# Mathematics Grade 12 Paper 1 GP Prelim 2023 Memo (English) > Grade 12 Mathematics Paper 1 memorandum from the GP Prelim 2023 (English), 22 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/mathematics-p1-english-memo/d9d63927-d3d4-4594-bd07-307226322b0c - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 12 - Language: English - Document type: Memorandum - Year: 2023 - Exam period: GP Prelim - Paper: 1 - Curriculum: CAPS - Pages: 22 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 GP Prelim 2023 Memo (English)](https://www.hlayiso.com/library/mathematics-p1-english-memo/d9d63927-d3d4-4594-bd07-307226322b0c) ## About this document Mathematics · Grade 12 · GP Prelim · 2023 · English. Memorandum, 22 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Matching paper and memo - [Mathematics P1 (English) QP](https://www.hlayiso.com/library/mathematics-p1-english-qp/11462bfe-d0db-4e5e-8cc4-7aa8f32251ac) — QP · 2023 · Eng ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com AMENDMENT TO MARKING GUIDELINES PREPARATORY EXAMINATION / VOORBEREIDENDE EKSAMEN 2023 FOR ATTENTION / VIR AANDAG: THE CHIEF INVIGILATOR / DIE HOOF TOESIGHOUER SUBJECT / VAK MATHEMATICS / WISKUNDE SUBJECT CODE / VAKKODE 10611 PAPER / VRAESTEL 1 DATE OF EXAMINATION / 8 SEPTEMBER 2023 DATUM VAN EKSAMEN The MATHEMATICS / WISKUNDE (Paper / Vraestel 1) written on 8 September 2023 has reference. It has come to our attention that there was an anomaly in the question paper which rendered Question / Vraag 9.2 ambiguous. To ensure that your candidates are not disadvantaged and prejudiced in way, you are advised to please ask Educators to ignore Question / Vraag 9.2 when marking. In other words, the paper must be marked out of a total of 145 instead of 150 and then the learners’ marks must be converted back to a mark out of 150. E.g. Should a learner have attained 65/145, then that mark is recalculated as 67/150. Use the a b formula: x 100 = b. Then, x 150 = c 145 100 c is the mark that is entered into SASAMS out of 150. DIRECTOR: EXAMINATIONS MANAGEMENT 9 September 2023 ### Page 2 Downloaded from hlayiso.com PREPARATORY EXAMINATION 2023 MARKING GUIDELINES MATHEMATICS (PAPER 1) (10611) 21 pages 2 ### Page 3 Downloaded from hlayiso.com 10611/23 INSTRUCTIONS AND INFORMATION: A – Accuracy CA – Continued Accuracy S – Statement R – Reason S and R – Statement and Reason NOTE: • If a candidate answered a question TWICE, mark only the FIRST attempt. • If a candidate crossed OUT an answer and did not redo it, mark the crossed-out answer. • Consistent accuracy applies to ALL aspects of the marking guidelines. • It is UNACCEPTABLE to assume values/answers in order to solve a question. 3 ### Page 4 Downloaded from hlayiso.com 10611/23 QUESTION 1 1.1 1.1.1 (2 x + 1) 2 − 4 = 0 4x2 + 4 x + 1 − 4 = 0 4x2 + 4 x − 3 = 0 ✓ standard form (2 x + 3)(2 x − 1) = 0 ✓ factors 3 1 ✓ answers x=− or x = 2 2 OR (2 x + 1) 2 − 4 = 0 2x +1 =  4 ✓ find root 2 x + 1 = 2 2 x = 2 − 1 ✓ simplification 2 x = 2 − 1 or 2 x = −2 − 1 2x = 1 or 2 x = −3 1 3 x= or x=− ✓ answers 2 2 NOTE: In option 2, the simplification mark must only be awarded if the candidate states  2 or demonstrates it later in the solution. (3) 1.1.2 4 x 2 − 11 = −12 x 4 x 2 + 12 x − 11 = 0 −12  (12) 2 − 4(4)( −11) ✓ correct substitution x= 2(4) x = −3, 74 or x = 0, 74 ✓✓ answers NOTE: Penalise 1 mark for rounding in this question only. Candidate must show the substitution to obtain full marks. (3) 1.1.3 15 x − 4  9 x 2 −9 x 2 + 15 x − 4  0 1 4  9 x 2 − 15 x + 4  0 ✓ standard form 3 3 (3x − 1)(3x − 4)  0 x ✓ factors 1 4 x  or x  3 3 ✓✓ answers NOTE: Penalise 1 mark for use of AND instead of OR (4) 4 ### Page 5 Downloaded from hlayiso.com 10611/23 1.1.4 2x − 2 − 7 − 2x = 1 ( 2 x − 2) 2 = ( 7 − 2 x + 1) 2 ✓ isolate surd and 2x − 2 = 7 − 2x + 2 7 − 2x +1 square both sides 4 x − 10 = 2 7 − 2 x (2 x − 5) 2 = ( 7 − 2 x ) 2 4 x 2 − 20 x + 25 = 7 − 2 x ✓ simplification 4 x 2 − 18 x + 18 = 0  2x2 − 9x + 9 = 0 ✓ standard form ( x − 3)(2 x − 3) = 0 ✓ factors 3  x = 3 or x = ✓ x=3 2 and rejection 3 reject : x  2 (5) 1.2 a 2b 2 − 2ab − 8 = 0 and log 2 ( a + 5) = 3 log 2 (a + 5) = 3 23 = a + 5 ✓ exponential form 8 = a+5 a=3 ✓ value of a  a 2b 2 − 2ab − 8 = 0 (3) 2 b 2 − 2(3)b − 8 = 0 ✓ substitution 9b 2 − 6b − 8 = 0 (3b − 4)(3b + 2) = 0 ✓ factors 4 2 b= or b = − ✓ both answers for b 3 3 OR OR a 2b 2 − 2ab − 8 = 0 and log 2 ( a + 5) = 3 log 2 (a + 5) = 3 23 = a + 5 ✓ exponential form 8 = a+5 a=3 ✓ value of a a 2b 2 − 2ab − 8 = 0 (ab − 4)(ab + 2) = 0 ✓ factors ab = 4 or ab = −2  3b = 4 3b = −2 ✓ substitution 4 2 b= or b = − ✓ both answers for b 3 3 (5) 5 ### Page 6 Downloaded from hlayiso.com 10611/23 1.3 x+2 p= 16 − x 2 x + 2  0  x  −2(1) ✓ inequality (1) 16 − x  0 2 ( 4 + x )( 4 − x )  0 ✓ factors  −4  x  4( 2) ✓ inequality (2) From (1) and (2) −2 x  4 ✓ answer (4) [24] QUESTION 2 2.1 2.1.1 1 − p; 2 p − 3; p + 5 d = T2 − T1 = T3 − T2 ✓ substitution  (2 p − 3) − (1 − p) = ( p + 5) − (2 p − 3) 2 p − 3 −1+ p = p + 5 − 2 p + 3 3p − 4 = − p +8 4 p = 12 p=3 ✓ answer (2) 2.1.2 (a) T1 = 1 − p T1 = 1 − 3 T1 = −2 ✓ answer (1) (b) T2 = 2 p − 3 T2 = 2(3) − 3  T2 = 3  d = T2 − T1 d = 3 − (−2) d =5 ✓ answer (1) 2.1.3 1 − p; 2 p − 3; p + 5;... = 1 − 3; 2(3) − 3; 3 + 5;... ✓ correct terms = −2; 3; 8; 13; 18;... All the terms except T1 end in either 3 or 8 while perfect squares end on 1; 4; 9; 6; 5; 0. ✓ explanation OR 6 ### Page 7 Downloaded from hlayiso.com 10611/23 T1 is not a perfect square. ✓✓ explanation OR Tn = 5n − 7 ✓✓ Tn = 5n − 7 (2)  not a perfect square 7 ### Page 8 Downloaded from hlayiso.com 10611/23 2.2 2.2.1 − 3; − 2; − 3; − 6 ; − 11; … First differences: +1; − 1; − 3; − 5 ✓ first differences Tn = a + (n − 1)d ✓ substitution Tn = 1 + (n − 1)(−2) Tn = 1 − 2n + 2 Tn = −2n + 3 ✓ answer (3) 2.2.2 Tn = −2n + 3 T35 = −2(35) + 3 T35 = −67 ✓ answer (1) 2.2.3 -3 -2 -3 -6 -11 1 -1 -3 -5 1st differences -2 -2 -2 2nd differences ✓ second differences 2a = −2 3a + b = 1 a + b + c = −3 ✓ a = −1 a = −1 3(−1) + b = 1 − 1 + 4 + c = −3 ✓ b=4 b=4 c = −6 ✓ c = −6 Tn = −n + 4n − 6 2 OR Tn = an2 + bn + c ✓ second differences but 2a = −2  a = −1  Tn = −n 2 + bn + c ✓ a = −1 T1 = −(1) 2 + b(1) + c − 3 = −1 + b + c ✓ b=4  −2 = b + c  (1) ✓ c = −6 T2 = −(2) 2 + b(2) + c − 2 = −4 + 2b + c  2 = 2b + c  (2) (2 − 1) b = 4 sub(1) − 2 = 4 + c  c = −6  Tn = −n 2 + 4n − 6 (4) 8 ### Page 9 Downloaded from hlayiso.com 10611/23 2.2.4 Tn = −n 2 + 4n − 6 T '(n) = −2n + 4 ✓ method 0 = −2n + 4 n=2 ✓ Tn (max) T (2) = −2  NO positive terms. OR Tn = −n 2 + 4n − 6 Tn = −[n 2 − 4n + 2 2 − 4 + 6] ✓ method Tn = −[( n − 2) 2 + 2] Tn = −(n − 2) 2 − 2 Tn (max) = −2 ✓ Tn (max)  NO positive terms. OR Max is given as -2 and is negative ✓✓ Max = -2 (2) [16] 9 ### Page 10 Downloaded from hlayiso.com 10611/23 QUESTION 3 3.1 S n = 4n 2 + 1 S 6 = 4( 6) 2 + 1  S 6 = 145 ✓ value of S 6 S 5 = 4(5) + 1 2  S 5 = 101 ✓ value of S 5  T6 = S 6 − S 5  T6 = 145 − 101  T6 = 44 ✓ answer (3) 3.2 (4 x − 3) + (4 x − 3) 2 + (4 x − 3)3  r = 4x − 3 ✓ r = 4x − 3 −1  r  1; r0 −1  4 x − 3  1; 4 x − 3  0 ✓ −1  r  1 3  2  4x  4 x 4 1 3 ✓ answer   x  1; x 2 4 3 NOTE: No penalty if candidate does not state: x  4 (3) 5 3.3 2  (−1) . k k =3 k 2 2 2 = (−1) 3 . + (−1) 4 . + (−1) 5 . 3 4 5 ✓ expansion 2 1 2 =− + − 3 2 5 17 =− ✓ answer 30 NOTE: Answer only, full marks. (2) [8] 10 ### Page 11 Downloaded from hlayiso.com 10611/23 QUESTION 4 4.1 x  −3 ✓ answer (1) 4.2 y2 ✓ answer (1) 4.3 4.3.1 6 units ✓ answer NOTE: Accept 6 units right, but not 6 units left. (1) 4.3.2 3,5 units ✓ answer NOTE: Accept 3,5 units upwards, but not 3,5 units downwards. (1) 4.4 x = −3 ✓ answer 3 y=− ✓ answer 2 NOTE: Do not accept any equating in terms of p and q. (2) 4.5 6 0= − 1,5 x+3 3 6 = 2 x+3 3( x + 3) = 6(2) 3x + 9 = 12 3x = 3 x =1 ✓ answer (1;0) NOTE: The answer does NOT need to be given in coordinate form. (1) 11 ### Page 12 Downloaded from hlayiso.com 10611/23 4.6 1 2 ✓ asymptotes -3 -3 1 ✓ intercepts ✓ shape 3 − 2 (3) 4.7 y = −x + k 3  − = −(−3) + k ✓ substitute point 2 3 − = 3+ k 2 9 k = − 2 ✓ answer NOTE: Award FULL marks for Answer only (2) 4.8 x  −3 ✓ answer (1) 4.9 −6 3 y= + x+3 2 OR 6 3 ✓ answer y= + −x − 3 2 OR 6 3 y=− + x+3 2 (1) [14] 12 ### Page 13 Downloaded from hlayiso.com 10611/23 QUESTION 5 5.1  7 81  E ;  2 4  ✓ answer (1) 5.2  7  81 2 f ( x) = −  x −  +  2 4 2  7  81 f (1) = − 1 −  + = 14 ✓ value of f (1)  2 4 2  7  81 f (5) = −  5 −  + = 18 ✓ value of f (5)  2 4 f (5) − f (1)  AG = 5 −1 18 − 14 ✓ substitution = 4  AG = 1 ✓ answer (4) 5.3 7 81 f ( x ) = −( x − ) 2 + 2 4  f ( x) = − x + 7 x + 8 2 f ( x) = g ( x) ✓ equating − x 2 + 7 x + 8 = −3x + 24 − x 2 + 10 x − 16 = 0  x 2 − 10 x + 16 = 0 ( x − 2)( x − 8) = 0  x = 2 or x = 8 ✓ x-values  xD = 2 ✓ answers/selection (3) 5.4 ST = f ( x) − g ( x) ST = − x 2 + 7 x + 8 − (−3x + 24) ✓ method ST = − x 2 + 7 x + 8 + 3x − 24 ST = − x 2 + 10 x − 16 ✓ answer (2) 5.5 d (ST ) For max: =0 dx d ( ST ) = −2 x + 10 = 0 ✓ derivative = 0 dx 2 x = 10 x = 5 ✓ value for x ST = −(5) 2 + 10(5) − 16 ✓ answer max:  ST = 9 (3) [13] 13 ### Page 14 Downloaded from hlayiso.com 10611/23 QUESTION 6 6.1 y = 3x Inverse x = 3y  y = log 3 x ✓ answer (1) 6.2 y = f −1 ( x) = log 3 x x-intercept: log 3 x = 0  x = 30 x =1 y-intercept: None y = f −1 ( x − 2) = log 3 ( x − 2) x-intercept: log 3 ( x − 2) = 0  x − 2 = 30 x = 3 y-intercept: None ✓ x-intercept of f −1 ( x) = log 3 x ✓ shape f −1 ( x) = log 3 x 1 ✓ x-intercept of 3 f −1 ( x − 2) = log 3 ( x − 2) ✓ shape f −1 ( x − 2) = log 3 ( x − 2) NOTE: If the candidate does not draw the graph but calculates both x-intercepts, award 2 marks. If the graph crosses the asymptote, DO NOT award the shape mark. (4) 6.3 log 3 ( x − 2)  1 ✓ critical values 2  x  5 ✓ correct notation (2) [7] 14 ### Page 15 Downloaded from hlayiso.com 10611/23 QUESTION 7 7.1 A = P (1 + i ) n 6 000 000 = 5 700 000(1 + i ) 7 ✓ substitution into correct formula 6 000 000 = (1 + i ) 7 5 700 000 7 1,0526... = 1 + i  i = 1,00735 − 1  i = 0,00735 ✓ value of i rate: 0,74% ✓ answer (3) 7.2 7.2.1 x[1 − (1 + i) − n ] Pv = i   0, 06 − n  30000 1 − 1 +     12   ✓ substitution into correct formula 4000000 = 0, 06 0,06 12 ✓ i= 12  0, 06  400000    −n  12  = 1 − 1 + 0, 06    30000  12  −n 1  0, 06  ✓ simplification = 1 +  3  12  1 log 0,06  = −n 1+   3 12  ✓ correct use of logs n = 220, 27  she will make 220 withdrawals. ✓ answer of 220 (5) 15 ### Page 16 Downloaded from hlayiso.com 10611/23 7.2.2   0, 06 − n  20 000 1 − 1 +     12   4 000 000 = 0, 06 12 −n  0, 06  0 = 1 +  ✓ valid method  12  She can make any number (an infinite ✓ simplification number) of withdrawals. Her interest earned equals her withdrawal ✓ explanation amount. She will only be drawing the interest amount. OR OR A = P(1 + i) n 1  0.06  ✓ valid method A = 4 000 000 1 +   12  A = R4 020 000 ✓ answer She can make any number (an infinite number) of withdrawals. ✓ explanation Her interest earned equals her withdrawal amount. She will only be drawing the interest amount. (3) 7.3 A = P (1 + i ) n 0,15 18 A = 1 000(1 + ) 12 A = R1 250,58 ✓ value of A x[(1 + i ) n − 1] Fv = i 0,15 18 700[(1 + ) − 1] Fv = 12 0,15 ✓ substitution in correct formula 12 Fv = R14 032,33 ✓ value for Fv Amount = R1 250,58 + R14 032,33  Amount = R15 282,91 ✓ answer (4) [15] 16 ### Page 17 Downloaded from hlayiso.com 10611/23 QUESTION 8 8.1 f ( x) = 3x 2 − 6 f ( x + h) = 3( x + h) 2 − 6 f ( x + h) = 3( x 2 + 2 xh + h 2 ) − 6 f ( x + h) = 3 x 2 + 6 xh + 3h 2 − 6 ✓ f (x + h) f ( x + h) − f ( x ) f '( x) = lim h →0 h 3 x + 6 xh + 3h 2 − 6 − 3 x 2 + 6 2 ✓ substitution f '( x) = lim h →0 h 6 xh + 3h 2 f '( x) = lim h →0 h h(6 x + 3h) f '( x) = lim ✓ factorisation (may be implied) h →0 h f '( x) = lim(6 x + 3h) h →0 f '( x) = 6 x ✓ answer NOTE: Penalise 1 mark for notation error in this question only. Award ZERO marks for Answer only (4) 8.2 1 f ( x) = (2 x − ) 2 x 1 1 f ( x) = 4 x − 4. + 2 ✓ simplification x x 1 1 − − f ( x) = 4 x − 4 x 2 + x −2 ✓ f ( x) = 4 x − 4 x 2 + x −2 3 − f '( x) = 4 + 2 x 2 − 2 x −3 ✓4 3 − 2 OR ✓ 2 x 2 or 2 2 x3 f '( x) = 4 + − 3 x3 x 2 ✓ − 2 x −3 or x3 (5) 8.3 f ( x) = 3x 3 − 3x 2 + 6 x − 2 f ( x) = 9 x 2 − 6 x + 6 ✓ f (x) f ( x) = 18 x − 6 ✓ f (x) for concave up 18 x − 6  0 ✓ correct condition of concavity 1 ✓ answer x  3 1   x  ;  3  (4) [13] 17 ### Page 18 Downloaded from hlayiso.com 10611/23 QUESTION 9 9.1 9.1.1 f ( x) = −6 x 2 − 6 x + 12 y-intercept x=0  f (0) = 12 A(0 ; 12) ✓ answer NOTE: Does not have to be in coordinate form. (1) 9.1.2 x-intercepts y = f ( x) = 0 −6 x 2 − 6 x + 12 = 0 ✓ f '( x)  x2 + x − 2 = 0 (= 0 may be implied) ( x + 2)( x − 1) = 0  x = −2 or x = 1 ✓ factors  B( − 2 ; 0) ; C(1 ; 0) ✓ both coordinates NOTE: Must be in coordinate form. (3) 9.1.3 Turning points ✓ answer (1) 9.1.4 f (x) increases where: m = f ( x)  0 ✓ m = f ( x)  0  −2  x  1 ✓✓ answers OR OR m = f ( x)  0 ✓ m = f ( x)  0  x  (– 2 ; 1) ✓✓ answers (3) 9.1.5 Point of inflection: f ( x) = 0 ✓ f ( x) = 0  −12 x − 6 = 0 − 12 x = 6 1 x = − ✓ value for x 2 f ( x) = −2 x 3 − 3x 2 + 12 x + 20 1 1 1 1  f (− ) = −2(− )3 − 3(− ) 2 + 12(− ) + 20 2 2 2 2 ✓ substitution 1  f (− ) = 13,5 1 2 ✓ value for f (− ) 2 (4) 9.2 DO NOT MARK THIS QUESTION. THIS QUESTION HAS BEEN REMOVED FROM THE EXAMINATION PAPER. [12] 18 ### Page 19 Downloaded from hlayiso.com 10611/23 QUESTION 10 10. x3 W ( x) = − + 3x 2 − 250 x − 2700 150  max : 3x 2 W ( x) = − + 6 x − 250 = 0 ✓ W (x) 150 − x 2 + 300 x − 12500 = 0  x 2 − 300 x + 12500 = 0 ✓ equate to 0 ( x − 250)( x − 50) = 0 ✓ factors  x = 250 or x = 50 ✓ values of x 6x W ( x) = − +6 150 x ✓ W (x) W ( x) = − + 6 25 50 ✓ W (50)  0 W (50) = − + 6  0 25 250 W (250) = − +60 25 ✓ W (250)  0  Maximum profit: 250 bicycles weekly. ✓ conclusion OR 19 ### Page 20 Downloaded from hlayiso.com 10611/23 x3 W ( x) = − + 3x 2 − 250 x − 2700 150  max : 3x 2 W ( x) = − + 6 x − 250 = 0 150 ✓ W (x) − x 2 + 300 x − 12500 = 0 ✓ equate to 0  x 2 − 300 x + 12500 = 0 ✓ factors ( x − 250)( x − 50) = 0  x = 250 or x = 50 ✓ values of x ✓ substitution 2503 W (250) = − + 3(250) 2 − 250(250) − 2700 150 ✓ W(250) W (250) = R18133,33 or 503 W (50) = − + 3(50) 2 − 250(50) − 2700 150 ✓ W(50) W (50) = − R8533,33  Maximum profit: 250 Bicycles weekly ✓ conclusion OR x3 W ( x) = − + 3x 2 − 250 x − 2700 150  max : 3x 2 ✓ W (x) W ( x) = − + 6 x − 250 = 0 150 − x 2 + 300 x − 12500 = 0 ✓ equate to 0  x 2 − 300 x + 12500 = 0 ( x − 250)( x − 50) = 0 ✓ factors  x = 250 or x = 50 ✓ values of x ✓ shape ✓ min at x = 50 ✓ max at x = 250  Maximum profit: 250 Bicycles weekly ✓ conclusion [8] 20 ### Page 21 Downloaded from hlayiso.com 10611/23 QUESTION 11 NOTE: Candidates can present solutions either in decimal or fraction form. Do not penalise for rounding-off. 11.1 11.1.1 For independent events: P ( A)  P ( B ) = P ( A and B )  ( x + 0,1)  (0, 4) = (0,1) ✓ correct substitution  ( x + 0,1) = 0, 25  x = 0,15 ✓ value for x x + 0,1 + 0,3 + y = 1 0,15 + 0,1 + 0,3 + y = 1 ✓ method  y = 0, 45 ✓ value for y (4) 11.1.2 0,15 + 0,3 = 0,45 ✓ answer NOTE: Answer only, full marks. (1) 11.2 F 2 5 F 3 5 4 F 3 7 P 5 P P F (unsuccessful)/ P (pass/succeed) 11.2.1 4 3 12 ✓ method  = 7 5 35 ✓ answer NOTE: Answer only, full marks. (2) 11.2.2 4 2 3 24 ✓ method   = 7 5 5 175 ✓ answer NOTE: Answer only, full marks. (2) [9] 21 ### Page 22 Downloaded from hlayiso.com 10611/23 QUESTION 12 12.1 12.1.1 1 ✓ answer NOTE: Accept 2. (1) 12.1.2 7!−(2!6! ) = 3 600 ✓ 7! ✓ –(2!6!) ✓ answer NOTE: Accept answer as 3 600 or (3) 7!–(2!6!) No CA if method is meaningless. 12.2 P(win) = 0,7  0,9 + 0,3  0,45 method = 0,765 = 76,5% answer NOTE: No CA if method is meaningless. 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