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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 ) = 3 sin ( 100 π t ) t 0  where t is measured in seconds.

The current i in this circuit is given by the equation

i ( t ) = 2 sin ( 100 π ( t + 0.003 ) ) .

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

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

p a v ( T ) = 1 T 0 T p ( t ) d t , 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 p a v (0.007).

 

[2]
e.

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

 

[2]
f.

Given that p ( t ) can be written as  p ( t ) = a sin ( b ( t c ) ) + d  where a b c d > 0, use your graph to find the values of a b c  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  2 3        A1

translation of  ( 0.003 0 )   (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.

p a v = 1 0.007 0 0.007 6 sin ( 100 π t ) sin ( 100 π ( t + 0.003 ) ) d t      (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 = 2 π 0.01

b = 628 ( = 200 π )        A1

c = 0.0035 0.01 4        (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.
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g.

Syllabus sections

Topic 2—Functions » SL 2.5—Modelling functions
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