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Date November 2019 Marks available 6 Reference code 19N.1.AHL.TZ0.H_2
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term Find Question number H_2 Adapted from N/A

Question

Given that 0 ln k e 2 x d x = 12 , find the value of k .

Markscheme

1 2 e 2 x seen       (A1)

attempt at using limits in an integrated expression ( [ 1 2 e 2 x ] 0 ln k = 1 2 e 2 ln k 1 2 e 0 )         (M1)

= 1 2 e ln k 2 1 2 e 0        (A1)

Setting their equation  = 12        M1

Note: their equation must be an integrated expression with limits substituted.

1 2 k 2 1 2 = 12        A1

( k 2 = 25 ) k = 5        A1

Note: Do not award final A1 for  k = ± 5 .

[6 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5 —Calculus » SL 5.10—Indefinite integration, reverse chain, by substitution
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