Mathematics Grade 11 Paper 1 Gauteng June 2017 Memo

Downloaded from hlayiso.com GAUTENG DEPARTMENT OF EDUCATION PROVINCIAL EXAMINATION JUNE 2017 GRADE 11 MATHEMATICS PAPER 1 MEMORANDUM 11 pages

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 GAUTENG DEPARTMENT OF EDUCATION PROVINCIAL EXAMINATION MATHEMATICS (Paper 1) MEMORANDUM INFORMATION A – Accuracy CA – Continued Accuracy 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 memorandum. • Assuming values/answers in order to solve a problem is UNACCEPTABLE. 2

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 QUESTION 1 MARKS: 19 𝑥 2 − 2𝑥 = 15 𝑥 2 − 2𝑥 − 15 = 0 ✓ std. form 1.1.1 (x – 5)(x + 3) = 0 ✓ correct factors x = 5 or x = -3 ✓ both solutions (3) 4𝑥 2 − 𝑥 − 5 < 0 (4𝑥 − 5)(𝑥 + 1) < 0 ✓ factors 1.1.2 + - + -1 1¼ −1 < 𝑥 < 14 1 ✓ critical values ✓ correct inequality (3) √2𝑥 − 1 + 2 = 𝑥 (√2𝑥 − 1 )2 = (𝑥 − 2)2 ✓ squaring both sides 2𝑥 − 1 = 𝑥 2 − 4𝑥 + 4 0 = 𝑥 2 − 6𝑥 + 5 ✓ std. form 1.1.3 0 = (𝑥 − 5)(𝑥 − 1) ✓ factors 𝑥 = 5 or 𝑥 = 1 ✓ both solutions Test: x = 5 only ✓ accepting x = 5 (5) 2 2𝑥 3 – 8 = 0 2 2𝑥 3 = 8 2 2 ✓ 𝑥3 = 4 𝑥3 = 4 𝟐 𝟑 𝟑 ✓ raising both sides to the (𝒙𝟑 )𝟐 = ± (𝟒)𝟐 power 3 2 𝑥 = ± 23 OR 1.1.4 ✓x = ±23 𝐎𝐑 𝑥 = ±8 𝑥 =±8 OR OR 2 2 𝑥3 = 4 ✓ 𝑥3 = 4  x   4 3 2 3 3 ✓ raising both sides to the power of 3 x2 = 64 x=±8 ✓ x = ±23 𝐎𝐑 𝑥 = ±8 (3) 3

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 𝑎+𝑏+𝑐 =0 𝑏 = −𝑎 − 𝑐 ✓ 𝑏 = −𝑎 − 𝑐 2 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0 −𝑏 ± √𝑏 2 − 4𝑎𝑐 𝑥= 2𝑎 −(−𝑎 − 𝑐) ± √(−𝑎 − 𝑐)2 − 4𝑎𝑐 = ✓ sub. in corr. formula 2𝑎 𝑎 + 𝑐 ± √𝑎2 + 2𝑎𝑐 + 𝑐 2 − 4𝑎𝑐 = 2𝑎 1.2 𝑎 + 𝑐 ± √𝑎2 − 2𝑎𝑐 + 𝑐 2 ✓ simplification to = 2𝑎 √𝑎2 − 2𝑎𝑐 + 𝑐2 𝑎 + 𝑐 ± √(𝑎 − 𝑐)2 = ✓ (𝑎 − 𝑐)2 2𝑎 𝑎 + 𝑐 ± (𝑎 − 𝑐) = 2𝑎 𝑐 ✓x=1 ∴ 𝑥 = 1 or 𝑥 = 𝑎 (5) QUESTION 2 MARKS: 15 3𝑦 + 𝑥 = 2 … … … … . . (1) 𝑦 2 + 𝑥 = 𝑥𝑦 + 𝑦.........(2) from (1) 𝑥 = 2 − 3𝑦 ✓ 𝑥 = 2 − 3𝑦 subst. 𝑥 = 2 − 3𝑦 in (2) 2 𝑦 + (2 − 3𝑦) = 𝑦(2 − 3𝑦) + 𝑦 ✓ sub. 𝑥 = 2 − 3𝑦 in........ (2) 𝑦 2 + 2 − 3𝑦 = 2𝑦 − 3𝑦 2 + 𝑦 2.1 4𝑦 2 − 6𝑦 + 2 = 0 ✓ std. form [÷ 2] 2𝑦 2 − 3𝑦 + 1 = 0 (2𝑦 − 1)(𝑦 − 1) = 0 ✓ factorisation 1 𝑦 = or 𝑦 = 1 ✓ both y values 2 1 𝑥= or 𝑥 = −1 ✓ both x values 2 (6) 4

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 L.H.S: √𝑏√𝑎 − 𝑏 . √𝑏√𝑎 + 𝑏 = √(𝑏√𝑎 − 𝑏)(𝑏√𝑎 − 𝑏) ✓ (𝑏√𝑎 − 𝑏)(𝑏√𝑎 − 𝑏) = √𝑏 2 𝑎 − 𝑏 2 2.2 ✓ simplification √𝑏 2 𝑎 − 𝑏 2 = √𝑏 2 (𝑎 − 1) ✓ factorisation ✓b = 𝑏√𝑎 − 1 = R.H.S (4) 3𝑎 = 21𝑏 (3𝑎 )𝑐 = (21𝑏 )𝑐 ✓ intro. power c on both sides 3𝑎𝑐 = (3𝑏 × 7𝑏 )𝑐 ✓ (3𝑏 × 7𝑏 )𝑐 2.3 3𝑎𝑐 = 3𝑏𝑐 × 3𝑎𝑏 ✓ deducing that 7𝑐 = 3𝑎 3𝑎𝑐 = 3𝑏𝑐 + 𝑎𝑏 𝑎𝑐 = 𝑏𝑐 + 𝑎𝑏 ✓ equating the indices 𝑎𝑐 = 𝑏(𝑐 + 𝑎) ✓ factorisation 𝑎𝑐 (5) ∴𝑏= 𝑎+𝑐 QUESTION 3 MARKS: 18 3.1.1 13 ✓ answer (1) 𝑇𝑛 = 𝑎𝑛 + 𝑏 3.1.2 −3 = 4(1) + 𝑏 −7 = 𝑏 ✓ b = −7 𝑇𝑛 = 4𝑛 − 7 ✓ answer ANY other valid method (2) 394 = 4𝑛 − 7 ✓ equation 401 = 4𝑛 3.1.3 100,25 = 𝑛 ✓ 100,25 = 𝑛 since 𝑛 ∉ ℕ, 394 is NOT a term ✓ any valid explanation (3) 5

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 𝑇𝑛 = 𝑛2 + 2𝑛 + 𝑐 0 = (1)2 + 2(1) + 𝑐 ✓ corr. sub. 0=3+𝑐 ✓ simplification −3 = 𝑐 3.2.1 OR OR T1 =a + b + c ✓ corr. sub. 0=1+2+c ✓ simplification 0=3+c –3 = c (2) 𝑇𝑛 = 𝑛2 + 2𝑛 + 𝑐 3.2.2 𝑇10 = (10)2 + 2(10) − 3 ✓ corr. sub. =117 ✓ answer (2) 𝑛2 + 2𝑛 − 3 > 360 𝑛2 + 2𝑛 − 363 > 0 ✓ 𝑛2 + 2𝑛 − 363 > 0 (−2)±√22 −4(1)(−363) 3.2.3 C.V: 𝑛= ✓ corr. sub. in corr. formula 2(1) 𝑛 = −20,08 or 𝑛 = 18,08 ✓ both critical values ∴ 𝑛 = 19 ✓ n = 19 𝑇19 > 360 (4) a = 10 ✓ answer 3.3.1 (1) b = 99 + 98 + 97 + 96 ........+ 50 +..........+ 3 + 2 + 1 ✓ sequence (99 + 1) = 100 (98 +2) = 100 3.3.2 : ∴ 𝑏 = (100) × 49 + 50 ✓ (100) × 49 b = 4950 ✓ answer (3) 6

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 QUESTION 4 MARKS: 6 𝑥−2≥0 ✓𝑥−2≥0 4.1 𝑥≥2 ✓ answer (2) 4.2 2 ✓ answer (1) √3 − 2 𝑘 =2+ 4 1 1 = 2 ✓𝑘=2 4 4 4.3 1 2 𝑔(𝑘) = (2 ) − 1 4 ✓ substitution 1 65 = 4 OR OR 4,06 16 16 ✓ answer (3) MARKS: 12 QUESTION 5 for x- intercept, make y = 0 3𝑥 − 1 = 0 3𝑥 = 1 3 𝑥 = 30 ✓ 1 = 30 5.1 x=0 ✓𝑥=0 for y- intercept, make x = 0 𝑦 = 30 − 1 y=0 ✓y=0 (3) 7

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 y ✓ shape x 5.2. 0 ✓ passing thr. origin ✓ asymptote at y = -1 -1 (3) (−1 ; ∞) 5.3 OR 𝑦 > −1 ✓ answer (1) 5.4.1 𝑥 = −2 ✓ answer (1) 5.4.2 𝑥 > −1 ✓ answer (1) 3ℎ(𝑥) = 726 ℎ(𝑥) = 242 ✓ 242 5.5 3𝑥 – 1 = 242 3𝑥 = 243 3 𝑥 = 35 ✓ 35 𝑥=5 ✓ answer (3) 8

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 QUESTION 6 MARKS: 10 𝑥=1 ✓𝑥=1 6.1 𝑦 = −2 ✓ 𝑦 = −2 (2) for x-intercept, make y = 0 3 ✓ y=0 0= −2 𝑥−1 3 2= 𝑥−1 𝑥 = 212 ✓ 𝑥 = 2½ 6.2 for y -intercept, make x = 0 3 𝑦= −2 0−1 𝑦 = −5 ✓ 𝑦 = −5 (3) y ✓ shape ✓ x and y-intercepts ✓ asymptotes 6.3 1 1 2 -2 2 -5 (3) ✓ reflection 6.4 Reflection about the y-axis. ✓ y-axis (2) 9

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 QUESTION 7 MARKS: 20 −𝑏 𝑥= 2𝑎 −(8) ✓ corr. sub. in corr. formula = 7.1.1 2(2) ✓ x -co-ord. - 2 = −2 ✓ y -co-ord. - 5 2 ( if not ordered pair) D(- 2 ; - 5) 3 (3) 𝑥 ≥ 0 or 𝑥 ≤ −4 ✓𝑥≥0 7.1.2 ✓ 𝑥 ≤ −4 1 ( incorrect notation) 2 (2) LM = 𝑘(𝑥) − 𝑝(𝑥) = 2𝑥 2 + 8𝑥 + 3 − (2𝑥 − 4) = 2𝑥 2 + 6𝑥 + 7 ✓ simplification 4𝑎𝑐 − 𝑏 2 ✓ formula MinLM = 4𝑎 ✓ corr. sub. in corr. formula 4(2)(7) − 62 = 4(2) ✓ answer 1 = 22 OR OR 7.1.3 LM = 𝑘(𝑥) − 𝑝(𝑥) = 2𝑥 2 + 8𝑥 + 3 − (2𝑥 − 4) = 2𝑥 2 + 6𝑥 + 7 ✓ simplification −𝑏 𝑥= 2𝑎 −(6) ✓ corr. sub. in corr. formula = 2(2) 3 =− 2 3 3 MinLM = 2(− 2)2 + 6 (− 2 ) + 7 ✓ sub. x in eqn. ✓ answer = 212 (4) 10

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 𝑘(−2) − 𝑘(3) Ave. grad = −2 − 3 ✓ −5 7.1.4 −5 − 45 ✓ −45 = −2 − 3 ✓ answer = 10 (3) 2𝑥 2 + 8𝑥 + 3 = 2𝑥 + 𝑡 2𝑥 2 + 6𝑥 + 3 − 𝑡 = 0 ✓ std. form ∆ = 𝑏 2 − 4𝑎𝑐 = (6)2 − 4(2)(3 − 𝑡) ✓ corr. sub. in formula = 39 − 24 + 8𝑡 = 12 + 8𝑡 ✓ value of ∆ 7.1.5 for = roots, ∆ = 0 12 + 8𝑡 = 0 ✓ ∆= 0 8𝑡 = −12 1 3 𝑡 = −1 or − or -1,5 2 2 ✓ answer (5) y ✓ shape 1½ x 7.2 –4 ✓ x-intercepts 1;–5 ✓ turning point in 3rd quadrant (3) 11