Mathematics Grade 11 P1 Gauteng November 2021 Memo (English)

Downloaded from hlayiso.com PROVINCIAL EXAMINATION NOVEMBER 2021 GRADE 11 MARKING GUIDELINES MATHEMATICS (PAPER 1) 17 pages

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 INSTRUCTIONS AND INFORMATION A – ACCURACY C.A. – CONSISTENT 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 guidelines. • Assuming values/answers in order to solve a question is UNACCEPTABLE. 2

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 QUESTION 1 1.1 1.1.1 (𝑥 + 2)(2𝑥 − 1) = 0  simplification 1  both values for x 𝑥 = −2 or 𝑥 = 2 (2) 1.1.2 3− x = 0  x=3  answer (1) 1.2 1.2.1 2 x 2 − 3x − 4 = 0 −b  b 2 − 4ac x= 2a  correct substitution −(−3)  (−3) − 4(2)(−4) 2 x=  answers 2(2) x = 2,35 or x = −0,85 (3) 1.2.2 (2 x + 3)(3 − x)  4 −2 x 2 + 3 x + 9  4  simplify −2 x 2 + 3 x + 5  0  standard form 2 x − 3x − 5  0 2  factors  critical values (2 x − 5)( x + 1)  0  answer 5 −1  x  2 NOTE: Candidates can write the answer as separate inequalities but must state AND between the inequalities, not OR. (5) 1.2.3 x+2+x =4 x+2 = 4− x  squaring both sides x + 2 = 16 − 8 x + x 2 𝑎2 + 𝑏 2 = 𝑐 2  standard form  factors 0 = x − 9 x + 14 2  critical values 0 = ( x − 7)( x − 2)  exclusion of x = 7 x  7 or x = 2 (5) 3

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 1.2.4 22 x +1 − 3(22 x −1 ) + 4 x = 12  3  22 x  2 − + 1 = 12  correct factorisation  2  3  simplification 2 2 x = 8 22 x   = 12  equating to like bases 2  answer 22 x = 8 2 2 x = 23 3 x= 2 (4) 1.3 1.3.1 f ( x) = x 2 + 8 x + 16 f (2) = 22 + 8(2) + 16 f (2) = 36  answer (1) 1.3.2 f ( x)  0 x 2 + 8 x + 16  0  factors ( x + 4)( x + 4)  0 ( x + 4) 2  0  critical value CV : − 4  answer x ; x  −4 NOTE: Both criteria must be presented for the answer mark. (3) 4

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 1.4 𝑦 − 2𝑥 + 1 = 0 and 𝑥𝑦 = 2𝑦 + 𝑥 2 + 3𝑥 − 10 … (2) 𝑦 = 2𝑥 − 1 … (1)  y as subject in (1) sub (2) into (1)  substitution  standard form 2 𝑥(2𝑥 − 1) = 2(2𝑥 − 1) + 𝑥 + 3𝑥 − 10  factors 2𝑥 2 − 𝑥 = 4𝑥 − 2 + 𝑥 2 + 3𝑥 − 10  values for x 𝑥 2 − 8𝑥 + 12 = 0 (𝑥 − 6)(𝑥 − 2) = 0  values for y 𝑥 = 6 or 𝑥 = 2 if 𝑥 = 6 if 𝑥 = 2 𝑦 = 2(6) − 1 𝑦 = 2(2) − 1 𝑦 = 11 𝑦=3 NOTE: Any other valid method. (6) 1.5 ( xp ) 2 + 3 = 2 x( p − x) x 2 p 2 + 3 = 2 px − 2 x 2  standard form x p + 3 − 2 px + 2 x = 0 2 2 2 ( p 2 + 2) x 2 − 2 xp + 3 = 0  substitution = b 2 − 4ac = ( −2 p) 2 − 4( p 2 + 2)(3)  value for ∆ = 4 p 2 − 12 p 2 − 24 = −8 p 2 − 24  complete explanation p 2  0 for p  −8 p 2  0 for p  −8 p 2 − 24  0 for p   roots are non-real. (4) [34] 5

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 QUESTION 2 1 2.1 2.1.1 −  16 x 2  2  −2   81x   interchange base AND 1 additive inverse of  81x −2  2 index = 2   16 x   answer 9 = 2 4x NOTE: Any correct alternative method. (2) 2.1.2 9 x + 32 x +1 18x.2− x  simplification of 9x and 32 x + 32 x +1 18x = 2x x −x 3 .2 .2  factorise 32 x (1 + 3)  answer = 2x 0 3 .2 =4 NOTE: Any correct alternative method. (3) 2.1.3 10 − 32 x = 3x − 2  squaring both sides 10 − 32 x = (3x − 2) 2  simplification 10 − 32 x = 32 x − 4.3x + 4  standard form 2.32 x − 4.3x − 6 = 0  factors 32 x − 2.3x − 3 = 0 (3x − 3)(3x + 1) = 0  value of x and rejection 3x = 3 or 3x = −1 x =1 no solution (3x  0) (5) 2.2 3 2a − 2 3a  surd and squaring = (3 2a − 2 3a ) 2  multiplication = 9. 4a 2 − 12 6a 2 + 4. 9a 2  simplification  simplification = 9.2a − 12 6a 2 + 4.3a 18a − 12a 6 + 12a = 30a − 12a 6 (4) [14] 6

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 QUESTION 3 3.1 3.1.1 A=9  answer B = 11  answer (2) 3.1.2 Tn = 2n + 1  answer (2) 3.1.3 Tn = 2n + 1 Tn = 2(100) + 1  substitution Tn = 201  answer NOTE : Answer only, award FULL marks. (2) 3.1.4 Tn = 2n + 1 305 = 2n + 1  substitution 304 = 2n  answer n = 152 (2) 3.2 𝑇3 − 𝑇2 = 𝑇2 − 𝑇1 3𝑥 2 + 4𝑥 − 3 − (2𝑥 2 + 5𝑥 + 2)  substitution = 2𝑥 2 + 5𝑥 + 2 − (2𝑥 2 + 5𝑥 + 5 ) 3𝑥 + 4𝑥 − 3 − 2𝑥 2 − 5𝑥 − 2 2  simplification = 2𝑥 2 + 5𝑥 + 2 − 2𝑥 2 − 5𝑥 − 5 𝑥 2 − 𝑥 − 5 = −3  standard form 𝑥2 − 𝑥 − 2 = 0 (𝑥 − 2)(𝑥 + 1)  answers ∴ 𝑥 = 2 or 𝑥 = −1 (4) [12] 7

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 QUESTION 4 4.1 -1 ; x ; 3 ; x+8 x-(-1) 3-x x+8-3  first differences 3-x-(x+1) x+5-(3-x)  second differences 3 − x − ( x + 1) = x + 5 − (3 − x)  equating 3 − x − x −1 = x + 5 − 3 + x 2 − 2x = 2x + 2  answer 4x = 0 x = 0 (4) 4.2 -1 ; 0 ; 3 ; 8  2nd difference 1 3 5  value for a 2 2  value for b 2a = 2 a =1  value for c 3a + b = 1  equation of Tn 3(1) + b = 1  complete the square of b = −2 Tn a + b + c = −1 1 − 2 + c = −1 c=0 Tn = n 2 − 2n Tn = n 2 − 2n + 1 − 1 Tn = (n − 1) 2 − 1 NOTE: If a candidate uses Tn = (n − 1) − 1 to 2 validate the quadratic pattern, award ZERO marks. (6) 8

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 4.3 Tn = (n − 1) 2 − 1  substitution T65 = (65 − 1) 2 − 1  answer T65 = 4095 (2) 4.4 (n − 1) 2 − 1  9800  setting up inequality n 2 − 2n − 9800  0  factors (n + 98)(n − 100) CVs -98 or 100  choice of n = 100  answer n  100  n = 101 (4) [16] QUESTION 5 5.1 5.1.1 A = P(1 − i ) n  correct substitution A = 250000(1 − 0.1)5 into correct formula A = R147622,50  answer (2) 5.1.2 𝐴 = 𝑃(1 + 𝑖)𝑛  correct substitution 𝐴 = 250 000(1 + 0.08)5 into correct formula 𝐴 = 𝑅367 332,02  answer (2) 5.1.3 95 × 147 622.50 100 = 𝑅140 241,375  R140 241, 375 New Machine - Sale of Old Machine  answer 𝑅367 332,02 − 𝑅140 241,375 𝑅227 090,65 (2) 5.2 5.2.1  i  m 1 + ieff = 1 + nom   m   correct substitution 12  0.115  into correct formula 1 + ieff = 1 +   12,13%  12   conclusion ieff = 12,13% Bank B offers a better interest rate. NOTE: Award no marks if candidate concludes only. (3) 9

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 5.2.2 𝐴 = 𝑃(1 + 𝑖)𝑛 0.115 24 0.115 36  value of i 𝑅280 000 = [𝑃 (1 + ) − 50 000] (1 + ) 12 12  n = 24 𝑃 = 𝑅197 759,51.  n = 36  setting up equation OR  answer 𝑛 𝐴 = 𝑃(1 + 𝑖) 0.115 60 0.115 36 OR 𝑅280000 = 𝑃 (1 + ) − 50 000 (1 + ) 12 12  value of i 𝑃 = 𝑅197 759,51  n = 60  n = 36 OR  setting up equation 𝐴 = 𝑃(1 + 𝑖)𝑛  answer 𝑅280 000 = [𝑃(1 + 0,1213)2 − 50 000](1 + 0,1213)3 OR 𝑃 = 𝑅197 729,98  value of i  n=2 OR  n=3 𝐴 = 𝑃(1 + 𝑖)𝑛  setting up equation 𝑅280 000 = 𝑃(1 + 0,1213)5 − 50 000(1 + 0,1213)3  answer 𝑃 = 𝑅197 727.98 OR  value of i  n=5  n=3  setting up equation  answer (5) [14] 10

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 QUESTION 6 6.1 3 f ( x) = −1 x+2 3  substitution 2= −1 k +2 3 3= k +2  answer 3k + 6 = 3 k = −1 (2) 6.2 x = −2  answer y = −1  answer (2) 6.3 3 f ( x) = −1 x+2 3 f (0) = −1 0+2 1 = 2  answer  1   0;   2 NOTE: Answer MUST be presented in coordinate form. (1) 6.4 3 f ( x) = −1 x+2 3 0= −1 x+2  simplification 3 1= x+2  answer x+2=3 x =1 (1;0) NOTE: Answer does NOT need to be in coordinate form. (2) 11

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 6.5  shape  asymptotes  x- and y- intercepts (3) 6.6 m = −1 and (-2;-1) is on the axis of symmetry  y = mx + c  substitution for m −1 = −1(−2) + c AND the point  c = −3  answer  y = −x − 3 NOTE: Any other valid method. (2) 6.7 x > –2  answer (1) 6.8 3 f ( x) = −1 x+2 3 −y = −1  make y negative x+2  3   subtract 4 y = − − 1  x+2  −3  answer y= +1 x+2−4 −3 y= +1 x−2 (3) [16] 12

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 QUESTION 7 7.1 q=5  answer (1) 7.2 f ( x) = − x 2 − 4 x + 5 x2 + 4x − 5 = 0  factors ( x + 5)( x − 1) = 0  critical values  answer x = −5 or x = 1  EP = 6 units (3) 7.3 −𝑏 𝑥= 2𝑎 −(−4) 𝑥=  x-value 2(−1) 𝑥 = −2 𝑓(−2) = 9  y-value ∴ 𝑁(−2; 9) OR OR 𝑓(𝑥) = −(𝑥 2 + 4𝑥) + 5  x-value 𝑓(𝑥) = −(𝑥 2 + 4𝑥 + 22 ) + 5 + 4  y-value 𝑓(𝑥) = −(𝑥 + 2)2 + 9 ∴ 𝑁(−2; 9) NOTE: Answer does NOT have to be in coordinate form. (2) 7.4 m of g = 1  m of h = −1  gradient of h = –1 y = mx + c 9 = −1(−2) + c  c=7 c=7  answer h( x ) = − x + 7 NOTE: Any other valid method. (3) 7.5 M(0;5) and N(-2;9) y −y mMN = 2 1 x2 − x1 9−5  correct substitution mMN = −2 − 0  answer mMN = −2 NOTE: NO marks for applying incorrect gradient formula. (2) 13

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 7.6 At point N: f (−2) = 9  g(–2) = 3 At point T: g (−2) = 3  method ∴ distance of NT = 9 − 3  answer = 6 units OR OR NT = f ( x) − g ( x) = − x 2 − 4 x + 5 − ( x + 5) = − x2 − 5x  method x = −2  NT = −(−2) 2 − 5(−2)  substitution NT = 6 units  answer NOTE: Any other valid method. (3) 7.7 By symmetry: 𝑆(−4; 5)  answer OR OR At point S: y = 5 ∴ 5 = −𝑥 2 − 4𝑥 + 5  y=5 𝑥 2 + 4𝑥 = 0  equating 𝑥(𝑥 + 4) = 0 𝑥 = 0 or 𝑥 = −4  selection of –4 ∴ 𝑆(−4; 5) (3) 7.8 7.8.1 x=–5  answer OR  answer x=0 (2) 7.8.2 x 1  answer (1) [20] 14

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 QUESTION 8 8.1 let x = 0 f ( x) = 3.2 x − 6 f (0) = 3.20 − 6 y = −3  answer  (0; −3) NOTE: Answer MUST be in coordinate form. (1) 8.2 let y = 0 0 = 3.2 x − 6 6 = 3.2 x  simplify both sides 2=2 x to base 2  answer x =1  (1;0) NOTE: Answer does NOT have to be in coordinate form. (2) 8.3  shape  asymptote  x- and y- intercepts (3) 8.4 𝑦 > −6 𝐎𝐑  answer 𝑦 ∈ [−6; ∞] (1) [7] 15

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 QUESTION 9 NOTE: Candidates may present answers as a fraction, decimal fraction or percentage. 9.1 9.1.1 120 120 P(male)=  236 236 P(male)= 0,51  answer (2) 9.1.2 67 𝑃(Female and plays sport) = 236 67  𝑃(Female and plays sport) = 0,28 236  answer (2) 9.2 51 𝑃(Male and does not play sport) = 236  0,22 𝑃(Male and does not play sport) = 0,22  conclusion ≠0 NOT mutually exclusive OR OR  statement There are 51 male learners who do not play sport.  conclusion Therefore the two are NOT mutually exclusive. NOTE: Accept alternative answers expressed as a percentage or in fraction form. (2) 9.3 𝑃(male and does not play sport) = 0,22  P(male) x 𝑃(male)×P(does not play sport) P(does not play sport) 120 100 = 0,22 = × 236 236 = 0,22  equating ∴ 𝑃(male and does not play sport)  conclusion = 𝑃(male)×P(does not play sport) ∴ the events ARE independent (correct to two decimal places) (4) [10] 16

Downloaded from hlayiso.com MATHEMATICS MEMORANDUM (Paper 1) GRADE 11 QUESTION 10 10.1 c = 84 − 75 c=9  answer (1) 10.2 f + 18 + x = 41  equating correctly f = 41 − x − 18  answer f = 23 − x (2) 10.3 f = 23 − x  x = 23 − f e = 17 − x  x = 17 − e  equating expressions of x  23 − f = 17 − e  equating to 75 but 40 + e + f + x = 75  simplification  e + f + x = 35  answer  17 − x + 23 − x + x = 35 40 − x = 35 x =5 (4) [7] TOTAL: 150 17