Mathematics Grade 12 Paper 1 KwaZulu Natal Mock 2019 Memo

Downloaded from hlayiso.com MATHEMATICS P1 SEPTEMBER 2019 PREPARATORY EXAMINATION MARKING GUIDELINE NATIONAL SENIOR CERTIFICATE GRADE 12 MARKS: 150 TIME: 3 hours This marking guideline consists of 13 pages. Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics P1 2 September 2019 Preparatory Examination NSC Marking Guideline QUESTION 1 1.1.1 𝑥 = 0 𝑜𝑟 𝑥 = 4 A✓ 0 A✓ 4 (2) 1.1.2 2 x2 + 5x − 1 = 0 A✓standard form − b  b 2 − 4ac x= 2a − (5)  (5) 2 − 4(2)(−1) = CA✓substitution in correct formula 2(2) = 0,19 or − 2,69 CA✓CA✓answers (4) (penalize 1 mark if rounding off is incorrect-once here for entire paper) 1.2.1 √𝑥 − 2 = 2 − 𝑥 (√𝑥 − 2)2 = (2 − 𝑥)2 A✓ squaring both sides 𝑥 − 2 = 4 − 4𝑥 + 𝑥 2 CA✓standard form 𝑥 2 − 5𝑥 + 6 = 0 CA✓factors (𝑥 − 2)(𝑥 − 3) = 0 CA✓answers and rejecting (4) 𝑥 = 2 𝑜𝑟 𝑥 = 3 n/a OR OR √𝑥 − 2 = 2 − 𝑥 A✓ A✓each inequality 𝑥 − 2 ≥ 0 𝑎𝑛𝑑 2 − 𝑥 ≥ 0 A✓both inequalities (4) 𝑥 ≥ 2 𝑎𝑛𝑑 𝑥 ≤ 2 CA✓answer 𝑥=2 Answer only FULL MARKS 1.2.2 𝑝2 − 𝑝 = 𝑥 letting 𝑝2 − 𝑝 = 𝑥 ∴ 𝑝2 − 𝑝 = 2 CA✓𝑝2 − 𝑝 = 2 𝑝2 − 𝑝 − 2 = 0 CA✓standard form (𝑝 + 1)(𝑝 − 2) = 0 CA✓factors 𝑝 = −1 𝑜𝑟 𝑝 = 2 CA✓answers MAX 3marks if four solutions (4) arrived at 1.3 −2𝑥 2 + 5𝑥 ≤ 0 5 2𝑥 (𝑥 − ) ≥ 0 AA✓✓ factors 2 5 5 𝑥≤0 or 𝑥≥ CA✓ 𝑥 ≤ 0 CA✓ 𝑥 ≥ 2 (4) 2 OR + OR + + If graphical Solution is used: 0 2,5 OR 0 2,5 AA 2 marks for graph - - CACA 2 marks for answer - (4)

Downloaded from hlayiso.com Mathematics P1 3 September 2019 Preparatory Examination NSC Marking Guideline 1.4 2𝑥+1 + 2𝑥 = 3𝑦+2 − 3𝑦 2𝑥 (2 + 1) = 3𝑦 (9 − 1) A✓factorising 2𝑥 (3) = 3𝑦 (8) CA✓simplifying 2𝑥−3 = 3𝑦−1 CA✓exponential form 𝑥 − 3 = 0 𝑎𝑛𝑑 𝑦 − 1 = 0 CA✓each exponent and equal to 0 𝑥 = 3 𝑎𝑛𝑑 𝑦 = 1 CA✓x and y value 𝑥+𝑦 =4 CA✓answer (6) If 𝑥 − 3 = 0 𝑎𝑛𝑑 𝑦 − 1 = 0 is missing then maximum 5/6 marks [24] QUESTION 2 2.1 48 ; 63 A✓answers (1) 2.2 8 15 24 35 1D 7 9 11 2D 2 2 A✓a value 2𝑎 = 2 𝑎=1 CA✓b value 3𝑎 + 𝑏 = 7 𝑏=4 𝑎+𝑏+𝑐 =8 𝑐=3 CA✓c value 2 𝑇𝑛 = 𝑛 + 4𝑛 + 3 CA✓answer (4) OR OR 2𝑎 = 2 𝑎=1 A✓a value 𝑇1 + 𝑑2 − 𝑑1 = 𝑐 8+2−7= 𝑐 3=c CA✓c value 𝑇𝑛 = 𝑛2 + 𝑏𝑛 + 3 8= 1+𝑏+3 𝑏=4 CA✓b value (4) 𝑇𝑛 = 𝑛2 + 4𝑛 + 3 CA✓answer

Downloaded from hlayiso.com Mathematics P1 4 September 2019 Preparatory Examination NSC Marking Guideline OR OR (n − 1(n − 2) A✓formula Tn = T1 + (n − 1)d1 + d2 2 A✓substitution into correct formula (n − 1(n − 2) = 8 + (n − 1)(7) + (2) 2 = 8 + 7n − 7 + n 2 − 3n + 2 CA✓simplifying = n 2 + 4n + 3 CA✓answer (4) OR OR A✓formula n −1 Tn = 2a + (n − 2)d  + T1 2 n −1 A✓substitution into correct formula = 2(7) + (n − 2)(2) + 8 2 CA✓simplifying n −1 = 14 + 2n − 4 + 8 2 (4) n −1 = 2n + 10 + 8 2 = (n − 1)(n + 5) + 8 = n 2 + 4n − 5 + 8 CA✓answer = n + 4n + 3 2 [5] QUESTION 3 3.1 T2 − T1 = T3 − T2 A✓ equating differences 𝑝 + 5 − 2𝑝 + 3 = 2𝑝 + 7 − 𝑝 − 5 CA✓simplifying −𝑝 + 8 = 𝑝 + 2 CA✓answer (3) 𝑝=3 3.2 Pattern is 3 ; 8 ; 13 ; . .. Sn = n 2a + (n − 1)d  2 CA✓a = 3 and d = 5 S120 = 120 2(3) + 119(5) CA✓substitution into formula 2 = 36060 CA✓answer (3) 3.3.1 x = k + 1 and y = k + 2 A✓x – value A✓y – value (2)

Downloaded from hlayiso.com Mathematics P1 5 September 2019 Preparatory Examination NSC Marking Guideline 3.3.2 𝑇𝑥 = 𝑎 + (𝑥 − 1)𝑑 = 3 + 5𝑘 CA✓substitution into nth term 𝑇𝑦 = 𝑎 + (𝑘 + 1)𝑑 CA✓3 + 5𝑘 = 3 + (𝑘 + 1)(5) = 8 + 5𝑘 CA✓8+5𝑘 𝑇𝑥 + 𝑇𝑦 = 11 + 10𝑘 CA✓answer (4) [12] QUESTION 4 4.1.1 15 ; 5 AA✓✓both terms (2) 4.1.2 a S = 1− r 15 CA✓substitution of common ratio = 1 1− 3 CA✓answer 45 = = 22,5 (2) 2 4.2 3 𝑠𝑖𝑛 3 0∘ ; 𝑐𝑜𝑠 3 0∘ ; 2 1 √3 3 ; ; A✓listing terms 2 2 2 1 𝑎 = ; 𝑟 = √3 2 𝑎𝑟 𝑛−1 = 40,5√3 1 𝑛−1 81 1 𝑛−1 81 (√3) = √3 CA✓2 (√3) = √3 2 2 2 𝑛−1 1 𝑛−1 1 CA✓3 2 = 34 . 32 3 2 = 34 . 32 𝑛−1 1 9 =4 = 𝑛−1 1 9 2 2 2 CA✓ = 42 = 2 𝑛−1=9 2 𝑛 = 10 CA✓answer (5) OR OR

Downloaded from hlayiso.com Mathematics P1 6 September 2019 Preparatory Examination NSC Marking Guideline 3 𝑠𝑖𝑛 3 0∘ ; 𝑐𝑜𝑠 3 0∘ ; 2 1 √3 3 A✓listing terms ; ; 2 2 2 1 𝑎 = ; 𝑟 = √3 2 𝑛−1 81 𝑎𝑟 = √3 2 1 𝑛−1 81 1 𝑛−1 81 (√3) = √3 CA✓2 (√3) = 2 √3 2 2 𝑛 (√3) = 81√3 √3 𝑛 243= (√3) 𝑛 CA✓243= (√3) 1 1 35 = 32𝑛 CA✓ 35 = 32𝑛 (5) n = 10 CA✓answer [9] QUESTION 5 5.1 𝑥 = 2 𝑎𝑛𝑑 𝑦 = −1 AA✓𝑥 = 2 ✓𝑦 = −1 (2) 5.2 𝑦 − intercept :(0 ; −3) A✓y – intercept −4 𝑥 − intercept: −1=0 −4 2−𝑥 A✓ −1=0 −4 1 2−𝑥 = 2−𝑥 1 −4 = 2 − 𝑥 𝑥=6 A✓x – intercept (3) (6 ; 0) (co-ordinate form not needed) 5.3 y CA✓x -intercepts f CA✓y - intercept CA✓both asymptotes O x A✓shape 2 6 (4) -1 -3 [9]

Downloaded from hlayiso.com Mathematics P1 7 September 2019 Preparatory Examination NSC Marking Guideline QUESTION 6 6.1 A(0; 6) A✓answer (Must be in coordinate (1) form) 6.2 𝑏 Using axis of symmetry (2) 𝑥=− = 2,5 2𝑎 S(5 ; 6) A✓x – value A✓y – value OR OR 𝑦 =𝑥+1=6 A✓equating equation to 6 (2) 𝑥=5 A✓x – value 𝑆(5 ; 6) OR OR A✓equating equation to 6 𝑦 = −𝑥 2 + 5𝑥 + 6 = 6 𝑥 2 − 5𝑥 = 0 𝑥(𝑥 − 5) = 0 (2) 𝑥 = 0 𝑜𝑟 𝑥 = 5 𝑛/𝑎 A✓x – values and rejection 𝑆(5 ; 6) OR −𝑥 2 + 5𝑥 + 6 = 𝑥 + 1 …………….. 6.3 −𝑥 2 + 5𝑥 + 6 = 0 𝑥 2 − 5𝑥 − 6 = 0 (𝑥 + 1)(𝑥 − 6) = 0 A✓factors 𝑥 = −1 𝑜𝑟 𝑥 = 6 (3) CACA✓✓x – values B(−1; 0), C(6; 0) 6.4 (−𝑥 2 + 5𝑥 + 6) − (𝑥 + 1) = 5 A✓subtraction of both graphs −𝑥 2 + 5𝑥 + 6 − 𝑥 − 1 = 5 A✓equating to 5 𝑥 2 − 4𝑥 = 0 𝑥(𝑥 − 4) = 0 CA✓factors 𝑥 = 0 𝑜𝑟 4 (4) OR = 4units CA✓OR value 6.5.1 −1+6 5 A✓midpoint formula 𝑥= =2 2 CA✓Axis of symmetry value 5 2 5 49 𝑦 = −( ) + 5( ) + 6 = = 12,25 CA✓substitution 2 2 4 5 ( ; 12,25) CA✓answer (4) 2 OR OR

Downloaded from hlayiso.com Mathematics P1 8 September 2019 Preparatory Examination NSC Marking Guideline 𝑏 5 5 A✓formula 𝑥 = − 2𝑎 = − 2(−1) = 2 CA✓Axis of symmetry value 5 2 5 49 𝑦 = −( ) + 5( ) + 6 = = 12,25 CA✓substitution 2 2 4 5 ( ; 12,25) CA✓answer (4) 2 OR OR 5 A✓derivative and equal to 0 𝑓 ′ (𝑥) = −2𝑥 + 5 = 0 ∴𝑥=2 5 2 5 49 CA✓Axis of symmetry value 𝑦 = −( ) + 5( ) + 6 = = 12,25 2 2 4 CA✓substitution 5 ( ; 12,25) (4) 2 CA✓answer 6.5.2 PQ = −𝑥 2 + 4𝑥 + 5 A✓ PQ in terms of x 𝑏 4 CA✓ substitution 𝑥 = − 2𝑎 = − 2(−1) = 2 CA✓x – value Max. PQ = −(2)2 + 4(2) + 5 = 9units CA✓answer (4) OR OR PQ = −𝑥 2 + 4𝑥 + 5 A✓PQ in terms of x CA✓derivative and equal to 0 PQ′ = −2𝑥 + 4 = 0 ∴ 𝑥 = 2 CA✓x – value (4) Max. PQ = −(2)2 + 4(2) + 5 = 9units CA✓answer [18] QUESTION 7 7.1 𝑦 = 5𝑥 AA✓✓ answer (2) 7.2 𝑦 > 0 or 𝑦 ∈ (0 ; ∞) A✓answer (1) 7.3 𝑙𝑜𝑔5 𝑥 = −4 A✓Equating log graph to ─4 1 𝑥 = 5−4 = A✓writing in exponential form 625 1 CA✓end points 0<𝑥≤ 625 A✓ interval Can be solved by log inequalities. (4) [7]

Downloaded from hlayiso.com Mathematics P1 9 September 2019 Preparatory Examination NSC Marking Guideline QUESTION 8 8.1 𝐴 = 𝑃(1 − 𝑖)𝑛 250000 = 𝑃(1 − 13,5%)5 A✓substitution into the correct 250000 formula 𝑃= (1 − 13,5%)5 CA✓making P the subject = 𝑅516249 CA✓answer (3) 8.2.1 𝑥[1 − (1 + 𝑖)−𝑛 ] A✓value of n 𝑃= 𝑖 A✓value of i 14,25% −240 CA✓substitution into correct 𝑥 [1 − (1 + ) ] 12 formula 950000 = 14,25% 12 (4) 𝑥 = 𝑅11986,33 CA✓answer 8.2.2 𝑥[1 − (1 + 𝑖)−𝑛 ] 𝑃= 𝑖 A✓Present value formula 14,25% −140 A✓value of n 11986,33 [1 − (1 + ) ] 12 CA✓substitution into correct = 14,25% formula 12 = 𝑅816048,67 CA✓answer (4) OR OR 𝐴 = 𝑃(1 + 𝑖)𝑛 14,25% 100 𝐴 = 950 000 (1 + ) 12 A✓Substitution into Compound =R3093215,766 Interest Formula 𝑥[(1 + 𝑖)𝑛 − 1] 𝐹= 𝑖 14,25% 100 11986,33 [(1 + 12 ) − 1] CA✓substitution into Future Value 𝐹= 14,25% Formula 12 =R2277167,107 Balance on Loan = R3093215,766 −R2277167,107 CA✓A -F (4) = R816048,67 CA✓answer

Downloaded from hlayiso.com Mathematics P1 10 September 2019 Preparatory Examination NSC Marking Guideline 8.2.3 A = P(1 + i) n 4  14,25%  = 816 048,671 +  CA✓substitution  12  = R855 506,92 𝑥[1−(1+𝑖)−𝑛 ] 855 506,92 = CA✓substitution of P and i 𝑖 A✓value of n 14,25% −136 𝑥 [1 − (1 + ) ] 12 = 14,25% 12 CA✓answer 𝑥 = 𝑅12711,51 (4) [15] QUESTION 9(penalize 1 mark once for incorrect notation in this question) 9.1 f (x + h ) − f (x ) A✓formula f ( x) = lim h →0 h (x + h )2 − 1 ( x + h) −  x 2 − 1 x  A✓substitution = lim 2  2  h →0 h 1 1 1 x 2 + 2 xh + h 2 − x − h − x 2 + x CA✓ simplification of numerator = lim 2 2 2 h →0 h (5)  1 h 2 x + h −  CA✓factorization = lim  2 h →0 h 1 = 2x − CA✓answer 2 OR OR 1 𝑓(𝑥 + ℎ) = (𝑥 + ℎ)2 − (𝑥 + ℎ) 2 A✓value of f(x+h) 1 1 CA✓ simplification 𝑓(𝑥 + ℎ) = 𝑥 2 + 2𝑥ℎ + ℎ2 − 𝑥 − ℎ 2 2 1 𝑓(𝑥 + ℎ) − 𝑓(𝑥) = 2𝑥ℎ + ℎ2 − ℎ 2 2 1 (5) 𝑓(𝑥 + ℎ) − 𝑓(𝑥) 2𝑥ℎ + ℎ − 2 ℎ = ℎ ℎ 1 CA✓factorization 𝑓(𝑥 + ℎ) − 𝑓(𝑥) ℎ(2𝑥 + ℎ − 2) = ℎ ℎ 1 A✓formula 𝑓 ′ (𝑥) = lim (2𝑥 + ℎ − ) ℎ→0 2 1 CA✓answer 𝑓 ′ (𝑥) = 2𝑥 − 2

Downloaded from hlayiso.com Mathematics P1 11 September 2019 Preparatory Examination NSC Marking Guideline 9.2.1 𝑑 5 [3𝑥4 + √𝑥 + 𝑎2 ] 𝑑𝑥 𝑑 1 [3𝑥4 + 𝑥5 + 𝑎2 ] A✓rewriting in exponential form 𝑑𝑥 A✓CA✓derivatives 1 4 Penalize 1 mark if a is included in (3) = 12𝑥 3 + 𝑥 −5 5 answer 9.2.2 𝑥𝑦 = 𝑥 + 𝑥 2 − 1 A✓ dividing by x (x ≠ 0) 𝑦 = 1 + 𝑥 − 𝑥 −1 A✓ 1 + 𝑥 − 𝑥 −1 𝑑𝑦 = 1 + 𝑥 −2 𝑑𝑥 CACA✓✓each derivative (4) [12] QUESTION 10 10.1.1 𝑥 3 + 5𝑥 2 − 8𝑥 − 12 = 0 (𝑥 + 1) is a factor 𝑓(−1) = 0 (𝑥 + 1)(𝑥 2 − 4𝑥 − 12) = 0 A ✓𝑓(−1) = 0 (6) A ✓(𝑥 + 1)(𝑥 2 − 4𝑥 − 12) = 0 (𝑥 + 6)(𝑥 + 1)(𝑥 − 2) = 0 A✓ all three factors 𝑥 = −6 𝑜𝑟 𝑥 = −1 𝑜𝑟 𝑥 = 2 CACACA✓✓✓each value Answer only 3/6 Marks 10.1.2 𝑓(𝑥) = 𝑥 3 + 5𝑥 2 − 8𝑥 − 12 𝑓 / (𝑥) = 3𝑥 2 + 10𝑥 − 8 = 0 A✓derivative and equal to 0 (3𝑥 − 2)(𝑥 + 4) = 0 CA✓factors 2 𝑥 = 𝑜𝑟 𝑥 = −4 CA✓x – values 3 2 2 3 2 2 2 400 𝑓 ( ) = ( ) + 5 ( ) − 8 ( ) − 12 = − 3 3 3 3 27 = −14,81 CA✓y – value 2 (4) 𝐵 ( ; − 14,81) 3 10.1.3 𝑓 ′′ (𝑥) = 6𝑥 + 10 = 0 CA✓second derivative and equal to 0 5 CA✓answer (2) 𝑥=− 3 OR OR 2 + (−4) 5 CA✓subst. into midpoint 𝑥=3 =− formula (2) 2 3 CA✓answer OR OR 𝑏 5 A✓formula (2) 𝑥=− =− 3𝑎 3 CA✓answer

Downloaded from hlayiso.com Mathematics P1 12 September 2019 Preparatory Examination NSC Marking Guideline 10.2.1 𝑓 / (0) = −8 CA✓gradient (2) 𝑦 = −8𝑥 − 12 CA✓answer 10.2.2 𝑓 ′ (𝑥). 𝑔′ (𝑥) > 0 Since 𝑔′ (𝑥) < 0 for all 𝑥 ∈ 𝑅 A✓𝑔′ (𝑥) < 0 (3𝑥 2 + 10𝑥 − 8) < 0 (3𝑥 − 2)(𝑥 + 4) < 0 CA✓factors 2 −4 < 𝑥 < 3 CA✓answer (3) OR OR 2 CACA ✓✓end points −4 < 𝑥 < 3 A✓interval (3) [14] QUESTION 11 11.1 𝛥𝑃𝑄𝐵: PQ = 𝑡𝑎𝑛 6 0∘ 2𝑦 A✓ 𝑡𝑎𝑛 6 0∘ ∴ PQ = 2√3𝑦 A✓setting up ratio QR = 𝑑 − 4𝑦 A✓value of PQ (4) A = 2√3𝑦(𝑑 − 4𝑦) A✓value of QR OR OR ∆𝐴𝑃𝑆 is equilateral AP = PS = AS = 𝑑 − 4𝑦 A✓value of PS A✓value of SC SC = 4𝑦 SR2 = (4𝑦)2 − (2𝑦)2 = 12𝑦 2 A✓use of theorem of Pythagoras SR = 2√3𝑦 A✓value of SR A = 2√3𝑦(𝑑 − 4𝑦) (4) 11.2 A = 2 3 y (d − 4 y ). = 2 3 yd − 8 3 y 2 A✓expression for Area A/ = 2 3d − 16 3 y = 0 CA✓derivative and equal to 0 d − 8y = 0 d y= CA✓y – value 8

Downloaded from hlayiso.com Mathematics P1 13 September 2019 Preparatory Examination NSC Marking Guideline d  d  CA✓substitution into original Max A = 2 3.  d − 4   equation 8  8  3d  d CA✓simplifying = d −  4  2 3d  d  3d 2 CA✓answer =  = (6) 4 2 8 [10] QUESTION 12 12.1 18 17 18 17 P(GG) = . A✓ A✓ 40 39 40 39 51 = 𝑜𝑟 0.1962 𝑜𝑟 19.62% A✓answer in any form (3) 260 12.2 12 10 10 12 12 10 10 12 P(B and R) = . + . A✓40 . A✓40 . 39 39 40 39 40 39 2 A✓answer in any form P(B and R) = 𝑜𝑟 0,1538 𝑜𝑟 15,38 % (3) 13 [6] QUESTION 13 13.1 9 x 9 x 9 = 729 A✓9 x 9 x 9 (2) A✓729 13.2 5 x 8 x 7 = 280 A✓5 x 8 x 7 A✓280 OR (2) OR 5 8 digits 7 digits 56 ways 6 8 digits 7 digits 56 ways 7 8 digits 7 digits 56 ways A✓table 8 8 digits 7 digits 56 ways 9 8 digits 7 digits 56 ways Total 280 ways A✓answer (2) 13.3 9 x 1 x 9 = 81 A✓9 x 1 x 9 (2) A✓81 [6] Total : 150