Mathematics Grade 11 P1 Gauteng November 2023 Memo (English)

Downloaded from hlayiso.com PROVINCIAL EXAMINATION NOVEMBER 2023 GRADE 11 MARKING GUIDELINES MATHEMATICS (PAPER 1) 18 pages

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

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 QUESTION 1 1.1 1.1.1 5 p K= + p+2 3 5 3 K= + 3+ 2 3 ✓ simplification K= 1 +1 ✓ answer K=2 (2) 1.1.2 5 p K= + p+2 3 p+2>0  p > –2 ✓ answer NOTE: Answer only, full marks. (1) 1.2 1.2.1 3x2 = 4x + 2 3x2 – 4x – 2 = 0 − b  b 2 − 4ac x= +1 2a − ( −4)  ( −4) 2 − 4(3)( −2) ✓ substitution x= +1 2(3) ✓ answers  x = 1,72 or x = –0,39 NOTE: Penalise with 1 mark for rounding- off in this question ONLY. Answers only, 1 mark. (2) 3

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 1.2.2 x2 + 10x > –21 x2 + 10x + 21 > 0 ✓ standard form (x + 7)( x + 3) > 0 ✓ factors x < –7 or x > –3 ✓✓ answers NOTE: Penalise with 1 mark for the use of AND instead of OR. (4) 1.2.3 x2 − 5 = 2 x ✓ squaring both sides ( x − 5 )2 = (2 x )2 2 ✓ simplification x2 – 5 = 4x ✓ standard form x2 – 4x – 5 = 0 ✓ factors (x – 5)(x + 1) = 0 ✓ both answers with exclusion  x = 5 or x = –1(N/A) (5) 1.2.4 10 3x+1 + 3x–1 = 9 10 ✓ factorisation 3x.31 + 3x.3–1 = 9 1 10 3x (3–1 + = 3 9 10 10 3x. = 3 9 1 3x = 3 ✓ simplification 3x = 3–1 ✓ same bases x = –1 ✓ answer (4) 4

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 1.3 1.3.1 −3 x + y = 2 and y + 1 = …(2) x −1 ✓ x as subject in (1) x = 2 – y …(1) sub (1) into (2) −3 y+1= ✓ substitution in (2) 2 − y −1 −3 y+1= 1− y (y + 1)(1 – y) = –3 1 – y2 = –3 y2 = 4 ✓ value of y 2 y = 2 … sub into (1) ✓ values for y x = 2 – 2 of x = 2 – (–2) x=0 x=4 ✓ values for x NOTE: Any other valid method. (5) 1.3.2  1 1   +  x y 1 1   +  ✓ correct selection for x 4 −2 1 and y values =− 4 ✓ answer (2) 1.4 1.4.1 2mx2 – (m – 2)x + m + 1 = 0  = b2 – 4ac ✓ substitution  = [–(m – 2)]2 – 4(2m) (m + 1)  = (m – 2)2 – 8m(m + 1) ✓ simplification  = m2 – 4m + 4 – 8m2 – 8m ✓ standard form  = 7m2 – 12m + 4 For real roots,   0 ✓ condition for  7m2 – 12m + 4  0 7m2 + 12m – 4  0 (7m – 2)(m + 2)  0 ✓ factors 2 –2m ✓ answers 7 (6) 5

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 1.4.2 For roots to have opposite signs, the value of ‘c’ must be positive. c > 0 ✓ condition of ‘c’ m + 1 > 0 m > – 1 Real roots opposite in sign for: –1<m<0 ✓ answer (2) [33] QUESTION 2 −1 2.1 2.1.1  512 x 3  3  −3   64 x  −1  29 x 3  3 ✓ prime bases =  6 −3  2 x  −1 = {23x6} 3 = {2–1x–2} 1 1 = x 2 ✓ simplification 2 x 1 = ✓ answer 2x2 NOTE: Any correct alternative method. (3) 2.1.2 x n+2 + x n +1 − x n − x n−1 x2 − 1 x n ( x 2 − 1) + x n ( x − x −1 ) = ✓ factorisation x2 − 1 x2 − 1 x n ( x 2 − 1) + x n ( ) = x x2 − 1 1 = xn + xn ) ✓ simplification x n n–1 =x +x ✓ answer NOTE: Any correct alternative method. (3) 6

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 2.1.3 a + 2a − 1 . a − 2a − 1 ✓ write as A single surd = (a + 2a − 1)(a − 2a − 1) = a 2 − ( 2a − 1) 2 ✓ quadratic expression = a 2 − 2a + 1 ✓ factors = ( a − 1) 2 ✓ answer =a–1 NOTE: Any correct alternative method. (4) 2.2 (1 – 2 )(x + y z ) = –3 + 2) ( −3 + 2 ) 1 + 2 ✓ isolating variables x+y z= x and rationalising 1− 2 1+ 2 ( −3 + 2 )(1 + 2 ) x+y z= 1− 2 −3− 2 2 + 2 x+y z= ✓ simplification −1 x+y z=1+2 2 ✓ simplification x=1 y=2 z=2 ✓ answers (4) [14] QUESTION 3 3.1 3.1.1 Multiply the previous term by cos x . ✓ answer (1) 3.1.2 sin x.cos2 x ✓ answer (1) 3.2 3.2.1 − 1; 2; 5; 8; … T2 – T1 = T3 – T2 T2 – T1 = 2 – (–1) = 3 T3 – T2 = 5 – 2 = 3 d = 3 ✓ value of d  Tn = dn + a  Tn = 3n + a ✓ substitution –1 = 3(1) + a a = –4 ✓ value of a  Tn = 3n – 4 ✓ answer NOTE: Answer only, full marks. (4) 7

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 3.2.2 Tn = 3n – 4 161 = 3n – 4 ✓ substitution 165 = 3n  n = 55 ✓ answer (2) [8] QUESTION 4 4.1 4.1.1 25 squares ✓ answer (1) 4.1.2 64 dots ✓ answer (1) 4.1.3 4 ; 9 ; 16 ; 25 5 7 9 2 2 ✓ second difference 2a = 2 a=1 ✓ value of a 3a + b = 5 b=2 ✓ value of b a+b+c=4 c=1 ✓ value of c  Tn = n2 + 2n + 1  Tn = (n + 1)2 (4) 4.2 4.2.1 17 ; 10 ; 5 ; 2 ; 1 ; 2 –7 –5 –3 –1 +1 2 2 2 2 r=1 ✓ answer s=2 ✓ answer (2) 8

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 4.2.2 2a = 2 ✓ 2 nd difference a=1 ✓ value of a d(n) = n2 +bn + c d(1) = 12 +b(1) + c 17 = 1 + b + c 16 = b + c …(2) d(2) = 22 +b(2) + c 10 = 4 +2b + c 6 = 2b + c …(1) (1) – (2)  b = –10 ✓ value of b 16 = –10 + c …(2)  c = 26 ✓ value of c NOTE: Any other valid method. (4) 4.2.3 d(n) = n2 – 10n + 26  d(8) = 82 – 10(8) + 26 ✓ substitution  d(8) = 10m ✓ answer (2) 4.2 4.2.4 d(n) = n2 – 10n + 26 Minimum turning point − (−10) n= 2(1) ✓ substitution n=5 ✓ value of n n>0…nN minimum value ✓ explanation and Mikayla is CORRECT choice of Mikayla NOTE: No marks are awarded for correct name only. The explanation and the choice in name must correlate. (3) [17] 9

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 QUESTION 5 5.1 A = P(1 – i)n 1 ✓ correct substitution = (1 – i)2 into correct formula 3 i = 0,42% ✓ answer NOTE: The substitution is for the correct formula and the interpretation of substituting for A and P. (2) 5.2 5.2.1  i  1 + ieff = 1 + nom   n  ✓ correct substitution 12 into correct formula  0,075  1 + ieff = 1 +   12  ieff = 0,07763… ✓ simplification ieff = 7,76% ✓ answer (3) 5.2.2 A = P(1 – i)n 0,075 12 0,075 12 0,078 8 ✓ 10 000(1 + )  A = 10 000(1 + ) (1 + ) + 12 12 4 0,078 8 0,078 6 ✓ x (1 + ) 5000(1 + ) 4 4 0,078 6  A = R18 191,03 ✓ 5000(1 + ) 4 ✓ answer YES. He will have sufficient funds. ✓ conclusion NOTE: The values must be substituted into the correct formula. Only award the conclusion mark if a valid calculation has been done. (5) 10

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 5.3 Mervin: A = P(1 – ni) A = 6 000(1 + 5 x 0,085) ✓ correct substitution into correct formula A = R8 550 ✓ value of A (Mervin) 5% x R6 000 = R300  total = R8 550 + R300 ✓ bonus value + A = R8 850 final value of A (Mervin) Haley: i = 0,08  4 = 0,02 n = 5 x 4 = 20 A = P(1 – i)n A = 6 000(1 + 0,02)20 ✓ correct substitution into correct formula A = R8 915,68 ✓ value of A (Haley) Haley will have the larger investment. ✓ conclusion NOTE: Only award the conclusion mark if a valid calculation has been done. (6) [16] QUESTION 6 6.1 x   ✓ answer OR x (–∞;∞) (1) 6.2 y = 0 ✓ answer (1) 6.3 f (x) = a.bx 2  3 144 = a.   ✓ substitution 4 a = 256 ✓ answer (2) 11

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 −x 6.4  3 g(x) =256   4 ✓ answer OR x 4 g(x) =256    3 (1) 6.5 Increasing function. ✓ answer If x increases, increases f(x). ✓ explanation (2) ✓ 6.6 [f (x)]2 – [f (–x)]2 = a.f (2x) – a.f (–2x) LHS [f (x)]2 – [f (–x)]2 3 3 [256( )x]2 – [256( )–x]2 ✓ substitution 4 4 3 3 = [2562 ( )2x] – [2562 ( )–2x] ✓ answer (LHS) 4 4 RHS a.f (2x) – a.f (–2x) 3 3 = 256 [256( )2x] – 256 [256( )–2x] 4 4 3 3 ✓ answer (RHS) = 2562 ( )2x – 2562 ( )–2x 4 4 (3) [10] 12

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 QUESTION 7 7.1 x = –4 ✓ answer y = –4 ✓ answer (2) 7.2 x-intercept: 2 f (x) = −1 x+4 2 ✓ make y = 0 0= −1 x+4 x+4=2  x = –2 ✓ x- value y-intercept: 2 f (x) = −1 x+4 2 y= −1 0+4 1 y=– ✓ y-value 2 (3) 7.3 y f ✓ shape ✓ asymptotes ✓ intercepts –2 O x –½ y = –1 x = –4 NOTE: The mark for shape is for the graph shifting from quadrants 1 towards 3. (3) 13

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 7.4 (–6 ; –2) ✓ x-value ✓ y-value (2) 7.5 y   and y  1 ✓ answer NOTE: Both conditions must be stated. (1) 7.6 Reflection in the y-axis ✓ answer (1) [12] QUESTION 8 8.1 y = a(x + p)2 + q  6 = a(0 –1)2 + 8 ✓ substitute points (1 ; 8) 6=a+8 and (0 ; 6) a = –2 ✓ value for a y = –2(x – 1) +8 2 y = –2(x2 – 2x + 1) +8 ✓ correct simplification y = –2x2 + 4x –2 +8  y = –2x2 + 4x + 6 NOTE: The mark for simplification is awarded either at the 3rd or 2nd last step. (3) 8.2 j(x) = –2x2 + 4x + 6  j(3) = 0 …(3;0)  j(1) = 8 …(1;8) ✓ coordinates of j(3) and y − y1 j(1) m= 2 x2 − x1 8−0 ✓ correct substitution m= 1− 3 ✓ answer  m = –4 NOTE: Candidates do not have to write the coordinates. (3) 14

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 8.3 j(x) = k(x) 1 ✓ equating –2x2 + 4x + 6 = x–1 3 –6x2 + 12x + 18 = x – 3 –6x2 + 11x + 21 = 0 ✓ standard form –6x2 – 11x – 21 = 0 (x – 3)(6x + 7) = 0 ✓ factors 7 x  3 or x = – ✓ x-values with exclusion 6  7 25 k−  = −  6 18  7 25  P  − ;−  ✓ y-value  6 18  NOTE: Answer need not be in coordinate form. (5) 1 8.4 VW = –2x2 + 4x + 6 – x + 1 3 ✓ method 11 VW = –2x2 + x+7 3  11 7 ✓ simplification VW = –2  x 2 − x −   6 2  11 121 121 7  VW = –2  x 2 − x + − −  ✓ method  6 144 144 2  2  11  625 VW = –2  x −  +  12  72 625 VWmax = of 8,68 units ✓ answer 72 OR OR 15

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 1  VW = –2x2 + 4x + 6 –  x − 1 ✓ method 3  1 VW = –2x2 + 4x + 6 – x + 1 3 11 VW = –2x2 + x+7 ✓ simplification 3 Turning point: 11 x= 3 ✓ method 2( −2) 11 x= 12 625 ✓ answer VWmax = of 8,68 units 72 (4) 8.5 8.5.1 x < 1 ✓ answer (1) 8.5.2 x  –1 … but … x  3 –1  x < 3… or … x > 3 ✓✓ answers NOTE: Both answers must be completely correct. (2) 8.6 k < 8 ✓✓ answer (2) [20] 16

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 QUESTION 9 ANSWERS CAN EITHER BE IN FRACTION OR DECIMAL FORM. 9.1 C 0,7 0,1 I C C 0,9 0,7 0,2 L 0,1 I ✓ Tree 1 0,1 I 0,7 C ✓ Tree 2 C 0,2 0,9 0,1 I L 0,1 I Tree 1 Tree 2 (2) 9.2 P(C and C) = 0,7 x 0,7 ✓ correct method P(C and C) = 0,49 ✓ answer NOTE: Answer only, full marks. (2) 9.3 P(C) or P(L and C) ✓ 0,7 = 0,7 + (0,2 x 0,9) ✓ +(0,2 x 0,9) = 0,88 ✓ answer NOTE: Answer only, full marks. (3) [7] 17

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 QUESTION 10 10.1 10.1.1 P(A and C) = 0 ✓ answer (1) 10.1.2 P(B and C) = P(B) x P(C) ✓ correct formula or = 0,4 x 0,2 correct substitution = 0,08 ✓ answer NOTE: Answer only, full marks. (2) 10.1.3 P(A and B) = P(A) x P(B) = 0,3 x 0,4 ✓ correct substitution = 0,12 ✓ value of P(A and B) P(A or B) = P(A) + P(B) – P(A and B) = 0,3 + 0,4 – 0,12 ✓ correct substitution P(A or B) = 0,58 ✓ answer (4) 10.2 10.2.1 8 + 10 + x + 65 + 3 + 5 + 5 + 2 = 103 x=5 ✓ answer (1) 10.2.2 n(any 2) P(any2) = n( S ) n(any 2) ✓ with 10 + 65 + 5 + 5 n( S ) P(any2) = 103 correct substitution 85 P(any2) = 103 ✓ answer (any shape) P(any2) = 0,83 (2) 10.3 9 (17 + 8) (17 + 8) 9 ✓ correct substitution x + x 80 79 80 79 9 25 25 9 ✓ simplification x + x 80 79 80 79 0,0712 ...  0,07 ✓ answer (3) [13] TOTAL: 150 18