Mathematics Grade 12 Paper 1 NSC June 2019 Memo

Downloaded from hlayiso.com SENIOR CERTIFICATE EXAMINATIONS/ SENIORSERTIFIKAAT-EKSAMEN NATIONAL SENIOR CERTIFICATE EXAMINATIONS/ NASIONALE SENIORSERTIFIKAAT-EKSAMEN MATHEMATICS P1/ WISKUNDE V1 MARKING GUIDELINES/NASIENRIGLYNE 2019 MARKS: 150 PUNTE: 150 These marking guidelines consist of 15 pages. Hierdie nasienriglyne bestaan uit 15 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 2 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • Consistent Accuracy applies in all aspects of the marking memorandum. LET WEL: • Indien ’n kandidaat ’n vraag TWEE keer beantwoord, merk slegs die EERSTE poging. • Volgehoue akkuraatheid is DEURGAANS op ALLE aspekte van die memorandum van toepassing. QUESTION/VRAAG 1 1.1.1 x 2 − 5x − 6 = 0 ( x − 6)( x + 1) = 0 x = 6 or x = –1  factors  both answers (2) OR/OF OR/OF x 2 − 5x − 6 = 0  correct subst into correct 5 ± (−5) 2 − 4(1)(−6) formula x= 2(1) 5 ± 49 x= 2 x = 6 or x = –1  both answers (2) 1.1.2 (3x − 1)(x − 4) = 16  standard form 3 x 2 − 13 x − 12 = 0 13 ± (−13) 2 − 4(3)(−12)  correct subst into correct x= 2(3) formula 13 ± 313 x= 6 x = 5,12 or x = −0,78   answers (4) OR/OF OR/OF 3 x 2 − 13 x − 12 = 0  standard form 13 x2 − x = 4 3 2 13  13  2  13  2  13  x2 − x + −  = 4 + −   adding  −  both 3  6  6  6 2 sides  13  313 x−  =  6 36 13 ± 313 x= 6 x = 5,12 or x = −0,78   answers (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 3 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne 1.1.3 x(4 − x ) ≥ 0  factorisation − x( x − 4 ) ≥ 0 or − x(x − 4 ) ≥ 0 x( x − 4 ) ≤ 0 0 4 0 4 0≤ x≤4 or x ∈ [0 ; 4]  0 ≤ x ≤ 4 (3) 1.1.4 5 −12x =4 5x + 1  factors in numerator (5 x + 1)(5 x − 1) =4 5x + 1 5x −1 = 4  5x −1 = 4 5x = 5 x =1  answer (3) OR/OF OR/OF 52x − 1 =4 5x + 1 5 2 x − 1 = 4.5 x + 4 5 2 x − 4.5 x − 5 = 0  standard form (5 − 5)(5 + 1) = 0 x x  factors 5 x = 5 or 5 x ≠ −1 x =1  answer (3) 1.2 x = 2 − 3 y ………………………..……..(1)  x = 2 − 3y 2 x + 4 xy − 5 = 0 ………………………(2) Substitute (1) in (2): (2 − 3 y ) 2 + 4 y (2 − 3 y ) − 5 = 0  correct subst into correct 4 − 12 y + 9 y 2 + 8 y − 12 y 2 − 5 = 0 formula − 3y 2 − 4 y −1 = 0 3y2 + 4 y +1 = 0  either standard form (3 y + 1)( y + 1) = 0 1 y = − or y = −1  y – values 3 x = 3 or x = 5  x – values (5) OR/OF OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 4 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne 2 x 2 x y= − ………………………..……..(1)  y= − 3 3 3 3 x 2 + 4 xy − 5 = 0 ………………………(2) Substitute (1) in (2): 2 x  correct subst into correct x 2 + 4 x −  − 5 = 0 3 3 formula 3x + 8 x − 4 x 2 − 15 = 0 2 − x 2 + 8 x − 15 = 0  either standard form x 2 − 8 x + 15 = 0 ( x − 5)( x − 3) = 0 x = 3 or x = 5  x – values 1 y=− or y = −1  y – values (5) 3 1.3 ab = 2 10 bc = 3 2 ac = 6 5  volume = abc ab.bc.ac = 2 10 .6 5.3 2  ab.bc.ac = 2 10 .6 5.3 2  (abc )2 = 36 100 (abc ) = 36 100 2 abc = 360 = 6 10  answer (5) OR/OF OR/OF 6 5 6 5 ac = 6 5 ∴ a =  a= c c 3 2 3 2 bc = 3 2 ∴ b =  b= c c ab = 2 10  6 5  3 2      c  c  = 2 10    18 10 = 2 10 .c 2 c2 = 9  value of c c=3  Volume = abc Volume = abc = 2 10 .3 = 360 = 6 10  answer (5) [22] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 5 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 2 2.1.1 59  answer (1) 2.1.2 15 29 41 51 14 12 10 –2 –2 2a = −2 second difference of – 2 a = −1 a 3(− 1) + b = 14 b = 17 b (− 1) + (17 ) + c = 15 c = −1 c Tn = −n 2 + 17n − 1 (4) 2.1.3 T27 = −(27 ) + 17(27 ) − 1  substitution 2  answer = −271 (2) 2.2.1 − 18 1 r= =−  answer 36 2 (1) 2.2.2 n −1  1 n −1  1  Tn = 36 −  Tn = 36 −   2  2 n −1 9  1 = 36 −  4096  2 n −1 n −1 1  1 1  1 = −   = −  16384  2  16384  2  14 n −1  1  1 −  = −   2  2 14 = n − 1  answer n = 15 (3) OR/OF OR/OF −9 9 −9 9 36 ; − 18 ; 9 ; ; ; ; ... ; 2 4 8 4096 If you look only at the denominator: 2 ; 4 ; 8 ; … ; 4096 2 k = 4096  2 k = 4096 2 k = 212 k = 12  k = 12 ∴ n = 15 terms  answer (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 6 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne 2.2.3 a S∞ = 1− r 36  correct subst into correct = formula with – 1 < r < 1  1 1− −   2 = 24  answer if – 1 < r < 1 (2) 2.2.4 250  1   − 18   − 1 1  4    r= and n = 250 S 250 even =   4 1 −1 4 = −24  S 250 even = −24   1  250  36   − 1  4   S 250 odd =   1 −1 4  S 250 odd = 48 = 48 S odd 48 = S even − 24 = −2  answer (4) OR/OF OR/OF T1 + T3 + T5 + T7 + ... + T499 T2 + T4 + T6 + T8 + ... + T500 a + ar 2 + ar 4 + ... + ar 498  a + ar 2 + ar 4 + ... + ar 498 = ar + ar 3 + ar 5 + ... + ar 499  ar + ar 3 + ar 5 + ... + ar 499 a + ar 2 + ar 4 + ... + ar 498 = ( r a + ar 2 + ar 4 + ... + ar 498 )  ( r a + ar 2 + ar 4 + ... + ar 498 ) 1 = r = −2  answer (4) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 7 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 3 3.1.1 p + 6 – (2p + 3) = p – 2 – (p + 6)  equating i.t.o p –p+3=–8  simplifying p = 11 (2) 3.1.2 Tn = 25 + (n − 1)(−8) = 33 − 8n  subst into T n formula 33 − 8n < −55 − 8n < −88  n > 11 n > 11  n = 12 ∴ Term 12 will be the first term smaller than –55 (3) ∴ Term 12 sal die eerste term kleiner as – 55 wees. 3.2 n 6 S 6 = [a + l ] = [( x − 3) + ( x − 18)] 2 2 = 6 x − 63  6 x − 63 n 9 S 9 = [a + l ] = [( x − 3) + ( x − 27)] 2 2 = 9 x − 135  9 x − 135 6 x − 63 = 9 x − 135 3 x = 72  24 x = 24 n 15 15 ∴ S15 = [a + l ] = [( x − 3) + ( x − 45)]  [( x − 3) + ( x − 45)] 2 2 2 15 = [2 x − 48] 2 15  substitution of x = [2(24) − 48] = 0 = RHS 2 (5) OR/OF OR/OF 9 ∑ (x − 3k ) = 0 k =7  expansion (x − 21) + (x − 24) + (x − 27 ) = 0  3 x − 72 = 0 ∴ 3x − 72 = 0 3x = 72  24 x = 24 15 ∑ (24 − 3k ) k =1 = 21 + 18 + 15 + ..... + −21.  substitution of x n S n = [a + l )] 2 15  sum of 15 terms = [21 − 21] 2 (5) = 0 = RHS OR/OF OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 8 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne ( x − 3) + ( x − 6) + ( x − 9) + ( x − 12) + ( x − 15) + ( x − 18)  expansion = ( x − 3) + ( x − 6) + ( x − 9) + ( x − 12) + ( x − 15) + ( x − 18) + ( x − 21) + ( x − 24) + ( x − 27)  3 x − 72 = 0 ∴ 3x − 72 = 0 3x = 72 x = 24  24 15 ∑ (24 − 3k ) k =1 = 21 + 18 + 15 + ..... + −21.  substitution of x n S n = [a + l ] 2 15 = [21 − 21] 2  sum of 15 terms (5) = 0 = RHS [10] QUESTION/VRAAG 4 4.1 y>0 answer (1) OR/OF OR/OF y ∈ (0 ; ∞ ) answer (1) 4.2 1 x g: y =   2 y −11 1 y g :x =  x =  2 2 1  equation (2) y = log 1 x or y = − log 2 x or y = log 2 2 x 4.3 Yes. The vertical line test cuts g–1 once  yes Ja. Die vertikale lyn toets sny g–1 slegs eenkeer.  valid reason (2) OR/OF OR/OF Yes. For every x-value there is a unique y-value  yes Ja. Vir elke x-waarde is daar ‘n unieke y-waarde  valid reason (2) OR/OF OR/OF Yes. g is a one-to-one function / Ja. g is ‘n een-tot-een funksie  yes  valid reason (2) OR/OF OR/OF Yes. The horizontal line cuts g only once  yes Ja. Die horisontale lyn sny g slegs een keer  valid reason (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 9 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne 4.4.1 y = − log 2 x 2 = − log 2 a  correct subst into correct 2 formula (a ; 2) 1 1 1 a = 2 −2 = or a=  = 4 2 4 answer (2) 4.4.2  1  answer M /  2 ;  or M / (2 ; a ) (1)  4 4.5  9  −1 M //  − 1;  9  4  4 (3) [11] QUESTION/VRAAG 5 5.1.1 x = −2  answer y =3  answer (2) 5.1.2  7 0 ;   answer  2 (1) 5.1.3 1 +3= 0 y=0 x+2 1 + 3( x + 2 ) = 0 3 x = −7 7 x=−  answer 3 (2)  7  x-intercept  − ; 0   3  5.1.4 y asymptotes at y = 3 and 3,5 x=–2 intercepts at 3 y = 3,5 and x = – 2,3 shape –2,3 x (reasonable – 2,3 –2 0 representation in correct quadrants) (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 10 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne 5.2.1 − 2x + 4 = 0 y=0 2x = 4 x=2 x=2 ∴ S(2 ; 0) (2) 5.2.2 Equation of k: y = a(x + 1) + 18  y = a( x + 1) + 18 2 2 0 = a(2 + 1) + 18 2 or 0 = a(− 4 + 1)2 + 18  substitute (2 ; 0) or (– 4 ; 0) 9a = −18 a = −2 a y = −2(x + 1) + 18 2 (3) 5.2.3 2 − 2 x − 4 x + 16 = −2 x + 4  equating − 2 x 2 − 2 x + 12 = 0 x2 + x − 6 = 0  standard form (x + 3)(x − 2) = 0  factors x = −3 or x = 2  choosing x = –3 y = −2(−3) + 4 = 10  answer T(–3 ; 10) (5) 5.2.4 x < −3 or x > 2  answer (2) OR/OF OR/OF (−∞ ; − 3) ∪ (2 ; ∞)  answer (2) 5.2.5(a) x < −1  answer (2) OR/OF OR/OF (−∞ ; − 1)  answer (2) 5.2.5(b) y  shape of cubic with local min tp moving to local max tp g  turning points at x = 2 and x = –4 –4 x  point of inflection at –1 0 2 x = –1 (3) [25] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 11 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 6 6.1.1 A = P(1 − i ) n 79866,96 = 180 000(1 − 0,15)n  substitution (1 − 0,15)n = 79866,96 180 000  79866,96  log   use of logs  180 000  n= log(1 − 0,15) n = 4,999… years n ≈ 5 years  answer (3) 6.1.2 A = P(1 + i ) n 20  values of i and n  0,1   substitution = 49 0001 +   4  = R80 292,21 The money will be enough to buy the car.  conclusion Die geld sal genoeg wees om die motor te koop. (consistent with answer) (3) 6.2.1 P= [ x 1 − (1 + i ) −n ] i 0,1025   0,1025  −234   n = 234  i = 7853,151 − 1 +   12   12    substitution in present P= 0,1025 value formula 12 P = R793 749,25  answer (4) OR/OF OR/OF Balance Outstanding / Uitstaande balans  0,1025  6   n = 6 in both 7853,151 +  − 1 0,1025  0 ,1025  6   12    i= = 800 0001 +  − 12  12  0,1025 A–F 12 = 841 885,56 − 48 136,62 = R793 748,94  R793 748,94 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 12 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne 6.2.2 A = P(1 + i ) n 3  0,1025  3 = 793749,251 +   0,1025   12   793749,251 +   12  = R814 263,3052 New instalment/Nuwe paaiement: P= [ x 1 − (1 + i ) −n ] i   0,1025  − 231   n = 231 x 1 − 1 +     12    substitution of new P 814 263,3052 =   0,1025  substitution of n and i into 12 formula x = R8 089,20  answer (5) [15] QUESTION/VRAAG 7 7.1 f ( x) = x 2 + 2 f ( x + h) = ( x + h ) + 2 2 = x 2 + 2 xh + h 2 + 2 ( f ( x + h) − f (x ) = x 2 + 2 xh + h 2 + 2 − x 2 + 2 )  x 2 + 2 xh + h 2 + 2 = 2 xh + h 2 f (x + h ) − f (x ) f / ( x) = lim h →0 h 2 xh + h 2 2 xh + h 2 = lim  lim h →0 h h →0 h h(2 x + h ) h(2 x + h ) = lim  lim h →0 h h →0 h = lim(2 x + h ) h →0 = 2x  answer (4) OR/OF OR/OF f (x + h ) − f (x ) f / ( x) = lim h →0 h = lim x + 2 xh + h + 2 − x 2 + 2 2 2 ( )  x 2 + 2 xh + h 2 + 2 h→0 h 2 2 xh + h 2 xh + h 2 = lim  lim h →0 h h →0 h h(2 x + h ) h(2 x + h ) = lim  lim h →0 h h →0 h = lim(2 x + h ) h →0 = 2x  answer (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 13 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne 7.2.1 y = 4 x 3 + 2 x −1  + 2 x −1 dy = 12 x 2 − 2 x − 2  12x 2 dx  − 2 x −2 (3) 7.2.2 y = 4 x + (3 x ) 3 3 2 1 1 6 = 4x + 9x 3 6  4x 3  9x 2 dy 4 − 3 2 4 − 5 = x + 54 x5  x 3  54x dx 3 3 (4) 7.3 Point of contact: (1 ; 5) m=2  m=2 y − y1 = m( x − x1 ) or y = 2x + c  substitution of (1 ; 5) y − 5 = 2( x − 1) 5= 2+c c=3 y = 2x + 3 y = 2x + 3  answer (3) [14] QUESTION/VRAAG 8 8.1 3 3 h( x) = −2( x + )( x − 1)( x + 3) – 2 ( x + )( x − 1)( x + 3) 2 2 ( h(x ) = −(2 x + 3) x 2 + 2 x − 3 ) correct simplification (3) h(x ) = −2 x 3 − 7 x 2 + 9 OR/OF OR/OF h( x) = −(2 x + 3)( x − 1)( x + 3)  − (2 x + 3)( x − 1)( x + 3) ( h(x ) = −(2 x + 3) x 2 + 2 x − 3 )  correct simplification (3) h(x ) = −2 x 3 − 7 x 2 + 9 8.2 h / ( x) = −6 x 2 − 14 x  first derivative − 6 x 2 − 14 x = 0 =0 − 2 x(3x + 7 ) = 0 7 x = 0 or x = −  both answers 3 (3) 8.3 7  answer x<− or x > 0 3 (2) OR/OF OR/OF  7 x ∈  − ∞ ; −  ∪ ( 0 ; ∞)  answer  3 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 14 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne 8.4 y = 4x + 7  y = 4x + 7 2 − 6 x − 14 x = 4  h / (x ) = 4 0 = 6 x 2 + 14 x + 4  standard form 0 = 3x 2 + 7 x + 2 0 = (3 x + 1)( x + 2 ) 1  both answers x=− or x = −2 3 (4) [12] QUESTION/VRAAG 9 9.1 Volume of Sphere 4 2048π = π (8)3 or = or = 2144,66 3 3  answer (1) 9.2 2 r +x =8 2 2 (Pythagoras)  substitution or reason r 2 = 64 − x 2 Pythagoras (1) 9.3 1 Vcone = πr 2 h 3 h=8+x 1 ( = π 64 − x 2 (8 + x ) 3 ) 1 ( )  π 64 − x 2 (8 + x ) 3 = π 3 (512 + 64 x − 8x − x ) 2 3  expansion dV 64π 16π 3π 2 = − x− x  dV = 64π − 16π x − 3π x 2 dx 3 3 3 dx 3 3 3 2 0 = 64 − 16 x − 3 x 0 = (8 − 3 x )( x + 8) 8 8 x= x ≠ −8 x= 3 3 1  512  32  π   Vcone 3  9  3   volume of the cone = Vsphere 2048π 3 8 = = 0,3 8 27  or 0,3 27 (7) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/Wiskunde V1 15 DBE/2019 SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 10 10.1 R R B R B B P(One Red and One Blue) 3 2 = P(Red, Blue) + P(Blue, Red)   ×   12   11  3 2 2 3 2 3 =  ×  +  ×    ×   12   11   12   11   12   11  1  addition of products = 11  answer (4) 10.2.1 a = 0,48 × 250 a = 120  answer (1) 10.2.2 b = 150  b P(S) × P(F) 200 150  P(S) × P(F) = × 250 250 200 150  and = 0,48 250 250 = P(S and F) These events are independent /  conclusion Hierdie gebeurtenisse is onafhanklik (with realistic probabilities) (4) [9] QUESTION/VRAAG 11 11.1 10 × 9  10 × 9 = 90 (2) 11.2.1 10!  10! =3 628 800 (1) 11.2.2 2! × 2! × 2! × 2! × 2! ×4!  2! × 2! × 2! × 2! × 2! = 768  4!  2! × 2! × 2! × 2! × 2!×4! or 768 (3) [6] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou