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, c∈R.
Hence, show that ∫excos2xdx=ex5sin2x+ex10cos2x+ex2+c, c∈R.
Find the x-coordinates of A and of C , giving your answers in the form a+arctanb, where a, b∈R.
Find the area enclosed by the curve and the x-axis between B and D, as shaded on the diagram.
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=ex2sin2xdx−∫ex2sin2xdx A1
= ex2sin2x−12(−ex2cos2x+∫ex2cos2x) M1A1
= ex2sin2x+ex4cos2x−14∫excos2xdx
∴54∫excos2xdx=ex2sin2x+ex4cos2x M1
∫excos2xdx=2ex5sin2x+ex5cos2x(+c) AG
METHOD 2
attempt at integration by parts with u=cos2x, dvdx=ex M1
∫excos2xdx=excos2x+2∫exsin2xdx A1
=excos2x+2(exsin2x−2∫excos2xdx) M1A1
=excos2x+2exsin2x−4∫excos2xdx
∴5∫excos2xdx=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+2exsin2x−4∫excos2xdx M1
∴5∫excos2xdx=excos2x+2exsin2x M1
∫excos2xdx=2ex5sin2x+ex5cos2x(+c) AG
[5 marks]
∫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]
f′(x)=excos2x−2exsinxcosx M1A1
Note: Award M1 for an attempt at both the product rule and the chain rule.
excosx(cosx−2sinx)=0 (M1)
Note: Award M1 for an attempt to factorise cosx or divide by cosx(cosx≠0).
discount cosx=0 (as this would also be a zero of the function)
⇒cosx−2sinx=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]
cosx=0⇒x=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)(=2e3π25−2eπ25) 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]