Date | May 2008 | Marks available | 3 | Reference code | 08M.3ca.hl.TZ2.1 |
Level | HL only | Paper | Paper 3 Calculus | Time zone | TZ2 |
Command term | Find | Question number | 1 | Adapted from | N/A |
Question
Find the value of limx→1(lnxsin2πx).
By using the series expansions for ex2 and cos x evaluate limx→0(1−ex21−cosx)..
Markscheme
Using l’Hopital’s rule,
limx→1(lnxsin2πx)=limx→1(1x2πcos2πx) M1A1
=12π A1
[3 marks]
limx→0(1−ex21−cosx)=limx→0(1−(1+x2+x42!+x63!+...)1−(1−x22!+x44!−...)) M1A1A1
Note: Award M1 for evidence of using the two series.
=limx→0((−x2−x42!−x63!−...)(x22!−x44!+...)) A1
EITHER
=limx→0((−1−x22!−x43!−...)(12!−x24!+...)) M1A1
=−112=−2 A1
OR
=limx→0((−2x−4x32!−6x53!−...)(2x2!−4x34!+...)) M1A1
=limx→0((−2−4x22!−6x43!−...)(1−4x24!+...))
=−21=−2 A1
[7 marks]
Examiners report
Part (a) was well done but too often the instruction to use series in part (b) was ignored. When this hint was observed correct solutions followed.
Part (a) was well done but too often the instruction to use series in part (b) was ignored. When this hint was observed correct solutions followed.