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Date November 2018 Marks available 6 Reference code 18N.1.AHL.TZ0.H_10
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term Find Question number H_10 Adapted from N/A

Question

The function f is defined by  f ( x ) = e x cos 2 x , where 0 ≤  x  ≤ 5. The curve  y = f ( x )  is shown on the following graph which has local maximum points at A and C and touches the x -axis at B and D.

Use integration by parts to show that e x cos 2 x d x = 2 e x 5 sin 2 x + e x 5 cos 2 x + c ,   c R .

[5]
a.

Hence, show that e x cos 2 x d x = e x 5 sin 2 x + e x 10 cos 2 x + e x 2 + c ,   c R .

[3]
b.

Find the x -coordinates of A and of C , giving your answers in the form  a + arctan b , where  a b R .

[6]
c.

Find the area enclosed by the curve and the x -axis between B and D, as shaded on the diagram.

[5]
d.

Markscheme

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

METHOD 1

attempt at integration by parts with  u = e x ,   d v d x = cos 2 x       M1

e x cos 2 x d x = e x 2 sin 2 x d x e x 2 sin 2 x d x       A1

= e x 2 sin 2 x 1 2 ( e x 2 cos 2 x + e x 2 cos 2 x )       M1A1

e x 2 sin 2 x + e x 4 cos 2 x 1 4 e x cos 2 x d x

5 4 e x cos 2 x d x = e x 2 sin 2 x + e x 4 cos 2 x       M1

e x cos 2 x d x = 2 e x 5 sin 2 x + e x 5 cos 2 x ( + c )     AG

 

 

METHOD 2

attempt at integration by parts with u = cos 2 x d v d x = e x       M1

e x cos 2 x d x = e x cos 2 x + 2 e x sin 2 x d x       A1

= e x cos 2 x + 2 ( e x sin 2 x 2 e x cos 2 x d x )       M1A1

= e x cos 2 x + 2 e x sin 2 x 4 e x cos 2 x d x

5 e x cos 2 x d x = e x cos 2 x + 2 e x sin 2 x       M1

e x cos 2 x d x = 2 e x 5 sin 2 x + e x 5 cos 2 x ( + c )    AG

 

METHOD 3

attempt at use of table      M1

eg

      A1A1 

Note: A1 for first 2 lines correct, A1 for third line correct.

e x cos 2 x d x = e x cos 2 x + 2 e x sin 2 x 4 e x cos 2 x d x       M1

5 e x cos 2 x d x = e x cos 2 x + 2 e x sin 2 x       M1

e x cos 2 x d x = 2 e x 5 sin 2 x + e x 5 cos 2 x ( + c )    AG

 

[5 marks]

a.

e x co s 2 x d x = e x 2 ( cos 2 x + 1 ) d x      M1A1

= 1 2 ( 2 e x 5 sin 2 x + e x 5 cos 2 x ) + e x 2       A1

= e x 5 sin 2 x + e x 10 cos 2 x + e x 2 ( + c )       AG

Note: Do not accept solutions where the RHS is differentiated.

 

[3 marks]

b.

f ( x ) = e x co s 2 x 2 e x sin x cos x       M1A1

Note: Award M1 for an attempt at both the product rule and the chain rule.

e x cos x ( cos x 2 sin x ) = 0       (M1)

Note: Award M1 for an attempt to factorise  cos x  or divide by  cos x ( cos x 0 ) .

discount  cos x = 0  (as this would also be a zero of the function)

cos x 2 sin x = 0

tan x = 1 2       (M1)

x = arctan ( 1 2 ) (at A) and  x = π + arctan ( 1 2 )  (at C)      A1A1

Note: Award A1 for each correct answer. If extra values are seen award A1A0.

 

[6 marks]

c.

 

cos x = 0 x = π 2 or  3 π 2       A1

Note: The A1may be awarded for work seen in part (c).

π 2 3 π 2 ( e x co s 2 x ) d x = [ e x 5 sin 2 x + e x 10 cos 2 x + e x 2 ] π 2 3 π 2       M1

= ( e 3 π 2 10 + e 3 π 2 2 ) ( e π 2 10 + e π 2 2 ) ( = 2 e 3 π 2 5 2 e π 2 5 )       M1(A1)A1

Note: Award M1 for substitution of the end points and subtracting, (A1) for  sin 3 π = sin π = 0 and  cos 3 π = cos π = 1  and A1 for a completely correct answer.

 

[5 marks]

d.

Examiners report

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

Syllabus sections

Topic 5 —Calculus » SL 5.7—The second derivative
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