NATIONAL SENI CERTIFICATE GRADE MATHEMATICS P FEBRUARY/MARCH 0 MEMANDUM MARKS: 50 This memorandum consists of 8 pages.
Mathematics/P DBE/Feb. Mar. 0 UESTION. Mean n n 000 9 R 44, 44 000 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 ()
Mathematics/P DBE/Feb. Mar. 0. YEAR 995 996 997 998 999 000 00 N (Number in millions) Log N ( to decimal place) 8 7 4 67 5 8 55 6,9 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 995 996 997 998 999 000 00 N (Number in 8 7 4 67 5 8 55 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 ()
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]
Mathematics/P 5 DBE/Feb. Mar. 0 UESTION. 40 40 (). Time, t, in minutes Frequenc for intervals in 0 t < 5 table 5 t <0 5 for first three 0 t < 5 0 5 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 5 0 5 0 5 0 5 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 5 4 + 7 + + 7 + 5 + c 5 7 c 5 c 7 e 4 g a d f b c e (7) [7]
Mathematics/P 6 DBE/Feb. Mar. 0 UESTION 5 5. 5. m AD AD 4 5 6 4 ( ) + ( ) ( 5 ) + ( 4) 6+ 6 5 5. + + M ; + 5 4 M ; M ( ; ) 5.4 m BC m AD Lines are parallel m ( ) ( + ) 9 + + 7 0 for substitution for answer for substitution 5 -value -value value m BC subst ( ; ) () () () equation () + c ( ) + c 7 c 7 + + 7 0 value m BC subst ( ; ) equation ()
Mathematics/P 7 DBE/Feb. Mar. 0 5.5. m AD tan β β 80 56, β,69 B( ; ) α E A( ; 4) β F D(5 ; ) tan β m AD,69 5.5. 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]
Mathematics/P 8 DBE/Feb. Mar. 0 UESTION 6 6. Coordinates of centre M ( ; ) + + 8 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 -
Mathematics/P 9 DBE/Feb. Mar. 0 ( a + ) + ( a ) 6 substitution ( a + ) + ( a + ) ( a + ) ( a + ) 6 6 8 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 + 4 + a + 4a + 4 6 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)
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 9 + 9 8 PS ( a ) a 4(8) 7 ( a ) a + ( b + ) a 5 0 + ( a + ) 4a 70 0 7...() 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
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] 7. 5 4 K N -5-4 - - - 4 5 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]
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)
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]
Mathematics/P 4 DBE/Feb. Mar. 0 UESTION 9 9.. r cos α 9.. () TÔR 80 (90 + α) 80 (90 + α) 90 α 90 α () 9.. 9. 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) []
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 ; ) -90-60 -0 0 60 90 0 50 80 f - - 0.5 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]
Mathematics/P 6 DBE/Feb. Mar. 0 UESTION.. sin 6 p sin 4 sin (80 + 6 ) 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 45 5.. 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) () ()
Mathematics/P 7 DBE/Feb. Mar. 0.. sin cos sin + sin 0 sin ( sin + ) 0 sin 0 or sin 0 + 80 k; k Z or {0 or 0 } + 60 k; k Z n.80 n.60 0 (n + ).80 + 0, n Z sin ( sin + ) 0 sin 0 or sin 0 + 80 k 0 0 + 60 k; k Z (6).4 tan tan tan tan 4. tan 87 tan 88 tan 89 sin sin sin 45 sin 88 sin 89...... 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 45... identit co-ratios simplification for answer identit co-ratios simplification for answer (4) (4) [9]
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