Mathematics Grade 11 P1 Gauteng June 2024 Memo (English)

Downloaded from hlayiso.com PROVINCIAL EXAMINATION JUNE 2024 GRADE 11 MARKING GUIDELINES MATHEMATICS (PAPER 1) 11 pages

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 INSTRUCTIONS AND INFORMATION A – ACCURACY CA. – CONTINUED ACCURACY NOTE: • If a candidate answerd 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 problem is UNACCEPTABLE. 2

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 QUESTION 1 1.1 1.1.1 x9 ✓ answer (1) 1.1.2 x=9 ✓ answer (1) 1.1.3 (1) 2 + 4 9 −1 5 ✓ answer 8  Irrational (1) 1.1.4 x=0 ✓ answer (1) 1.2 1.2.1 2 x 2 − 3x − k = 0 2 x 2 − 3x − 5 = 0 (2 x − 5)( x + 1) = 0 ✓ factors 5 x = of x = −1 ✓ answer 2 (2) 1.2.2 2 x 2 − 3x − 4 = 0 −(−3)  (−3) 2 − 4(2)(−4) ✓ substitution x= 2(2)  x = 2,35 of x = −0,85 ✓✓ answer NOTE: Penalise 1 mark for incorrect rounding in this question ONLY. (3) 1.3 1.3.1 x−4= − x−2 16 − 8 x − x 2 = x − 2 ✓ square both sides x 2 − 9 x + 18 = 0 ✓ standard form ( x − 6)( x − 3) = 0 ✓ factors x = 3 of x  6 ✓ critical values ✓ rejection (5) 3

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 1.3.2 x 2 + 4  3x + 2 x 2 − 3x + 2  0 ✓ standard form ( x − 2)( x − 1)  0 ✓ factors x  1 or x  2 OR ✓✓ answers x  (−;1)  (2; ) (4) 1.4 y + 2 = 2x  y = 2x − 2 (1) ✓ y as subject 2x2 = 2 − y 2 (2)  2 x 2 = 2 − (2 x − 2) 2 ✓ substitution 2 x 2 = 2 − (4 x 2 − 8 x + 4) 2 x2 = 2 − 4 x2 + 8x − 4  6x2 − 8x + 2 = 0  3x 2 − 4 x + 1 = 0 ✓ standard form (3 x − 1)( x − 1) = 0 1 ✓ factors x = of x = 1 3 ✓ both x- values 4 y =− of y = 0 3 ✓ both y- values OR y + 2 = 2x OR y+2 x = (1) 2 ✓ x as subject 2x2 = 2 − y 2 (2) y+2 2  2( ) = 2 − y2 ✓ substitution 2 y2 + 4 y + 4 2( ) = 2 − y2 4 y + 4y + 4 2 ( ) = 2 − y2 2  y + 4 y + 4 = 4 − 2 y2 2 3y2 + 4 y = 0 ✓ standard form y (3 y + 4) = 0 ✓ factors 4 y =0 of y=− 3 ✓ both y-values 1 x =1 of x= ✓ both x-values (6) 3 1.5 1.5.1 For unequal roots:   0 ✓ condition of  4

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11  2 x 2 − 4 x + (3 − p) = 0  = b 2 − 4ac   = (−4) 2 − 4(2)(3 − p) ✓ substitution into   = 16 − 24 + 8 p (4)   = −8 + 8 p ✓ expression for   −8 + 8 p  0 8p  8  p 1 ✓ answer 1.5.2 From 1.5.1 3 − p  0 (a  0 and  b  0) ✓ c0  − p  −3 ✓ answer 1  p  3 NOTE: Any other valid method (2) 1.6 ( p 2 + 1) x 2 = −2 pqx − q 2 ✓ standard form  ( p 2 + 1) x 2 + 2 pqx + q 2 = 0  = (2 pq) 2 − 4( p 2 + 1)q 2 ✓ substitute into   = 4 p q − 4 p q − 4q 2 2 2 2 2 ✓ simplification   = −4q 2 ✓ expression for   q 2  0, q  ; q  0  −4q 2  0 ✓ explanation  Roots non-real (5) [35] 5

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 QUESTION 2 2.1 2.1.1 (2 −1 + 3−1 ) 2 1 1 ✓ converting exponents = ( + )2 2 3 3+ 2 =( ) 6 5 ✓ simplification = ( )2 6 25 = ✓ answer 36 (3) 2.1.2 2 .4 x + 2 − 4 x − 3 2 x .2 x 2 .4 x .4 2 − 4 x .4 − 3 = 4x 1 4 x (32 − ) 64 ✓✓ factors = x 4 1 = 32 − 64 2 048 − 1 = 64 2 047 = ✓ answer 64 63 NOTE: Accept 31 (3) 64 1 2.1.3 − 3 2 [ 12 + 3 (3 3)] 1 1 − = 3 2 [2 3 + (3 3) 3 ] ✓ simplify brackets 1 1 1 1 − = 3 [2.3 + 3 .3 ] 2 2 3 6 ✓ surds as rational 1 1 1 − exponents = 3 [2.3 + 3 ] 2 2 2 1 1 − = 3 .3 (2 + 1) 2 2 ✓ factors = 30 (3) =3 ✓ answer (4) 6

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 2.2.1 3 2 x = 16 4 3 x4 = 8 3 4 4   x 4 3 = 83 ✓ method  x = (3 8 ) 4 ✓ x as subject  x = 16 ✓ answer (3) NOTE: Any other valid method. 2.2.2 5 x +1 + 5 x = 150 5 x.51 + 5 x = 150  5 x (51 + 1) = 150 ✓ factors  5 (6) = 150 x 5 x = 25 ✓ simplification 5 = 5 x 2 ✓ 25 as base of 5 x = 2 ✓ answer (4) NOTE: Any other valid method. 2.3 9 − 54 6 2 (9 − 9.2.3 ) 2 =  ✓  2 6 2 2 2 9 2 − 3 2 . 3. 2 = 6.2 9 2 − 3.2 3 ✓ simplification = 12 3(3 2 − 2 3 ) ✓ factorisation = 12 3 2 −2 3 = 4 (3) [20] 7

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 QUESTION 3 3.1 T4 = −17 ✓ answer (1) 3.2 Tn = −7n + 11 ✓ answer (1) 3.3 Tn = −7n + 11  −227 = −7n + 11 ✓ substitute Tn − 238 = −7n ✓ answer  n = 34 (2) 3.4 2a = −7 7 ✓ value for a a=− 2 ✓ conclusion  maximum (2) NOTE: Any other valid method. [6] QUESTION 4 4.1 69 0 − 63 T4 − 69 − 63 − 57 6 6 next term (T4 ) is: − 120 ✓ answer (1) 4.2 Tn = an2 + bn + c T2 = a(2) 2 + b(2) + c = 4a + 2b + c T1 = a(1) 2 + b(1) + c = a + b + c T2 − T1 = 3a + b 2a = 6 ✓ 2 nd difference a = 3 ✓ value for a 3a + b = −69 3(3) + b = −69 9 + b = −69  b = −78 ✓ value for b a + b + c = 69 3 + (−78) + c = 69 ✓ value for c  c = 144 Tn = 3n 2 − 78n + 144 (4) 8

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 4.3 Smallest term (minimum) is at the turning point of the quadratic equation: − b − (−78) 78 ✓ substitute into A.O.S n = = = = 13 2a 2(3) 6 ✓ value for n Tn = 3n − 78n + 144 2 T13 = 3n 2 − 78n + 144 T13 = 3(13) 2 − 78(13) + 144 ✓ substitution T13 = −363 ✓ answer (4) [9] QUESTION 5 5.1 The points A( − 4 ; 0) and B(0 ; 8) lie on g. 8−0 ✓ substitute for mAB m AB = 0 − (−4) ✓ value of mAB m AB = 2 y = mx + c  8 = 2(0) + c  A(0;8) ✓ value of c c = 8  g ( x) = 2 x + 8 NOTE: Any other valid method. (3) 5.2 f ( x) = a( x − x1 )( x − x2 ) f ( x) = a( x + 4)( x − 2) ✓ substitute points A subst (0;8) and C 8 = a (0 + 4)(0 − 2) 8 = a (−8) a = −1 ✓ value of a f ( x) = −( x + 4)( x − 2) f ( x) = −( x 2 + 2 x − 8) f ( x) = − x 2 − 2 x + 8 ✓ standard form (A)  a = −1 b = −2 ✓ value of b c =8 ✓ value of c (5) 9

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 5.3 f ( x) = − x 2 − 2 x + 8 − b − (−2) ✓ correct substitution x = = 2a 2(−1)  x = −1  f (−1) = −(−1) 2 − 2(−1) + 8 ✓ correct substitution  f (−1) = 9 NOTE: Any other valid method. (2) 5.4 y9 ✓ answer (1) 5.5 mg = 2 1 ✓  mp = − mp 2 y − y1 = m( x − x1 ) 1 ✓ substitute point C  y − 0 = − ( x − 2) 2 1  y = − x +1 2 ✓ answer (3) 5.6 f (−1) = 9 ✓ f (−1) = 9 and g (−1) = 6 g (−1) = 6  f (−1) − g (−1)  PQ = 9 − 6  PQ = 3 ✓ answer OR OR DP(−1;9) ✓ using TP and g (−1) g (−1) = 6  PQ = 9 − 6 ✓ answer  PQ = 3 (2) 5.7 x  2 ; x  −4 ✓✓ answer (A) (2) [18] QUESTION 6 6.1 x = −3 ✓ answer y =1 ✓ answer (2) 10

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 1) GRADE 11 6.2 −4 f ( x) = +1 x+3 −4 0= +1 x+3 −4 −1 = x+3 − x − 3 = −4 − x = −1 x =1 ✓ x-intercept −4 y= +1 0+3 −4 y= +1 3 1 ✓ y- intercept y =− 3 (2) 6.3 y = −4 ✓ answer (1) 6.4 ✓ shape of k ✓ k and h intercepts ✓ asymptotes k and h ✓ shape of h (4) 6.5 p ( x) = 3h( x)  p( x) = 3(2 − x − 4)  p( x) = 3.2 − x − 12 asymptote: y = −12 ✓ answer NOTE: Answer only, full marks. (1) 6.6 q( x) = 2 x − 4 ✓ answer (1) 6.7 x=0 ✓ answer (1) [12] TOTAL: 100 11