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Date May 2018 Marks available 3 Reference code 18M.2.sl.TZ1.10
Level SL only Paper 2 Time zone TZ1
Command term Hence and Write Question number 10 Adapted from N/A

Question

Let \(f\left( x \right) = 12\,\,{\text{cos}}\,x - 5\,\,{\text{sin}}\,x,\,\, - \pi  \leqslant x \leqslant 2\pi \), be a periodic function with \(f\left( x \right) = f\left( {x + 2\pi } \right)\)

The following diagram shows the graph of \(f\).

There is a maximum point at A. The minimum value of \(f\) is −13 .

A ball on a spring is attached to a fixed point O. The ball is then pulled down and released, so that it moves back and forth vertically.

The distance, d centimetres, of the centre of the ball from O at time t seconds, is given by

\(d\left( t \right) = f\left( t \right) + 17,\,\,0 \leqslant t \leqslant 5.\)

Find the coordinates of A.

[2]
a.

For the graph of \(f\), write down the amplitude.

[1]
b.i.

For the graph of \(f\), write down the period.

[1]
b.ii.

Hence, write \(f\left( x \right)\) in the form \(p\,\,{\text{cos}}\,\left( {x + r} \right)\).

[3]
c.

Find the maximum speed of the ball.

[3]
d.

Find the first time when the ball’s speed is changing at a rate of 2 cm s−2.

[5]
e.

Markscheme

−0.394791,13

A(−0.395, 13)      A1A1 N2

[2 marks]

a.

13      A1 N1

[1 mark]

b.i.

\({2\pi }\), 6.28      A1 N1

[1 mark]

b.ii.

valid approach      (M1)

eg recognizing that amplitude is p or shift is r

\(f\left( x \right) = 13\,\,{\text{cos}}\,\left( {x + 0.395} \right)\)   (accept p = 13, r = 0.395)     A1A1 N3

Note: Accept any value of r of the form \(0.395 + 2\pi k,\,\,k \in \mathbb{Z}\)

[3 marks]

c.

recognizing need for d ′(t)      (M1)

eg  −12 sin(t) − 5 cos(t)

correct approach (accept any variable for t)      (A1)

eg  −13 sin(t + 0.395), sketch of d′, (1.18, −13), t = 4.32

maximum speed = 13 (cms−1)      A1 N2

[3 marks]

d.

recognizing that acceleration is needed      (M1)

eg   a(t), d "(t)

correct equation (accept any variable for t)      (A1)

eg  \(a\left( t \right) =  - 2,\,\,\left| {\frac{{\text{d}}}{{{\text{d}}t}}\left( {d'\left( t \right)} \right)} \right| = 2,\,\, - 12\,\,{\text{cos}}\,\left( t \right) + 5\,\,{\text{sin}}\,\left( t \right) =  - 2\)

valid attempt to solve their equation   (M1)

eg  sketch, 1.33

1.02154

1.02      A2 N3

[5 marks]

e.

Examiners report

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

Syllabus sections

Topic 6 - Calculus » 6.6 » Kinematic problems involving displacement \(s\), velocity \(v\) and acceleration \(a\).
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