Mathematics Grade 11 P1 Gauteng June 2023 Memo (English)

Downloaded from hlayiso.com PROVINCIAL EXAMINATION JUNE 2023 GRADE 11 MARKING GUIDELINES MATHEMATICS (PAPER 1) 10 pages

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRAD QUESTION 1 1.1 1.1.1 x  {4 ; 5}  answer  answer (2) 1.1.2 x  {0 ; 3}  answer  answer (2) 1.1.3 x  {1 ; 2}  answer  answer (2) 1.2 1.2.1 3x 2  4 x  0  x ( 3 x  4)  0  factors 4 x  0 or  x   answers 3 NOTE: Any other valid method. (3) 1.2.2 3x – 14 = –6x2  standard form  6x2 + 3x – 14 = 0  (3)  (3) 2  4(6)( 14) x=  substitution 2(6)  x = 1,29 or x = –1,79  answers (4) 1.2.3 ( x  1)( x  3)  12 x 2  2 x  3  12  standard form x 2  2 x  15  0 -3 5  factors ( x  5)( x  3)  0  x  5or  x  3  answers (4) 1.2.4 2 x 2 x 2 x  x2 ( 2  x ) 2  ( x  2) 2  squaring both sides 2  x  x2  4x  4 0  x 2  3x  2  standard form 0 = (x – 2)(x – 1)  factors  x = 2 ... or ... x = 1(NA)  answers with rejection (4) 2

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRAD 1.2.5 x6  0 x  6  value of x  x  2y  3 6  2y  3  substitution 2y  3 3 y   value of y 2 (3) 1.3 y – 1 = 2x  y = 2x + 1 ………(1)  expression for y x2 + xy – 3x – y + 2 = 0 ………(2)  x2 + x(2x + 1) – 3x – (2x + 1) + 2 = 0  substitution x2 + 2x2 + x – 3x – 2x – 1 + 2 = 0  3x2 – 4x + 1 = 0  standard form (3x – 1)(x – 1) = 0 1 x= ... or ... x = 1  x-values 3 5 y= ... or ... y = 3  y-values 3 (5) 1.4 x 2  px  p 2  2  x 2  px  p 2  2  0   b 2  4ac   ( p ) 2  4(1)( p 2  2)  substitute into    p2  4 p2  8   5 p2  8    5 p2  8  p 2  0 p    5 p 2  0 p    p 2  0 , 5 p 2  0 and 2  5 p  8  0 p   5 p2  8  0  roots are real and unequal. (3) 3

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRAD 1.5 The 100 m fencing will make up three sides because the existing wall will make one side. Wall � � � fencing = 2x + y = 100  y = –2x + 100  A  xy  expression for y A = x(–2x + 100) A = –2x2 + 100x  expression for A A = –2(x2 – 50x)  A = –2(x2 – 50x – 625 – 625)  complete square A = –2(x – 25)2 –625) Maximum area: x  25  value of x A  2( 25) 2  100( 25)  value of A  A  1 250  1 250  25 y  y  50  value of y (7) [39] 4

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRAD QUESTION 2 2.1 2.1.1 3n2.9n1 27n1 3n2.32n2  32 n  2 and 33n 3 33n3 33n4 33n3 33n43n3  33 n  4  3 n  3 37  answer (3) 2.1.2 x2 1 x (1  3 ) 2   substitution 11 3 1 2 3  3   simplification 2 3 42 3  2 3 2(2  3 ) 2  factorisation  2 3 =2  answer (4) 5 2.1.3 ( a 2  b 2 )  ( a  b) 2 1 ( a  b) 2 5 (a  b)(a  b)  (a  b) 2  ( a  b) 2 1  factorisation 1 1 5 ( a  b) 2 ( a  b) 2  ( a  b) 2  1  simplification ( a  b) 2 1 5 ( a  b) 2  ( a  b) 2  simplification ( a  b) 3 a 3  3a 2b  3ab2  b3  answer (4) 5

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRAD 2 8 2.2 RTP: – = 2 1 2 8 2 8 – 1 2 8 2 8  2 2  – 1 2 2 2 2(2 2 )  8(1  2 )   2 2(1  2) 2 2 (1  2 ) 4 2 88 2  2 2 2 4 4 2 88 2  2 2 4 4 2 8   simplification 2 2 4 4 2 2   factorisation 2 2 2  2 (4) 2.3 MK  LM  2  MK= 2 JM 2  (2  3) 2  22 pythag  substitution JM 2  4  4 3  3  4  simplification JM 2  4 3  3 JM  4 3  3  JM 1 A  (4) 4 3  3  substitution into area formula 2 A  6,3 units2  answer (6) [21] 6

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRAD QUESTION 3 3.1 T2  T1  T3  T2 4 x  5  x  10 x  5  ( 4 x  5)  method 3x  5  10 x  5  4 x  5 3x  5  6 x  10  3x  15 x  5  answer (2) 3.2 3.2.1  3n  20  106  equating 3n  126  n  42  answer (2) 3.2.2  3n  20  0  Tn  0 20  3n 20  n 3 n  7  answer (2) 3.2.3 Odd valued terms: 17; 11; 5; ………..  sequence General term: Tn  6n  23  Tn Tn  6( 20)  23  Tn  97  answer (3) 3.3 3; a; 10; b; 21  1st differences a  3; 10  a; b  10; 21  b 1st difference 2nd differences: 10  a  a  3  1  equating 2nd difference in  2a  13  1 terms of a, then b.  2a  12 a  6  value of a and 21  b  b  10  1  2b  31  1  2b  30  b  15  value of b (4) [13] 7

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRAD QUESTION 4 y D(6 ; 7) C(–1 ; 0) B x O A(2 ; y) 4.1 B(5; 0)  answer NOTE: Must be in coordinate form. (2) 4.2 y  a ( x  x1 )( x  x2 )  7  a (6  (1))(6  5)  substitute roots and 7  7a point D(6;7) a  1  value for a  y  1( x  1)( x  5) y  x 2  5x  x  5  y  x 2  5x  x  5  y  x2  4x  5 (3) 4.3 B(5 ; 0) C(0 ;  5) 50 mBC  05 5 mBC   mBC 5  mBC  1  mh   1  mh  y  y1  m( x  x1 ) y  (1)  1( x  0) pt (1;0)  y  x 1  answer (3) 8

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRAD 4.4 1  x  0  all critical values (independent) x5  answers OR x  [–1 ; 0] or [5 ; ]  all critical values NOTE: Deduct 1 mark if brackets are incorrect in the (independent) alternative solution.  answers (2) [10] QUESTION 5 5.1  asymptotes 2  intercepts  shape NOTE: If the candidate calculates the intercepts and lists the asymptotes but does not sketch the graph, award 2 marks. (3) 5.2 tan 135 = –1  m  1 y – y1 = m(x – x1) y – (–1) = –1(x – (–2))  subs. point (2;–1) y + 1 = –x – 2 y = –x – 3  answer (3) 5.3 x  2  answer (1) 5.4 5.4.1 5 units right  answer NOTE: Accept an answer of 5 units. (1) 5.4.2 3 units up  answer NOTE: Accept an answer of 3 units. (1) [9] 9

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRAD QUESTION 6 6.1 6.1.1 y  6  answer (1) 6.1.2 h( x)  3.2 x  6 0  3.2 x  6  equate to 0 6  3.2 x 2  2x x 1  answer (2) 6.1.3 h( x )  3.2 x  6 y = 3.2 – 6 y = –3  answer (1) 6.1.4 x >1  answer (1) 6.2 y  x-intercept x O  asymptote –1  shape (3) [8] TOTAL: 100 10