Date | November 2020 | Marks available | 7 | Reference code | 20N.2.AHL.TZ0.H_6 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Prove | Question number | H_6 | Adapted from | N/A |
Question
Use mathematical induction to prove that dndxn(xepx)=pn-1(px+n)epx for n ∈ ℤ+, p ∈ ℚ.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
n=1: LHS=d(xepx)dx=xpepx+epx=(px+1)epx, RHS=p0(px+1)epx
LHS=RHS so true for n=1: A1
Note: Award A1 if n=0 is proved.
assume proposition true for n=k, i.e. dkdxk(xepx)=pk-1(px+k)epx M1
Notes: Do not award M1 if using n instead of k.
Assumption of truth must be present.
Subsequent marks are not dependent on this M1 mark.
dk+1dxk+1(xepx)=ddx(dkdxk(xepx)) (M1)
=ddx(pk-1(px+k)epx) M1
=pk-1(px+k)pepx+epx(pk)
=pk(px+k)epx+epx(pk) A1
Note: Award A1 for correct derivative.
=pk(px+k+1)epx A1
=p((k+1)-1)(px+(k+1))epx
Note: The final A1 can be awarded for either of the two lines above.
hence true for n=1 and n=k true ⇒n=k+1 true R1
therefore true for all n ∈ ℤ+
Note: Only award the final R1 if the three method marks have been awarded.
[7 marks]