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Date May 2018 Marks available 2 Reference code 18M.1.AHL.TZ1.H_7
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 1
Command term Find Question number H_7 Adapted from N/A

Question

Let  y = arccos ( x 2 )

Find  d y d x .

[2]
a.

Find 0 1 arccos ( x 2 ) d x .

[7]
b.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

y = arccos ( x 2 ) d y d x = 1 2 1 ( x 2 ) 2 ( = 1 4 x 2 )     M1A1

Note: M1 is for use of the chain rule.

[2 marks]

a.

attempt at integration by parts     M1

u = arccos ( x 2 ) d u d x = 1 4 x 2

d v d x = 1 v = x      (A1)

0 1 arccos ( x 2 ) d x = [ x arccos ( x 2 ) ] 0 1 + 0 1 1 4 x 2 d x       A1

using integration by substitution or inspection      (M1)

[ x arccos ( x 2 ) ] 0 1 + [ ( 4 x 2 ) 1 2 ] 0 1       A1

Note: Award A1 for  ( 4 x 2 ) 1 2  or equivalent.

Note: Condone lack of limits to this point.

attempt to substitute limits into their integral     M1

= π 3 3 + 2      A1

[7 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 5 —Calculus » SL 5.3—Differentiating polynomials, n E Z
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Topic 5 —Calculus » SL 5.6—Differentiating polynomials n E Q. Chain, product and quotient rules
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Topic 5 —Calculus

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