Date | November 2019 | Marks available | 7 | Reference code | 19N.1.AHL.TZ0.H_6 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 0 |
Command term | Prove | Question number | H_6 | Adapted from | N/A |
Question
Consider the function f(x)=xe2x, where x∈R. The nth derivative of f(x) is denoted by f(n)(x).
Prove, by mathematical induction, that f(n)(x)=(2nx+n2n−1)e2x, n∈Z+.
Markscheme
f′(x)=e2x+2xe2x A1
Note: This must be obtained from the candidate differentiating f(x).
=(21x+1×21−1)e2x A1
(hence true for n=1)
assume true for n=k: M1
f(k)(x)=(2kx+k2k−1)e2x
Note: Award M1 if truth is assumed. Do not allow “let n=k”.
consider n=k+1:
f(k+1)(x)=ddx((2kx+k2k−1)e2x)
attempt to differentiate f(k)(x) M1
f(k+1)(x)=2ke2x+2(2kx+k2k−1)e2x A1
f(k+1)(x)=(2k+2k+1x+k2k)e2x
f(k+1)(x)=(2k+1x+(k+1)2k)e2x A1
=(2k+1x+(k+1)2(k+1)−1)e2x
True for n=1 and n=k true implies true for n=k+1.
Therefore the statement is true for all n(∈Z+) R1
Note: Do not award final R1 if the two previous M1s are not awarded. Allow full marks for candidates who use the base case n=0.
[7 marks]