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Date May 2019 Marks available 2 Reference code 19M.2.AHL.TZ1.H_10
Level Additional Higher Level Paper Paper 2 Time zone Time zone 1
Command term Write down Question number H_10 Adapted from N/A

Question

The voltage v in a circuit is given by the equation

v(t)=3sin(100πt)t0 where t is measured in seconds.

The current i in this circuit is given by the equation

i(t)=2sin(100π(t+0.003)).

The power p in this circuit is given by p(t)=v(t)×i(t).

The average power pav in this circuit from t=0 to t=T is given by the equation

pav(T)=1TT0p(t)dt, where T>0.

Write down the maximum and minimum value of v.

[2]
a.

Write down two transformations that will transform the graph of y=v(t) onto the graph of y=i(t).

[2]
b.

Sketch the graph of y=p(t) for 0 ≤ t ≤ 0.02 , showing clearly the coordinates of the first maximum and the first minimum.

[3]
c.

Find the total time in the interval 0 ≤ t ≤ 0.02 for which p(t) ≥ 3.

 

[3]
d.

Find pav(0.007).

 

[2]
e.

With reference to your graph of y=p(t) explain why pav(T) > 0 for all T > 0.

 

[2]
f.

Given that p(t) can be written as p(t)=asin(b(tc))+d where abcd > 0, use your graph to find the values of abc and d.

 

[6]
g.

Markscheme

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

3, −3       A1A1 

[2 marks]

a.

stretch parallel to the y-axis (with x-axis invariant), scale factor 23       A1

translation of (0.0030)  (shift to the left by 0.003)      A1

Note: Can be done in either order.

[2 marks]

b.

correct shape over correct domain with correct endpoints       A1
first maximum at (0.0035, 4.76)       A1
first minimum at (0.0085, −1.24)       A1

[3 marks]

c.

p ≥ 3 between t = 0.0016762 and 0.0053238 and t = 0.011676 and 0.015324       (M1)(A1)

Note: Award M1A1 for either interval.

= 0.00730       A1

[3 marks]

d.

pav=10.0070.00706sin(100πt)sin(100π(t+0.003))dt     (M1)

= 2.87       A1

[2 marks]

e.

in each cycle the area under the t axis is smaller than area above the t axis      R1

the curve begins with the positive part of the cycle       R1

[2 marks]

f.

a=4.76(1.24)2       (M1)

a=3.00       A1

d=4.76+(1.24)2

d=1.76       A1

b=2π0.01

b=628(=200π)       A1

c=0.00350.014       (M1)

c=0.00100       A1

[6 marks]

g.

Examiners report

[N/A]
a.
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b.
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c.
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d.
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e.
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f.
[N/A]
g.

Syllabus sections

Topic 2—Functions » SL 2.5—Modelling functions
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Topic 2—Functions » AHL 2.8—Transformations of graphs, composite transformations
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Topic 2—Functions
Topic 5—Calculus

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