NATIONAL SENIOR CERTIFICATE GRADE 12

Transcription

1 NATIONAL SENI CERTIFICATE GRADE MATHEMATICS P FEBRUARY/MARCH 0 MEMANDUM MARKS: 50 This memorandum consists of 8 pages.

2 Mathematics/P DBE/Feb. Mar. 0 UESTION. Mean n n R 44, answer (). Standard deviation n ( ) n R4 460,97. Value of one standard deviation above mean R 44,44 + R4 460,97 R5 805,4 Onl one person earned a commission of more than R 5 805,4. Therefore onl person received a rating of good. answer adding mean and std. dev. deduction () () [6] UESTION. at least four points all points (). Eponential (The increase in growth is showing a virtual doubling for each ear). eponential ()

3 Mathematics/P DBE/Feb. Mar. 0. YEAR N (Number in millions) Log N ( to decimal place) ,9 7, 7,8 8, 8,4 8,7 at least four values all values () (if onl log of values in table taken in account) YEAR N (Number in millions) Log N ( to decimal place) 0,9,,5,8,,4,7 at least four values all values ().4 at least 4 points l plotted all points ()

4 Mathematics/P 4 DBE/Feb. Mar. 0 (if onl log of values in table taken in account) at least 4 points l plotted all points ().5 The graph representing log N is a straight line. That is, log N m + c m+ c N 0 Therefore eponential graph. linear reason () [9]

5 Mathematics/P 5 DBE/Feb. Mar. 0 UESTION (). Time, t, in minutes Frequenc for intervals in 0 t < 5 table 5 t <0 5 for first three 0 t < t < 0 5 frequencies 0 t < 5 7 for last two frequencies (). 5 Frequenc 0 first three bars last two bars no gaps between bars Time intervals () [7] UESTION 4 a 7 b 5 c 7 d e 4 f 7 g 4 each answer (7) g 4 ; a 7 ; d ; f 7 ; b c 5 7 c 5 c 7 e 4 g a d f b c e (7) [7]

6 Mathematics/P 6 DBE/Feb. Mar. 0 UESTION m AD AD ( ) + ( ) ( 5 ) + ( 4) M ; M ; M ( ; ) 5.4 m BC m AD Lines are parallel m ( ) ( + ) for substitution for answer for substitution 5 -value -value value m BC subst ( ; ) () () () equation () + c ( ) + c 7 c value m BC subst ( ; ) equation ()

7 Mathematics/P 7 DBE/Feb. Mar m AD tan β β 80 56, β,69 B( ; ) α E A( ; 4) β F D(5 ; ) tan β m AD, m BD 5 ( ) 8 tanα 8 α 80 0,56 α 59,44 FED ˆ 80 59,44 0,56 EFD ˆ,69 FDE ˆ 80 (0,56 +,69 ) 5,75 m BD 59,44 8 0,56,69 5,75 () (5) 5.6 Co-ordinates of centre M ( ; ) Radius of circle: of AD ( ) 5 Equation of the circle is: ( ) + ( ) value of radius substitution into equation of circle centre form () r ( ) + ( 4) Equation of the circle is: + ( ) ( ) value of r substitution into equation of circle centre form () 5.7 M( ; ) B( ; ) MB ( + ) + ( ) ) MB 6 Point B lies outside the circle because MB > radius M( ; ) B( ; ) MB + 6 Radius of the circle < 6 Point B lies outside the circle because MB > radius substitution outside () substitution outside () [0]

8 Mathematics/P 8 DBE/Feb. Mar. 0 UESTION 6 6. Coordinates of centre M ( ; ) r ( ) ( ) Radius 8 or coordinates of centre calculation value (4) 6. mms mms mrs mrs m ( ) + ( ) tangent radius gradient MS gradient RS subst ( ; ) equation (4) m m MS MS m RS mrs + c + c c gradient MS gradient RS subst ( ; ) equation (4) 6. MS MP MP MS MP 9MS ( a + ) + ( b ) 9( + ) 6 () MS SR and PS SR m PS m MS b + a b + a + b a () Subst () into() MP MS equation equal gradients gradient b -a -

9 Mathematics/P 9 DBE/Feb. Mar. 0 ( a + ) + ( a ) 6 substitution ( a + ) + ( a + ) ( a + ) ( a + ) a + 9 or 9 a 7 or b a 8 P(7 ; 8) a 7 b -8 (8) MS MP MP MS MP 9MS ( a + ) + ( b ) 9( + ) 6 () MS SR and PS SR m PS m MS b + a b + a + b a () Subst () into() MP MS equation equal gradients gradient b -a - a + 4a a + 4a a + 8a 54 0 a + 4a 77 0 ( a + )( a 7) 0 a 7 or But a > 0 a 7 b a 8 P(7 ; 8) substitution a 7 b -8 (8)

10 Mathematics/P 0 DBE/Feb. Mar. 0 P(a ; b) MSP is a straight line (MS SR) m PM b a + b a b a...() PS MS PS ( a ) a 4(8) 7 ( a ) a + ( b + ) a ( a + ) 4a () 7 ( a 7)( a + 5) 0 a 7 or a / 5 b 7 8 P(7 ; 8) ( a ) ( a ) 7 6 a 6 or 6 a 7 or 5 a 7 b 8 P(7 ; -8) MSP a straight line m PM b a + equation equation substitution of equation into equation coordinates (8) M( ; ) diagram ( ; 8) ( ; ) ( ; ) S( ; ) (; 8) P(7 ; 8) 6 (8) ( ; 8) (; 8) 6 P(a ;b) division of line segment into

11 Mathematics/P DBE/Feb. Mar. 0 P(a ; b) S P b a + 9 b b 8 9 a + a 7 M M P(7 ; 8) UESTION 7 S P M M given ratio substitution equation equation coordinates (8) [6] K N N / - L K / - M / M - L / -4 For coordinates and label of each image: K L M N 7.. Transformation is not rigid, because the area is not preserved under enlargement. -5 (4) not rigid size not preserved () 7.. // N ( ; ) coordinates of // N () 7. ( ; ) ( ; ) ( ; ) (4) 7.4 Area of KLMN : area of // // // // K L M N : 4 answer () 7.5 If the point that is furthest awa from the origin is sent into the circle, the whole quadrilateral is sent into the circle. K is furthest awa. KO + 8 p. KO, p 8 K furthest KO 8 answer () [7]

12 Mathematics/P DBE/Feb. Mar. 0 UESTION 8 8. cosθ + sinθ cos5 + ( ) sin5 cosθ sinθ or 5 cos5 ( ) sin5 + 5 ; 5 cosθ sinθ cosθ + sinθ or or,54 cos( 5 ) ( ) sin( 5 ) 5 or cos( 5 ) + ( ) sin( 5 ) + 5 ; 5,54-0,7-0,7 subst and into formula for using 5 coordinate (in an format) subst and into formula for for coordinate (in an format) (5) subst and into formula for using 5 -coordinate (in an format) subst and into formula for for -coordinate (in an format) (5) cosθ sinθ cos5 sin5 cosθ + sinθ cos5 + sin5 + Solving () and () simultaneousl: 5 5 and () () subst and 5 into formula for simplification subst and 5 into formula for -coordinate -coordinate (5)

13 Mathematics/P DBE/Feb. Mar. 0 Using first principles: ( r cosα; r sinα) r 5 α θ r θ / ( ; ) tanθ r + θ 56, α 5 56, 78, 69 ( r cosα; r sinα) ( 0,7;,54) tan θ r θ 56, ( r cosα; r sinα) answer (5) [5]

14 Mathematics/P 4 DBE/Feb. Mar. 0 UESTION r cos α 9.. () TÔR 80 (90 + α) 80 (90 + α) 90 α 90 α () TR cos TÔR OT cos(90 α) OT OT cos(90 α) OT sinα OT 5 OT 9,5 sin(rtˆ O) OT OT sinα OT 5 OT 9,5 cos.cos ( tan ) LHS cos sin cos. cos sin RHS cos( 90 α) 5 sinα 9,5 sin( Rˆ O) sinα 5 9,5 cos tan sin cos answer O O (4) (4) (4) []

15 Mathematics/P 5 DBE/Feb. Mar. 0 UESTION 0 0. Period 0º 0 0. sin - 0 or 90 () 0 90 () 0. Maimum value of f () is Maimum value of h() is 0 ma of f() answer () 0.4 g 90 ; 90 ( 0 ;) ( 80 ; ) f sin cos 0 sin cos 0 sin cos - sin cos () There are solutions where graphs f and g are equal 0.6 f().g() < 0 ( 60º ; 0º) or (60º ; 90º) or (0º ; 80º) 60 < < 0 or 60 < < 90 or 0 < < 80 answer () for each interval brackets or smbols (4) [4]

16 Mathematics/P 6 DBE/Feb. Mar. 0 UESTION.. sin 6 p sin 4 sin ( ) sin 6 p 6 p p sin 6 answer ().. cos6 sin 6 identit answer p ().. cos cos (6 ) double angle cos 6 epansion ( p ) ( p) p p..4 cos 7 cos5 + sin 7.sin5 cos(7 5 ) cos 58 (cos 80 ) (cos ) ( p) p.. (cos + sin ) (cos sin ) LHS (cos sin )(cos + sin ) cos + cos sin + sin (cos sin cos + sin (cos sin )(cos + sin ) 4cos sin cos sin sin cos tan RHS.. cos sin or cos sin sin cos sin sin sin sin sin + sin 0 ) answer cos(7-5 ) (cos ) answer (cos sin )(cos + sin ) () () (cos + sin ) (cos sin ) numerator 4 cos sin cos sin sin cos for answer sin (6) () ()

17 Mathematics/P 7 DBE/Feb. Mar. 0.. sin cos sin + sin 0 sin ( sin + ) 0 sin 0 or sin k; k Z or {0 or 0 } + 60 k; k Z n.80 n.60 0 (n + ) , n Z sin ( sin + ) 0 sin 0 or sin k k; k Z (6).4 tan tan tan tan 4. tan 87 tan 88 tan 89 sin sin sin 45 sin 88 sin cos cos cos 45 cos88 cos89 sin sin sin 45 sin(90 ) sin(90 ) cos cos cos 45 cos(90 cos(90 ) sin sin sin 45 cos cos cos cos cos 45 sin sin tan 45 tan 89 cot tan 88 cot... product is (tan.cot )(tan.cot )...(tan 44.cot 44 ).tan identit co-ratios simplification for answer identit co-ratios simplification for answer (4) (4) [9]

18 Mathematics/P 8 DBE/Feb. Mar. 0 UESTION In Δ CBG and ΔCDH: CG² ² + ² Pthagoras CH² ² + ² Pthagoras In ΔFAE AE² ² + ² ² GH² In Δ CGH GH² CG² + CH² - CG.CH. cos GCH CG + CH GH cos GĈH CG. CH cos GĈH +. + cos GĈH ( + cos GĈH + ) CG² CH² AE² AE² GH² use of cos rule manipulation of formula substitution cosgĉh ( + ) (8) [8] TOTAL: 50