# Mathematics Grade 12 Paper 1 NSC June 2017 Memo > Grade 12 Mathematics Paper 1 memorandum from the NSC June Exam 2017, 20 pages. Read online or get the free PDF download, question paper linked when available. - Page: https://www.hlayiso.com/library/mathematics-p1-may-june-2017-memo-afr-eng/3c63f2c2-50f3-4098-93bc-7fb8b5283ef8 - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 12 - Document type: Memorandum - Year: 2017 - Exam period: NSC June Exam - Paper: 1 - Curriculum: CAPS - Pages: 20 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Mathematics](https://www.hlayiso.com/library/grade-12/mathematics) › [Mathematics Grade 12 Paper 1 NSC June 2017 Memo](https://www.hlayiso.com/library/mathematics-p1-may-june-2017-memo-afr-eng/3c63f2c2-50f3-4098-93bc-7fb8b5283ef8) ## About this document Mathematics · Grade 12 · NSC June Exam · 2017. Memorandum, 20 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com SENIOR CERTIFICATE EXAMNATIONS SENIORSERTIFIKAAT-EKSAMEN MATHEMATICS P1/WISKUNDE V1 2017 MARKING GUIDELINES/NASIENRIGLYNE MARKS: 150 PUNTE: 150 These marking guidelines consist of 20 pages. Hierdie nasienriglyne bestaan uit 20 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 2 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 2 DBE/2017 SCE/SSE – 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 3x 2 + 10 x + 6 = 0 − b ± b 2 − 4ac x= 2a  substitution into correct formula − 10 ± (10) − 4(3)(6) 2 = 2(3)  x = −2,55 x = −2,55 or x = − 0,78  x = − 0,78 (3) OR/OF 100 for adding on 10 100 100 36 x2 + x + = −2 + 3 36 36 both sides 2  5 7 x +  =  3 9 5 7 x+ =± 3 3 −5± 7  x = −2,55 x= 3  x = − 0,78 x = −0,78 or x = −2,55 (3) 1.1.2 2 6 x − 15 = x + 1 6 x 2 − 15 = (x + 1) 2  concept of squaring 2 2 6 x − 15 = x + 2 x + 1 both sides 5 x 2 − 2 x − 16 = 0  standard form (5 x + 8)(x − 2) = 0 (accurate) 8  factors x=− or x = 2  both answers 5 ∴x = 2  correct selection (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 3 DBE/2017 SCE/SSE – Marking guidelines/Nasienriglyne 1.1.3 x 2 + 2 x − 24 ≥ 0 (x + 6)(x − 4) ≥ 0  factors y • • ‒6 4 x OR + –6 – 4 +  x ≤ −6 or x ≥ 4 x ≤ −6 or x ≥ 4 (3) 1.2 y = −5 x + 3  y subject of formula 3 x − 2 x(− 5 x + 3) = (− 5 x + 3) − 105 2 2  substitution 2 2 2 3x + 10 x − 6 x = 25 x − 30 x + 9 − 105  simplification − 12 x 2 + 24 x + 96 = 0 x 2 − 2x − 8 = 0 (x − 4)(x + 2) = 0  factors x = −2 or x = 4  values of x  values of y y = 13 or y = −17 (6) OR/OF 3− y  x subject of formula x= 5 2 3− y  3− y  2  substitution 3  − 2 y  = y − 105  5   5   9 − 6y + y  2 3− y  3  − 2 y 2  = y − 105  25   5  27 − 18 y + 3 y 2 − 30 y + 10 y 2 = 25 y 2 − 2625  simplification 12 y 2 + 48 y − 2652 = 0 y 2 + 4 y − 221 = 0 ( y − 13)( y + 17 ) = 0  factors y = −17 or y = 13  values of y 3 + 17 3 − 13 x= or x = 5 5 x=4 or x = −2  values of x (6) 1.3.1 2 p − 48 p − 49 = 0  factors ( p − 49)( p + 1) = 0  p = −1 p = −1 or p = 49  p = 49 (3) 1.3.2 x 7 = −1 or x 7 = 49 x  7 = −1 or 7 = 49x no solution x=2  no solution  x=2 (3) [23] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 4 DBE/2017 SCE/SSE – Marking guidelines/Nasienriglyne QUESTION/VRAAG 2 2.1.1 3; 2; k ; ... 2 2 r=  r= / 0,67 3 3 (1) 2.1.2 T3 r= T2 T3 = r × T2 2 2  ×2 = ×2 3 3 4 = 3 4 4 Thus k =  / 1,34 3 3 (2) 2.1.3 Tn = a.r n −1 n −1 n −1 128 2 128 2  = 3×   = 3×   729 3 729 3 n −1 n −1 2 128 2 128   =   = 3 2187 3 2187 n −1 7 2 2 2 7   =    3 3 3 n −1 = 7 n=8 answer OR / OF OR / OF n −1 Tn = a.r n −1 128 2 128 n −1 = 3×   2 729 3  = 3×   729 3 n −1 2 128 n −1   = 2 128 3 2187   = 3 2187 128 n − 1 = log 2 128 3 2189  n − 1 = log 2 n −1 = 7 3 2189 answer n=8 (4) 2.2.1 Tn = a + (n − 1)d  substitution of n, T18 = 100 + (18 − 1)(150 ) a and d into AS  2 650 = R 2 650 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 5 DBE/2017 SCE/SSE – Marking guidelines/Nasienriglyne 2.2.2 n Sn = [2a + (n − 1)d ] 2 n 30 500 = [2(100 ) + (n − 1)(150 )] substitute 30 500, 2 a and d into sum 61 000 = n(150n + 50 ) formula for AS 61 000 = 150n 2 + 50n 3n 2 + n − 1 220 = 0  simplification  factors or (3n + 61)(n − 20) = 0 quad formula 61 n=− or n = 20  n = 20 3 N/A x = 100 + (20 − 1)(150)  substitution Tn of = R 2 950 AS  2 950 (6) [15] QUESTION/VRAAG 3 3.1 First differences: 17; 15 17; 15 Second difference: –2 Tn = an 2 + bn + c second difference − 2 a= = = −1 value of a 2 2 3a + b = 17 3(− 1) + b = 17 b = 20  value of b a+b+c = 0 − 1 + 20 + c = 0 c = −19 value of c Tn = −n 2 + 20n − 19 OR / OF First differences: 17; 15 17; 15 Tn = T1 + (n − 1)d1 + (n − 1)(n − 2) d 2 2 value of a = (0) + (n − 1)(17) + (n − 1)(n − 2) (− 2) 2  value of b 2 = 17 n − 17 − n + 3n − 2 value of c = −n 2 + 20n − 19 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 6 DBE/2017 SCE/SSE – Marking guidelines/Nasienriglyne 3.2 56 = −n 2 + 20n − 19  Tn = 56 2 n − 20n + 75 = 0 (n − 15)(n − 5) = 0 factors n = 5 or n = 15 both answers (3) 3.3 10 15 ∑T − ∑T n =5 n n =11 n = T5 + T6 + T7 + T8 + T9 + T10 − T11 − T12 − T13 − T14 − T15 = (T5 − T15 ) + (T6 − T14 ) + ... + (T9 − T13 ) + T10 = T10  symmetry of terms because by symmetry T5 = T15 ; T6 = T14 ........... T10 = −(10 ) + 20(10 ) − 19 2  T10 = 81  81 (4) OR/OF T5 T 10 T 15 0; 17; 32; 45; 56; 65; 72; 77; 80; 81; 80; 77; 72; 65; 56  writing out the symmetry of Hence, terms 10 15 ∑ Tn − ∑ Tn n =5 n =11  = (56 + 65 + 72 + 77 + 80 + 81) − (80 + 77 + 72 + 65 + 56) 56 + 65 + 72 + 77 + 80 + 81 = 81  80 + 77 + 72 + 65 + 56  81 (4) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 7 DBE/2017 SCE/SSE – Marking guidelines/Nasienriglyne QUESTION/VRAAG 4 4.1 A (4; 3) (4; 3) (1) 4.2 6 y= +3 x=0 −4 3 = 3 2 y= 2  3 (2) B 0 ;   2 4.3 6 0= +3 y = 0 x−4 6 −3 = x−4 − 3(x − 4 ) = 6 − 3 x + 12 = 6 x=2 x=2 (2) C(2 ; 0) 4.4 3 3 0− 0− Average gradient = 2  2 2−0 2−0 3 3 =− − 4 4 (2) 4.5 y = −x + 7  m = −1  y = −x + 7 OR/OF OR/OF m = −1  m = −1 ∴ y − 3 = −( x − 4)  y = −x + 7 y = −x + 7 (2) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 8 DBE/2017 SCE/SSE – Marking guidelines/Nasienriglyne Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 9 DBE/2017 SCE/SSE – Marking guidelines/Nasienriglyne QUESTION/VRAAG 5 5.1 f f:  x-intercepts y-intercept ( ̶ 2; 0) 0 (7; 0)  shape TP g:  x-intercept and y-intercept  shape (0; ̶ 14) (6) g (2½; ̶ 20¼) 5.2 1 1 y = −20  y = −20 / 4 4 81 − 4 (2) 5.3 1 1 − 20 < k < −14  − 20 < k 4 4  k < −14 (2) 5.4 Reflecting in the x-axis: y = −2 x + 14  y = −2 x + 14 y = −2( x + 7 ) + 14 Shifting 7 units to the left: = −2 x − 14 + 14  y = −2 x = −2 x (2) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 10 DBE/2017 SCE/SSE – Marking guidelines/Nasienriglyne QUESTION/VRAAG 6 6.1 f : y = bx  interchange x and y f −1 : x = b y  answer y = logb x (2) 6.2 y = x  answer (1) 6.3 P(0; 1)  answer (1) 6.4 T(1; 0)  coordinates of T y = mx + c y = −x + 1  y = −x +1 (2) 6.5 At point R, PT and OR intersect: − x +1 = x 2x = 1  − x +1 = x 1 x= 1 2  x= 2 1 1 y=  y= 2 2 1 1 Substitute  ;  into the equation of f: 2 2 y = bx 1 1 = b2 2 2  substitution 1 1 b=  = 1 2 4  b= 4 OR/OF (5) At point R, PT and OR intersect: − x +1= x 2x = 1  − x +1 = x 1 x= 1 2  x= 1 2 y= 1 2  y= 2 1 1 Substitute  ;  into the equation of g: 2 2 y = logb x 1 1 = logb   2 2  substitution 1 1 b = 2 2 2 1 1 1  b= b=  = 4 2 4 (5) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 11 DBE/2017 SCE/SSE – Marking guidelines/Nasienriglyne QUESTION/VRAAG 7 7.1 A = P(1 − i ) n 331527 = 500000(1 − i ) 3  substitution of A, P & (1 − i )3 = 331527 n in correct formula 500000 331527 331527 1− i = 3  1− i = 3 or 500000 500000 i = 0,12800... 1 − i = 3 0,663054 = 12,8%  answer (3) 7.2 [ x 1 − (1 + i ) P= −n ] i 0,24 1   0,24  − n  i = / 0,02 / 19001 − 1 +   12 50   12    substitution of 46 000 = 0,24 P , x and i in correct 12 formula −n 46  0,24  = 1 − 1 +  95  12  −n  0,24  49 1 +  =  12  95 49  0,24  49 n = − log 0 , 24  OR/OF − n log1 +  = log  1+  12   95  12  95 = 33,43276544... months  33,43 It will take him 34 months to pay back the loan.  answer (4) 7.3 x[(1 + i ) − 1] n F= i 0,075  0,075  4×6,5  i= / 0,01875 35001 +  − 1 4  4    n = 4 × 6,5 = 26 = 0,075 substitution into correct 4 formula = R 115 902,69  115 902,69 A = P(1 + i ) n 4×3, 5  0,075  substitution into correct = 115 902,691 +  formula  4   150 328,12 = R 150 328,12 (6) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 12 DBE/2017 SCE/SSE – Marking guidelines/Nasienriglyne QUESTION/VRAAG 8 8.1 f (x + h ) = 3 − 2(x + h ) 2 = 3 − 2 x 2 − 4 xh − 2h 2 f (x + h ) − f (x ) = 3 − 2 x 2 − 4 xh − 2h 2 − 3 + 2 x 2  3 − 2 x 2 − 4 xh − 2h 2 = −4 xh − 2h 2  − 4 xh − 2h 2 f (x + h ) − f (x ) f ′(x ) = lim h →0 h  f ′(x ) = lim f (x + h ) − f (x ) h →0 h − 4 xh − 2h 2 = lim h →0 h = lim h(−4 x − 2h)  lim(− 4 x − 2h ) h →0 h →0 h –4x (5) = lim(− 4 x − 2h ) h →0 = −4 x OR/OF f (x + h ) − f (x )  f ′(x ) = lim h →0 h f (x + h ) − f (x ) f ′( x ) = lim h →0 h = lim ( 3 − 2( x + h ) − 3 − 2 x 2 2 )  3 − 2 x 2 − 4 xh − 2h 2 h →0 h 3 − 2 x − 4 xh − 2h 2 − 3 + 2 x 2 2  − 4 xh − 2h 2 = lim h →0 h 2 − 4 xh − 2h = lim h →0 h  lim(− 4 x − 2h ) h →0 h(−4 x − 2h) –4x (5) = lim h →0 h = lim(− 4 x − 2h ) h →0 = −4 x 12 x 2 + 2 x + 1 8.2 y= 6x 1 1 = 2x + + 12 x 2 2 x 1 3 6x  + + 6x 6x 6x 1 1 = 2 x + + x −1 1 3 6  x −1 6 dy 1 = 2 − x −2 dx 6 2 1 1 = 2− 2  − x −2 6x 6 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 13 DBE/2017 SCE/SSE – Marking guidelines/Nasienriglyne 8.3 y = x 3 + bx 2 + cx − 4 y / = 3 x 2 + 2bx + c  y / = 3 x 2 + 2bx + c y // = 6 x + 2b  y // = 6 x + 2b At point of inflection: y // = 6 x + 2b = 0  y // = 0 Substitute x = 2: 6(2) + 2b = 0  sub x = 2 into y // = 0 2b = −12 b = −6  value of b 3 2 y = x − 6 x + cx − 4 Substitute (2 ; 4 ) : 4 = 23 − 6(2 ) + c(2 ) − 4 2  substitute (2; 4 ) 2c = 24 c = 12  value of c y = x 3 − 6 x 2 + 12 x − 4 (7) [16] QUESTION/VRAAG 9 9.1 (0 ; 1)  answer (1) 9.2 3 2 f ( x) = x − x − x + 1 f ( x) = x 2 ( x − 1) − ( x − 1)  (x – 1) ( f ( x) = ( x − 1) x − 1 2 ) (  x2 −1 ) f ( x) = ( x − 1)( x − 1)( x + 1)  (x − 1)(x − 1)(x + 1) f ( x) = 0 (x − 1)(x − 1)(x + 1) = 0 (–1; 0) x-intercepts: (–1; 0); (1; 0) (1; 0) (5) OR f ( x) = x 3 − x 2 − x + 1  (x – 1) ( f ( x) = ( x − 1) x − 1 2 ) (  x2 −1 ) f ( x) = ( x − 1)(x − 1)(x + 1)  (x − 1)(x − 1)(x + 1) f ( x) = 0 (x − 1)(x − 1)(x + 1) = 0 (–1; 0) x-intercepts: (–1; 0); (1; 0) (1; 0) (5) OR Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 14 DBE/2017 SCE/SSE – Marking guidelines/Nasienriglyne f ( x) = x 3 − x 2 − x + 1  (x + 1) ( f ( x) = (x + 1) x 2 − 2 x + 1 ) (  x2 − 2x + 1 ) f ( x) = (x + 1)(x − 1)(x − 1)  (x − 1)(x − 1)(x + 1) f ( x) = 0 (x − 1)(x − 1)(x + 1) = 0 (–1; 0) x-intercepts: (–1; 0); (1; 0) (1; 0) (5) 9.3 3 f ( x) = x − x − x + 1 2 f / (x ) = 3x 2 − 2 x − 1  f / (x ) = 3x 2 − 2 x − 1  f / (x ) = 0 f / (x ) = 0 (3 x + 1)( x − 1) = 0 factorisation 1 x =− or x = 1 3 x value 32 x value y= y=0 32 27 y = 27  1 32   − ;  (1 ; 0 ) (6)  3 27  9.4  1 32  − ;   3 27  (0 ; 1) y– and x–intercepts ( ̶ 1 ; 0) 0 (1 ; 0) shape turning points (3) 9.5 f ′( x) < 0 1 1 − < x <1  x>− 3 3  x <1 OR/OF (2) 1 1 (– ; 1)  (– 3 3  1) (2) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 15 DBE/2017 SCE/SSE – Marking guidelines/Nasienriglyne QUESTION/VRAAG 10 1 1 10.1 60 = 2b + 2r + (2πr )  60 = 2b + 2r + (2πr ) 2 2 2b = 60 − 2r − πr 1 1  b = 30 − r − πr b = 30 − r − πr 2 2 (2) 10.2 Area = area of rectangle + area of semicircle 1 A(r ) = length × breadth + (area of circle) 2  1  1  1  (2r ) 30 − r − πr   ( ) = (2r ) 30 − r − πr  + πr 2 2  2  2   1 1 ( )  πr 2 2 = 60r − 2r 2 − πr 2 + πr 2 2 1 1 = 60r − 2r 2 − πr 2  60r − 2r 2 − πr 2 2 2  1  = 60r −  2 + π r 2  2  For a maximum, A / (r ) = 0  A / (r ) = 0  1  60 − 2 2 + π r = 0 1  2   60 − 2 2 + π r  2  60 − (4 + π )r = 0 60 r= 4+π = 8,40 m answer (6) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 16 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 16 DBE/2017 SCE/SSE – Marking guidelines/Nasienriglyne QUESTION/VRAAG 11 11.1 8! 8! 8× 7× 6×5× 4 or  8× 7 × 6× 5× 4 / 3! 3! = 6720  6720 (2) 11.2 P(A and B) = P(A )× P(B) = 0,4 × 0,35  0,4 × 0,35 = 0,14  0,14 P(A or B) = P(A ) + P(B) − P(A and B) = 0,4 + 0,35 − 0,14  substitution = 0,61  answer (4) pass 0,35 School A 0,65 fail 0,2 pass 0,65 0,3 School B 0,35 fail 0,5 pass 0,9 School C 0,1 fail 11.3.1 100 % − 20% 1 − 0,2  100 % − 20% or 1 – 0,2 or/of 80% or 0,8 = 80% = 0,8 OR/OF 30% + 50% or 0,3 + 0,5 30% + 50% = 80% or/of 0,3 + 0,5 = 0,8  80% or 0,8 (2) 11.3.2 0,3 × 0,35 = 0,105  0,3 = 10,5%  0,35  0,105 = 10,5% (3) 11.3.3 (0,2 × 0,35) + (0,3 × 0,65) + (0,5 × 0,9)  0,2 × 0,35 = 0,715  0,3 × 0,65  0,5 × 0,9 = 71,5% answer (4) [15] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 17 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 17 DBE/2017 SCE/SSE – Marking guidelines/Nasienriglyne SENIOR CERTIFICATE EXAMINATIONS/ SENIORSERTIFIKAAT-EKSAMEN MATHEMATICS P1/WISKUNDE V1 2016 MARKING GUIDELINE (ADDENDUM) MARKS/PUNTE: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 18 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 18 DBE/2017 SCE/SSE – 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, sien slegs die EERSTE poging na. • Volgehoue akkuraatheid is op ALLE aspekte van die memorandum van toepassing. Once a candidate has reached 2 errors related to marks: stop marking. QUESTION/VRAAG 1 1.1.1 • incorrect rounding 2/3 – only rounding penalization • use of calculator 2/3 – this is where use of calculator for factors get used • answer in surd form 2/3 ( at least simplified under square root) 1.1.2 • CA mark only if quadratic equation • check answers • if 6 x 2 − 15 = x + 1 breakdown 0/3 • both answer must be seen before selection if no factors are shown • if in the context of their incorrect sum, both of the answers are NA, both need to be shown as NA 1.1.3 ( x + 6)( x − 4) ≥ 0 • x ≥ 4 or / and x ≥ −6 , award 1/3 marks (factors) • x ≤ 4 or / and x ≤ −6 , award 1/3 marks (factors) • − 6 ≤ x ≤ 4 , award 1/3 marks (factors) • x ≤ −6 and x ≥ 4 , award 2/3 marks • equal is left out: –1 Answer only 3/3 1.2 NB: At the second error related to a mark (two skills) – no further marking. If incorrect algebra leads to the equation being linear: max 2/6 These marks will be the changing of the formula and the substitution mark. 1.3.2 CA from 1.3.1 • If 7 x = p can award 1 mark for the concept • If answer x = 2 only 2/3 QUESTION/VRAAG 2 2.1.2 CA from 2.1.1 Answer only 2/2 2.1.3 Answer only 1/4 • If n = 7 2/4 • Incorrect working that leads to use of logs and an not a natural number max 2/4 2.2.1 Answer only 2/2 2.2.2 • Answer only 1/6 Copyright reserved/Kopiereg voorbehou ### Page 19 Downloaded from hlayiso.com Mathematics/P1/Wiskunde/V1 19 DBE/2017 SCE/SSE – Marking guidelines/Nasienriglyne • S n has to equal 30 500 otherwise a BD QUESTION/VRAAG 3 3.2 n = 5 only 1/3 3.3 Answer only 1/4 QUESTION/VRAAG 4 4.1 x = 4 ; y = 3 1/1 4.3 y = 0 can be implied 4.4 CA from 4.2 and 4.3 QUESTION/VRAAG 5 5.1 Only working out, but no sketch max 4/6 – loose shape mark per graph not sketched 5.2 CA from turning point in 5.1 5.3 CA from sketch (TP to y-intercept) 5.4 Answer only 2/2 QUESTION/VRAAG 6 6.1 Answer only 2/2 If answer not in terms of b max 1/2 6.3 Coordinate from not needed QUESTION/VRAAG 7 Penalise candidates a maximum of one mark (overall) for notation error in 7.1 and 7.2 7.1 • Interchange A and P – breakdown 0/3 • Wrong formula 0/3 • Early rounding: answer is 12,93% – 2/3 • i and n incorrect – learner can still get the substitution mark 1/6 7.3 • If quarterly is taken as monthly consistently in both parts 5/6 A = P(1 + i ) n 4×10  0,075  • If 10 years is used: = 115 902,691 +  5/6  4  = R 243 667,94 Copyright reserved/Kopiereg voorbehou Text truncated here. 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