MINISTRY OF EDUCATION

REPUBLIC OF NAMIBIA MINISTRY OF EDUCATION NAMIBIA SENIOR SECONDARY CERTIFICATE MATHEMATICS SPECIMEN PAPERS 1, MARK SCHEMES AND ANALYSIS HIGHER LEVEL GRADES 11 1 THESE PAPERS AND MARK SCHEMES SERVE TO EXEMPLIFY THE SPECIFICATIONS IN THE LOCALISED NSSC MATHEMATICS HIGHER LEVEL SYLLABUS 006

Ministry of Education National Institute for Educational Development (NIED) Private Bag 034 Okahandja Namibia Copyright NIED, Ministry of Education, 005 NSSCH Mathematics Specimen Paper Booklet Grades 11-1 ISBN: 99916-69-0-7 Printed by NIED Publication date: 005

TABLE OF CONTENTS Paper 1: Specimen Paper...1 Paper 1: Mark Scheme...9 Paper 1: Paper Analysis...11 Paper : Specimen Paper...1 Paper : Mark Scheme...18 Paper : Paper Analysis...

MINISTRY OF EDUCATION Namibia Senior Secondary Certificate (NSSC) MATHEMATICS: HIGHER LEVEL PAPER 1: SPECIMEN PAPER TIME: 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. 1

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. (). (a) Solve the inequality 10 6.5x < 3. () (b) Write down the smallest integer that satisfies this inequality. (1) 3. The diagram shows a rope AB of length 0 m, correct to the nearest metre. NOT TO SCALE A piece, CB, is cut from the rope. Given that the length CB is.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. () (b) Find the extension of the string when the stored energy is 135 joules. (1)

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, () (b) angle BEC, () (c) angle BDC. () 6. The length of a vehicle is 4.5 metres. A model of the vehicle is made to the scale 1 : 0. NOT TO SCALE (a) Find the length of the model in centimetres. (1) (b) The volume of the interior of the model is 600 cm 3. Find the volume of the interior of the vehicle, in cubic metres. (3) 7. Amakoe has been invited to spend her 004 Christmas in Britain. Her uncle gives her N$400 to take as pocket money. The exchange rate in 004 is 1 = N$11.50 (a) Convert N$400 to, giving your answer to the nearest. (1) (b) The number of N$ obtained for 1 in 004 is 30% less than the number of N$ obtained for 1 in 001. Find the exchange rate in 001. () 3

8. The diagram shows the velocity-time graph of a cyclist for a journey. v (ms 1 ) 10 NOT TO SCALE 0 0 30 t (s) The magnitude of the deceleration of the cyclist is twice that of the acceleration. The maximum velocity of 10 ms 1 is reached after 0 seconds. (a) Find the acceleration of the cyclist in the first 0 seconds. () (b) Find the total time taken by the cyclist for the journey. () (c) Find the total distance travelled by the cyclist for the journey. () 4

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. o. (a) Calculate the distance AC. (3) (b) Given that B is due west of C, find the bearing of A from C. (3) 5

11..X(, ) The diagram above shows a triangle ABC in which A is (5, 8), B is (, 4), C is (9, 6). The point X is (, ). The following transformations can be applied to the triangle ABC P reflection in the line y = x 3 Q translation by the vector 7 R rotation clockwise through 90 0 about X(, ). (a) Find the image of point A, when transformation Q is applied, followed by transformation P. () (b) Find the image of point C when transformation R is applied to triangle ABC. () 6

1. 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 1.5 cm and the sector angle is 100 0. Find (a) the arc length of the sector, () (b) the radius of the cone, () (c) the height of the cone. () 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 0 < M 100 100 < M 00 00 < M 300 300 < M 400 400 < M 500 500 < M 600 600 < M 700 700 < M 800 Number of weeks 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. () (c) Calculate an estimate ofthe mean number of meals served per week. () 7

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, () (b) she hits a straight drive. () 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. () 15. The functions f and g are defined by 6 f : x a, x R, x 1. 5 x + 3 g : x a 4x, x R (a) Express fg(x) in terms of x. () (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. () 17. The graph of the curve x + y = 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) 8

MINISTRY OF EDUCATION Namibia Senior Secondary Certificate (NSSC) MATHEMATICS: HIGHER LEVEL PAPER 1: MARK SCHEME 1. (a) 10 hours 57 min 380 (b) = 34.7 km/h 10.95. (a) 6.5x < 13 x > (b) x = 1 3 3 for t d for collecting terms 3. AC = 0.5.45 = 18.05 m 3 for using limits for subtraction of correct limits 4. (a) E = kx k =.4 (b) x = 7.5 3 5. (a) 7 0 (5 equal angles around a point) (b) 36 0 (angle at centre equals twice angle at circumference) (c) 36 0 (angles in the same segment) 6. (a).5 cm (b) 600 x (0) 3 = 480000 cm 3 = 4.8 m 3 6 4 Accept any valid reason 7. (a) 09 11.5 (b) = N$ 16.43 0.7 3 10 8. (a) = 0.5 ms 0 (b) time for decelaration = 0.5 x x 10 total time = 30 + 10 = 40 sec. (c) 0.5 x 10 x 0 + 10 x 10 + 0.5 x 10 x 10 = 50 m 6 9. y 0; x 4; y < x + 1; y x + 3 B3,, 1, 0 9 3

10. (a) AC = 45 + 67 x 45 x 67 x cos 115. 0 = 95.3 km sin ACB sin 115. (b) = 45 95.3 angle ACB = 5.3 0 Bearing = 95.3 0 11. (a) (5, 8) (, 15) (15, ) 6 for correct formula, for correct input, for final answer (b) (9, 6) (6, 5) 4 one for each coordinate 100 1. (a) x x π x 1.5 = 1.8 cm 360 (b) (1.8) : π = 3.47 cm (c) h = 1.5 3.47 h = 1.0 cm 13. (a) correct calculation of cumulative frequency accurate plotting of points and drawing of curve (b) 30 330 (c) (50 x + 150 x 7 + 50 x 11 + 350 x 17 + 450 x 7 + 550 x 4 + 650 + 750) : 50 = 33 14. (a) 0.7 x 0.8 = 0.56 (b) (a) + 0.3 x 0.4 = 0.68 (c) (1 (b)) = 0.104 6 A, 1, 0 7 6 if it is shown on curve where reading is taken accept (50.5 x + 150.5 x 7 + 50.5 x 11.) : 50 = 33.5 Accept solution by tree diagram or any other valid method 15. (a) 6 6 = ( 4x) + 3 7 8x (b) 6 = 3 7 8x 5 x = 0.65 (or ) 8 5 y 6 5 16. (a) m AB = Equation of AB = 6 x + 3 6 5x + 6y = 1 (b) M AB (3, 1) Equation y 1 6 = x 3 5 6x 5y = 13 4 Accept Accept 5 1 y = x + 3 or equivalent 6 6 3 y = x or equivalent 5 5 (c) Substitute (13,13) in 6x 5y =13 or equivalent Show LHS = RHS (d) AC = ( 13 + 3) + (13 6) = 17.5 9 17. x + (x + 1) = 9 x + x 8 = 0 x + x 4 = 0 1 ± 1+ 16 x = x 1 = 1.6 x =.6 y 1 =.6 y = 1.6 5 Accept (y 1) + y = 9 an subsequent solution of y first for applying formula for a 3 term quadratic for solving for x for solving for y 10

MINISTRY OF EDUCATION Namibia Senior Secondary Certificate (NSSC) MATHEMATICS: HIGHER LEVEL PAPER 1: PAPER ANALYSIS Q Syll. Ref. Topic Context Target Grades Total 4 1 Time and speed Localized 3 3 1 (a)3 1(j)4 5(d)7 Algebra and number Linear inequalities and 3 3 1(a)1 types of number 3 1(i) 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 Scales and similar Practical 1 4 4(a)3 figures Volumes 3 7 1(l)1 Money and finance Currency 1 3 8 6(a)4 Kinematics Velocity-time graph 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 4 1 3(a) Mensuration Arc length, cone 4 6 13 10(a), 6,7,8 Data handling Cumulative frequency, mean 5 Localized 14 10(b)5 Probability Localized 6 6 15 6(c) Functions Composite 3 3 16 7(a)1, Coordinate geometry Application without 3 9 4(e)3 Locus sketch 4 17 5(a)6 Algebra: simultaneous Points of intersection of 5 5 equations graphs Total for paper 7 31 80 Percentage 33 39 8 100 11

MINISTRY OF EDUCATION Namibia Senior Secondary Certificate (NSSC) MATHEMATICS: HIGHER LEVEL PAPER : SPECIMEN PAPER TIME: 3 hours MARK: 10 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 10. 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. 1

dy 6 1. A curve is such that dx = x +. 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). 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 sec x + tan x 5 = 0 can be written as a quadratic equation in tan x. () (b) Hence solve the equation sec x + tan x 5 = 0, for 0 radians x π radians. (5) 4. (a) Sketch, on the same diagram, the graph of y = x 5 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 x 5 + x = 4. (3) 13

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 t 1 s = 10 10e t. 0 Calculate (a) the initial velocity of the particle, (3) (b) the value of t when the particle is instantaneously at rest, () (c) the acceleration when the particle is instantaneously at rest. (3) 6. The diagram below shows part of the curve y = 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 360 o about the x- axis. (3) 7. The points A, B and C have position vectors, relative to an origin O, of 3i + k, i j + 5k and i + 10j + 7k respectively. (a) Show that BA = i + j 3k. (1) (b) Use a scalar product to find angle ABC. (6) 1 (c) Find the coordinates of the point D, such that OD = AC. () 5 14

8. (a) Show that (x + 3) is a factor of x 3 + x 7x 1 and hence solve the equation x 3 + x 7x 1 = 0 giving your answers correct to two decimal places where necessary. (5) (b) The polynomial f(x) = x 3 + ax + bx 3 has a factor of (x + 1). When f(x) is divided by (x + ), it has the same remainder as when it is divided by (x ). Find the value of a and of b. (4) 9. The function f is defined by 10. f : x a 5 4cos x, for 0 o x 360 o. (a) State the amplitude and period of f. () (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) x cm h cm NOT TO SCALE x cm A solid rectangular block has a base, which measures x cm by x cm. The height of the block is h cm and the volume is 7 cm 3. (a) Express h in terms of x. () (b) Show that the total surface area, A cm, of the block, in terms of x, is 16 A = 4x +. () x (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) 15

11. (a) Solve the equation log x 3log =. (4) x (b) The variables x and y are related by the equation y = ae bx, 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 (, 1.6), find the value of a and of b. (4) (iii) Find the value of y when x = 4. (1) 1. The function f is defined by f : x a x + 8x 10. (a) Express x + 8x 10 in the form a(x + b) + 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. () (c) Find the value of the constant k for which the equation x + 8x 10 = k has real, distinct roots. (3) The function g is defined by g : x a x + 8x 10 for the domain x. (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. () 16

3 13. (a) Evaluate (45 3r ). (5) r= 0 (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 0% 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. () 17

MINISTRY OF EDUCATION Namibia Senior Secondary Certificate (NSSC) MATHEMATICS: HIGHER LEVEL PAPER : MARK SCHEME 6 1. (a) y = + x + C x 10 = + 6 + C 6 y = + x + 6 x (b) At x = 3, m = 3 8 y = x + c 3 10 = 8 + c 3y 8x = 6 6 or equivalent. dy 5 = dx x when y = ln 8, ln 8 = 5 ln x ln 4 x = 5 gradient of curve = 5 for attempt at differentiation 3. (a) (tan x + 1) + tan x 5 = 0 tan x + tan x 3 = 0 4. (b) tan x = 1 or tan x = x = 0.785, 3.93 x =.16, 5.30 3 Pts of intersection on y-axis (5,0) and (4,0) Pts of intersection on x-axis (.5,0) and (4,0) (b) x 5 + x = 4 x = 3 or 5 x + x = 4 x = 1 7 7 for use of correct identity if in degrees, max. of 3 marks (45 0, 5 0, 13.7, 303.7 0 ) for y + x = 4 for pts of intersection with axes for shape of modulus graph for pts of intersection with axes 18

ds t 1 5. (a) v = = 10e dt 0 when t = 0, v = 9.95 ms 1 t 1 (b) when v = 0, 10 e = 0 0 t = ln 00 = 5.9 dv t (c) a = = 10 e dt 6. (a) x = 4 A when t = ln 00, = 4 0 ( 4 x) 1 dx 4 1 a = ms 0 3 = (4 x ) 3 0 3 3 = (4 3) (4 0) 3 14 = 3 4 (b) V = π ( 4 x) dx = π 0 = 8π 4 1 4 x x 0 7. (a) BA = OA OB 8 9 for differentiation for differentiation for minus sign for rest of expression correct accept 5.1 or equivalent (b) BC = 4i + 1j + k cos ABC = = = BA.BC BA BC 4 + 4 6 1+ 4 + 9 10 + 144 + 4 14 14 164 angle ABC = 73.0 0 (c) OD = i + j + k D( 1,, 1) 9 for knowing the formula for knowing the scalar product for knowing how to find the modulus (either one) for dividing AC by 5 19

8. (a) f( 3) = 7 +18 +1 1 = 0 (x + 3)(x x 4) = 0 x = 3 or 1± 17 x = x =.56 or x = 1.56 (b) f() = f( ) 8 + 4a + b 3 = 8 + 4a b 3 b = 4 f( 1) = 0 1 + a b 3 = 0 a = 0 9. (a) Amplitude = 4 Period = 180 0 or π (b) Sketch graph cycles between 1 and 9 proper position (i.e. not a sin curve) Max points (90 0,9), (70 0, 9) (c) 5 4cos x = 3 cos x = 0.5 x = 60 0 x = 30 0 10. (a) 7 = x x h 36 h = x (b) A = x x + 6x h = 4x 36 + 6x x (c) da 16 = 8x dx x 16 8x 0 x = x = 3 (d) A = 108 cm d A 43 = 8 + > 0 when x = 3 3 dx x A = 108 cm is a minimum 9 10 10 for division for correct use of formula allow ± 4 for 90 0 and 70 0, for 9 for reducing to cos x allow for any other correct method 0

3 11. (a) log x log = x (log x) log x 3 = 0 log x = 3 or log 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 = 1.8 1.6 0.6 1 b = = 0 (iii) y = 1.8 e = 13.4 1. (a) (x + ) 18 a =, b =, c = 18 (b) from (a) x = and y = 18 (c) x + 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 = (x + ) - 18 g 1 : x = (y + ) 18 1 x + 18 g : x a (f) domain: x 18 range: y 13. (a) 45 + 4 + 39.. AP where a = 45, d = 3 n = 4 S 4 = 1(90 69) = 5 (b) Geometric progression with r = 0.8 7000 S = = 35000 kg 1 0.8 (c) 7000(0.8) n 1 > 1000 1 log n < 7 log 0.8 n = 10 years 1 14 or by differentiation or by formula of axis of symmetry accept any correct convincing explanation (d) S 9 7000(1 0.8 ) = = 30300 kg 1 0.8 9 14 1

MINISTRY OF EDUCATION Namibia Senior Secondary Certificate (NSSC) MATHEMATICS: HIGHER LEVEL PAPER : PAPER ANALYSIS Q Syll. Ref. Topic Context Target Grades Total 4 1 1 0.1 19.4 Integration and differentiation Equation of curve Equation of tangent 4 6 15.1 19.3 Logarithms Differentiation Gradient of curve involving natural logs 3 5 3 17.3,5 Trigonometry Identities and quadratic 4 3 7 1.3 equation 4 16.3 Absolute values Graph sketching and 4 1 7 16. solving equation 5 19.6 Differentiation Application of rate of 5 3 8 change 6 0.3 Integration Areas and volumes 3 6 9 7 13 Vectors in 3 dimensions Scalar product 5 4 9 8 11 Polynomials and Applications of 3 3 3 9 1. remainder theorem remainder and factor theorem 9 17.4 Trigonometry Graphs 3 5 10 17.5 10 3.1 19.5 11 15.1 15.4 1 1.4,7 14.1,3 13 18.1 5(e)4 Mensuration Differentiation Logarithms Theory of quadratics Functions Sigma notation Geometric progression Application of stationary points to volume and area Change of base and transformation to linear form Application of theory of quadratics and inverse functions, domain, range Apply sigma notation, solve GP problem 5 5 10 4 4 4 1 5 5 4 14 3 9 14 Total for paper 4 45 33 10 Percentage 35 38 7 100

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