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Date November 2018 Marks available 3 Reference code 18N.2.SL.TZ0.S_10
Level Standard Level Paper Paper 2 Time zone Time zone 0
Command term Find Question number S_10 Adapted from N/A

Question

All lengths in this question are in metres.

 

Consider the function f ( x ) = 4 x 2 8 , for −2 ≤ x  ≤ 2. In the following diagram, the shaded region is enclosed by the graph of f and the x -axis.

A container can be modelled by rotating this region by 360˚ about the x -axis.

Water can flow in and out of the container.

The volume of water in the container is given by the function g ( t ) , for 0 ≤ t ≤ 4 , where t is measured in hours and g ( t ) is measured in m3. The rate of change of the volume of water in the container is given by g ( t ) = 0.9 2.5 cos ( 0.4 t 2 ) .

The volume of water in the container is increasing only when  p  < t  < q .

Find the volume of the container.

[3]
a.

Find the value of  p and of  q .

[3]
b.i.

During the interval  p  < t  < q , he volume of water in the container increases by k  m3. Find the value of k .

[3]
b.ii.

When t = 0, the volume of water in the container is 2.3 m3. It is known that the container is never completely full of water during the 4 hour period.

 

Find the minimum volume of empty space in the container during the 4 hour period.

[5]
c.

Markscheme

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

attempt to substitute correct limits or the function into formula involving  f 2       (M1)

eg       π 2 2 y 2 d y ,   π ( 4 x 2 8 ) 2 d x

4.18879

volume = 4.19,  4 3 π   (exact) (m3)      A2 N3

Note: If candidates have their GDC incorrectly set in degrees, award M marks where appropriate, but no A marks may be awarded. Answers from degrees are p = 13.1243 and q = 26.9768 in (b)(i) and 12.3130 or 28.3505 in (b)(ii).

 

[3 marks]

 

 

a.

recognizing the volume increases when g is positive      (M1)

eg    g ( t ) > 0,  sketch of graph of g indicating correct interval

1.73387, 3.56393

p = 1.73,  p = 3.56      A1A1 N3

 

[3 marks]

 

 

b.i.

valid approach to find change in volume      (M1)

eg    g ( q ) g ( p ) ,   p q g ( t ) d t

3.74541

total amount = 3.75  (m3)      A2 N3

 

[3 marks]

b.ii.

Note: There may be slight differences in the final answer, depending on which values candidates carry through from previous parts. Accept answers that are consistent with correct working.

 

recognizing when the volume of water is a maximum     (M1)

eg   maximum when  t = q ,   0 q g ( t ) d t

valid approach to find maximum volume of water      (M1)

eg    2.3 + 0 q g ( t ) d t ,   2.3 + 0 p g ( t ) d t + 3.74541 ,  3.85745

correct expression for the difference between volume of container and maximum value      (A1)

eg    4.18879 ( 2.3 + 0 q g ( t ) d t ) ,  4.19 − 3.85745

0.331334

0.331 (m3)      A2 N3

 

[5 marks]

c.

Examiners report

[N/A]
a.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.

Syllabus sections

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