NATIONAL SENIOR CERTIFICATE GRADE 12

Transcription

1 NATIONAL SENI CERTIFICATE GRADE MATHEMATICS P NOVEMER 009() MEMANDUM MARKS: 50 This memorandum consists of 5 pages.

2 Mathematics/P DoE/November 009() QUESTION. 5,5 Mean 43, 5 ANSWER ONLY: Full marks. Ordered Data 9,3 4,9 5 3,6 6, 8 3,5 60,9 65,7 7,9 76,4 98, 8 + 3,5 Median 30, ,6 Lower quartile 9,3 65,7 + 7,9 Upper quartile 68,8 The five number summary is (9,3 ; 9,3 ; 30,5 ; 68,8 ; 98,) If they use the formula: Ordered Data 9,3 4,9 5 3,6 6, 8 3,5 60,9 65,7 7,9 76,4 98, + P50 6,5 Position of median: 8 + 3,5 Q 30,3 5,5 answer No penalty for Rounding: Accept 43,54 ; 44 9,3 9,3 30,3 68,8 98, If indicated on the bo and whisker diagram in.3 5 marks () (5) 3 Position of lower quartile: P 5 4 Q 5 + (0,5(3,6 5)) 7.5 Position of upper quartile: P 0,75 9, ,7 + (0,75(7,9 65,7)) 70,35 Q Min 9,3 Ma 98, Accept any one of these five number summaries: (9,3 ; 9,3 ; 30,3 ; 68,8 ; 98,) (9,3 ; 5 ; 30,3 ; 7,9 ; 98,) (9,3 ; 7, ; 30,3 ; 70,4 ; 98,)

3 Mathematics/P 3 DoE/November 009() Note: If just a bo and whisker without any reference to the numbers: /3 minimum and maimum values quartiles and median whiskers with median line.4 The data is skewed to the right (positively skewed). This suggests that there was a large difference between the median and the maimum rainfall (some months had eceptionally high rainfall in that year). Die data is skeef na regs (positief skeef) Dit dui daarop dat daar n groot verskil is tussen die mediaan en die maksimum reënval (sommige maande het ongewoon hoë reënval gehad gedurende die jaar..5 y using the calculator, σ 8, 9. (8,905856) Pen and Paper method (not recommended) Mean 43,54 (43, ) ( ) 60,9 7,36 30,3696 4,9-8,64 80,496 9,3-34,4 7,378 8,0-5,54 4,496 7,9 8,36 804,896 76,4 3,86 079,78 98, 54,66 987,76 65,7,6 49,0656 6, -7,44 304,536 3,5 -,04,886 3,6-9,94 397,6036 5,0-8,54 84,536 Sum 9536, ,509 σ 8,9 (8,9059..) comment about rainfall. () Note: Skewed to right / verwysing na reënval () answer Accept: 8 ; 8, ; 8, headings correct sum of the squares of the mean deviations answer [5]

4 Mathematics/P 4 DoE/November 009() QUESTION. Linear or Eponential answer. Time taken (in seconds) Scatter plot of times taken by winners of men's 00 m freestyle at Olympic Games Year () line of best fit () Scatter Plot of time taken by the winner of 00m Freestyle at Olympic Games Time taken (in seconds) Year For this set of data we will accept the straight line.

5 Mathematics/P 5 DoE/November 009().3 The scatter plot shows an overall decrease in the time taken by the winner since 97. Die spreidiagram dui n algehele afname in tye aangeteken deur die wenners vanaf 97. Times are faster. Tye is vinniger. Negative correlation between year and time. Negatiewe korrelasie tussen jaar en tyd..4 The top athletes of the world have turned professional. This allows them to train at the best facilities and receive the best coaching available. Also, equipment manufacturers are in competition with each other. In this case, manufacturers are designing swimsuits that assist swimmers Swimmers train harder and put in more effort. Die top atlete van die wêreld het professionele atlete geword. Dit laat hulle toe om by die beste fasiliteite te oefen en die beste afrigting te ontvang. Vervaardigers van voorraad is in kompetisie met mekaar. Hul onwerp dus swembroeke wat die swemmers help. Swemmers oefen harder en gebruik meer tyd om te oefen..5 In the contet of the times around these two observations, one can consider the efforts of 976 and 988 to be outliers. This shows that these athletes were eceptionally good swimmers at the time. inne die konteks van tye gedurende hierdie twee waarnemings, kan die poging van 976 and 988 gesien word as uitskieters. Dit dui daarop dat hierdie atlete uitstekende swemmers was daardie tyd..6 Winning time of 008 is epected to be about 47,6 seconds. Accept answer from candidate s graph. decrease/afname () any acceptable reason relating to the trend () enige aanvaarbare rede wat verband hou met die neiging. () acceptable reason in contet () aanvaarbare rede binne die konteks () answer from graph () [8] QUESTION answer () 3. Cut off mark of 56% (37 students)or 58% (38 students) Accept interval: 55% - 60% answer read off from ogive ()

6 Mathematics/P 6 DoE/November 009() 3.3 QUESTION 4 Marks (out of 00) Frequency ( f ) 0 marks <0 0 marks <0 3 0 marks < marks <40 40 marks <50 50 marks < marks < marks < marks <90 90 marks <00 0 class intervals Accept 0 0 ; 0 0 Or 0 < marks 0 Or etween 0 and 0 Or From 0 to 0 If the intervals not in tens, the mark for intervals not given method accuracy of five answers [5] tan 45 ma m A t t 3 0 tan 45 t t 3 t y m + c 3 ()() + c c y + ( t ;0) in y m + 0 t + t tan 45 answer Answer only: full marks equating value c value Answer only: full marks () () ()

7 Mathematics/P 7 DoE/November 009() 4.3 ( p) + ( 3 + 4) ( p) + ( 3 + 4) 4.4 p + p p ( p) + ( 3 + 4) ( p) ( p) p or p 0 Let p AC p p 0 p( p ) ( ) which is true midpoint of C midpoint of C p p 0 50 p + (3 + 4) 50 or p substitution into distance formula epansion factors answer Note: If an answer was not chosen: 3/4 substitution into distance formula (4) epansion factors answer (4) If gradient of C assumed as - and p calculated correctly: 0/4 Answer only: /4 substitution into distance formula 50 which is true(justification) answer (4) If equating to 50 from the start, then 3/ t + p ; -value ( ) ( 0 ; ) y-value () 4.5 Gradient of line m A Equation of line is: y + 4 ( ) y 6 y m + c y p 4 gradients are equal substitution of (p;-4) equation in any form [3]

8 Mathematics/P 8 DoE/November 009() QUESTION 5 D(0 ; 8) A( 8 ; ) M 0 ( ; 6) Midpoint D ; ( ; ) 5. y 7( 8) + 58 Alieson theline. 5.3 The line y is a tangent to the circle at A. -coordinate y-coordinate substitution () () Substitute both at the same time with justification () relationship m m line AM 7 8 ( ) 7 mline mam 7 7 AM to the line NOTE: m line 7 and CA gradient of AM then no relationship: 4/5 m AM 8 ( ) 7 m 7 line product (5)

9 Mathematics/P 9 DoE/November 009() 5.3 contd m m D line 7 7 line // diameter ( + ) + ( y ) y y (7 + 58) (7 + 58) ( + 8) 0 8 y y is a tangent to the circle m D 7 m 7 line conclusion (5) Note: Only lines parallel 4/5 circle equation substitution of y standard form answer tangent (5) 5.4 AD A (8 ) ( + 6) (0 + 8) + ( 8 + ) substitution answer substitution answer (4) Note: Answers 0 then 3/4

10 Mathematics/P 0 DoE/November 009() 5.5 m AD m AD m A 8 () 0 ( 8) 3 4 ( 6) 8 ( ) m. m A AD DÂ gradient of AD gradient of A PRODUCT D (8 + 6) 00 AD D A 90 + (0 + ) + A distance formula Pythagoras conclusion a b + d ( b)( d)cos A (0)(0) cos A 0 00cos A A 90 cos rule substitution conclusion ( AD) ( A) D ( 0) + ( 6 8) D AD + A DA ˆ 90 (Pyth) D 00 D AD + A conclusion A ˆ 90 (angles in semi - circle) reason 5.6 θ 45 answer ()

11 Mathematics/P DoE/November 009() D(0 ; 8) T A( 8 ; ) N Z M 0 ( ; 6) 5.7 Let the radius of circle TNM be r N M (properties of a kite) AN TZ r (TZNA is a square) N 0 r D M (8 ( 6)) + (0 ( )) 0 (0 r) 00 (0 r) (0 r) r 0 5,93 N M AN TZ r N 0 r D M D 00 answer (6) ZM 90 M 00 7,07 ZM tan, 5 M ZM 7, 07 tan, 5, 93 tan radius theorem M tan,5 answer (6)

12 Mathematics/P DoE/November 009() 5.7 contd M ( + ) + ( + 6) M 50 ZM tan, 5 M ZM 7, 07 tan, 5, 93 M tan,5 answer (6) y a well known formula Area AD r (semi perimeter) 0 0 r (0 + 00) 50 r(0 + 5 r,93 ) M 50 (radius of circle) N 50 (adjacent sides of kite) A 0 AN 0 50,93 ut TANZ is a square AN ZN radius,93 formula 00 answer M N AN,93 square answer (6) (6)

13 Mathematics/P 3 DoE/November 009() QUESTION squared units answer 5 / If 5 0 0/ 6.. ( ; y) ( ; y) Note: If candidate state: coordinates times two / y If ( k ; ky) :/ () () y 6..3 coordinates ( ;8) 0 9 A / 8 ( 8;8) / ( ;4) A O C C / If ( ; y) : / / A / coordinates / coordinates C If diagram not drawn but coordinates correctly given: /3 If coordinates correctly plotted but not joined: / Not rigid. The shape remains the same, whilst the size is changed /enlarged 6. Note: Shape remains the same: / Only the shape remains the same: / Reflection about the line y : ( ; y) ( y ; ) Rotate clockwise about the origin: ( y ; ) ( ; y) Translate left and 3 down: ( ; y) ( ; y 3) General rule: ( ; y) ( ; y 3) same shape and different size () not rigid only / just enlarged 0/ Mark per coordinate reflection rotation translation (6) Answer only: Full marks [5]

14 Mathematics/P 4 DoE/November 009() The first transformations in the given order is the same as the reflection in the -ais i.e. ( ; y) ( ; y) Then the translation gives us ( ; y) ( ; y) ( ; y 3) NOTE: If just given: ( ; y) ( ; y 3) : /6 If using ( ; y) (y ; ) ( ; y) ( y ; - ) ( ; y) ( ; y 3) throughout :4/6 If learner starts with ( ; y) and continue to use ( ; y) for the second and third transformation 4/6 QUESTION 7 7. T / ( cosθ y sinθ ; y cosθ + sinθ ) coordinate y coordinate () Clock-wise formula: 0/ 7. A / ( p cos35 q sin35 ; q cos35 + psin35 ) If clockwise rotation: A / ( p cos35 + q sin35 ; q cos35 p sin35 ) coordinate y coordinate () CA from p cos(35 ) q sin(35 ) p cos 45 q sin 45 p p q and y y cos(35 ) + p sin(35 ) q q...() q cos 45 + p sin 45 q + + p p...() equating substitution equating substitution () + (): q q solving simultaneously

15 Mathematics/P 5 DoE/November 009() Substitute q into..() answer for q p p p () Note: If not left in surd form: 6/7 answer for p A ( ;) (7) p cos(35 ) q sin(35 ) p cos 45 q sin 45 p p q and y y cos(35 ) + p sin(35 ) q...() q cos 45 + p sin 45 q p + 0,4 0,7q + 0,7p...() () + (): q q Substitute q into..(),4 0,7p 0,7q (),4 p p,4 A ( ;) Note: If not left in surd form: 6/7 equating substitution equating substitution solving simultaneously answer for q answer for p (7)

16 Mathematics/P 6 DoE/November 009() p + q p + q and ( p + q) ( + ( p q) p q p + q p p q + ) A(p ; q) is obtained from A / by a rotation through 35 in a clockwise direction p ( ( q ( ( ) cos( 35 ) ( ) ) A ( ;) ( ) cos( 35 ) + ( + ( ) ) ) sin( 35 ) )sin( 35 ) ( p + q) substitution ( p q) substitution solving simultaneously answer for q answer for p (7) substituting ( ) substitution equating substitution substituting ( ) answer for q answer for p (7)

17 Mathematics/P 7 DoE/November 009() QUESTION 8 8. sin α 8 7 ( 5 ; 8) sin α > 0 in second quadrant y 8 r 7 α α 5 tanα α 8 5 ( Pythagoras) sin(90 + α) cosα 5 7 cos α sin α 7 α 5 answer For drawing the radius vector in the correct quadrant /3 Without a sketch but correct values: 3/3 reduction answer () Answer only: full marks Cannot accept decimal values epansion cos α cos α cos α cos α sin α substitution any further calculation or answer epansion substitution any further calculation or answer epansion substitution any further calculation or answer [8]

18 Mathematics/P 8 DoE/November 009() QUESTION 9 NOTE: Only penalise once in the question for leaving out the Penalise once in this question for treating as an equation 9. sin(90 ).cos(80 ) + tan.cos( ).sin(80 + ) cos ( cos ) + tan (cos )( sin ) cos cos (cos sin cos sin + sin cos sin ) sin90 cos 5 tan 390 cos00 sin35 sin0 ( cos 45 ) tan 30 sin0 sin or tan 30 If using cos 80 : no penalty If the candidate stop at /7 sin + cos sin + ( sin sin sin ) + sin + 0 sin 0 (sin + )(sin ) 0 sin 90 + k.360 ; k Z Or sin( 90 ) cos cos( 80 ) cos cos( ) cos sin( 80 + ) sin sin tan cos simplification answer sin 90 sin 0 cos 5 cos 45 tan 390 tan 30 cos 00 sin 0 sin 35 sin 45 or cos 45 o substitution substitution of identity standard form factorisation sin ;sin 90 + k.360 ; k Z answers (any two answers) (7) (7) If k Z not included: 6/7 (7) Also ± k. 360 ; k N0 or Z

19 Mathematics/P 9 DoE/November 009() sin sin + cos sin cos sin cos [ sin(90 ) ] sin 80 + (90 ) + k k k0 k Z sin + cos sin cos sin cos cos(90 ) cos 80 (90 ) + k k360 k Z 0 + k.360 ; k Z or k.360 ; k Z 50 + k.360 ; k Z or k.360 or or 0 + k.360 or 30 + k k.360 ; k Z or 30 + k (90 ) + k k (90 ) + k k k0 manipulation substitution of identity co ratios 80 + (90 ) + k k0 360 (90 ) + k k360 If k Z not included: 6/7 manipulation substitution of identity co ratios (7) 80 (90 ) + k k (90 ) + k k0 (7) If k Z not included: 6/7 [0]

20 Mathematics/P 0 DoE/November 009() QUESTION 0 0. sin( A + ) sin A.cos + cos A.sin cos( A + ) cos A.cos sin A.sin sin A.cos + cos A.sin cos A.cos cos A.cos sin A.sin cos A.cos sin A.cos cos A.sin + cos A.cos cos A.cos cos A.cos sin A.sin cos A.cos cos A.cos tan A + tan tan A.tan epansions divisions tana and tan 0. tan A + tan RHS tan A. tan sin A sin + cos cos cos A.cos A sin A sin cos A.cos cos A cos sin Acos + sin cos A cos Acos sin Asin sin sin( A + ) cos( A + ) tan( A + ) LHS tan C tan(80 ( A + )) tan C tan( A + ) tan A + tan tan C tan A.tan tan C( tan A.tan ) (tan A + tan ) tan C tan A.tan.tan C tan A tan tan A + tan + tan C tan A.tan.tan C sin A cos A multiplication epansions C tan( A + ) substitution into formula multiplication with LCD If no conclusion: 3/4 (4)

21 Mathematics/P DoE/November 009() C ˆ 80 ( Aˆ + ˆ) (angles in a triangle) tan C tan(80 ( A + )) tan C tan((80 A) + ( )) tan(80 A) + tan( ) tan C tan(80 A).tan( ) tan C( tan(80 A).tan( )) tan(80 A) + tan( ) tan C tan C tan Atan tan A tan tan A + tan + tan C tan A.tan.tan C C rearrange angle substitution into formula epansion (4) QUESTION NOTE: Penalty of one for early rounding off once in this question.. Dˆ A sin D A sin sin D A 0, Dˆ A 30,58 earing of Ship A from Ship 80 ( ) + 30, 58 58, 58 Dˆ A sin DA ˆ sin sin DA ˆ 0, DA ˆ 30,58 then ND ˆ (refle angles) ND ˆ 5 but MD ˆ + ND ˆ 80 MD ˆ 8 then MA ˆ MD ˆ + DA ˆ 30, ,58 (co - interior angles/ angles around a point) Dˆ C 4 sine rule substitution ˆ 30, 58 method or M ˆ D 8 answer (6) Dˆ C 4 sine rule substitution N D ˆ 5 M ˆD 8 answer (6)

22 Mathematics/P DoE/November 009().. ˆ 30, 58 E A D 8 30,58 0km EA sin(8 + 30,58 ) 0 EA 0sin(8 + 30,58 ) EA 0,4 km definition substitution 0 A answer P 58,58 Q R Let Q, then AQ 0 PQ QR sin 58,58 sin 58,58 0 PQ.sin 58,58 QR (0 )sin 58,58 PQ + QR.sin 58,58 + (0 )sin 58,58 P AR 0sin 58,58 0,4 (assume ships move at same speed) trigonometeric ratios sum answer trigonometeric

23 Mathematics/P 3 DoE/November 009() PQ RAQ Q QA 60 km PQ sin58,58 60 PQ 60sin 58,58 PR PQ 5,0 km 0,4 km (angle, angle, side) ratios 5,0 km answer A M cos3,4 0 M 0 cos3,4 0,4 8 30,58 3,4 0 M trigonometeric ratios substitution. A C a c b a + c ac cos b a + a a a cos b b a a b cos a b cos a b sin a a cos ( cos ) cos sin b a b a a b / b / answer equal sides cos rule substitution simplification (4) sin b sin a formula substitution (4) [3]

24 Mathematics/P 4 DoE/November 009() a + c b cos ac cos rule equal sides but a c a + a b cos a. a a b a b a substitution simplification (4) QUESTION. y ( 0 ; 0) or ( 60 ; 0) A ( 30 ; ) or ( 0 ; ) ()

25 Mathematics/P 5 DoE/November 009() QUESTION. cos( 30 ) cos( 30 ) See points A and on the graph Note: If drawn the line y and put A and on the graph: 0/ manipulation answer () A and in the correct place on the graph: full marks If A and on the -ais: / If A 30 and 90 : /.3 cos( 30 ) 0, g ( ) 0 is at maimum and minimum values of graph 30 ; 0.5 [ 90 ; 60 ) (0 ; 70 ] 90 < 60 or 0 < 70 If < 60 or > 0 /3 60 (ref angle) Answer only: 3/3 one for each -value () notation critical values []