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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 xR. The nth derivative of f(x) is denoted by f(n)(x).

 

Prove, by mathematical induction, that f(n)(x)=(2nx+n2n1)e2xnZ+.

Markscheme

f(x)=e2x+2xe2x      A1

Note: This must be obtained from the candidate differentiating f(x).

=(21x+1×211)e2x      A1

(hence true for n=1)

 

assume true for n=k:      M1

f(k)(x)=(2kx+k2k1)e2x

Note: Award M1 if truth is assumed. Do not allow “let n=k”.

consider n=k+1:

f(k+1)(x)=ddx((2kx+k2k1)e2x)

attempt to differentiate f(k)(x)       M1

f(k+1)(x)=2ke2x+2(2kx+k2k1)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]

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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