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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)=excos2x, 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 excos2xdx=2ex5sin2x+ex5cos2x+c,  cR.

[5]
a.

Hence, show that excos2xdx=ex5sin2x+ex10cos2x+ex2+c,  cR.

[3]
b.

Find the x-coordinates of A and of C , giving your answers in the form a+arctanb, where abR.

[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=ex,  dvdx=cos2x      M1

excos2xdx=ex2sin2xdxex2sin2xdx      A1

= ex2sin2x12(ex2cos2x+ex2cos2x)      M1A1

ex2sin2x+ex4cos2x14excos2xdx

54excos2xdx=ex2sin2x+ex4cos2x      M1

excos2xdx=2ex5sin2x+ex5cos2x(+c)    AG

 

 

METHOD 2

attempt at integration by parts with u=cos2xdvdx=ex      M1

excos2xdx=excos2x+2exsin2xdx      A1

=excos2x+2(exsin2x2excos2xdx)      M1A1

=excos2x+2exsin2x4excos2xdx

5excos2xdx=excos2x+2exsin2x      M1

excos2xdx=2ex5sin2x+ex5cos2x(+c)    AG

 

METHOD 3

attempt at use of table      M1

eg

      A1A1 

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

excos2xdx=excos2x+2exsin2x4excos2xdx      M1

5excos2xdx=excos2x+2exsin2x      M1

excos2xdx=2ex5sin2x+ex5cos2x(+c)    AG

 

[5 marks]

a.

excos2xdx=ex2(cos2x+1)dx     M1A1

=12(2ex5sin2x+ex5cos2x)+ex2      A1

=ex5sin2x+ex10cos2x+ex2(+c)      AG

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

 

[3 marks]

b.

f(x)=excos2x2exsinxcosx      M1A1

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

excosx(cosx2sinx)=0      (M1)

Note: Award M1 for an attempt to factorise cosx or divide by cosx(cosx0).

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

cosx2sinx=0

tanx=12      (M1)

x=arctan(12) (at A) and x=π+arctan(12) (at C)      A1A1

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

 

[6 marks]

c.

 

cosx=0x=π2 or 3π2      A1

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

3π2π2(excos2x)dx=[ex5sin2x+ex10cos2x+ex2]3π2π2      M1

=(e3π210+e3π22)(eπ210+eπ22)(=2e3π252eπ25)      M1(A1)A1

Note: Award M1 for substitution of the end points and subtracting, (A1) for sin3π=sinπ=0 and cos3π=cosπ=1 and A1 for a completely correct answer.

 

[5 marks]

d.

Examiners report

[N/A]
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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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Topic 3— Geometry and trigonometry
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