Grade 12 Mathematics: Memorandum Paper 1 - …

1 mathematics (NSC)/ Grade 12/P1 89 ExemplarMEMORANDUM Grade 12 mathematics : Memorandum Paper 1 x = 9 x = 012 xD21 ? 6 + 8 + 10 + 12 + 14 = 50 15)51(255 S D10)41(244 S D510155 ?T a = 64 and r = 1,5 DD729)23(6467 ? n)087,1(2 D3,82log087,1 ?nDThus during the 9th )1(4)1()(2 xxxxfD)2)(2)(1( xxxDThusx = 1 or x = 2 or x = -2 x = 5 y- intercept is y = -0,2 x 5 = -1 DThusx = y = x 5 OR y = -x + x = 2y 4 DThusy = (x + 4)/2 DThus22)( 3)(xxf DD(may include a constant) Distance = 80604341u uDD = 15 + 60 = 75 km t)15,1(0005000100 DDt)15,1(20?43,21 ?tDThus 21 hours and 26 minutes 92,192112021121110118888 2112111021110 fnnSSD11966,105,0log0005,0log0005,02101, 021202112020 ?

2 !?!? ? ? From AP: b a = (a b) bD?3b = 2a?b = 23aDFrom GP: a ba = 1 a bD?(a b) = a?(a 23a) = aD?(13a) = a?19a a = 0 ?a 9a = 0 ? a(a 9) = 0D?a = 0 or a = 9 ? b = 23 9 ?b = )5441(!3!5!4!3u DD= 150 Net salary = 0,7582506187,50Ru Bond repayments: 0,36187,501856,25Ru iD2401220 u nD iiAn)1( ,742153RA DDThus you can afford the flat. D6 Copyright reserved mathematics (NSC)/ Grade 12/P1 90 ExemplarMEMORANDUM iD iin1)1(30020000D914)1( ?niDnii ? 914)(logDThusn = 53,2 months Thus need 54 months 22)0(0 ? u abafD362)1(1 ? bbf )( ? 22)0(0 ? u abag D92,25,8172)2(22 ? ? bbbgDxxg)92,2.(2)( ? f(2,3) = 25,03 f(6) = 1458Dg(2,3) = 23,52 g(6) = 1239, f(x)D is the closer approximation as the values of f(2,3) and f(6) are closer to the collected data than those of g(2,3) and g(6).

3 1)(3()( xxxfD3 ?xD or 1 E = 3 6 = 3 E (0 ; -3) )(xg)12)(34(2 xxxD3272)(23 ? )12)(34()34(22 xxxxxD0)34)(112(2 ?xxxD1 ?xD)8;1(K? 02146)(2 cxxxgD6377r ?xD18,2 ?x or 15,0 xDAxis of symmetry of f(x) is x = -2 DThus F does not lie on the axis of symmetry of f (x). )()(xgxfc cD2146422 ?xxxD021262 ?xxD01632 ?xx66)1(34366 uu r ? ?xD or Line 5 DCan t by ( a b) because a b = 0 )352)(2()(2 xxxxfDBy factor theorem or inspection )1)(32)(2()( ?xxxxfD2 ?xD or x = -1,5 D or x = -1 Thus125,1222 xxx thus DDD15,04 hxfhxfxfh)()(lim)(0 coDhxhxxfh2121lim)(0 c?oDhhxxhxxxfh)2)(2()2()2(lim)(0 c?oD)2)(2(1lim)(0 c?ohxxxfhD2)2(1)( c? ? 2)0( cf thus gradient of tangent is 2 Increasing implies that D0)(!cxf0)1)(2(022 ?! ?xxxxD12 ? cm 23 Dmc23 ?)

4 23(mmxy Thusx intercept is when y = 0 Dmmx23 ? Area = 0,5 (x-intercept)(y-intercept) D = )23(2321mmm M ultiply out above expression to get Area = mm26129 DThus02292 mdxdAD 492 mD 23r mDBut m < 0 thus 23 mD5 Copyright reserved mathematics (NSC)/ Grade 12/P1 91 ExemplarMEMORANDUM Twice as much labour as bootleg, thus 500 2 = 250 straight leg jeans. 150dxD250dyD5002d yxDCorrect feasible region yxP58 DPoint A (125; 250) D and B(150 ; 200) DBy substituting points into profit function get maximum profit at A D = R 2 250 yxP511 DThus to maximize profit now use point B (by substitution) Thus 150 straight leg and 200 bootleg D220406080100120140160180200220240260280 300100200300400500xy Feasible region is not shadedABCopyright reserved)