JUNE EXAMINATION GRADE 12 2024 MARKING GUIDELINES MATHEMATICS (PAPER 1) 12 pages MARKING GUIDELINES MATHEMATICS (PAPER 1) GR12 0624 NOTE: Read the following instructions carefully before answering the questions. • • • • If a candidate answers a question TWICE, only mark the first attempt. If a candidate has crossed OUT an answer and did not redo it, mark the crossed-out answers. Consistent accuracy applies in ALL aspects of the marking guidelines Assuming values/answers in order to solve a question is UNACCEPTABLE. QUESTION 1 1.1 1.1.1 2๐ฅ๐ฅ(3๐ฅ๐ฅ + 4) = 0 4 ๐ฅ๐ฅ = 0 or ๐ฅ๐ฅ = − 3 ๏ผ ๐ฅ๐ฅ = 0 4 ๏ผ ๐ฅ๐ฅ = − 3 1.1.2 2๐ฅ๐ฅ 2 − 4๐ฅ๐ฅ + 1 = 0 −(−4) ± ๏ฟฝ(−4)2 − 4(2)(1) ๐ฅ๐ฅ = (2)(2) 4 ± √8 ๐ฅ๐ฅ = 4 ๐ฅ๐ฅ = 1,71 or ๐ฅ๐ฅ = 0,29 ๏ผ standard form ๏ผ substitute into correct formula 1.1.3 (๐ฅ๐ฅ − 2)2 ≥ 1 ๐ฅ๐ฅ 2 − 4๐ฅ๐ฅ + 4 ≥ 1 ๐ฅ๐ฅ 2 − 4๐ฅ๐ฅ + 3 ≥ 0 (๐ฅ๐ฅ − 3)(๐ฅ๐ฅ − 1) ≥ 0 ๐ถ๐ถ๐ถ๐ถ ๐ฅ๐ฅ = 3 or ๐ฅ๐ฅ = 1 ๐ฅ๐ฅ ≤ 1 ๐๐๐๐ ๐ฅ๐ฅ ≥ 3 ๏ผ standard form ๏ผ factors ๏ผ critical values ๏ผ answer Or 1.2 ๏ผ ๐ฅ๐ฅ = 1,71 ๏ผ ๐ฅ๐ฅ = 0,29 [-1 for incorrect rounding only in this question] (๐ฅ๐ฅ − 2) ≤ −1 ๐๐๐๐ (๐ฅ๐ฅ − 2) ≥ 1 ๐ฅ๐ฅ ≤ −1 + 2 ๐๐๐๐ ๐ฅ๐ฅ ≥ 1 + 2 ๐ฅ๐ฅ ≤ 1๐๐๐๐ ๐ฅ๐ฅ ≥ 3 1.2.1 √๐ฅ๐ฅ − 2 = 4 − ๐ฅ๐ฅ ๐ฅ๐ฅ − 2 ≥ 0 ๐๐๐๐๐๐ 4 − ๐ฅ๐ฅ ≥ 0 ๐ฅ๐ฅ ≥ 2 ๐๐๐๐๐๐ ๐ฅ๐ฅ ≤ 4 2 ≤ ๐ฅ๐ฅ ≤ 4 1.2.2 √๐ฅ๐ฅ − 2 = 4 − ๐ฅ๐ฅ ๐ฅ๐ฅ − 2 = 16 − 8๐ฅ๐ฅ + ๐ฅ๐ฅ 2 ๐ฅ๐ฅ 2 − 9๐ฅ๐ฅ + 18 = 0 (๐ฅ๐ฅ − 6)(๐ฅ๐ฅ − 3) = 0 ๐ฅ๐ฅ ≠ 6 or ๐ฅ๐ฅ = 3 (2) 2 ๏ผ๏ผ correct inequalities with square rooting both sides ๏ผ simplifying ๏ผ answer ๏ผ ๐ฅ๐ฅ − 2 ≥ 0 ๏ผ 4 − ๐ฅ๐ฅ ≥ 0 ๏ผ square both sides ๏ผ standard form ๏ผ factors ๏ผ selecting ๐ฅ๐ฅ = 3 (4) (4) (2) (4) MARKING GUIDELINES 1.3 MATHEMATICS (PAPER 1) GR12 0624 ๏ผ subject of equation ๏ผ substitution ๏ผ standard form ๏ผ factors ๏ผ x-values ๏ผ y-values 3๐ฅ๐ฅ + ๐ฆ๐ฆ = 2 and ๐ฆ๐ฆ 2 = 2๐ฅ๐ฅ 2 − 1 Equation 1.๐ฆ๐ฆ = 2 − 3๐ฅ๐ฅ (2 − 3๐ฅ๐ฅ)2 = 2๐ฅ๐ฅ 2 − 1 4 − 12๐ฅ๐ฅ + 9๐ฅ๐ฅ 2 = 2๐ฅ๐ฅ 2 − 1 7๐ฅ๐ฅ 2 − 12๐ฅ๐ฅ + 5 = 0 (7๐ฅ๐ฅ − 5)(๐ฅ๐ฅ − 1) = 0 5 ๐๐๐๐ ๐ฅ๐ฅ = 1 ๐ฅ๐ฅ = 7 1 ๐ฆ๐ฆ = − 7 of ๐ฆ๐ฆ = −1 OR 2−๐ฆ๐ฆ Euqation 1 ๐ฅ๐ฅ = 3 2 − ๐ฆ๐ฆ 2 ๏ฟฝ −1 ๐ฆ๐ฆ = 2 ๏ฟฝ 3 4 − 4๐ฆ๐ฆ + ๐ฆ๐ฆ 2 ๏ฟฝ−1 ๐ฆ๐ฆ 2 = 2 ๏ฟฝ 9 9๐ฆ๐ฆ 2 = 8 − 8๐ฆ๐ฆ + 2๐ฆ๐ฆ 2 − 9 7๐ฆ๐ฆ 2 + 8๐ฆ๐ฆ + 1 = 0 (7๐ฆ๐ฆ + 1)(๐ฆ๐ฆ + 1) = 0 1 ๐๐๐๐ ๐ฆ๐ฆ = −1 ๐ฆ๐ฆ = − 7 5 ๐ฅ๐ฅ = ๐๐๐๐ ๐ฅ๐ฅ = 1 7 2 1.4 ๐๐ + 2๐ ๐ = ๐๐ ๐๐ − 2๐ ๐ = ๐๐ 2๐๐ = ๐๐ + ๐๐ ๐๐+๐๐ ๐๐ = 2 ๐๐2 − ๐๐ 2 ๐๐๐๐ = 8 ๐๐2 −๐๐2 (6) ๐๐ + 2๐ ๐ = ๐๐ ๐๐ − 2๐ ๐ = ๐๐ 4๐ ๐ = ๐๐ − ๐๐ ๐๐−๐๐ ๐ ๐ = 4 ๏ผ 2๐๐ = ๐๐ + ๐๐ ๏ผ 4๐ ๐ = ๐๐ − ๐๐ ๏ผ r and s subject of equation ๏ผ multiplication OR OR ๏ผ substitution of a and b ๏ผ ๏ผ expand ๏ผ simplify RHS = 8 (๐๐ + 2๐ ๐ )2 − (๐๐ − 2๐ ๐ )2 = 8 (๐๐ 2 + 4๐๐๐๐ + 4๐ ๐ 2 ) − (๐๐ 2 − 4๐๐๐๐ + 4๐ ๐ 2 ) = 8 8๐๐๐๐ = 8 = ๐๐๐๐ (4) [26] 3 MARKING GUIDELINES MATHEMATICS (PAPER 1) GR12 0624 QUESTION 2 2.1 2.1.1 2.1.2 2.2 85 ; 82 ; 79 ; 76 ๐๐ = 85 ๐๐ = −3 ๐๐๐๐ = ๐๐ + (๐๐ − 1)๐๐ ๐๐๐๐ = 85 + (๐๐ − 1)(−3) ๐๐๐๐ = 85 − 3๐๐ + 3 ๐๐๐๐ = 88 − 3๐๐ ๏ผ d value ๏ผsubstitute a and d ๏ผ answer ๐๐๐๐ = 88 − 3๐๐ < 0 −3๐๐ < −88 88 ๐๐ > 3 ∴ ๐๐30 will be the first negative number. ๐๐๐๐ − ๐๐๐๐−1 = 4๐๐ − 3 ∴ ๐๐1 = 4(2) − 3 = 5 ๐๐2 = 4(3) − 3 = 9 3 = 13 = 2๐๐ = 4 3๐๐ + ๐๐ = 5 First difference Second difference = ๐๐ = 2 ๐๐11 = 190 5 ; 9 ; 13 ๐๐3 = 4(4) − (3) ๏ผ Tn<0 ๏ผ simplify ๏ผ answer Answer only full marks (3) ๏ผ first difference ๏ผ value of a 4 ; 4 ๏ผ value of b 3(2) + ๐๐ = 5 ๏ผ value of c ๐๐ = −1 ๏ผ value of ๐๐1 ๐๐๐๐ = 2๐๐2 − 1๐๐ + ๐๐ 190 = 2(11)2 − 1(11) + ๐๐ 190 = 242 − 11 + ๐๐ ๐๐ = −41 ๐๐๐๐ = 2๐๐2 − ๐๐ − 41 ๐๐1 = 2(1)2 − (1) − 41 = −40 OR OR (5) 4 MARKING GUIDELINES MATHEMATICS (PAPER 1) GR12 0624 ๏ผ expanding ๐๐๐๐−1 ๐๐๐๐ = ๐๐๐๐2 + ๐๐๐๐ + ๐๐ ๐๐๐๐ − 1 = ๐๐(๐๐ − 1)2 + ๐๐(๐๐ − 1) + ๐๐ ๐๐๐๐ − ๐๐๐๐ − 1 = ๐๐๐๐2 + ๐๐๐๐ + ๐๐ − [๐๐(๐๐2 − 2๐๐ + 1) + ๐๐๐๐ − ๐๐ + ๐๐] = ๐๐๐๐2 + ๐๐๐๐ + ๐๐ − ๐๐๐๐2 + 2๐๐๐๐ − ๐๐ − ๐๐๐๐ + ๐๐ − ๐๐ = 2๐๐๐๐ − ๐๐ + ๐๐ 2๐๐๐๐ − ๐๐ + ๐๐ = 4๐๐ − 3 2๐๐ = 4 ๐๐ = 2 −๐๐ + ๐๐ = −3 −2 + ๐๐ = −3 ๐๐ = −1 121(2) + 11(−1) + ๐๐ = 190 ๐๐ = −41 ๐๐๐๐ = 2๐๐2 − ๐๐ − 41 ๏ผ value of a ๏ผ value of b ๏ผ value of c ๏ผ value of ๐๐1 ๐๐1 = 2(1)2 − (1) − 41 = −40 2.3 ๐๐ [2๐๐ + (๐๐ − 1)๐๐] 2 50 [2๐๐ + (50 − 1)๐๐] 1275 = 2 51 = [2๐๐ + 49๐๐] ๐๐25 + ๐๐26 = ๐๐ + 24๐๐ + ๐๐ + 25๐๐ ๐๐25 + ๐๐26 = 2๐๐ + 49๐๐ ๐๐25 + ๐๐26 = 51 ๐๐๐๐ = ๏ผ substitute in formula ๏ผ expanding ๐๐25 + ๐๐26 ๏ผ answer (3) [14] QUESTION 3 3.1 ∞ ∑ ( 4 x − 1) k ๏ผr k =1 (4๐ฅ๐ฅ − 1)1 + (4๐ฅ๐ฅ − 1)2 + (4๐ฅ๐ฅ − 1)3 … .. ๐๐ = (4๐ฅ๐ฅ − 1) −1 < ๐๐ < 1 −1 < 4๐ฅ๐ฅ − 1 < 1 0 < 4๐ฅ๐ฅ < 2 1 1 0 < ๐ฅ๐ฅ < 2 ๐ฅ๐ฅ ≠ 4 3.2 ๏ผ condition ๏ผ answer 1 ๏ผ excluding ๐ฅ๐ฅ ≠ 4 3.2.1 ๐๐1 = 3 and ๐๐5 = 48 ๐๐๐๐ = ๐๐๐๐ ๐๐−1 48 = 3. ๐๐ 4 16 = ๐๐ 4 ∴ ๐๐ = 2 (4) ๏ผ sub into formula ๏ผ simplify ๏ผ answer 5 (3) MARKING GUIDELINES 3.2.2 Sum of radii for 5 circles ๐๐(๐๐ ๐๐ − 1) ๐๐๐๐ = ๐๐ − 1 3(25 − 1) ๐๐5 = = 93 ๐ข๐ข๐ข๐ข๐ข๐ข๐ข๐ข๐ข๐ข 2−1 ๐ฟ๐ฟ = 93 × 2 = 186 ๐ข๐ข๐ข๐ข๐ข๐ข๐ข๐ข๐ข๐ข MATHEMATICS (PAPER 1) GR12 0624 ๏ผ subt in formula ๏ผ simplify ๏ผ answer OR Sum of diameter for 5 circles ๐๐(๐๐ ๐๐ − 1) ๐๐5 = ๐๐ − 1 6(25 − 1) ๐๐5 = 2 ๐๐5 = 186 ๐ฟ๐ฟ = 186 units OR 6 + 12 + 24 +48 + 96 = 186 (3) 3.2.3 ๐๐32 + ๐๐. 62 + ๐๐. 122 + โฏ ๐ก๐ก๐ก๐ก 10 terms ๐๐. 62 ๐๐ = ๐๐. 32 ๐๐ = 4 9๐๐(410 − 1) ๐๐10 = 4−1 ๐๐10 = 3๐๐(1 048 575) ๐๐10 = 3 145 725๐๐ ๏ผ area of circle ๏ผ r value ๏ผ subt into formula ๏ผ answer (4) [16] QUESTION 4 4.1 ๐ฆ๐ฆ = ๐ฅ๐ฅ + 1 and ๐ฆ๐ฆ = −๐ฅ๐ฅ + 3 ๐ฅ๐ฅ + 1 = −๐ฅ๐ฅ + 3 2๐ฅ๐ฅ = 2 ๐ฅ๐ฅ = 1 ๐ฆ๐ฆ = 1 + 1 ๐ฆ๐ฆ = 2 ๐๐ = −1 ๐๐ = 2 ๏ผ ๐ฅ๐ฅ + 1 = −๐ฅ๐ฅ + 3 ๏ผ ๐ฅ๐ฅ-value ๏ผ ๐ฆ๐ฆ-value (3) 6 MARKING GUIDELINES 4.2 −4 +2 ๐ฅ๐ฅ − 1 ๏ผ ๐ฆ๐ฆ = 0 −2๐ฅ๐ฅ + 2 = −4 −2๐ฅ๐ฅ = −6 ๐ฅ๐ฅ = 3 ๏ผ ๐ฅ๐ฅ = 3 ๐ฆ๐ฆ = −4 +2 ๐ฅ๐ฅ − 1 −4 −2 = ๐ฅ๐ฅ − 1 0= 4.3 MATHEMATICS (PAPER 1) GR12 0624 (2) ๏ผ horizontal asymptote ๐ฅ๐ฅ = 1 ๏ผ vertical asymptote ๏ผ y-intercept ๏ผ shape ๐ฆ๐ฆ = 2 (4) 4.4 ๐ฅ๐ฅ < 1 ๐๐๐๐ ๐ฅ๐ฅ > 3 ๏ผ notation ๏ผ critical values 7 (2) [11] MARKING GUIDELINES MATHEMATICS (PAPER 1) GR12 0624 QUESTION 5 5.1 5.1.1 5.1.2 5.1.3 5.1.4 5.1.5 5.1.6 5.2 ๐๐(๐ฅ๐ฅ) = −(๐ฅ๐ฅ − 2)2 + 9 0 = −๐ฅ๐ฅ 2 + 4๐ฅ๐ฅ − 4 + 9 0 = ๐ฅ๐ฅ 2 − 4๐ฅ๐ฅ − 5 (๐ฅ๐ฅ − 5)(๐ฅ๐ฅ + 1) = 0 ๐ฅ๐ฅ = 5 or ๐ฅ๐ฅ = −1 ๐ด๐ด๐ด๐ด = 6 units ๏ผ let ๐ฆ๐ฆ = 0 ๏ผ standard form ๏ผ ๐ฅ๐ฅ-values ๏ผ 6 units OR OR (๐ฅ๐ฅ − 2)2 = 9 ๐ฅ๐ฅ − 2 = ±3 ๐ฅ๐ฅ − 2 = 3 ๐๐๐๐ ๐ฅ๐ฅ − 2 = −3 ๐ฅ๐ฅ = 5 ๐๐๐๐ ๐ฅ๐ฅ = −1 ๐ด๐ด๐ด๐ด = 6 units ๏ผ let ๐ฆ๐ฆ = 0 ๏ผ (๐ฅ๐ฅ − 2)2 = 9 ๏ผ ๐ฅ๐ฅ-values ๏ผ 6 units D(2:9) ๐ฆ๐ฆ = ๐๐ ๐ฅ๐ฅ 9 = ๐๐ 2 ๐๐ = 3 ๏ผ sub in ๐๐(๐ฅ๐ฅ) ๐๐(๐ฅ๐ฅ) = −(๐ฅ๐ฅ − 2)2 + 9 ๐ฆ๐ฆ = −(๐ฅ๐ฅ + 2 − 2)2 + 9 − 9 ๐ฆ๐ฆ = −๐ฅ๐ฅ 2 ๐ฅ๐ฅ ≥ 2 ๐ฅ๐ฅ ≤ 0 or ๐ฅ๐ฅ ≥ 0 1 Prove: ๐๐ ๏ฟฝ๐ฅ๐ฅ + 2๏ฟฝ = √3๐๐(๐ฅ๐ฅ) ๐๐(๐ฅ๐ฅ) = 3๐ฅ๐ฅ 1 1 ๐๐ ๏ฟฝ๐ฅ๐ฅ + ๏ฟฝ = 3๐ฅ๐ฅ+2 2 1 = 3๐ฅ๐ฅ . 32 = √3๐๐(๐ฅ๐ฅ) (4) ๏ผ ๐๐ = 3 (2) ๏ผ answer (1) ๏ผ subst ๏ผ answer Answer only full marks (2) ๏ผ answer (1) (accuracy mark) ๏ผ subt ๏ผ use of exp law (2) ๐ฆ๐ฆ = ๐๐(๐ฅ๐ฅ − ๐ฅ๐ฅ1 )(๐ฅ๐ฅ − ๐ฅ๐ฅ2 ) ๐ฆ๐ฆ = ๐๐(๐ฅ๐ฅ + 3)(๐ฅ๐ฅ − 2) ๐ฆ๐ฆ = ๐๐๐ฅ๐ฅ 2 + ๐๐๐๐ − 6๐๐ ๐ฆ๐ฆ = ๐๐๐๐ + ๐๐ 0 = ๐๐(−6) + ๐๐ ๏ผ subt in formula ๏ผ simplifying ๏ผ subt in formula 0 = ๐๐(−6) − 6๐๐ 6๐๐ = −6๐๐ ๏ผ subt c in formula ๐๐ = −๐๐ ๏ผ ๐๐ = −๐๐ OR 8 (5) MARKING GUIDELINES −3+2 MATHEMATICS (PAPER 1) GR12 0624 1 ๐ฅ๐ฅ = 2 = − 2 ๐๐ ′ (๐ฅ๐ฅ) = 2๐๐๐๐ + ๐๐ 1 0 = 2 ๏ฟฝ− ๏ฟฝ ๐๐ + ๐๐ 2 ๐๐ = ๐๐ (−6; 0) (0; ๐๐) ๐๐ ๐๐ = 6 ๐๐ = 6๐๐ 0 = 4๐๐ + 2๐๐ + 6๐๐ 0 = 4๐๐ + 2๐๐ + 6๐๐ −6๐๐ = 6๐๐ ๐๐ = −๐๐ [17] QUESTION 6 6.1 6.2 6.3 ๏ผ ๐ฅ๐ฅ = 1 ๏ผ answer (1) ๐ฆ๐ฆ = ๐๐๐๐๐๐1 ๐ฅ๐ฅ ๏ผ swop x and y ๏ผ answer (2) ๐ฅ๐ฅ > 0 ๐ฅ๐ฅ = 6.4 (1) ๐ด๐ด(1: 0) 2 log 1 ๐ฆ๐ฆ 2 ๐ฅ๐ฅ Answer only full marks 1 ๐ฆ๐ฆ = ๏ฟฝ ๏ฟฝ 2 ๏ผ shape ๏ผ y-intercept ๏ผ one other point (3) 6.5 ๏ผ answer ๐ฆ๐ฆ = log 1 ๐ฅ๐ฅ 2 1 22 ๐ฆ๐ฆ = log 1 ๐ฆ๐ฆ = 1 (1) 9 MARKING GUIDELINES 6.6 MATHEMATICS (PAPER 1) Reflection in the y-axis and translated one unit down GR12 0624 ๏ผ reflection in yaxis ๏ผ translate one down (2) [10] QUESTION 7 7.1 ๐๐๐๐๐๐๐๐ ๐๐ ๏ฟฝ 1 + ๐๐๐๐๐๐๐๐ = ๏ฟฝ1 + ๐๐ ๐๐๐๐๐๐๐๐ = ๏ฟฝ1 + 7.2 ๐๐ = 7,71% ๏ subt into formula ๏ simplify 4 0,075 ๏ฟฝ −1 4 ๏ answer (accept i=0.0771) ๏ n-value ๐ด๐ด = ๐๐(1 − ๐๐)๐๐ ๏ sub in form 4 200 = 60 000(1 − ๐๐)42 ๏ simplify 4 200 = (1 − ๐๐)42 60 000 42 ๏ฟฝ 42 7.3 ๏ r = 6,14% (accept in i form) 4 200 = 1 − ๐๐ 60 000 ๐๐ = 1 − ๏ฟฝ ๐๐ = 6,14% 4 200 60 000 ๐ด๐ด = ๐๐ ๏ฟฝ1 + (3) (4) ๏ substitution (3 years) ๐๐ ๐๐× ๐๐ ๏ฟฝ ๐๐ ๏ answer 0.054 7×12 ๏ฟฝ 17 614,76 = ๐๐ ๏ฟฝ1 + 12 ๐๐ = R12 080,41 12 080,41 + ๐ฅ๐ฅ = 27 000 ๏ฟฝ1 + 12 080,41 + ๐ฅ๐ฅ = 31 736,69 ๐ฅ๐ฅ = R19 656,28 ๏ substitutions 0,054 3×12 ๏ฟฝ 12 OR ๏ simplify ๏ answer Or 10 (5) MARKING GUIDELINES 0,054 17 614,76 = 27 000(1+ 12 )120 − ๐ฅ๐ฅ(1 + 12 120 0,054 0,054 84 ๏ฟฝ ) = 27 000 ๏ฟฝ1 + 12 12 ๐ฅ๐ฅ = R19 656,28 ๐ฅ๐ฅ(1 + 0,054 84 ) − 17 614,76 MATHEMATICS (PAPER 1) GR12 0624 0,054 ๏ i= 12 ๏ n=120 ๏-๐ฅ๐ฅ ๏n=84 ๏answer [12] QUESTION 8 - Only penalise 1 mark for incorrect notation 8.1 8.1 ๐๐(๐ฅ๐ฅ) = 2 − 3๐ฅ๐ฅ 2 ๐๐(๐ฅ๐ฅ + โ) = 2 − 3(๐ฅ๐ฅ + โ)2 ๐๐(๐ฅ๐ฅ + โ) = 2 − 3๐ฅ๐ฅ 2 − 6๐ฅ๐ฅโ − 3โ2 2 − 3๐ฅ๐ฅ 2 − 6๐ฅ๐ฅโ − 3โ2 − (2 − 3๐ฅ๐ฅ 2 ) ′ (๐ฅ๐ฅ) ๐๐ = lim โ→0 โ −6๐ฅ๐ฅโ − 3โ2 = lim โ→0 โ โ(−6๐ฅ๐ฅ−3โ) = lim โ→0 โ = lim (−6๐ฅ๐ฅ − 3โ) โ→0 8.2 = −6๐ฅ๐ฅ 8.2.1 8.2.2 ๏ผ substitution ๏ผ factors ๏ผ simplify ๏ผ answer (5) ๐๐(๐ฅ๐ฅ) = 2๐ฅ๐ฅ 4 − 3๐ฅ๐ฅ + ๐๐2 ๐๐ ′ (๐ฅ๐ฅ) = 8๐ฅ๐ฅ 3 − 3 ๏ผ 8๐ฅ๐ฅ 3 ๏ผ -3 ๏ผ 0 (implied) ๏ฃฎ 2 x3 − x ๏ฃน Dx ๏ฃฏ ๏ฃบ x ๏ฃป ๏ฃฐ ๏ฃฎ 2 − ๏ฃน ๏ผ Dx ๏ฃฏ 2 x − x 2 ๏ฃบ ๏ฃฐ๏ฃฏ ๏ฃป๏ฃบ ๏ผ 4๐ฅ๐ฅ (3) 1 1 ๐ท๐ท๐ฅ๐ฅ ๏ฟฝ2๐ฅ๐ฅ 2 − ๐ฅ๐ฅ −2 ๏ฟฝ 8.3 ๏ผ f (x + h) 1 3 ๏ผ 2 ๐ฅ๐ฅ −2 1 3 = 4๐ฅ๐ฅ + ๐ฅ๐ฅ −2 2 (3) ๏ผ gradient ๏ผ subt ๏ผ equation ๐ฆ๐ฆ = ๐๐๐๐ + ๐๐ ๐ฆ๐ฆ = 7๐ฅ๐ฅ + ๐๐ 5 = 7(4) + ๐๐ ๐๐ = −23 ๐ฆ๐ฆ = 7๐ฅ๐ฅ − 23 OR ๐ฆ๐ฆ − 5 = 7(๐ฅ๐ฅ − 4) ๐ฆ๐ฆ − 5 = 7๐ฅ๐ฅ − 28 ๐ฆ๐ฆ = 7๐ฅ๐ฅ − 23 (3) 11 MARKING GUIDELINES MATHEMATICS (PAPER 1) GR12 0624 QUESTION 9 9.1 ๐๐(๐ฅ๐ฅ) = 2๐ฅ๐ฅ 3 + ๐๐๐ฅ๐ฅ 2 + ๐๐๐๐ + 3 ๏ผ sub in f N(2;-9) −9 = 2(2)3 + ๐๐(2)2 + ๐๐(2) + 3 −9 = 16 + 4๐๐ + 2๐๐ + 3 −28 = 4๐๐ + 2๐๐ … …๏ช ๏ผ ๐๐(๐ฅ๐ฅ) = 0 ๏ผ sub in f’ ๏ผ solve for q ๐๐ ′ (๐ฅ๐ฅ) = 6๐ฅ๐ฅ 2 + 2๐๐๐๐ + ๐๐ 0 = 6(2)2 + 2๐๐(2) + ๐๐ 0 = 24 + 4๐๐ +q −24 = 4๐๐ + ๐๐ … … ๏ซ ๏ผ solve for p Equation ๏ช – ๏ซ ∴ −4 = q 9.2 9.3 9.4 9.5 −24 = 4๐๐ − 4 4๐๐ = −20 ๐๐ = −5 (5) G(0;3) ๏ผ y-value 3 ๏ผ x-value 0 ๐๐(๐ฅ๐ฅ) = 2๐ฅ๐ฅ 3 − 5๐ฅ๐ฅ 2 − 4๐ฅ๐ฅ + 3 0 = (๐ฅ๐ฅ − 3)(2๐ฅ๐ฅ 2 + ๐ฅ๐ฅ − 1) 0 = (๐ฅ๐ฅ − 3)(2๐ฅ๐ฅ − 1)(๐ฅ๐ฅ + 1) 1 ๐ฅ๐ฅ = 3 ๐๐๐๐ ๐ฅ๐ฅ = ๐๐๐๐ ๐ฅ๐ฅ = −1 2 AB = 1,5 units ๏ผ (2๐ฅ๐ฅ 2 + ๐ฅ๐ฅ − 1) ๏ผ factors ๏ผ roots ๏ผ 1,5 (4) ๏ผ ๐๐ ′ (๐ฅ๐ฅ) ๏ผ factors ๏ผ ๐ฅ๐ฅ value ๐๐ ′ (๐ฅ๐ฅ) = 6๐ฅ๐ฅ 2 − 10๐ฅ๐ฅ − 4 0 = 3๐ฅ๐ฅ 2 − 5๐ฅ๐ฅ − 2 (3๐ฅ๐ฅ + 1)(๐ฅ๐ฅ − 2) = 0 1 ๐ฅ๐ฅ = − 3 ๐๐ ′′ (๐ฅ๐ฅ) = 12๐ฅ๐ฅ − 10 0 = 12๐ฅ๐ฅ − 10 10 5 ๐ฅ๐ฅ = = 12 6 OR ๐ฅ๐ฅ = (2) (3) ๏ผ ๐๐ ′′ ๏ผ=0 5 ๏ผ6 1 −3 + 2 5 ๐ฅ๐ฅ = 6 2 OR (3) 12 MARKING GUIDELINES 9.6 9.7 MATHEMATICS (PAPER 1) GR12 0624 ๐ฆ๐ฆ ′′ = 6๐๐๐๐ + 2๐๐ = 0 6๐๐๐๐ = −2๐๐ 2๐๐ ๐ฅ๐ฅ = − 6๐๐ −2(−5) 5 ๐ฅ๐ฅ = = 6(2) 6 ๐๐ ′′ > 0 6๐ฅ๐ฅ − 5 > 0 5 ๐ฅ๐ฅ > 6 ๐ฅ๐ฅ < −1 ๐๐๐๐ − 5 ๏ผ ๐ฅ๐ฅ > 6 1 1 < ๐ฅ๐ฅ < ๐๐๐๐ 2 < ๐ฅ๐ฅ < 3 3 2 ๏ผ ๐ฅ๐ฅ < −1 1 1 ๏ผ − 3 < ๐ฅ๐ฅ < 2 ๏ผ 2 < ๐ฅ๐ฅ < 3 13 (1) (3) [21] MARKING GUIDELINES MATHEMATICS (PAPER 1) GR12 0624 QUESTION 10 10.1 10.1.1 P(S and T) = P(not S) = 1 ๐๐(๐๐) = 4 3 4 1 6 ๏ผ ๐๐(๐๐) ๏ผ subst in formula ๐๐(๐๐ ๐๐๐๐๐๐ ๐๐) = ๐๐(๐๐) × ๐๐(๐๐) 10.1.2 10.2.1 1 1 = × ๐๐(๐๐) 6 4 2 ๐๐(๐๐) = 3 ๐๐(๐๐ ๐๐๐๐ ๐๐) = ๐๐(๐๐) + ๐๐(๐๐) − ๐๐(๐๐ ๐๐๐๐๐๐ ๐๐) 1 2 1 ๐๐(๐๐ ๐๐๐๐ ๐๐) = + − 4 3 6 3 ๐๐(๐๐ ๐๐๐๐ ๐๐) = 4 0,30 0,35 0,65 10.2.2 B 0,70 0,30 (3) ๏ผ sub into formula (2) ๏ผ answer BB ๏ผ Branch B or C 30% and 65% B C BC B CB C CC ๏ผ Branch B or C 30% and 70% ๏ผ outcomes C 0,70 10.2 ๏ผ ๐๐(๐๐) P(same meal) = (0,35)(0,30) + (0,65)(0,70) P(same meal) = 0,105 + 0,455 = 0,56 Number of people = 200 x 0.56 =112 (3) ๏ผ (0,35)(0,30) + (0,65)(0,7) ๏ผ 0,56 ๏ผ 112 (3) [11] TOTAL: 14 150
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