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REPUBLIC OF NAMIBIA
MINISTRY OF EDUCATION
NAMIBIA SENIOR SECONDARY CERTIFICATE
THESE PAPERS AND MARK SCHEMES SERVE TO EXEMPLIFY THE SPECIFICATIONS IN THE LOCALISED
NSSC MATHEMATICS HIGHER LEVEL SYLLABUS
2006
MATHEMATICS SPECIMEN PAPERS 1 – 2, MARK SCHEMES
AND ANALYSIS
HIGHER LEVEL
GRADES 11 – 12
Ministry of Education National Institute for Educational Development (NIED) Private Bag 2034 Okahandja Namibia © Copyright NIED, Ministry of Education, 2005 NSSCH Mathematics Specimen Paper Booklet Grades 11 - 12 ISBN: 99916-69-02-7 Printed by NIED Publication date: 2005
TABLE OF CONTENTS Paper 1: Specimen Paper .........................................................................................................1
Paper 1: Mark Scheme.............................................................................................................9
Paper 1: Paper Analysis .........................................................................................................11
Paper 2: Specimen Paper .......................................................................................................12
Paper 2: Mark Scheme...........................................................................................................18
Paper 2: Paper Analysis .........................................................................................................22
1
MINISTRY OF EDUCATION
Namibia Senior Secondary Certificate (NSSC)
MATHEMATICS: HIGHER LEVEL
PAPER 1: SPECIMEN PAPER
TIME: 2 hours MARK: 80 marks Additional Materials: Answer Book / Paper Electronic calculator Geometrical instruments One sheet of graph paper READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name in the spaces provided on the answer paper. Write in dark blue or black pen on both sides of the paper. 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 and working on the separate answer paper provided. All working must be clearly shown. It should be done on the same sheet as the rest of the answer. Marks will be given for working which shows that you know how to solve a problem even if you get the answer wrong. 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. Non-programmable scientific calculators should be used. If the degree of accuracy is not specified and if the answer is not exact, give the answer to three significant figures. Give answers in degrees to one decimal place.
This question paper consists of 8 printed pages.
2
1. A train leaves Windhoek at 18:40 on Monday evening and arrives at Swakopmund at 05:37 on Tuesday.
(a) Find the time of the train journey in hours and minutes. (1)
(b) The distance between Windhoek and Swakopmund is 380 km.
Calculate the average speed of the train in km/h. (2)
___________________________________________________________________ 2. (a) Solve the inequality
10 – 6.5x < 23. (2)
(b) Write down the smallest integer that satisfies this inequality. (1)
___________________________________________________________________ 3. The diagram shows a rope AB of length 20 m, correct to the nearest metre.
NOT TO SCALE
A piece, CB, is cut from the rope.
Given that the length CB is 2.5 metres, correct to 1 decimal place, find the maximum possible length of the remaining piece, AC. (3)
___________________________________________________________________ 4. The stored energy, E joules, in an elastic string is directly proportional to the square of
the extension, x, in centimetres. The stored energy is 60 joules when the extension is 5 cm.
(a) Find the equation relating E to x. (2)
(b) Find the extension of the string when the stored energy is 135 joules. (1)
___________________________________________________________________
3
5. The diagram shows a regular pentagon ABCDE inscribed in a circle, centre O.
NOT TO SCALE
Find the value of each of the following angles, giving a reason for each answer:
(a) angle BOC, (2)
(b) angle BEC, (2)
(c) angle BDC. (2)
___________________________________________________________________ 6. The length of a vehicle is 4.5 metres. A model of the vehicle is made to the scale 1 : 20.
NOT TO SCALE
(a) Find the length of the model in centimetres. (1)
(b) The volume of the interior of the model is 600 cm3. Find the volume of the interior of the vehicle, in cubic metres. (3)
___________________________________________________________________ 7. Amakoe has been invited to spend her 2004 Christmas in Britain. Her uncle gives her
N$2400 to take as pocket money. The exchange rate in 2004 is £1 = N$11.50
(a) Convert N$2400 to £, giving your answer to the nearest £. (1)
(b) The number of N$ obtained for £1 in 2004 is 30% less than the number of
N$ obtained for £1 in 2001. Find the exchange rate in 2001. (2)
___________________________________________________________________
4
8. The diagram shows the velocity-time graph of a cyclist for a journey.
NOT TO SCALE
The magnitude of the deceleration of the cyclist is twice that of the acceleration. The maximum velocity of 10 ms─1 is reached after 20 seconds.
(a) Find the acceleration of the cyclist in the first 20 seconds. (2)
(b) Find the total time taken by the cyclist for the journey. (2)
(c) Find the total distance travelled by the cyclist for the journey. (2)
___________________________________________________________________
v (ms─1) 10 0 20 30 t (s)
5
9.
NOT TO SCALE
Write down the four inequalities which define the shaded region in the diagram above. (3)
___________________________________________________________________ 10. The diagram shows three waterholes A, B and C in the Etosha National Park.
NOT TO SCALE
The distance between A and B is 45 km and the distance between A and C is 67 km. Angle ABC is 115.2o.
(a) Calculate the distance AC. (3)
(b) Given that B is due west of C, find the bearing of A from C. (3)
___________________________________________________________________
6
11.
.X(2, 2)
The diagram above shows a triangle ABC in which A is (5, 8), B is (2, 4), C is (9, 6). The point X is (2, 2).
The following transformations can be applied to the triangle ABC
P reflection in the line y = x
Q translation by the vector
−73
R rotation clockwise through 900 about X(2, 2).
(a) Find the image of point A, when transformation Q is applied, followed by transformation P. (2)
(b) Find the image of point C when transformation R is applied to triangle ABC. (2)
___________________________________________________________________
7
12. The diagram shows a sector of a circle which Eluid is using to form a hollow cone, without overlap.
NOT TO SCALE
The radius of the sector is 12.5 cm and the sector angle is 1000.
Find
(a) the arc length of the sector, (2)
(b) the radius of the cone, (2)
(c) the height of the cone. (2)
___________________________________________________________________ 13. A café owner in Oshakati sells take-away meals at lunch times.
He kept a record of the number of meals, M, that were sold per week over a period of 50 weeks and summarised the information in the following table.
Number of meals (M) served per
week
Number of weeks
0 < M ≤ 100 100 < M ≤ 200 200 < M ≤ 300 300 < M ≤ 400 400 < M ≤ 500 500 < M ≤ 600 600 < M ≤ 700 700 < M ≤ 800
2 7 11 17 7 4 1 1
(a) Draw, on graph paper, a cumulative frequency curve for the data. (3)
(b) Use your graph to estimate the median number of meals served over the
period of 50 weeks. (2)
(c) Calculate an estimate ofthe mean number of meals served per week. (2)
___________________________________________________________________
8
14. A lady golfer from Walvis Bay plays a particular hole many times in a year. On 30% of all days, there is a wind blowing across the course. If the wind is blowing, the probability that she hits a straight drive is 0.4, but if the wind is not blowing, the probability that she hits a straight drive, is 0.8. Find the probability that on a particular day
(a) the wind is not blowing and she hits a straight drive, (2)
(b) she hits a straight drive. (2)
She plays the hole on two successive days. Find the probability that
(c) she does not hit a straight drive on either of the two days. (2)
___________________________________________________________________ 15. The functions f and g are defined by
326:f+x
x a , ∈x ℝ, 5.1−≠x
xx 42:g −a , ∈x ℝ
(a) Express fg(x) in terms of x. (2)
(b) Hence, solve
fg(x) = 3. (1)
___________________________________________________________________ 16. The points A and B have the coordinates (─3, 6) and (9, ─4) respectively.
(a) Find the equation of the line AB. (3)
(b) Find the equation of the locus of all points equidistant from A and B. (3)
(c) Verify that the point C(13, 13) lies on this locus. (1)
(d) Find the distance AC. (2)
___________________________________________________________________ 17. The graph of the curve x2 + y2 = 9 meets the graph of the straight line x – y + 1 = 0
at two points. Find the coordinates of these points, giving your answers correct to 1 decimal place. (5)
___________________________________________________________________
9
MINISTRY OF EDUCATION
Namibia Senior Secondary Certificate (NSSC)
MATHEMATICS: HIGHER LEVEL
PAPER 1: MARK SCHEME
1. (a) 10 hours 57 min
(b) 95.10
380 = 34.7 km/h
B1 M1 A1√ 3
for td
2. (a) – 6.5x < 13 x > ─ 2 (b) x = ─ 1
M1 A1 B1√ 3
for collecting terms
3. AC = 20.5 – 2.45 = 18.05 m
M1 M1 A1 3
for using limits for subtraction of correct limits
4. (a) E = kx2 k = 2.4 (b) x = 7.5
M1 A1 A1√ 3
5. (a) 720 (5 equal angles around a point) (b) 360 (angle at centre equals twice angle at circumference) (c) 360 (angles in the same segment)
M1 A1 M1 A1 M1 A1 6
Accept any valid reason
6. (a) 22.5 cm (b) 600 x (20)3 = 480000 cm3 = 4.8 m3
B1 M1 A1 A1√ 4
7. (a) £ 209
(b) 7.05.11 = N$ 16.43
B1 M1 A1 3
8. (a) 2010 = 0.5 ms─2
(b) time for decelaration = 0.5 x 2 x 10 total time = 30 + 10 = 40 sec.
(c) 0.5 x 10 x 20 + 10 x 10 + 0.5 x 10 x 10 = 250 m
M1 A1 M1 A1 M1 A1√ 6
9. y ≥ 0; x ≤ 4; y < x + 1; y ≥ – x + 3
B3, 2, 1, 0 3
10
10. (a) AC2 = 452 + 672 – 2 x 45 x 67 x cos 115.20 = 95.3 km
(b) 3.95
2.115sin45
sin=
ACB
angle ACB = 25.30 Bearing = 295.30
M1 A1 A1 M1 A1 A1√ 6
M1 for correct formula, A1for correct input, A1 for final answer
11. (a) (5, 8) → (2, 15) → (15, 2) (b) (9, 6) → (6, ─ 5)
B1 B1√ B1 B1 4
one for each coordinate
12. (a) 360100 x 2 x π x 12.5 = 21.8 cm
(b) (21.8) : 2π = 3.47 cm
(c) h2 = 12.52 – 3.472 h = 12.0 cm
M1 A1 M1 A1 M1 A1√ 6
13. (a) correct calculation of cumulative frequency accurate plotting of points and drawing of curve (b) 320 – 330 (c) (50 x 2 + 150 x 7 + 250 x 11 + 350 x 17 + 450 x 7 + 550 x 4 + 650 + 750) : 50 = 332
M1 A2, 1, 0 M1 A1 M1 A1 7
M1 if it is shown on curve where reading is taken accept (50.5 x 2 + 150.5 x 7 + 250.5 x 11 ….) : 50 = 332.5
14. (a) 0.7 x 0.8 = 0.56 (b) (a) + 0.3 x 0.4 = 0.68 (c) (1 – (b))2 = 0.1024
M1 A1 M1 A1 M1 A1√ 6
Accept solution by tree diagram or any other valid method
15. (a) xx 87
63)42(2
6−
=+−
(b) 387
6=
− x x = 0.625 (or
85 )
M1 A1 M1 A1 4
16. (a) mAB = 65− Equation of AB
65
36 −=
+−
xy
5x + 6y = 21
(b) MAB(3, 1) Equation 56
31=
−−
xy
6x – 5y = 13
(c) Substitute (13,13) in 6x – 5y =13 or equivalent Show LHS = RHS
(d) 22 )613()313( −++=AC = 17.5
M1 M1 A1 B1 M1 A1 B1 M1 A1 9
Accept 213
65
+−
= xy or equivalent
Accept 532
56
−= xy or equivalent
17. x2 + (x + 1)2 = 9 2x2 + 2x – 8 = 0 x2 + x – 4 = 0
2
1611 +±−=x
x1 = 1.6 x2 = ─2.6 y1 = 2.6 y2 = ─1.6
M1 A1 M1 A1 A1√ 5
Accept (y – 1)2 + y2 = 9 an subsequent solution of y first for applying formula for a 3 term quadratic for solving for x for solving for y
11
MINISTRY OF EDUCATION
Namibia Senior Secondary Certificate (NSSC)
MATHEMATICS: HIGHER LEVEL
PAPER 1: PAPER ANALYSIS
Q Syll.
Ref. Topic Context Target
Grades Total
4 2 1 1 2(a)3
1(j)4 Time and speed Localized 3 3
2 5(d)7 1(a)1
Algebra and number Linear inequalities and types of number
3 3
3 1(i)2 Limits of accuracy Practical 3 3 4 1(j)6 Variation Science 3 3 5 4(d)9,10 Geometry Angles in circles 6 6 6 1(j)3
4(a)3 Scales and similar figures
Practical Volumes
1 3
4
7 1(l)1 Money and finance Currency 1 2 3 8 6(a)4 Kinematics Velocity-time graph 2 4 6 9 6(b)1 Linear programming Determining shaded
region 3 3
10 8(a)5 Trigonometry and bearing
Localized 3 3 6
11 9(b)5 Transformations Use and combine 2 2 4 12 3(a)2 Mensuration Arc length, cone 2 4 6 13 10(a),
6,7,8 Data handling Cumulative frequency,
mean Localized
2 5
14 10(b)5 Probability Localized 6 6 15 6(c)2 Functions Composite 3 3 16 7(a)1,2
4(e)3 Coordinate geometry Locus
Application without sketch
2 3 4
9
17 5(a)6 Algebra: simultaneous equations
Points of intersection of graphs
5 5
Total for paper 27 31 22 80 Percentage 33 39 28 100
12
MINISTRY OF EDUCATION
Namibia Senior Secondary Certificate (NSSC)
MATHEMATICS: HIGHER LEVEL
PAPER 2: SPECIMEN PAPER
TIME: 3 hours MARK: 120 marks Additional Materials: Answer Book / Paper Electronic calculator Geometrical instruments READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name in the spaces provided on the answer paper. Write in dark blue or black pen on both sides of the paper. 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 and working on the separate answer paper provided. All working must be clearly shown. It should be done on the same sheet as the rest of the answer. Marks will be given for working which shows that you know how to solve a problem even if you get the answer wrong. 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 120. Non-programmable scientific calculators should be used. If the degree of accuracy is not specified and if the answer is not exact, give the answer to three significant figures. Give answers in degrees to one decimal place.
This question paper consists of 6 printed pages.
13
1. A curve is such that 26dd
2 +=xx
y .
Given that the curve passes through the point P(3, 10), find
(a) the equation of the curve, (3)
(b) the equation of the tangent to the curve at P. (3)
___________________________________________________________________ 2. Find the gradient of the curve
y = 5 ln x – ln 4 at the point where y = ln 8. (5)
___________________________________________________________________ 3. (a) Show that the equation
2 sec2 x + tan x – 5 = 0 can be written as a quadratic equation in tan x. (2)
(b) Hence solve the equation
2 sec2 x + tan x – 5 = 0, for 0 radians ≤ x ≤ 2π radians. (5)
___________________________________________________________________ 4. (a) Sketch, on the same diagram, the graph of 52 −= xy and the graph of
y + x = 4, showing, on your diagram, the coordinates of the points of intersection of each graph with the coordinate axes. (4)
(b) Solve the equation 452 =+− xx . (3)
___________________________________________________________________
14
5. A particle moves in a straight line, so that t seconds after leaving a fixed point O, its displacement, s metres, from O is given by
tes t
2011010 −−= − .
Calculate
(a) the initial velocity of the particle, (3) (b) the value of t when the particle is instantaneously at rest, (2)
(c) the acceleration when the particle is instantaneously at rest. (3)
___________________________________________________________________ 6. The diagram below shows part of the curve y2 = 4 – x.
NOT TO SCALE
(a) Find the area of the shaded region. (6) (b) Find the volume generated when the shaded region is rotated through 360o about
the x- axis. (3)
___________________________________________________________________ 7. The points A, B and C have position vectors, relative to an origin O, of
3i + 2k, 2i – 2j + 5k and –2i + 10j + 7k respectively.
(a) Show that BA = i + 2j – 3k. (1)
(b) Use a scalar product to find angle ABC. (6)
(c) Find the coordinates of the point D, such that ACOD51
= . (2)
___________________________________________________________________
15
8. (a) Show that (x + 3) is a factor of x3 + 2x2 – 7x – 12 and hence solve the equation
x3 + 2x2 – 7x – 12 = 0
giving your answers correct to two decimal places where necessary. (5)
(b) The polynomial f(x) = x3 + ax2 + bx – 3 has a factor of (x + 1). When f(x) is divided by (x + 2), it has the same remainder as when it is divided by (x – 2). Find the value of a and of b. (4)
___________________________________________________________________ 9. The function f is defined by
xx 2cos45:f −a , for 0o ≤ x ≤ 360o.
(a) State the amplitude and period of f. (2)
(b) Sketch the graph of y = f(x) and write down the coordinates of the
maximum points. (5)
(c) Find the smallest value of x that satisfies the equation f(x) = 3. (3)
___________________________________________________________________ 10. NOT TO SCALE x cm h cm 2x cm
A solid rectangular block has a base, which measures 2x cm by x cm. The height of the block is h cm and the volume is 72 cm3.
(a) Express h in terms of x. (2)
(b) Show that the total surface area, A cm2, of the block, in terms of x, is
xxA 2164 2 += . (2)
(c) Given that x can vary, calculate the value of x for which A has a stationary
value. (3) (d) Find this value of A and determine whether it is a maximum or a minimum. (3)
___________________________________________________________________
16
11. (a) Solve the equation
22log3log2 =− xx . (4)
(b) The variables x and y are related by the equation y = aebx, where a and b are constants.
(i) Show that the graph of ln y against x is a straight line and express the
gradient and the intercept on the ln y-axis in terms of a and/or b. (3)
(ii) Given that this line passes through (0, 0.6) and (2, 1.6), find the value of a and of b. (4)
(iii) Find the value of y when x = 4. (1)
___________________________________________________________________ 12. The function f is defined by
1082:f 2 −+ xxx a .
(a) Express 2x2 + 8x – 10 in the form a(x + b)2 + c, where a, b and c are
constants. (3) (b) Hence, or otherwise, state the least value of y and the corresponding value
of x. (2) (c) Find the value of the constant k for which the equation
2x2 + 8x – 10 = k
has real, distinct roots. (3)
The function g is defined by
1082:g 2 −+ xxx a for the domain x ≥ – 2.
(d) Explain why g has an inverse but f has not. (1)
(e) Find an expression for g–1. (3)
(f) State the domain and range of g–1. (2)
___________________________________________________________________
17
13. (a) Evaluate ∑=
−23
0)345(
rr . (5)
(b) Copper is extracted from its ore obtained from a mine in Namibia. During the first
year of operation, the ore obtained yielded 7 000 kg of copper. With the increased difficulty of mining, the production of copper in each subsequent year shows a decrease of 20% on the previous year’s production. Assuming that mining continues in the same way for an infinite period of time,
(i) calculate the maximum amount of copper which could possibly be
extracted. (4)
For economic reasons, mining is abandoned once the annual yield falls below 1 000 kg. Calculate
(ii) the number of years the mine is in operation, (3)
(iii) the total yield of copper during this time. (2)
___________________________________________________________________
18
MINISTRY OF EDUCATION
Namibia Senior Secondary Certificate (NSSC)
MATHEMATICS: HIGHER LEVEL
PAPER 2: MARK SCHEME
1. (a) Cxx
y ++−= 26
10 = ─2 + 6 + C
626++−= x
xy
(b) At x = 3, m = 322
cxy +=38
10 = 8 + c 3y – 8x = 6
M1 M1 A1 M1 M1 A1 6
or equivalent
2. xdx
dy 5=
when y = ln 8, ln 8 = 5 ln x – ln 4 x = 2
gradient of curve = 25
M1 A1 M1 A1 A1 5
M1 for attempt at differentiation
3. (a) 2(tan2x + 1) + tan x – 5 = 0 2tan2x + tan x – 3 = 0
(b) tan x = 1 or tan x = 23
−
x = 0.785, 3.93 x = 2.16, 5.30
M1 A1 M1 A1 A1√ A1 A1√ 7
for use of correct identity if in degrees, max. of 3 marks (450, 2250, 123.7, 303.70)
4.
Pts of intersection on y-axis (5,0) and (4,0) Pts of intersection on x-axis (2.5,0) and (4,0)
(b) 2x – 5 + x = 4 x = 3
or 5 – 2x + x = 4 x = 1
B1 A1 M1 A1 B1 M1 A1 7
for y + x = 4 for pts of intersection with axes for shape of modulus graph for pts of intersection with axes
19
5. (a) 20110 −== −te
dtdsv
when t = 0, v = 9.95 ms─1
(b) when v = 0, 020110 =−−te
t = ln 200 = 5.29
(c) tedtdva −−== 10
when t = ln 200, 2
201 −−= msa
M1 A1 A1√ M1 A1 M1A1 A1 8
M1 for differentiation M1 for differentiation
6. (a) x = 4
∫ −=4
021
)4( xA dx
=
4
0
23
)4(32
−− x
=
−−−− 2
323
)04()34(32
= 3
14
(b) ∫ −=4
0)4( dxxV π
= 4
0
2
214
− xxπ
= 8π
B1 M1 A1 A1 M1 A1 M1 A1 A1 9
for minus sign for rest of expression correct
accept 25.1
7. (a) OBOABA −= (b) BC = ─4i + 12j + 2k
cos ABC = BCBA
BCBA.
= 414410941
6244++++
−+−
= 16414
14
angle ABC = 73.00 (c) OD = ─i + 2j + k D(─1, 2, 1)
B1 B1 M1 M1 M1 A1√ A1√ M1 A1 9
or equivalent for knowing the formula for knowing the scalar product for knowing how to find the modulus (either one) M1 for dividing AC by 5
20
8. (a) f(─3) = ─27 +18 +21 ─ 12 = 0 (x + 3)(x2 – x – 4) = 0
x = ─3 or 2
171±=x
x = 2.56 or x = ─1.56
(b) f(2) = f(─2) 8 + 4a + 2b – 3 = ─8 + 4a – 2b – 3 b = ─4 f(─1) = 0 ─1 + a – b – 3 = 0 a = 0
B1 M1 M1 A1 A1 M1 A1 M1 A1 9
for division for correct use of formula
9. (a) Amplitude = 4 Period = 1800 or π
(b) Sketch graph 2 cycles between 1 and 9 proper position (i.e. not a sin curve) Max points (900,9), (2700, 9)
(c) 5 – 4cos 2x = 3 cos 2x = 0.5 2x = 600
x = 300
B1 B1 B1 B1 B1 B1 B1 M1 M1 A1 10
allow 4± B1 for 900 and 2700, B1 for 9 for reducing to cos 2x
10. (a) hxx ××= 272
h = 2
36x
(b) A = hxxx ×+×× 622
= 4x2 + 2
366x
x×
(c) 2
2168x
xdxdA
−=
021682=−
xx
x = 3
(d) A = 108 cm2
32
2 4328xdx
Ad+= > 0 when x = 3
A = 108 cm2 is a minimum
M1 A1 M1 M1 M1 M1 A1 A1 M1 A1 10
allow for any other correct method
21
11. (a) 2log
3log2
2 =−x
x
(log2 x)2 – 2log2 x – 3 = 0 log2 x = 3 or log2 x = ─1 x = 8 or x = 0.5
(b) (i) ln y = ln a + bx ln e ln y = ln a + bx (ln e = 1)
Gradient = b Intercept on ln y- axis = ln a (ii) ln a = 0.6 a = 1.82
b = 21
026.06.1=
−−
(iii) y = 1.82 e2 = 13.4
M1 M1 A1 A1 B1 B1 B1 M1A1 M1A1 A1 12
12. (a) 2(x + 2)2 – 18 a = 2, b = 2, c = ─18
(b) from (a) x = ─2 and y = ─18 (c) 2x2 + 8x + (─10 – k) = 0 ∆ > 0 64 + 80 + 8k > 0 k > ─18
(d) g is a one-one function(turning point excluded) but f is not
(e) g: y = 2(x + 2)2 - 18 g─1: x = 2(y + 2)2 – 18
2218:1 −
+− xxg a
(f) domain: x ≥ ─18 range: y ≥ ─2
B1 B1 B1 B1√ B1√ M1 M1 A1 B1 M1 M1 A1 B1√ B1√ 14
or by differentiation or by formula of axis of symmetry accept any correct convincing explanation
13. (a) 45 + 42 + 39 ….. AP where a = 45, d = ─3 n = 24 S24 = 12(90 – 69) = 252
(b) Geometric progression with r = 0.8
350008.01
7000=
−=∞S kg
(c) 7000(0.8)n – 1 > 1000
8.0log
71log
<n
n = 10 years
(d) 8.01
)8.01(7000 9
9 −−
=S = 30300 kg
M1 A1 B1 M1 A1 M1 A1 M1 A1 M1 M1 A1 M1 A1 14
22
MINISTRY OF EDUCATION
Namibia Senior Secondary Certificate (NSSC)
MATHEMATICS: HIGHER LEVEL
PAPER 2: PAPER ANALYSIS
Q Syll.
Ref. Topic Context Target
Grades Total
4 2 1 1 20.1
19.4 Integration and differentiation
Equation of curve Equation of tangent
4 2 6
2 15.1 19.3
Logarithms Differentiation
Gradient of curve involving natural logs
2 3 5
3 17.3,5 12.3
Trigonometry Identities and quadratic equation
4 3 7
4 16.3 16.2
Absolute values Graph sketching and solving equation
4 1 2 7
5 19.6 Differentiation Application of rate of change
5 3 8
6 20.3 Integration Areas and volumes 3 6 9 7 13 Vectors in 3 dimensions Scalar product 5 4 9 8 11
12.2 Polynomials and remainder theorem
Applications of remainder and factor theorem
3 3 3 9
9 17.4 17.5
Trigonometry Graphs 3 2 5 10
10 3.1 19.5
Mensuration Differentiation
Application of stationary points to volume and area
5 5 10
11 15.1 15.4
Logarithms Change of base and transformation to linear form
4 4 4 12
12 12.4,7 14.1,3
Theory of quadratics Functions
Application of theory of quadratics and inverse functions, domain, range
5 5 4 14
13 18.1 5(e)4
Sigma notation Geometric progression
Apply sigma notation, solve GP problem
2 3 9 14
Total for paper 42 45 33 120 Percentage 35 38 27 100
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