NATIONAL SENIOR CERTIFICATE GRADE 12

Transcription

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

2 Mathematics/P DoE/November 009 QUESTION B( ; 5) y O // M A(5 ; ) // C ( ; 5) D(9 ; 7). m AC m AC ( 5) 5 ( ). Equation of AC is: y ( 5) y M( ; y) ; ( ; ) Substitute into equation of AC: y M lieson line AC m BM m m BM AC 5 + m. BM AC BM AC or AMˆ B 90.5 BM ( 5 + ) + ( ) BM 7 AC (5 + ) AC 7 Area of ABC + ( + 5) ) 6 square units Δ ( 7)( 7 into gradient formula () into straight line equation () midpoint substitution of y into gradient formula m BM m m BM AC into distance formula BM 7 AC 7 formula for area of () () (5) [6]

3 Mathematics/P DoE/November 009 QUESTION A(0 ; ) B(7 ; ) θ O C D( ; 5) E Midpoint BD : ; (; ) CA CB ( 0) ( 7) CA CB CD + ( ) + ( ).. r 5 and centre( ; ) Equation of circle is.. ( ) + ( y + ) ( 5) m BD 7 ( ) substitution into midpoint formula answer () CA CD CB radius and centre () equation of circle () gradient m BD tan θ 6, θ Let E ( ; y) E(6 ; 6) y + y 6 tan θ 6 y 6 () () E( + ; ) (6 ; 6) with transformation ()

4 Mathematics/P DoE/November The diagonals AE and BD bisect each other + 5 m AB m AD Â 90 ABED is a rectangle (diagonals bisect each other and adjacent sides are perpendicular) bisect (from..) gradients 90 ()..7 Show that Aˆ Bˆ Cˆ Dˆ 90 Gradient of tangent is. Equation of a tangent at B is: y ( 7) 8 y + + y + gradient substitution into straight line formula ().. y 6 ( ) (7) + c c y () ( + ) But ( ; y) ( ; y + ) ( ; ) ( ; + ) ( ; 6) ( + ) ( + ) + y ( + ) + ( y ) + ( y 6) + y ( + ) + ( y ) + 6 y ( y 6) 0 y + ( ) ( y ) 5 0 gradient into straight line formula completing the square equation equation () () replacing with + and replacing y with y. completing the square equation ()

5 Mathematics/P 5 DoE/November Distance from origin to centre ( 0) + (6 0) 5 Since 5 > 0, the origin lies outside the circle. distance from origin to centre conclusion () [] QUESTION. Enlargement with a scale factor of enlargement scale factor (). // // R ( ; ) R ( + ; + ) R ( ; 7) 7 (). ( ; y) ( y ; ) (y ; ) (). Reflection about the line y P( 5 ; ) P //// ( ; 5) ( ; y) ( y ; ) θ 90 K rotation in an anticlockwise direction P( 5 ; ) P //// ( ; 5) ( ; y) ( y ; ) θ rotation in a clockwise direction reflection line y () for any new correct coordinate 90 anti-clockwise () for any new correct coordinate 70 clockwise () [0]

6 Mathematics/P 6 DoE/November 009 QUESTION. h y f θ φ β tan( φ) 6 φ 80 80, φ 99,6 tan β β 6, θ φ β 99,6 6, θ 6,0 method () NOTE: The following solution is beyond the bounds of the curriculum, but is mathematically correct. tanθ tan( φ β ) tanφ tan β tanθ + tanφ.tan β 6 8 tanθ + ( 6)() θ 6,0 8 formula epansion simplification ()

7 Mathematics/P 7 DoE/November 009. For clockwise rotation: / cosθ + ysinθ / cos50 sin50 formula simplification / y y / / sinθ + y cosθ sin50 cos50 / y + ; + + ; + / P // P formula (8) For anticlockwise rotation: cosθ ysinθ cos( 50 ) ( )sin( 50 ) ( cos0 ) sin 0 y sinθ + y cosθ / P // P sin( 50 ) + ( )cos( 50 ) sin 0 ( cos0 ) + + ; + + ; + formula simplification formula (8) []

8 Mathematics/P 8 DoE/November 009 QUESTION tanθ r r 5 ( ) + (5) 69 r () cosθ sinθ sin(90 ) tan(60 ) cos(80 ) cos tan (cos )( tan ) cos cos tan cos( 60 ) + tan5 tan 5 + cos 660 cos60 tan 5 simplification tan 5 + cos60 denominator simplification numerator + sin 0, sin 0,86 Ref angle 9, , k.60 09,08 + k.60, k Z or or 60 9, k.60 0,9 + k.60, k Z -0,86 ref angle 09,08 + k. 60 0,9 + k. 60 k Z () () () 5.5 k ,9 or k.60 9,08 sin cos tan 60 sin sin or 0 equation (5) simplification (any one) () []

9 Mathematics/P 9 DoE/November 009 QUESTION cos p p diagram p p () 6.. sin cos sin( 66 ) tan 0 (sin cos ) + sin 66 tan sin + cos tan p p p + p p + p p ( ) sin cos tan sin cos substitution (6) sin cos sin( 66 ) tan 0 sin + cos tan sin + sin sin p 6. sin + sin LHS + cos cos sin + sin cos + cos cos + sin ( + cos ) 5 + cos cos sin ( + cos ) (5 cos )( + cos ) sin 5 cos sin cos tan sin simplification (6) cos cos sin sin.cos common factor factors () []

10 Mathematics/P 0 DoE/November 009 QUESTION Q RS ˆ 80 ( α) ( angles of triangle) α In triangle QRS: QR sin(50 α) sin(50 α) QR (sin50 cos50 ) + 6( + ) QRS ˆ α into sine rule rewriting in terms of QR epansion (5) 7.. PQ QR tanα 6( + ) PQ tanα PQ PQ 6 6 PQ + PQ tanα ratio of QR tan α simplification simplification (5)

11 Mathematics/P DoE/November 009 In triangle QPR: Q P ˆR 80 (90 + α ) 90 α ( angles triangle) PQ QR sin(90 α) QR PQ QR tanα PQ QRtanα sin(50 α) tanα (sin50 cos50 ) tanα ( + tanα tanα) of QR simplification tan α simplification (5) tanα ) 6.tanα 7 7 tanα 6 tanα, α 58,56 7. Area of red part Area of bigger triangle area smaller triangle.(80)(80).sin 60 (50)(50).sin ,75 cm equating simplification to 7 tan α 6 () method area formula ()

12 Mathematics/P DoE/November 009 a a a a a Area of equilateral triangle of side a (80 50 ) 975 cm q. a. a method area formula () Area of red part (80 50 ) ,75 cm method area formula () [7]

13 Mathematics/P DoE/November 009 QUESTION 8 8. a tan 5 a 8. y reading from graph () h f - amplitude shape endpoints -intercepts () () 8. f (,5 ) cos 59, 5 0,5075 h (,5 ) tan, 5 0,57 > f (,5 ) calculation of f (,5 ) and h(,5 ) From the graph,,5 lies to the RIGHT of the point of intersection. θ <, 5 h(,5 ) > f (,5 ) θ <, 5 () [0]

14 Mathematics/P DoE/November 009 QUESTION 9 9. Scatter plot of new AIDS infections for the period 00 to 007 New infections (in millions) Year labelling of aes and title plotting points labels () 9. There is more or less a linear decrease in the number of new infections year after year. 9. There is greater awareness about the AIDS pandemic through advertising in the media. Education in lifestyle has happened with many people. Any valid reason in the contet of the situation QUESTION decrease two valid reasons () () [6] 0. (Calculator used) σ 8, σ 8, 0. The times are more closely spread around the mean because out of 0 travelling times from the data set falls within one standard deviation from the mean. So the teacher s observation is acceptable. () () comment on spread use of SD in eplanation () [6]

15 Mathematics/P 5 DoE/November 009 QUESTION arranging in ascending order. Petrol Median 76 Lower quartile 69 Upper quartile 85 median lower quartile upper quartile () There are 9 data points. The lower quartile (600) is at position 5 and the upper quartile (800) is at position 5. There are data points from 600 to 800. Therefore there are 9 data points strictly between 600 and ( ) 9 bo whiskers () data points from 600 to data points () data points () NOTE: If a learner uses 9,5 data points from 600 to 800, award mark only.[8] QUESTION ().. About 050 ().. IQR Q Q method () light bulbs 0 60 () light bulbs 80 Therefore the cost will be: R5,00 80 R00 R00 () [9] TOTAL: 50