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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+, p.

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+


Note: Only award the final R1 if the three method marks have been awarded.


[7 marks]

Examiners report

[N/A]

Syllabus sections

Topic 1—Number and algebra » AHL 1.15—Proof by induction, contradiction, counterexamples
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Topic 1—Number and algebra

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