Mathematics Grade 12 Paper 1 NSC Supplementary 2017 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12/GRAAD 12 MATHEMATICS P1/WISKUNDE V1 MARCH 2017 MEMORANDUM MARKS: 150 PUNTE: 150 This memorandum consists of 18 pages. Hierdie memorandum bestaan uit 18 bladsye. Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/ Wiskunde V1 2 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum 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  3( x  1)  0  answer x  3 or x  1  answer (2) 1.1.2 x 3  512 3 3 2  x x  512 2 2  8  2  32  3   2 3 3  x   83 3     x  64  answer (3) OR x 3  512  squaring both x 3  262144 sides x 3  218  x3  218 x  26 x  64  answer (3) 1.1.3 x( x  4)  0 + – + OR / 0 0 4 OF 4 0  x  4 OR / OF x  0; 4 critical values inequality or interval (2) Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/ Wiskunde V1 3 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum 1.2.1 x 2  5x  2  0  subst correct 5  52  412 formula x 21 5  17 x 2  answer x = 0,44 or x = 4,56  answer (3) OR x 2  5x  2  0 x 2  5 x  2 2 2  5  5  5  17 2 x 2  5 x      2      x    2  4  2 4 2  5 17 x     2 4 5  17 5  17  answer x or x  2 2  answer x = 0,44 or x = 4,56 (3) 1.2.2 f ( x)  x 2  5 x  2 x 2  5x  2  c x 2  5x  2  c  0  standard form b 2  4ac  0  b 2  4ac  0  52  412  c   0  substitution 25  8  4c  0 4c  17 17 c  answer 4 (4) 1.3 x  2y  2 x 2  2 xy  3 y 2  4  substitution 2 y  22  2 y2 y  2  3 y 2  4  simplification 4 y2  8y  4  4 y2  4 y  3y2  4  standard form 3y2  4 y  0  factors y 3 y  4   0 4  y 0; y 4 y0 or y 3 3 2  x-values (ca on x2 x 3 both x-values) (6) Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/ Wiskunde V1 4 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum OR / OF x  2y  2 1 y  x 1 2 x  2 xy  3 y 2  4 2 2 1  1  x  2 x x  1  3 x  1  4 2  substitution 2  2  1  x 2  x 2  2 x  3 x 2  x  1  4 4   simplification 3 2 x  x 2  3x  3  4 4  standard form 3x  4 x  4  0 2  factors 3x  2x  2  0 2  x2 ; x2 x2 or x 3 3 4  y-values (ca on y0 y 3 both y-values) (6) 6 1.4 S x 2 2 For S to be a maximum the denominator needs to be at a minimum.  Minimum of Vir S om ’n maksimum waarde te hê, moet die deler ’n minimum x 2  2 is 2 waarde h Minimum of x 2  2 is 2 6 M aximum of S  x 2 2 3 6 (2)  2 3 [22] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/ Wiskunde V1 5 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum QUESTION/VRAAG 2 For geometric: 2.1 1  ; b ;  1 ; ...... b 1 4   b 1 1 b   1 b 4  4 1 b2   b 1 4 2 1 b b 1 2 2 (3) OR  1 b      1  1  b      1  4  4 1 b  b 1 2 2 1 b 2 (3) 2.2 1 1  ; ;  1 ; ...... 4 2 r = – 2 r  2 T19  ar 18  1 subst. into correct    (2)18 formula  4  218     2   2   216 –65536 / –216  65536 (3) OR / OF T19  ar 18  1 r = – 2    (2)18  4 subst. into correct      2  2 218 formula  2 16 –65536 / –216  65536 (3) Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/ Wiskunde V1 6 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum 1 1 2.3 The series is:  ; ;  1 ; 2 ;  4 ; 8 ; ........ 4 2 1 The new positive term series: ; 2 ; 8 ; 32 ; 128 ; 1 2 a  ……… 2 1 r  4 a r4 2 20 19  20  1  n 1  4  n 1 or  p 0 n 1  2   correct formula OR/OF (4) 19 1   2 4 p p 0 etc. 2.4 No, the series is not convergent / Nee, die reeks konvergeer no nie r = 4 and for convergence –1 < r < 1 r = 4 en vir konvergering –1 < r < 1 reason (2) [12] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/ Wiskunde V1 7 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum QUESTION/VRAAG 3 3.1.1 24  24 (1) 3.1.2 6 6 9 15 0 3 6 3 3 3 2a  3 3a  b  0 abc  6  a 2 3 9 9 a b c9 b 2 2 2 3 9  c9 Tn  n 2  n  9 2 2 3 9  Tn  n 2  n  9 2 2 OR/OF (4) (n  1)(n  2)d 2 formula Tn  T1  (n  1)d1  2 substitution (n  1)(n  2)(3)  6  (n  1)(0)  2 simplifying n  3n  2  3  2  6   1 2 3 9  6  n2  n  3 3 2 9 2 2  Tn  n  n9 3 9 2 2  n2  n  9 (4) 2 2 3.1.3 3 2 9 n  n  9  3249  equating general 2 2 term to 3249 3n  9n  18  6498 2 3n 2  9n  6480  0 n 2  3n  2160  0  standard form n  45n  48  0  factors n  45 or n  48  n  45 or n  48 (4) 3.2  1 ; 2 sin 3 x ; 5 ; ....... 2 sin 3 x  1  5  2 sin 3 x  2 sin 3x  1  5  2 sin 3x 4 sin 3 x  4  sin 3x  1 sin 3 x  1  3x  90  3 x  90   x  30  x  30  (4) [13] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/ Wiskunde V1 8 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum QUESTION/VRAAG 4 4.1 U(1; 0)  (1 ; 0) (1) 4.2 x 1  x 1 y 1  y 1 (2) 4.3 2 1  0  y=0 x 1 2  x  1 x  1  x  1 T 1 ; 0  (2) 4.4 f  x   log 5 x h : x  log 5 y  change x and y y  5x  y  5x (2) 4.5 y=0  answer (1) 4.6  V 2  1; 2  1   2  1 V2,41; 2,41  2  1 (4) OR / OF 2 2 x 1 x  1 x 1 x 1 x2  x  2  x 1 x 2  2x  1  0 2  4  41 1 subs into correct x 2 formula 2 8  2 22 2   x  2 1 2  y  2 1  1 2  V 1 2 ; 1 2  (4) OR / OF 2 2 x 1   x 1  x 1 x 1 x  1  2 2  x  1  2 2 x  1 2  x  2 1  V 1 2 ; 1 2   y  2 1 (4) 4.7 T 3 ; 2 /  x3  y2 (2) [14] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/ Wiskunde V1 9 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum QUESTION 5 5.1.1 C(0 ; –3)  C(0 ; –3) (1) 5.1.2 f x   x  2 x  3 2  factors x  3x  1  0 x  1 or x  3  x-value AB = 3 – (–1)  other x-value AB 4 units  answer (4) 2 1 3 5.1.3 x or 2x  2  0 or x  21 2 1  x 1 y  1  21  3 2  4 D1 ; 4  y value (2) 5.1.4 C(0 ; –3) D(1 ; – 4) Average gradient / Gemiddelde gradient 43 3 4 43 3 4  or  or 1 0 0 1 1 0 0 1  1 –1 (2) 5.1.5 OC = OB = 3 equal lengths OĈB = 45 isosceles right angled triangle  45 Gelykbenige reghoekige driehoek (2) OR / OF tan   m g  tan   1 tan   1   45  45 OB̂C  45 (2) OĈB  45 5.1.6  4  k  3 OR (–4 ; –3) –4  –3  notation (3) 5.1.7 f x  . f x   0 / //  2x  2 2 x  2. 2  0 2 2x  2  0 x 1  x 1 (3) Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/ Wiskunde V1 10 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum 5.2 y (3 ; 4) TP  x=3 f  y=4 x O (1 ; 0) (5 ; 0)  x – intercepts  y-intercept  shape (0 ; 5) (5) f ( x)  a x  1 x  5 4  a3  13  5 4  4a a  1 f x    x 2  6 x  5 [22] QUESTION/VRAAG 6 6.1.1 A  150 0001  0,2 n=2 2  150 000 in correct formula  R 96 000  96 000 (3) 6.1.2 150 000(1  0,2) n  49 152  150 000(1  0,2)  49 152 n 0,8n  1024 3125 n log 0,8  log 1024  n log 0,8  log 1024 3125 3125 n5  n5 The machine will need to be replaced at the beginning of 2020 / Masjien moet aan die begin van 2020  2020 vervang word (4) OR / OF 150 000(1  0,2) n  49 152  150 000(1  0,2) n  49 152 0,8n  1024 3125 1024 1024  n  log 0,8 n  log 0,8 3125 3125  n5 n5 The machine will need to be replaced at the beginning of 2020 / Masjien moet aan die begin van 2020  2020 vervang word (4) Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/ Wiskunde V1 11 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum 6.1.3 R280 000 – R49 152  R230 848 =R230 848  0,085  20  i= 0,085 = 0,02125 x 1    1 4  4   230 848  and n = 20 0,085  subs into correct formula 4 x  R 9 383,26  R 9 383,26 (4) n x[1  (1  i ) ] 6.2 Pv  0,11 i  i=   0,11  180  12 9 000 1  1     n = 180   12    substitution correct formula  0,11 12  R 791 837,43  R 791 837,43 Lerato qualifies for a loan of R 791 000 under the given conditions / Lerato kwalifiseer vir ’n lening van  R791 000 (5) R 791 000 gegewe die kondisies [16] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/ Wiskunde V1 12 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum QUESTION/VRAAG 7 PENALISE ONLY ONCE for incorrect notation in this question. 7.1 f ( x  h)  ( x  h) 2  5  ( x 2  2 xh  h 2 )  5  x 2  2 xh  h 2  5  f ( x  h)  f ( x)  x 2  2 xh  h 2  5  x 2  5  simplifying  2 xh  h 2 f ( x  h)  f ( x ) f ( x)  lim h 0 h formula 2 xh  h 2  lim subst. into formula h 0 h h( 2 x  h)  lim factorisation h 0 h  lim 2 x  h  h 0 answer  2x (5) OR/OF f ( x  h)  f ( x ) f ( x)  lim formula h 0 h  lim  x  2 xh  h 2  5  x 2  5 2  subst. into formula h 0 h 2 xh  h 2  lim simplifying h 0 h h( 2 x  h) factorisation  lim h 0 h  lim 2 x  h  h 0 answer  2x (5) 2x 7.2 g ( x)  5 x 2  x3  5 x 2  2 x 2  5x 2  2 x 2 g /  x   10 x  4 x 3  10 x 4 4  10 x  3  4 x 3 or x x3 (3) Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/ Wiskunde V1 13 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum 7.3 h( x)  ax 2 , x  0 h 1 : x  ay 2 y 0 x x y y a a h 1 8  8 8 a  a h  x   2ax / h / 4   2a 4   h 4  8a /  8a 8 8  8a   8a a a 8 64a 2  a 1 a3  1  a3  8 8 1 a 1 a  2 2 (6) [14] QUESTION/VRAAG 8 8.1 f / ( x)  0 6 x 2  10 x  4  0  derivative 3x 2  5 x  2  0  derivative = 0 3x  2x  1  0  factors 2 x or x 1  x-values 3 3 2 2 2 2 y  2   5   4  y  21  51  41 3 2 3 3 3 28  y-values y or y 1 27  2 28  Turning points are  ;  and (1 ; 1)  3 27  (5) Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/ Wiskunde V1 14 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum 8.2 2 x3  5 x 2  4 x  0  x 2 x2  5x  4  0    x 2 x 2  5x  4  0 5  25  42 4  x = 0 x  0 or x  4 5 7 5 7   4 4 No real roots / Geen reële wortels (3) OR / OF 2 x 3  5x 2  4 x  0   x 2 x 2  5x  4  0    x 2 x 2  5x  4  0 x=0 x0 or b 2  4ac  25  424 f  7  0  b 2  4ac  0 No real roots / Geen reële wortels (3) 8.3 f ( x)  2 x 3  5 x 2  4 x   x 2 x2  5x  4  0 y    2 ; 28  f   3 27  (1 ; 1) (0 ; 0) x turning points 0 shape (3) Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/ Wiskunde V1 15 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum 8.4 f ( x)  2 x 3  5 x 2  4 x f / ( x)  6 x 2  10 x  4 f // ( x)  12 x  10  12 x  10 f ( x)  0 //  f // ( x)  0 12 x  10  0 5  answer x 6 (3) OR b Point of inflection: x 3a x  5  x  5 32  32 5 x 5 6  x 5 6 The function is concave up for x  since a  0  f // ( x)  0 6 (3) OR 2 2 1 1  x 3 Point of inflection: x 3 2 2 5 x  x 5 6 6 5  f ( x)  0 // The function is concave up for x  since a  0 6 (3) [14] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/ Wiskunde V1 16 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum QUESTION/VRAAG 9 9. Length of one side of the square / lengte van sy van vierkant x  x 4  4 Length of the rectangle / lengte van die reghoek : x 2l  x   6 6 5x 4 4 24  5 x  or 5x 2 8 6 l 4 2 24  5 x  2 8  x     x 2 x  24  5 x  4 A     4 4 8  x  24  5 x  x 2 24 x  5 x 2      4 8  16 32 24 x  3 x 2  32 24 x  3 x 2 A . 32 dA  0 dA dx For minimum area / Vir min imum oppervlaktte 0 24  6 x dx  dA 24  6 x 32  dx 32  x4 6 x  24 x4 (7) [7] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/ Wiskunde V1 17 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum QUESTION/VRAAG 10 10.1.1 PS and T   PS  PT  1 1     PT   PS and T  PS PT 6 4  PT   2 PT   2 3 3 (2) 10.1.2 PS or T   PS  PT   PS and T  1 2 1    1 2 1 4 3 6      3 4 3 6  3 4  4 (2) 10.2.1 5! 5 = 120  5! or 120 (2) 10.2.2 55  3125  5 5 or 3 125 (1) 10.3 nE   5!  2!  2!  5! nS  7!  2!  2! 5!  2!  2! 5!  2!  2! PE    7! 7! 2 2   21 21 (4) [11] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1/ Wiskunde V1 18 DBE/Feb.–Mar./Feb.–Mrt. 2017 NSC/NSS – Memorandum QUESTION/VRAAG 11 11 0,85 W P(F and W) = 0,595 F 0,7 0,15 L P(F and L) = 0,105 0,55 W P(not F and W) = 0,165 0,3  0,3 not F 0,45 L P(not F and L) = 0135 P(Win) = P(F and W) + P(not F and W)  P(F and W) = 0,7  0,85 = 0,595 = 0,7  0,85 + 0,3  0,45  P(not F and W) = 0,3  0,45 = 0,595 + 0,165 = 0,165  0,595 + 0,165 = 0,76 19  0,76 / 76% / = 76% 25 (5) 19 = 25 [5] TOTAL/TOTAAL: 150 Copyright reserved/Kopieregvoorbehou