Date | November 2021 | Marks available | 7 | Reference code | 21N.1.AHL.TZ0.11 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 0 |
Command term | Prove | Question number | 11 | Adapted from | N/A |
Question
Prove by mathematical induction that dndxn(x2ex)=[x2+2nx+n(n-1)]ex for n∈ℤ+.
Hence or otherwise, determine the Maclaurin series of f(x)=x2ex in ascending powers of x, up to and including the term in x4.
Hence or otherwise, determine the value of limx→0[(x2ex-x2)3x9].
Markscheme
For n=1
LHS: ddx(x2ex)=x2ex+2xex(=ex(x2+2x)) A1
RHS: (x2+2(1)x+1(1-1))ex(=ex(x2+2x)) A1
so true for n=1
now assume true for n=k; i.e. dkdxk(x2ex)=[x2+2kx+k(k-1)]ex M1
Note: Do not award M1 for statements such as "let n=k". Subsequent marks can still be awarded.
attempt to differentiate the RHS M1
dk+1dxk+1(x2ex)=ddx([x2+2kx+k(k-1)]ex)
=(2x+2k)ex+(x2+2kx+k(k-1))ex A1
=[x2+2(k+1)x+k(k+1)]ex A1
so true for n=k implies true for n=k+1
therefore n=1 true and n=k true ⇒n=k+1 true
therefore, true for all n∈ℤ+ R1
Note: Award R1 only if three of the previous four marks have been awarded
[7 marks]
METHOD 1
attempt to use dndxn(x2ex)=[x2+2nx+n(n-1)]ex (M1)
Note: For x=0, dndxn(x2ex)x=0=n(n-1) may be seen.
f(0)=0, f'(0)=0, f''(0)=2, f'''(0)=6, f(4)(0)=12
use of f(x)=f(0)+xf'(0)+x22!f''(0)+x33!f'''(0)+x44!f(4)(0)+… (M1)
⇒f(x)≈x2+x3+12x4 A1
METHOD 2
' Maclaurin series of ' (M1)
(A1)
A1
[3 marks]
METHOD 1
attempt to substitute into M1
(A1)
EITHER
A1
OR
A1
THEN
so A1
METHOD 2
M1
(A1)
attempt to use L'Hôpital's rule M1
A1
[4 marks]