Mathematics Grade 11 P1 Gauteng June 2022 Memo (English)

Downloaded from hlayiso.com PROVINCIAL EXAMINATION JUNE 2022 GRADE 11 MARKING GUIDELINES MATHEMATICS PAPER 1 12 pages

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (Paper 1) GRADE 11 QUESTION 1 1.1 x  −3 ; x  −1  answer ✓ exclusion (2) 1.2.1 ( x + 4)( x − 1) = 0 x = −4 or x = 1 ✓ answers (1) 1.2.2 3x 2 − 2 x = 14 3x 2 − 2 x − 14 = 0  standard form −(−2)  (−2) 2 − 4(3)(−14)  correct x= 2(3) substitution 1  43 x= 3 x = −1,85 or x = 2,52  answer  answer NOTE: Penalise 1 mark for incorrect rounding in this question ONLY. (4) Any other valid method. 1.2.3 45 x 2 − 2 x = 18 − 2 x − 2x restrictio n  x 2 − 2 x  0  method let : k = x 2 − 2 x 45  k = 18 − k  k = 18k − 45 2  k 2 − 18k + 45 = 0  (k − 15)( k − 3) = 0  values of k  k = 15 or  k = 3 substitute x 2 − 2 x  back  int o  k .  x 2 − 2 x = 15 or  x 2 − 2 x = 3  equating  x 2 − 2 x − 15 = 0 or  x 2 − 2 x − 3 = 0  ( x − 5)( x + 3) = 0 or  ( x − 3)( x + 1) = 0  x = 5 or  x = −3 or  x = 3 or  x = −1  all values of x (4) NOTE: Any other valid method. 2

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (Paper 1) GRADE 11 1.2.4 5 − x +1 = −x 5 − x = −x −1 5 − x = x2 + 2 x + 1  method 0 = x 2 + 3x − 4  standard form 0 = ( x + 4)( x − 1)  factors  x = −4 or x  1  critical values  rejection (5) NOTE: Any other valid method. 1.2.5 32 x +1 − 4.3x = −1 3.32 x − 4.3x + 1 = 0  standard form (3.3 − 1)(3 − 1) = 0 x x  factors 1 3x = or 3x = 1  equating 3 3x = 3−1 or 3x = 30 x = −1 x=0  answers OR 32 x +1 − 4.3x = −1 let 3x = k  3k 2 − 4k = −1 3k 2 − 4k + 1 = 0  standard form (3k − 1)(k − 1) = 0  factors 1 k= or k = 1  equating 3 3x = 3−1 or 3x = 30  answers x = −1 x=0 (4) 3

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (Paper 1) GRADE 11 1.2.6(a) x 2 − 3x  0 x( x − 3)  0  factors CVs 0 ; 3 0 x3  answer (2) 1.2.6(b) 3 x  6 ✓ critical values NOTE: Answer can be written as separate ✓ answer (2) inequalities. 1.3 x + y = −10 xy = −600...(2)  setting up x = −10 − y...(1) equations sub (1) in (2)  substitution (−10 − y ) y = −600 −10 y − y 2 = − 600 y 2 + 10 y − 600 = 0  factors ( y − 20)( y + 30) = 600 y = 20 or of y = −30  values of y as y = 20 as y = −30 x = −30 x = 20  values of x NOTA: Enige ander geldige metode. (5) [29] 4

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (Paper 1) GRADE 11 QUESTION 2 2.1.1 2 8 − 4 32 + 3 50 Mark = 2 4.2 − 4 16.2 + 3 25.2 allocated  simplify surds = 2.2 2 − 4.4 2 + 3.5 2 to 1st step.  simplification = 4 2 − 16 2 + 15 2  answer =3 2 (3) 1 2.1.2 − 3 2 [ 12 + 3 (3 3) ] 1 1 1 1 − 3 2 [( 2 2.3) 2 + (3.3 2 ) 3 ]  simplify surds 1 1 3 1 − 3 [21.3 + (3 ) ] 2 2 2 2 1 1 1 − 3 2 [2.3 2 + 3 2 ] 2.30 + 30 ✓ simplification 2 +1 ✓ simplification 3 (4)  answer 2.1.3 52006 − 52004 + 24 52004 + 1 52004(52 − 1) + 24 52004 + 1  factorise 52004(24) + 24  simplify 52004 + 1 24(52004 + 1)  factorise 52004 + 1  answer 24 (4) 2.2.1 x = 6 + 6 + 6 + .........  method  x 2 = 6 + 6 + 6 + ...  x2 = 6 + x  answer (2) 2.2.2 x2 − x − 6 = 0 ✓ standard form ( x − 3)( x + 2) = 0 ✓ factors x = 3 x  −2 ✓ answer ✓ exclusion (4) 5

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (Paper 1) GRADE 11 2.3 d 2 = ( 8 − 1)2 + ( 8 + 1)2 Pythag  method d 2 = 8 − 2 8 +1+ 8 + 2 8 +1 d 2 = 18  value of d 2 d =3 2  answer (3) [20] 6

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (Paper 1) GRADE 11 QUESTION 3 3.1.1 -7 0 9 20 7 9 11 2 2 ✓ second difference 2a = 2 a = 1  value of a 3(1) + b = 7 b = 4  value of b 1 + 4 + c = −7  c = −12 (4)  value of c Tn = n + 4n − 12 2 3.1.2 Tn = n 2 + 4n − 12 128 = n 2 + 4n − 12 ✓ equating 0 = n + 4n − 140 2 0 = (n − 10)( n + 14)  factors  n = 10  n  −14  selection of n = 10 (3) 3.1.3 Tn = 2n + 5 599 = 2n + 5 ✓ equating 594 = 2n n = 297  value of n This difference will be between term number 297  conclusion and 298. (3) 3.2.1 93  answer (1) 3.2.2 Tn = 10n − 3 ✓ Tn = 10n − 3 Tn = (10n − 2) − 1 ✓ method Tn = 2(5n − 1) − 1 Thus : (10n − 2) = 2(5n − 1)  equating 10n − 2 EVEN for any value of n.  explanation Tn is ODD when : Tn = (10n − 2) − 1 (4) [15] 7

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (Paper 1) GRADE 11 QUESTION 4 4.1 1 − p;2 p − 3; p + 5 T2 − T1 = T3 − T2 ✓ method  (2 p − 3) − (1 − p) = ( p + 5) − (2 p − 3)  2 p − 3 −1+ p = p + 5 − 2 p + 3 3 p − 4 = − p + 8 ✓ simplification  4 p = 12 ✓ answer p=3 (3) 4.2.1 T1 = 1 − p T1 = 1 − 3 ✓ answer T1 = −2 (1) NOTE: Answer only: full marks 4.2.2 T2 = 2 p − 3 T2 = 2(3) − 3 T2 = 3 d = T2 − T1 d = 3 − (−2) ✓ answer d =5 (1) NOTE: Answer only: full marks 4.3 1- p ; 2p - 3 ; p + 5 = 1 - 3 ; 2(3) - 3 ; 3 + 5 = -2 ; 3 ; 8 ; 13 ; 18; … All the terms except T1 end in either a 3 or an 8  explanation while perfect squares end on 0; 1; 4; 5; 6; 9 . (2) [7] 8

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (Paper 1) GRADE 11 QUESTION 5 5.1 −3 x  2 ✓ answer NOTE: Candidates can present the answer as separate inequalities. (1) 5.2 f ( x) = a( x − x1 )( x − x2 ) f ( x) = a( x + 3)( x − 2) − 8 = a(1 + 3)(1 − 2) (1;−8) ✓ sub. point D and − 8 = a(4)( −1) x-intercepts a = 2 f ( x) = 2( x + 3)( x − 2) ✓ value of a f ( x) = 2( x 2 + x − 6) ✓ simplification of binomial f ( x) = 2 x 2 + 2 x − 12  a = 2 b = 2 c = −12 ✓ values of a, b, c (4) 5.3 b x=− 2a 2 x=− 2(2) 2 x=− 4 1 ✓ value of x x=− 2 1 1 1  f (− ) = 2( ) 2 + 2( ) − 12 ✓ sub for x into 2 2 2 f(x) 1 1 f (− ) = − 1 − 12 2 2 1 1 f (− ) = −12 ✓ value of y 2 2 1 1  turning  po int( − ;−12 ) 2 2 NOTE: Answer does not have to be in coordinate (3) form. 9

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (Paper 1) GRADE 11 1 5.4 x = − +7 2 1 x=6 2 ✓ answer NOTE: Answer only: full marks. (1) 5.5 In  g  m = 6 1 ✓ gradient of q  In  q  m = − 6  y = mx + c 1 0 = − (−3) + c 6 1 0= +c 2 1 c=− ✓ value of c 2 1 1 ✓ answer y =− x− 6 2 (3) NOTE: Answer only: FULL marks [12] 10

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (Paper 1) GRADE 11 QUESTION 6 6.1 ✓ shape of p with y-intercept = 1 ✓ shape of h ✓ x- and y-intercepts of h ✓ turning point of h (4) 6.2 See sketch in Q6.1. ✓ shape of f(x) with y-intercept = 1 (1) 6.3 y = 2( x − 1 + 2) 2 − 8 y = 2( x + 1) 2 − 8 ✓ answer (1) 6.4 p( x) = 4 x 1 1 ( x+ )  p( x + ) = 4 2 2 1 1 ✓ expansion p ( x + ) = 4 x.4 2 2 1 1 p ( x + ) = 4 x.(22 ) 2 ✓ simplification 2 1 p ( x + ) = 4 x.2 ✓ simplification 2 1  p ( x + ) = 2 p( x) 2 (3) [9] 11

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (Paper 1) GRADE 11 QUESTION 7 7.1 p=4  value of p  value of q q=2 a  f ( x) = +2 x−4 a  sub. point (5 ; 3) 3= + 2 T (5;3) 5−4 3=a+2 a = 3− 2 a =1  answer (4) 7.2 1 ✓ answer h( x) = − +2 x−4 (1) 7.3 𝑦∈ℝ; y2  answer NOTE: Both conditions must be stated. (1) 7.4 Axis of Symmetry of f is: y =x-p+q and y=-x-p+q y =x-4+2 y=-x+4+2 ✓ method y =x-2 y=-x+6 Selection of: y = - x + 6 ✓ selection c=6 (2) [8] TOTAL: 100 12