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FP2 Mock - March 24 - MS

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0% found this document useful (0 votes)
37 views21 pages

FP2 Mock - March 24 - MS

Mock exam marking scheme

Uploaded by

suren2006
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
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FURTHER PURE MATHEMATICS : F2


International Advanced Subsidiary /Advanced Level

Mock Exam – March 2024 Paper Reference


Duration: 1 hour 30 minutes WFM02/01

Additional materials required Total Marks

Mathematical Formulae and Statistical Tables (Lilac), Calculator

Instructions
• Use black ink or ball-point pen.
• Fill in the boxes at the top of this page with your name, and class.
• Answer the questions in the spaces provided – there may be more space than you need

Information
• The total mark for this paper is 75.
• The marks for each question are shown in brackets.
- use this as a guideline as to how much time spend on each question.
Advice
• Read each question carefully before you start to answer it.
• Try to answer every question.
• Check your answers if you have time at the end.

This document consists of 21 pages


Turn over
2

1.
2
(a) Express 𝑟𝑟(𝑟𝑟+2)
in partial fractions.
(2)
(b) Hence show that

𝑛𝑛
2 𝑛𝑛(3𝑛𝑛 + 5)
� =
𝑟𝑟(𝑟𝑟 + 2) 2(𝑛𝑛 + 1)(𝑛𝑛 + 2)
𝑟𝑟=1
(4)
20
2
(c) Find the value of � , to 4 decimal places.
𝑟𝑟=11 𝑟𝑟(𝑟𝑟+2)
(2)
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2. Using algebra, find the set of values of 𝑥𝑥 for which

𝑥𝑥 2
<
𝑥𝑥 + 1 𝑥𝑥 + 3
(7)
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3.
(a) Determine the general solution of the differential equation

𝑑𝑑𝑑𝑑
+ 2𝑦𝑦 cot 𝑥𝑥 = 2𝑒𝑒 2𝑥𝑥 cosec 2 𝑥𝑥
𝑑𝑑𝑑𝑑

giving your answer in the form 𝑦𝑦 = 𝑓𝑓(𝑥𝑥).


(5)
𝜋𝜋
(b) Determine the particular solution for which 𝑦𝑦 = 0 at 𝑥𝑥 = 2 .
(2)
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8

4. The complex number 𝑧𝑧 = cos 𝜃𝜃 + 𝑖𝑖 sin 𝜃𝜃, where 𝜃𝜃 is real.

(a) Use de Moivre’s theorem to show that

1
𝑧𝑧 𝑛𝑛 + = 2 cos 𝑛𝑛𝑛𝑛
𝑧𝑧 𝑛𝑛

where 𝑛𝑛 is a positive integer.

(2)
(b) Show that

1
cos 4 𝜃𝜃 = (cos 4𝜃𝜃 + 4 cos 2𝜃𝜃 + 3)
8
(5)
(c) Hence show that
𝜋𝜋
3
7√3
�(8 cos4 𝜃𝜃 − 3)𝑑𝑑𝑑𝑑 =
8
0
(3)
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5.
(a) Given that 𝑥𝑥 = 𝑒𝑒 𝑡𝑡 , determine, in terms of 𝑦𝑦 and 𝑡𝑡,

𝑑𝑑𝑑𝑑
(i) 𝑑𝑑𝑑𝑑

𝑑𝑑2 𝑦𝑦
(ii) 𝑑𝑑𝑥𝑥 2
(4)
(b) Hence show that the transformation 𝑥𝑥 = 𝑒𝑒 𝑡𝑡 , transforms the differential equation

𝑑𝑑 2 𝑦𝑦 𝑑𝑑𝑑𝑑
𝑥𝑥 2 2
− 4𝑥𝑥 + 6𝑦𝑦 = 𝑥𝑥 3 𝑥𝑥 > 0 (𝐼𝐼)
𝑑𝑑𝑥𝑥 𝑑𝑑𝑥𝑥

Into the differential equation

𝑑𝑑 2 𝑦𝑦 𝑑𝑑𝑑𝑑
2
−5 + 6𝑦𝑦 = 𝑒𝑒 3𝑡𝑡 (𝐼𝐼𝐼𝐼)
𝑑𝑑𝑡𝑡 𝑑𝑑𝑑𝑑
(2)

(c) Solve differential equation (𝐼𝐼𝐼𝐼) to determine a general solution for 𝑦𝑦 in terms of 𝑡𝑡.
(7)
(d) Hence determine the general solution of differential equation (𝐼𝐼).

(1)
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6. The complex number 𝑧𝑧 on an Argand diagram is represented by the point 𝑃𝑃 where

|𝑧𝑧 − 2 + 6𝑖𝑖| = 2|𝑧𝑧 − 5 + 3𝑖𝑖|

Given that the locus of 𝑃𝑃 is a circle,

(a) determine the centre and radius of this circle, and sketch the locus of 𝑃𝑃.
(5)
(b) Find the complex number 𝑧𝑧 which satisfy both |𝑧𝑧 − 2 + 6𝑖𝑖| = 2|𝑧𝑧 − 5 + 3𝑖𝑖| and
𝜋𝜋
arg(𝑧𝑧 − 6 + 4𝑖𝑖) = − 4 .
(4)
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7. Given that 𝑦𝑦 = tan 𝑥𝑥

(a) Show that

𝑑𝑑3 𝑦𝑦
= sec 2 𝑥𝑥 (𝑝𝑝 sec 2 𝑥𝑥 + 𝑞𝑞)
𝑑𝑑𝑥𝑥 3

where 𝑝𝑝 and 𝑞𝑞 are integers to be determined.


(4)
𝜋𝜋
(b) Hence determine the Taylor series expansion about 4
of tan 𝑥𝑥 in ascending
𝜋𝜋 𝜋𝜋 3
powers of �𝑥𝑥 − 4 �, up to and including the term in �𝑥𝑥 − 4 � , giving each
coefficient in simplest form.
(3)
(c) Use the answer to part (b) to determine, to four significant figures, an approximate
5𝜋𝜋
value of tan �12 �.
(2)
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8.

Figure 1

The curve 𝐶𝐶, shown in Figure 1, has polar equation

𝑟𝑟 = 2𝑎𝑎(1 + cos 𝜃𝜃) , 0 ≤ 𝜃𝜃 < 𝜋𝜋

The tangent to 𝐶𝐶 at the point 𝐴𝐴 is parallel to the initial line.

(a) Find the polar coordinates 𝐴𝐴.


(6)
The finite region 𝑅𝑅, shown shaded in Figure 1, is bounded by the curve 𝐶𝐶, the initial line
and the line 𝑂𝑂𝑂𝑂.

(b) Use calculus to find the area of the shaded region 𝑅𝑅, giving your answer in the form
𝑎𝑎2 �𝑝𝑝𝑝𝑝 + 𝑞𝑞√3 �, where 𝑝𝑝 and 𝑞𝑞 are rational numbers to be found.
(5)
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