Mathematics Grade 11 P1 Gauteng November 2022 Memo (English)

Downloaded from hlayiso.com PROVINCIAL EXAMINATION NOVEMBER 2022 GRADE 11 MARKING GUIDELINES MATHEMATICS (PAPER 1) 19 pages

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 QUESTION 1 1.1 1.1.1 When x = 2 and when x = 3 3− x 3−3 3− 2 0 and 1 =0 =1  x = 2 or x = 3 ✓ ✓ answers (2) 1.1.2 3 − x = 2x – 3 3 – x  0 and 2x – 3  0 ✓ setting up correct inequalities – x  –3 and 2x  3 3  x  3 and x  ✓✓ answers 2 3  x3 2 NOTE: Answer can be written as separate inequalities. OR NOTE: If the candidates solve the equation, the 3 answers are: x = or x = 2, the answer 4 3 3 of x = must be rejected to obtain , but 4 3 2 if not rejected, award . 3 (3) 1.2 (3a – 8)(2b + 7) = 0  3a – 8 = 0 or 2b + 7 = 0 8 a= ✓ value of a 3 7 b=– ✓ value of b 2 (2) 2

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 1.3 1.3.1 4x2 – 20x + 1 = 0 − b  b 2 − 4ac x= 2a − ( −20)  ( −20) 2 − 4(4)(1) ✓ substitution x= 2(4) 20  384 x= 8  x = 4,95 or x = 0,05 ✓✓ answers NOTE: Penalise one mark for incorrect rounding-off in this question ONLY. (3) 1.3.2 (x + 1)(x – 3) > 12 x2 – 2x – 3 > 12 x2 – 2x – 15 > 0 ✓ standard form (x – 5) (x + 3) > 0 ✓ factors  x > 5 or x < –3 ✓✓ answers (4) 1.3.3 x – 5+ x = 7 x–7= 5+ x (x – 7)2 = ( 5 + x ) 2 ✓ squaring both sides x2 – 14x + 49 = 5 + x ✓ simplification x2 – 15x + 44 = 0 ✓ standard form (x – 4)(x – 11) = 11 ✓ factors x = 11 or x  4 ✓ answers with rejection (5) 3

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 1.4 1.4.1 x = 3 and y = a x–y=1 3–a=1 a=2 but x2 – 3xy + by2 = –5 ✓ value of a  32 – 3(3)(2) + b (2)2 = –5 ✓ substitute for x and y  9 –18 + 4b = –5  4b = 4 ✓ simplification b=1 ✓ value of b (4) 1.4.2 x – y = 1  x = y + 1……(1) ✓ x as subject x2 – 3xy + y2 = –5……(2)  (y + 1)2 – 3y(y + 1) + y2 = –5 ✓ substitute into (2)  y2 + 2y + 1 – 3y2 – 3y + y2 = –5  – y2 – y + 6 = 0 ✓ standard form  y2 + y – 6 = 0 ✓ factors (y + 3)(y – 2) = 0  y = – 3 or y = 2 ✓ y-values  x = – 2 or x = 3 ✓ x-values The other solution: (– 2;– 3) NOTE: Candidates do not have to write the answer in coordinate form. (6) 4

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 1.5 1.5.1 (p + 1)x2 + 2px + (p + 2) = 0  = b2 – 4ac  = (2p)2 – 4(p + 1)(p + 2) ✓ substitution for   = 4p2 – 4p2 – 12p – 8  = – 12p – 8 ✓ expression for  For equal roots, =0 ✓ condition for  – 12p – 8 = 0 8 p=– 12 2 ✓ answer p=– 3 (4) 1.5.2  > 0 ✓ condition of  – 12p – 8 > 0 2 p<– 3 ✓ answer (2) [35] 5

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 QUESTION 2 2.1 1 1 1 2.1.1 ( n−1 . n+1 ) n 3 3 1 = (3−n+1.3−n−1 ) n ✓ simplification 1 = ( 3− 2 n ) n = 3−2 ✓ simplification 1 = ✓ answer 9 NOTE: Any other valid method. (3) 5 2 2 2.1.2 – 3 27 2 − 2 + − −2 8 3 4 5 1 1 2 25 ✓ simplification = [(33 ) 2 ] 3 − + −2 −2 (2 3 ) 3 (2 2 ) 5 1 2 25 = 3 − −2 + 4 2 ✓ simplification 2 − 2 5 = 9 – 23 + 21 ✓ simplification =3 ✓ answer NOTE: Any other valid method. (4) 6

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 2.2 32–x + 8 = 3x 32.3–x + 8 = 3x 9 + 8 = 3x ✓ simplification 3x 9 + 8.3x = 32x 32x + 8.3x – 9 = 0 ✓ standard form (3x – 9)(3x + 1) = 0 ✓ factors 3x = 9 or 3x = –1 3x = 32 or NA ✓ answer with rejection x=2 NOTE: Any other valid method. (4) [11] QUESTION 3 3.1 1 ; 2 ; 5 ; 10 +1 +3 +5 +2 +2 2 nd difference 2a = 2 3a + b = 1 a+b+c=1 ✓ value of a a=1 3(1) + b = 1 1–2+c=1 ✓ value of b b = –2 c=2 ✓ value of c  Tn = n2 – 2n +2 NOTE: No penalty if not written as an equation. (4) 3.2 T50 = 502 – 2(50) + 2 ✓ substitution T50 = 2402 ✓ answer (2) 3.3 ROW 11 ✓✓ answer (2) 7

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 3.4 1 ; 3 ; 5 ; 7 +2 +2 +2 Tn = 2n – 1 ✓ correct equation 241 = 2n – 1 ✓ Tn = 241 n = 121 ✓ answer (3) [11] QUESTION 4 4.1 4 ; 9 ; x ; 37… 5 ; x − 9 ; 37 − x ✓ 1st differences x − 9 − 5 ; 37 − x − ( x − 9) ✓ 2nd differences x − 14 ; 46 − 2 x x – 14 = 46 – 2x ✓ equating  3x = 60  x = 20 ✓ answer (4) 4.2 5 ; 11 ; 17 ;… (first differences) Tn = 6n – 1 ✓ correct equation 599 = 6n – 1 ✓ equating 600 = 6n n = 100 ✓ value of n  Between 100th and 101th terms ✓ conclusion (4) 8

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 4.3 4.3.1 4 ; 9 ; 20 ; 37 5 11 17 6 6 ✓ 2 nd difference 2a = 6 a=3  Tn = 3n2 + bn +c ✓ value of a Since the a-value is positive, Tn has a MINIMUM value. ✓ conclusion (3) 4.3.2 y = 3n2 – 4n +5 − (−4) x= 2(3) 4 x= ✓ value of x 6 2 x= 3  2  11 2  f  = ✓ value of f    3 3  3 11  range: y  ✓ answer 3 OR 11 y [ ; ) 3 NOTE: Correct brackets must be used to obtain the answer mark in option 2. This is a theoretical solution as it is understood that a number pattern is composed of terms which are indicated as natural numbers. (3) 9

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 4.4 For n to be common to both patterns: 6n – 1 = 3n2 – 4n + 5 ✓ equating  0 = 3n2 – 10n + 6 − b  b 2 − 4ac n= 2a − ( −10)  ( −10) 2 − 4(3)(6) n= 2(3) 5 7 n= ✓ simplification 3  NO, n is not a Natural Number. ✓ conclusion (3) [17] QUESTION 5 5.1 1 + ieffective = (1 + inominal)n 0,135  inominal = ✓ inominal 12 0,135 12 ✓ correct substitution into  1 + ie = (1 + ) 12 correct formula  1 + ie = 1,14  ie = 0,14 ✓ simplification ie  ie = 14% ✓ answer (4) 10

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 5.2 A = P(1 + in)  7 500 = 6 800(1 + 0,075n) ✓ substitution into correct formula 7 500  – 1 = 0,075n 6 800 0,075n = 0,10  n = 1,333 yrs ✓ value of n  n = 1,333(12)  n = 15,996  n  16 mnths ✓ answer NOTE: The answer mark is for 16 months. (3) 5.3 Compounded semiannually for 6 years.  n = 12 ✓ value of n and i 0,05 i= = 0,025 2 A = P(1 + i)n A = 5 000(1 + 0,025)12 ✓ substitution into correct formula  A = R6 724,44 ✓ answer (3) 5.4 5.4.1 A = P(1 + in) A = P + Pin  linear function ✓ answer (1) 11

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 5.4.2 m = Pi P>0 i>0 m>0 A ✓m>0 ✓ shape (c > 0) P n 0 (2) 5.4.3 n = 0: A = P + Pi(0) A=P n = 1: A = P + Pi(1) A = P + Pi  an INCREASE in Pi ✓ answer NOTE: Answer only, full marks. (1) [14] 12

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 QUESTION 6 6.1 x=2 ✓ answer y=1 ✓ answer (2) 6.2 y = –1 ✓ answer (1) 6.3 x g f ✓ shape of f x=1 ✓ intercepts of f x ✓ shape of g O ✓ intercept of g x = –1 x=2 (4) 6.4 x  R; x  2 or x  (–;2) or (2; ) ✓ answer NOTE: Must state both conditions in option 1. (1) 6.5 y > –1 or y  (–1;) ✓ answer (1) 6.6 Point of intersection of asymptotes: (2;1) y – y1 = m(x – x1) y – 1 = –1(x – 2) ✓ substitute m and pt (2 ; –1) y = –x +3 ✓ answer (2) 6.7 2 units to the right and 1 unit upwards ✓ units right ✓ units up (2) 13

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 6.8 d = ( x2 − x1 ) 2 + ( y 2 − y1 ) 2 d = (2 − 0) 2 + (1 − 0) 2  (0;0)(2;1) ✓ correct substitution d= 5 d = 2,24 ✓ answer NOTE: Answer only, full marks. No penalty for rounding-off incorrectly. (2) 6.9 (3 ; 0) ✓ x-value ✓ y-value (2) 6.10 x (–;2) or x<2 ✓ answer (1) [18] QUESTION 7 7.1 y = a( x − x1 )( x − x2 ) 4 = a(0 − 1)(0 − 3) ✓ substitute roots and point (0 ; 4) 4 = 3a 4 a = ✓ value for a 3 4 y= ( x − 1)( x − 3) 3 4 2 y= ( x − 4 x + 3) 3 4 2 16 y= x − x+4 ✓ answer 3 3 (3) 14

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 4 2 16 7.2 f ( x) = x − x+4 3 3 −b x= 2a 16 − (− ) x= 3 ✓ substitution into correct 4 formula 2( ) 3 x=2 ✓ x-value 4 2 16  f (2) = (2) − (2) + 4 3 3 4  f (2) = − ✓ y-value 3 4  D(2 ; − ) 3 NOTE: Answer does not have to be in coordinate form. (3) 0−4 7.3 mg = (0;4)( −8;0) ✓ substitute correctly into −8−0 gradient formula 1  mg = ✓ value of mg 2  m p = −2 ✓ value of mp 4 4  − = −2(2) + c  D(2;− ) ✓ substitute m and point D 3 3 4  c =4− 3 8  c= ✓ value of c 3 8  p( x ) = −2 x + 3 NOTE: Answer does not have to be an equation. (5) 15

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 7.4 m = tan 1 1 = tan ✓ tan = 2 2   = 26,57° ✓ answer (2) 7.5 4 2 16 1 x – x+4= x+4 ✓ equating 3 3 2 4 2 35 x – x =0 3 6 4 35  x x −  = 0 ✓ factors 3 6 35 x = 0 of x = ✓ x-answers 8 1  35  y=   +4 2  8  99 y= ✓ y-answer from correct selection 16 of x-value  35 99  K ;   8 16  NOTE: Answer does not need to be in coordinate form. (4) 7.6 7.6.1 1 < x < 3 ✓ answer NOTE: Can be written as separate inequalities. (1) 7.6.2 x(–8 ; 1] or [3 ; ) ✓ answer ✓ answer NOTE: Penalise 1 mark if answer is not with correct brackets. (2) 16

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 4 7.7 TP(2 ; – ) from Q. 7.2 3 4 4  f (x) = (x – 2)2 – ✓ equation of f 3 3 4 13 ✓ equation of h  h(x) = (x – 4)2 – 3 3 (2) 7.8 7.8.1 j(x) = ax –8  0 = –4a –8……pt(–4 ; 0) ✓ substitution  4a = –8  a = –2 ✓ answer (2) 7.8.2 The graph of g to j is a 90° anticlockwise ✓ answer rotation about the origin. NOTE: Accept, graphs are perpendicular and if g is shifted anti-clockwise by 90°, j is obtained. (1) [25] QUESTION 8 8.1 Total = 13 + 7 + 24 + 3 Total = 47 ✓ answer (1) 13 + 7 8.2 8.2.1 P(M) = 47 20 P(M) = …or…0,43 ✓ answer 47 (1) 7 8.2.2 P(M and S) = or 0,15 ✓ answer 47 (1) 3 8.2.3 P(not M or S) = or 0,06 ✓ answer 47 (1) 17

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 13 + 7 + 24 8.2.4 P(M or S) = ✓ method 47 44 P(M or S) = or 0,94 ✓ answer 47 OR P(M or S) = P(M) + P(S) – P(M and S) 20 31 7 = + – ✓ method 47 47 47 44 P(M or S) = or 0,94 ✓ answer 47 (2) 8.2.5 13 + 24 P(M or only S) = ✓ method 47 37 P(M or only S) = or 0,79 ✓ answer 47 NOTE: Answer only, full marks. (2) 8.3 8.3.1 P(A or B) = P(A) + P(B) – P(A and B) = 0,95 + 0,98 – 0,94 ✓ correct substitution P(A or B) = 0,99 ✓ answer NOTE: Answer only, full marks. (2) 8.3.2 P(Not Detected) = 1 – 0,99 P(Not Detected) = 0,01 ✓ answer NOTE: Answer only, full marks. (1) [11] 18

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 QUESTION 9 9.1 9.1.1 NO. The survey was done on ONE day of the ✓ NO month and only in the morning. ✓ reasonable motivation NOTE: Answer must be NO. (2) 9.1.2 The survey should be done: • At different TIMES of the day. ✓ answer • On different DAYS of the month, especially at the end of the month when most people do shopping. NOTE: Any other valid reason. (1) 9.2 26 25 24 26 25 24 P(black) =   ✓   52 51 50 52 51 50 2  P(black) = or 0,118 ✓ answer P(black) 17 26 25 24 P(red) =   52 51 50 2  P(red) = or 0,118 ✓ answer P(red) 17  P(3 black or 3 red) = 0,118 + 0,118 ✓ method (+) 59  P(3 black or 3 red) = 0,236 or ✓ answer P(3 black or 3 red) 250 (5) [8] TOTAL: 150 19