+…) A1
=a0+a1x+(a2-12a0)x2+(a3-12a1+12a0)x3+… (A1)
attempts to equate coefficients,
a0=0, M1
A1
so
METHOD 2
uses to form M1
A1
(A1)
attempts to expand the up to and including the term M1
A1
Note: Accept use of long division.
[5 marks]