2009 prelim espring s4e am p1
TRANSCRIPT
East Spring Secondary SchoolTowards Excellence and Success
Name: ( )
S 4E
Preliminary Examinations 2009Sec 4 Express
Additional Mathematics 4038/01Paper 1
Thursday 27th August 2009 2 hours12.40 -2.40 pm
Additional Materials:6 Writing Papers1 Graph Paper
READ THESE INSTRUCTIONS FIRST
Write your name, index number and class on all the work you hand in.Write in dark blue or black pen.You may use a soft pencil for any diagrams or graphs.Do not use staples, paper clips, highlighters, glue or correction fluid.
Answer all questions.Write your answers on the separate Answer Paper provided.Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the cases of angles in degrees, unless a different level of accuracy is specified in the question.The use of a scientific calculator is expected, where appropriate.You are reminded of the need for clear presentation in your answers.
At the end of the examination, fasten all your work securely together.The number of marks is given in brackets [ ] at the end of each question or part question.The total number of marks for this paper is 80.
This question paper consists of 5 printed pages including the cover page.
2009 Prelims AM P1
East Spring Secondary SchoolMathematics DepartmentDo It Right. Always
Mathematical Formulas
1. ALGEBRA
Quadratic EquationsFor the equation
Binomial expansion
where n is a positive integer and
2. TRIGONOMETRY
Identities
Formulas for
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1. Find all the angles between 0o and 360o which satisfy the equation . [4]
2. Calculate the coordinates of the points of intersection of the curve
and the straight line . [5]
3. Find the first four terms in the expansion of and of . Hence,
obtain the coefficient of in the expression of . [5]
4. Given that , where is an acute angle measured in degrees, obtain an
expression, in terms of p, for
(a) ,
(b) ,
(c) . [5]
5. A curve has an equation .
(a) Find an expression for . [2]
(b) Given that y is increasing at a rate of 0.2 units per second when x = 5, find the corresponding rate of change of x. [3]
6. A closed cylinder has a base radius of cm and a volume of
3.
Find in surd form and in terms of the exact values of
(a) the base area of the cylinder,
(b) the height of the cylinder,
(c) the total surface area of the cylinder.
[6]
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7. The function is defined by .
(a) State the amplitude and period of . [2]
(b) Find the -coordinates of the points where the graph of
crosses the x-axis. [3]
(c) Sketch the graph of . [2]
8. (a) Given that , find the value of a and of b, and Q(x). [4]
(b) The polynomial leaves the same remainder when divided by or where . Express in terms of . [3]
9. (a) Find the range of values of for which for all real values of . [4]
(b) Find the range of values of p for which the line meets the curve . [4]
10. A circle, , passes through the points (5, 0) and (3, 4).
(a) Find
(i) the midpoint of , [1]
(ii) the equation of the perpendicular bisector of . [2]
Given that the centre of the circle lies on the line ,
(b) (i) show that the centre of is (2, 1), and [2]
(ii) hence find the equation of the circle . [3]
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11. A particle, moving in a straight line, passes through a fixed point O with
velocity 14 ms–1. The acceleration, a ms–2, of the particle, t seconds after
passing through O, is given by a = 2t – 9. The particle subsequently comes to
instantaneous rest, first at A and later at B. find
(a) the acceleration of the particle at A and at B, [4]
(b) the greatest speed of the particle as it travels from A to B. [2]
(c) the distance AB. [4]
12. Answer the whole of this question on the graph paper provided. Variables and are related by the equation where a and n are constants.
x 2 3 5 6 8y 22 200 350 600 1450
The table shows experimental values of and , but an error has been made in recording one of the values of .
(a) On graph paper, plot against lg , using a scale of 2 cm to represent 0.1 unit on the horizontal axis and 2 cm to represent 0.5 unit on the vertical axis.Draw a straight line graph to represent the equation . [3]
(b) Use your graph to (i) correct the reading of for which an error has been made, [2](ii) estimate the value of a and of n. [3]
(c) On the same diagram, draw the straight line representing and hence find the value of for which . [2]
End of Paper
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Additional Mathematics Paper 1 (4038/01)Secondary 4 Express Marking Scheme
Qn Solution Mks
1.
or
(A2-for all correct answers, A1 for 2-3 correct answers)
B1
M1
A2
2. … (1)
… (2)
Subst (2) into (1):
The points of intersection are (2, 4) and (1, 2).
M1
M2M1
A1
3.
Coef of = = 8960
A1
M1
A1
M1
A1
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For correct substitution
For correct Quadratic equation
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Qn Solution Mks
4.
(a)
(b)
(c)
A1
M1, A1
M1, A1
5.
(a)
(b) unit/s
When ,
unit/s
M1, A1
M1
M1, A1
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p
121 p
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Qn Solution Mks
6. (a)
(b)
(c)
Base area of cylinder
cm2
Height of cylinder
cm
Total surface area of cylinder
cm2
A1
M1
M1
A1
M1
A1
7.
(a)
(b)
Amplitude = 2; Period = A1, A1
M1
M1
A1
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For Rationalising
For Factorising 841
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Qn Solution Mks
7. (c) Shape -
A1
Points -
A1
8.(a)
Let
Comparing the coefficients:
::
: :
M1
A1A1
A1
8.(b) Let
or (reject)
M1
M1
A1
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o30
6
o60
3
o90
2
o120
3
2
o150
6
5
o180
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Qn Solution Mks
9.(a)
M1
M1
M1
A1
9.(b) … (1)… (2)
Subst (1) into (2):
(1) meets (2)
M1
M1
M1
A1
10.(a) (i)
(ii)
(b) (i)
(ii)
Midpoint of AB =
Gradient of AB = Gradient of bisector of AB
=
So,
Equation of bisector of AB is … (1)
… (2)
Subst (1) into (2):
Centre = (2, 1) (shown)
Radius = units
Equation of circle is
A1
M1
A1
M1
A1
A1
M1
A1
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Qn Solution Mks
11(a) Integrating, v = t2 – 9t + c
At t = 0, v = 14, so c = 14.
Hence v = t2 – 9t + 14 = (t – 7)(t – 2).
When particle is at rest, v = 0; t = 2 or 7
At A, t = 2, a = –5m/s2,At B, t = 7, a = 5m/s2.
B1
B1
A1A1
(b) At greatest speed, a = 0 and this occur at t = 4.5
v = 4.52 – 9(4.5) + 14 = –6.25max speed = 6.25m/s
B1
A1
(c) Distance, s = t3 – t2 + 14t (constant = 0)
At A, s = 12 ,
At B, s = –8 .
AB = 20 m
(Alt: by definite integral)
B1
B1
B1
A1
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Qn Solution Mks
12. (a)
Plotting of points
Line of best fit
M1
A1
A1
(b) (i)
(ii)
From the graph:
Error reading of = 200
From the graph:
correct the reading of = 79.4 [70.8---89.1]
From the graph:
a = 2 [ ± 0.3 ]
[± 0.3]
M1
A1
M1
A1
A1
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lg x0.30.50.70.80.9lg y1.32.32.52.83.2
12
log y
0.18
1.9
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Qn Solution Mks
12(c)
Plot lg y against lg x.
lg x 0 0.9
lg y 0.7 1.6
From the graph:
[1.48-1.55]
A1
A1
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