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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 ) = x e 2 x , where  x R . The  n th  derivative of  f ( x ) is denoted by  f ( n ) ( x ) .

 

Prove, by mathematical induction, that  f ( n ) ( x ) = ( 2 n x + n 2 n 1 ) e 2 x n Z + .

Markscheme

f ( x ) = e 2 x + 2 x e 2 x       A1

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

= ( 2 1 x + 1 × 2 1 1 ) e 2 x       A1

(hence true for n = 1 )

 

assume true for  n = k :      M1

f ( k ) ( x ) = ( 2 k x + k 2 k 1 ) e 2 x

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

consider  n = k + 1 :

f ( k + 1 ) ( x ) = d d x ( ( 2 k x + k 2 k 1 ) e 2 x )

attempt to differentiate  f ( k ) ( x )        M1

f ( k + 1 ) ( x ) = 2 k e 2 x + 2 ( 2 k x + k 2 k 1 ) e 2 x       A1

f ( k + 1 ) ( x ) = ( 2 k + 2 k + 1 x + k 2 k ) e 2 x

f ( k + 1 ) ( x ) = ( 2 k + 1 x + ( k + 1 ) 2 k ) e 2 x       A1

     = ( 2 k + 1 x + ( k + 1 ) 2 ( k + 1 ) 1 ) e 2 x

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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