Mathematics Grade 12 P1 KwaZulu Natal November 2015 Memo

Basic Education Pd : ** Fo MATHEMATICS P1 ; * _ PREPARATORY EXAMINATION woe SEPTEMBER 2015 * MEMORANDUM : Merrie terre er ere eer NATIONAL SENIOR CERTIFICATE MARKS: 150 This memorandum consists of 11 pages. KwaZulu-Natal Department of Education REPUBLIC OF SOUTH AFRICA

2015 Oct 22 3:04PM HP Fax Downloaded fromhlayiso.com Mathematics P1 2 September 2015 Preparatory Examination NSC-MEMORANDUM(KZN) QUESTION 1 L111) | x(x-5)=0 2 x=0 or x= ¥v Answers 1.12 | 4x? ~5x=3 4x? — $x —3=0 ¥ equation in std. form yal (-5) £ J (-5)* - 4(4)(-3) Y substitution 2(4) x=1,69 or x=-044 vv answers 4 113 | 2° @x+1)<0 2 >Oforalixe R ¥2" >Oandxek “3x+1<0 ¥ 3x+1<0 1 x<~— ~ answer 3 1.1.4 a x-3x? =4 1 x-4=3Vx ¥ isolating 3Vx or 3x? 9x =x? ~ 8x +16 ¥ squaring both sides x’ ~17x +16=0 Y trinomial (1X - 16)=0 v factors x=1 or x=16 ~ both answers ¥ selection 6 nia ! OR i | L x-3x? —-4=0 ¥ standard form 1 1 ¥ factors [#-4]e-1 =< ' a a Varad = = 1 x?=4 or x? =-1 ¥ va-] x=16 no solution Y answer ¥ rejecting 6 1.2 fg 2028 fg7030 _ 7s /g2028 yor (yor -1 ¥ factorising denominator vs v simplifying 80 2 Y answer 3 80 Copyright reserved Please turn over

2015 Oct 22 3:04PM HP Fax Downloaded frori’hlayiso.com Mathematics Pt 3 September 2015 Preparatory Examination NSC-MEMORANDUM(KZN) 13 24” and x 4+2y=29 3xtl=2y > Y for 2y as subject w4+2y=29 > (2) Subst. (1) into (2) x? 43x41=29 ¥ substitution of 2y x? +3x—-28=0 ¥ std. form (x+7)(—4)=0 ve? or x=4 vx values y=-10 of yee vy values 6 [24 QUESTION 2 211]3:;2;6:2;9.:,2., 12; 2 vw answers 2 2.1.2 | a=3,d=3,n=50 ¥ for a, d and n values 50 50 So = ray +150] or Ze) 3(50-1)] Y for substitution into formula = 3825 a=2 and is a constant sequence “Ss = 50x 2 = 100 Swoy = 3825 +100 = 3925 5 2.2 §,=at+artart+..tar* > () ¥ for equation (1) rS, =artar+.. tar" — (2) ¥ for equation (2) (2)- a): 7S, -S, = ar" -a ¥ subtraction on LHS and RHS S,¢-= alr’ . 1) | ~ factorising alr” = 1) 1S St in rl 4 [ Copyright reserved Please turn over

2015 Oct 22 3:04PM HP Fax Downloaded frori*hlayiso.com Mathematics P1 4 September 2015 Preparatory Examination NSC-MEMORANDUM(KZN) QUESTION 3 3.1 24 Sy 1124 6x4 3x? tee a=6r=2 Ya value 1s T value S= of2 ~1) “substitution inte formula 2-1 = 196602 answer 4 3.2 -l<r<l ¥ for condition 1 -teltyey Y for r=—x 2 2 -2<x<2 ¥ answer 3 3.3 r>l1 test vr>l 2 2 v4 x> answer 4 19] QUESTION 4 4.1 2a=4 a=2 ¥ avalue 3a+b=2 b=-4 vb value 4 atbht+e=5 c=7 ¥c value T,=2n? -4n+7 ¥ general term OR 2a=4 a=2 ¥ a value 3at+b=2 hea-4 %b value T) =e=7 ¥c value LT, =2n? -4n+7 4 ¥ general term OR Copyright reserved Please turn over

2015 Oct 22 3:04PM HP Fax Mathematics P1 5 NSC-MEMORANDUM(KZN) Downloaded frori’hlayiso.com September 2015 Preparatory Examination _ (nn - 2) 7. =T+(- Nd}, ¥ formula n-1\n—2 = 5+ (n—1X2)+ {a- he 2) (4) ¥ substituting first and second difference values =5+2n-2+2n? —6n+4 =2n? -4n+7 ¥ simplifying ¥ general term 4 42 | First Difference n™ term = (4n-2) 4n-2= 2018 v4n—2= 2018 4n = 2020 n=505 ¥n=505 Toys = 2(506)’ — 4(506) + 7 = 510055 Fo, = 2(505)? - 4(505)+ 7 = 508037 Between the 505" and the $06" terms. “answer 3 OR 7, -T,) =2n* —4n+7-[a(n-1)? -4(n—-1)+7] | 2 2 ¥T,-T._, substitution =2n? —4n+7-[2n* —4n+2-4n+4+7] =4n-6 4n—-6=2018 v4n-6=2018 4n = 2024 n=506 Between the 505” and the S06" terms. Vanswer 3 [7] Copyright reserved Please turn over

2015 Oct 22 3:04PM HP Fax Downloaded from hlayiso.com Mathematics PL 6 September 2015 Preparatory Examination NSC-MEMORANDUM(KZN) QUESTION 5 5.1 A=P(1+i)" ¥ ¥ correct 4 substitution into 2x = x{l + 0,085) correct formula . log2 ¥ making 7 the log(1 + 0,085) subject =85 or years or 8 years and 6 months ¥ answer \ 5.2.11 4 = P(+i)" 2 ¥ correct = 350 ovo + 23) substitution into correct formula =R357919,30 answer e 3.2.2 ~ jy" pal 497] i 0135)" xy 1-—|14+-4 ( 12 } ¥ “correct 35791930 = —435 substitution into —— correct formula 12 0,135 357919,30x 2" 12 238 (1-285) | 2 x= v x as subject x =R4328.60 v answer 4 5.2.3 | Balance on Loan after the 180” payment 178 in 43 2300 28) - 1 = 457919 a 285) _ ¥ ~ substitution , 0,135 into formulae 12 = R188118,54 ¥ R188118,54 Total Payments = R4328,60« 60 = R259716 v R1030206,80 Savings = R259716 - R188118,54 = R71597,46 v answer 8 Copyright reserved Please tum over

2015 Oct 22 3:06PM HP Fax Downloaded from hlayiso.com Mathematics Pt 7 September 2015 Preparatory Examination NSC-MEMORANDUM(KZN) QUESTION 6 6.1 a x*)=——+q. P(-2;4) isa point on f= + © PC24) isa point onf p=l v pvalue gel ¥ qvalue a = 4] » xt+t Now subst. P(-2;4): 4=—4 4) ¥ sub d point P 24] subst. p, g and point .a@=-3 ¥a value 4 62 3 iso xt1 4 =o) x+1=3 wl=3 x=2 T@;0) “answer 3 63 y=(x+p)+¢@ y=(e+D41 , “substitution of p and q values into equation of line of symmetry y=xt2 “c=2 ~ answer 2 OR yexte Point of intersection of asymptotes (—1 ; 1 l= ym ( ) ~ substitution of p and q values ~lelte into equation of line of symmetry c-2 ¥ answer 2 im ka Copyright reserved Please turn over

2015 Oct 22 3:06PM HP Fax Downloaded frori’hlayiso.com Mathematics P! 8 September 2015 Preparatory Examination NSC-MEMORANDUM(KZN) QUESTION 7 val a(x)=—x? +2x+3 and ACx)-axt+q. CD = 6 units. -x? 4+2e+3=0 vy=0 (x+1(x-3)=0 ¥ factors x=-1 or x=3 B(3;0) ¥ coordinates of B C(0;3) ¥ coordinates of C 4 72 DO; -3) v¥ answer 2 73 a=1 and g=-3 v¥ answers 2 74 ~x° 42x43-x-3 ~ equating both equations x -~x¥-6=0 (x+2(x-3)=0 © factors x=-2 or x=3 ¥ x values y=-5 or y=0 vy values E(-2;-5) answer 5 75 gyal +4 VAxis of symmetry value ; _lex<l or x>3 0% ond points and inequality 4 [17] QUESTION 8 8.1] p@)=log, x x= log, y Diiy=3* vv answer 2 8.2 ¥¥ graph of p ¥ graph of p@ 4 8.3 log, x=2 x=9 O<x<9 ¥vfor end points and inequality 2 8.4 x intercept of p is (1 ; 0) x intercept of A is (-1 ; 0) vv answer 2 {10} Copyright reserved Please turn over

2015 Oct 22 3:06PM HP Fax Downloaded fromhlayiso.com Mathematics PI 9 September 2015 Preparatory Examination NSC-MEMORANDUM(KZN) QUESTION 9 91 fQx)=x? +3x 1°) =tim flx+h)- 4) h0 h =i (x+AY +3(x+A)- (2 +3x) v f(x +h)= (+h) +3(4h) am h ¥ substitution into correct formula _ x +2xh+h? +3x4+3h—x? -3x an = lim h ¥ simplifying hao xh +h? 43h hoo h _ AQQx+h43) = tim a ¥ factors hoo =2x+3 answer 5 921 | a ity 35? Yaxt dx 16 0 y=6x3 Y6% x3 7 * =20x3 or 20Vx7 ¥ answer 3 9.2.2 bons x 5x? 44x S(t f= 4 ¥ factorising f= xx? -5x+4) ¥ factors (x4) = x(e~1fx-4) % simplification of f (x-4) “answer =x -x Sx) = 2x-1 4 | {12] Copyright reserved Please turn over

2015 Oct 22 3:06PM HP Fax Downloaded fromhlayiso.com Mathematics P1 10 September 2015 Preparatory Examination NSC-MEMORANDUM(KZN) QUESTION 10 1 10 y= fe the gradient of the tangent to the x curve is — 1 ya4yt ¥ exponential form dy a _74 ¥ derivati ok =-~4y? = 7 derivative a4} Y derivative =—1 x v=4 “x=-2 or x=2 ¥ x~ values y=-2 or y=2 (-2;-2) (2;2) ~y —values 5 10.2.1) y= (x 1e- 4)’ ¥ (e—Dr-4y? =(x- 1)? ~ 4x44) ¥ squaring binomial =x ~9x? 424y—-16 2 10.2.2) y= x3 -9x? +24x-16 & = 3x? —18x+24=0 ¥ derivative and equal to 0 x” ~6x+8=0 (x—-2)x—4)=0 “factors x=2 or x=4 ya4 or y=0 v y—values B(; 4) ¥ coordinates of B 4 10.23 4&<-16 v inequality | W16 2 10.2.4] f(x) =6x-18>0 ¥ 6x-18>0 x>3 answer 2 [15] Copyright reserved Please tum over

2015 Oct 22 3:07PM HP Fax Downloaded from hlayiso.com Mathematics Pt 1 September 2015 Preparatory Examination NSC-MEMORANDUM(KZN) QUESTION 11 11.1 Velxbxh ¥ formula = 5x(9 ~2x\(x) Y substitution = 45x? —10x? 2 2 | 790% — 30x? v derivative 90x ~ 30x? =0 ¥ derivative equal to 0 vw 30x(3 — x)= 0 factors vx values x=0 or x=3 Therefore the box will have a maximum atx =3 | ¥ choosing x =3 § [7] QUESTION 12 12.1 | (x+0,2)x0,5= 0,2 ¥P(A) x P(B) = P(A and B) x+0,2=0,4 x= 0,2 vx value... P(A only) 0,24+0,.24+03+y=1 v y=1-07 y= 9,3 vy value 4 12.2 15! 2t5! P(X and Y together) = = ¥ 7 al 3 vi 3 1 P(X and Y not together) = 1 — 3 =2 3 Y answer 3 12.3.1 | 4 digit numbers(with repetitions) ¥ product =9x10x10x10 ¥ answer 2 =9000 12.3.2 | 4 digit numbers(without repetitions) =9x9x8x7 ¥ 9x9 =4536 v 8x7 v answer 3 12.3.3 | 4 digit numbers(with repetitions and last digit = 0) =9x10x10x1 ¥ product =900 ¥ answer 2 [14] TOTAL MARKS: 150 Copyright reserved