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.
Show that the coefficient of x3 in the Maclaurin series for f(x) is zero.
Using the Maclaurin series for arctanx and e3x−1, find the Maclaurin series for arctan(e3x−1) up to and including the x3 term.
Hence, or otherwise, find limx→0f(x)−1arctan(e3x−1).
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(cos2x−sinx) (or equivalent) A1
attempting to find f(0), f′(0) and f″(0) M1
f(0)=1; f′(0)=(cos0)esin0=1; f″(0)=esin0(cos20−sin0)=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]
METHOD 1
attempting to differentiate f″(x) M1
f‴(x)=(cosx)esinx(cos2x−sinx)−(cosx)esinx(2sinx+1) (or equivalent) A2
substituting x=0 into their f‴(x) M1
f‴(0)=1(1−0)−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+(x−x33!+…)+(x−x33!+…)22!+(x−x33!+…)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]
substituting 3x into the Maclaurin series for ex M1
e3x=1+3x+(3x)22!+(3x)33!+… A1
substituting (e3x−1) into the Maclaurin series for arctanx M1
arctan(e3x−1)=(e3x−1)−(e3x−1)33+(e3x−1)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+9x22−9x32 A1
[6 marks]
METHOD 1
substitution of their series M1
limx→0x+x22+…3x+9x22+… A1
=limx→01+x2+…3+9x2+…
=13 A1
METHOD 2
use of l’Hôpital’s rule M1
limx→0(cosx)esinx3e3x1+(e3x−1)2 (or equivalent) A1
=13 A1
[3 marks]