Mathematics Grade 12 P1 Gauteng Prelim 2024 Memo (English)

Downloaded from hlayiso.com PREPARATORY EXAMINATION 2024 MARKING GUIDELINES MATHEMATICS (PAPER 1) (10611) 22 pages

Downloaded from hlayiso.com 10611/24 INSTRUCTIONS AND INFORMATION A – Accuracy CA – Continued Accuracy S – Statement R – Reason S and R – Statement and Reason 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 in ALL aspects of the marking guidelines. • It is UNACCEPTABLE to assume values/answers in order to solve a question. 2

Downloaded from hlayiso.com 10611/24 QUESTION 1 1.1 1.1.1 2k = ( x − 5)( x − k ) ∴ 2k = (2 − 5)(2 − k ) 2k = (−3)(2 − k ) 2k = −6 + 3k  simplification ∴k = 6  answer (2) 1.1.2 2k = ( x − 5)( x − k ) ∴ 2(2) = ( x − 5)( x − 2) 4 = x 2 − 7 x + 10  simplification 2 ∴0 = x − 7x + 6  standard form 0 =( x − 1)( x − 6)  factors  answers (4) ∴ x 1 or = = x 6 1.2 1.2.1 2x2 + 3 =8x  standard form ∴ 2 x2 − 8x + 3 =0 −b ± b 2 − 4ac x= 2a −(−8) ± (−8) 2 − 4(2)(3) ∴x =  substitution into correct 2(2) formula 8 ± 64 − 24 x= 4 =∴ x 3,58 or = x 0, 42  answers NOTE: Penalise 1 mark for rounding in this question ONLY. Candidates must show substitution to get full marks. (4) 1.2.2 2( x + 10) − 10 =x − 12 2( x + 10) = x−2 2( x + 10) = x 2 − 4 x + 4  isolate surd and square both 2 x + 20 = x 2 − 4 x + 4 sides 0 = x 2 − 6 x − 16  standard form 0= ( x + 2)( x − 8)  factors x ≠ −2 or x =8  critical values with exclusion (4) 1.2.3 3 x ( x − 5) < 0 3 x > 0 for x ∈   3x > 0 x −5 < 0 ∴x < 5  answer (2) 3

Downloaded from hlayiso.com 10611/24 1.3 3 = x .9 y 27 x+ = 4 y 2 6......(2) . x 3 2 .32 y = 33  prime bases of 3 1 x + 2y 3= 2 x + 4y 6 =  simplification (linear) x= 6 − 4 y....(1)  expression for x sub (1) into (2) 6 − 4 y + 4 y2 = 6 4 y2 − 4 y = 0 4 y ( y − 1) = 0  factors =y 0 or = y 1  both y-values  both x-values =x 6=x 2 (6) 1.4 1.4.1 2p +5 = 0 5 ∴p=−  answer 2 (1) 1.4.2 2p +5 < 0 5 ∴p<−  answer 2 (1) [24] 4

Downloaded from hlayiso.com 10611/24 QUESTION 2 2.1 2.1.1 T4 = 27  answer (1) 2.1.2 0 5 14 27 44 5 9 13 17 1st differences 4 4 4 2nd differences  2 nd differences 2a = 4 ∴a = 2  value of a 3a + b =5 3(2) + b =5 ∴ b =−1  value of b a+b+c = 0 2 − 1 + c =0 ∴ c =−1  value of c ∴Tn= 2n 2 − n − 1 (4) 2.1.3 860= 2n 2 − n − 1  equating correctly/or correct standard form ∴ 0= 2n 2 − n − 861 −(−1) ± (−1) 2 − 4(2)(−861) n=  substitution 2(2) =∴ n 21 or n ≠ −20,5 There are 21 terms in the sequence.  answer with rejection/ selection NOTE: Candidate must reject a negative answer or decimal answer to obtain full marks. (3) 2.2 Series: 110 + 121 +…… + 990  series with T1 = 110 and ∴a = 110 and d = 11 Tn = 990 ∴110 + (n − 1)11 = 990  substitution into Tn (n − 1)11 =880 formula n −1 = 80 ∴n = 81  value of n  substitution into S n 81 formula ∴ S= 81 [110 + 990]  answer 2 44 550 ∴ S81 = OR (5) 5

Downloaded from hlayiso.com 10611/24 81 S81 = [2(110) + (81 − 1)(11)] 2  substitution into S n S81 = 44550 formula  answer [13] 6

Downloaded from hlayiso.com 10611/24 QUESTION 3 3.1 3.1.1 4( x − 2) 3 r= 8( x − 2) 2 x−2  expression for r ∴r = 2 x−2  −1 < r < 1 −1 < <1 2 ∴ −2 < x − 2 < 2  answer ∴0 < x < 4 (3) 3.1.2 a = 8( x − 2) 2 a = 8(2,5 − 2) 2 ∴a = 2  value of a x−2 r= 2 2,5 − 2 r= 2 1  value of r ∴r = 4 2 S∞ =  substitution into correct 1 1− formula 4 8 S∞ =  answer 3 OR 8(2,5 − 2)  value of a (numerator) S∞ =  2,5 − 2   value of r 1−    2  (denominator)  substitution into correct 8 S∞ = formula 3  answer (4) 3.2 3.2.1 n = (12 − 3) + 1 ∴ n = 10  answer (1) 7

Downloaded from hlayiso.com 10611/24 3.2.2 a = 3(−2) 3− 2 ∴ a = −6  value of a and r r = −2 a (1 − r n ) ∴ Sn = 1− r − 6(1 − (−2)10 )  correct substitution in ∴ Sn = 1+ 2 correct formula ∴ S n = 2046  answer (3) [11] 8

Downloaded from hlayiso.com 10611/24 QUESTION 4 4.1 f ( x) = ax 2 x≤0 − 8 = a (−6) 2  substitute point P − 8 = 36a 2 ∴a = −  answer 9 NOTE: Value of a must be in simplified form. (2) 4.2 2 2 x=− y ; y≤0 9  interchange x and y 9 ∴ y2 = − x y ≤ 0 2 9  answer ∴y =− − x 2 NOTE: No penalty for not writing constraints of y. (2) 4.3 y≤0  answer (1) 4.4  shape  point on f −1 (2) 4.5 9 −y=− − x  setting equation for 2 reflection across 9 y = x and x-axis ∴y = − x 2  answer NOTE: Answer only, award FULL marks. (2) [9] 9

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Downloaded from hlayiso.com 10611/24 QUESTION 5 5.1 P( − 3 ; 2)  answer NOTE: If a candidate draws a sketch indicating understanding of a rotation of 900 anticlockwise and shows the point in the second quadrant, award 1 mark. If a candidate has a negative x-coordinate, indicating understanding of the point being in the second quadrant, award 1 mark. (2) 5.2 5.2.1 a 0 = 1 for a ∈  ; a ≠ 0 ∴ h(0) = a 0  making x = 0 ∴ h(0) = 1  value of h(0) (A) ∴ Q(0 ; 1) NOTE: Any other valid answer (2) 5.2.2 h( x ) = a x 1 h(−1) = a −1 =  substitute point A 2 1 1 ∴ = a 2 ∴a = 2  answer (2) 5.2.3 h: y = 2x x = 2y  interchange x and y −1 ∴h :  answer ∴ y = log 2 x NOTE: Answer only, award FULL marks. Accept an answer in terms of a: y = log a x (2) 11

Downloaded from hlayiso.com 10611/24 5.2.4  shape (increasing log function)  x-int  asymptote (3) 5.2.5 log 2 x > −1  log to exponential ∴ x > 2 −1 form 1 ∴x > 2  answer NOTE: Answer only, full marks (2) 5.2.6 100.3x = 2 x 2x 100 = 3x  simplification x 2 100 =   3  express correctly in log 2 100 = x terms of logs 3 x = −11,36  answer OR (3) 12

Downloaded from hlayiso.com 10611/24 h( x ) = g ( x )  express correctly in ∴ 2x = 100.3x terms of logs 2x 100 ∴ x = 3 x  x as subject 2 ∴   100 = 3 2 x  answer ∴ log   = log100 3 2 ∴ x log = log100 3 log100 x= 2 log 3 ∴ x =−11,36 5.3.1  q = p log 10 +   2  1980  = p log 10 +   2   p = log1000 p = log(1000) 3 ∴p=  value of p  total price Total Price R3 ×1980 = R5940 (3) 5.3.2  q = p log 10 +   2  q = ∴ 2 log10 10 +   2 q  log to exponential ∴102 =10 + equation 2 ∴ 200 =20 + q ∴q = 180  answer (2) [21] 13

Downloaded from hlayiso.com 10611/24 QUESTION 6 6.1 x ∈  ; x ≠ −1  answer NOTE: Both conditions must be stated. (1) 6.2 x = −1  answer y=2  answer NOTE: Do not accept answers written in terms of p and q. (2) 6.3 2 4 − x += +2  equating x +1 ∴ (− x + 4)( x + 1) = 2 + 2( x + 1) ∴− x 2 + 3 x + 4= 2 + 2 x + 2  simplification ∴0 = x − x 2  standard form ∴ 0= x( x − 1) =∴x 0 = or x 1 ∴ xD 1 =  x-values with ∴ yD =−1 + 4 =3 selection (A)  y-value ∴ D(1 ; 3) NOTE: Does not have to be in coordinate form. (5) [8] 14

Downloaded from hlayiso.com 10611/24 QUESTION 7 7.1 A = P (1 + i ) n i 24 ∴ 2 x = x(1 + )  substitution into 4 correct formula i ∴ (1 + ) 24 = 2 4 i ∴1 + = 24 2 4 i ∴ = 0,0293...  value of i 4 ∴ i = 0,1172...  answer ∴ Annual % interest rate = 11,72% (3) 7.2.1 A = P(1 + i ) n  correct i and n ∴ A = 10000(1 + 0,0079...) 5  substitution into ∴ A = 10000(1,0079...) 5 correct formula ∴ A = R10 402,15  answer (3) 7.2.2. x[1 − (1 + i ) − n ] P= i 450[1 − (1, 0079) − n ] ∴ 10402,15 =  substitution into 0, 0079 correct formula ∴[1 − (1, 0079) − n ] =0,183 ∴1 − 0,183 =(1, 0079) − n  [1 − (1,0079) −n ] = 0,183 ∴ 0,8169 (1, = 0079) − n  correct use of logs ∴−n = log1,0079 0,8169 ∴−n = −25, 63... ∴n = 26  answer NOTE: Answer must be a natural number. (4) 7.2.3 Balance of the loan after the 25th payment: = value of loan - value of annuity at that time 450[(1,0079) 25 − 1]  10402,15(1, 0079) 25 = 10 402,15(1,0079) 25 − 0,0079 450[(1, 0079) 25 − 1] = 12 668,89 − 12 386,53  0, 0079 = R 282,36  answer OR (3) 15

Downloaded from hlayiso.com 10611/24   0.095  −0,6315128   correct substitution in 450 1 − 1 +   Pv formula   12   Pv = 0.095  n = 0, 6315128 12  answer Pv = R 282,36 [13] 16

Downloaded from hlayiso.com 10611/24 QUESTION 8 8.1 −2 x 2 + 3 x f ( x) = −2( x + h) 2 + 3( x + h) f ( x + h) = −2( x 2 + 2 xh + h 2 ) + 3 x + 3h f ( x + h) = −2 x 2 − 4 xh − 2h 2 + 3 x + 3h f ( x + h) = f ( x + h) − f ( x ) f '( x) = lim h →0 h −2 x − 4 xh − 2h 2 + 3 x + 3h − (−2 x 2 + 3 x) 2  substitution f '( x) = lim h →0 h −2 x − 4 xh − 2h + 3 x + 3h + 2 x 2 − 3 x 2 2 f '( x) = lim h →0 h 2 −4 xh − 2h + 3h f '( x) = lim  simplification h →0 h h(−4 x − 2h + 3)  correct factorisation f '( x) = lim h →0 h f '( x) = lim(−4 x − 2h + 3) h →0 f '( x) =−4 x + 3  answer NOTE: Penalise ONE mark for incorrect notation in this question only. (4) 8.2 3x 2 f ( x) = − 24 x 2 1 3x 2 ∴ f ( x) = − 24 x 2  simplification 2 12 ∴ f ′( x) = 3x − 1  derivatives x2 12 ∴ f ′(9) = 3(9) − 1  substitution 92 ∴ f ′(9) = 23  answer (5) 17

Downloaded from hlayiso.com 10611/24 8.3 Minimum value at x = 4 ∴ g ′( x) = 0 at x = 4  ∴ g ′(4) 0 = 2 −1 g ( x) = ax + bx ∴ g ′( x) = 2ax − bx − 2  derivative ∴ 2a (4) − b(4) − 2 = 0 b ∴ 8a − = 0 16 ∴128a − b = 0 (1)  equation 1 Function value at x = 4 is 96 ∴ g (4) = 96 ∴ a (4) 2 + b(4) −1 = 96 b ∴16a + = 96 4 ∴ 64a + b = 384 (2)  equation 2 (1) + (2) ∴192a = 384  value of a ∴a = 2 (1)∴ b = 256  value of b (6) [15] 18

Downloaded from hlayiso.com 10611/24 QUESTION 9 9.1 9.1.1 E(0 ; − 4)  answer (1) 9.1.2 y = a ( x + 2)( x − 6) roots are: − 2 and 6 − 4 = a (2)(−6) substitute point (0 ; − 4)  substitute roots and ∴12a = 4 point 1 ∴a = 3  value of a equation of g ′ : 1 y = ( x + 2)( x − 6) 3 1  substitute a and y = ( x 2 − 4 x − 12) 3 simplification 1 4 ∴ y = x2 − x − 4  answer 3 3 (4) 9.1.3 At the turning point of g: g ′( x) = 0 ∴ x = −2 and x = 6  answers (2) 9.1.4 x=2  answer (1) 9.1.5 For x < −2 , g ′( x) > 0 ∴ the gradient of g is positive and for the values of x immediately right of − 2 .  explanation For g ′( x) < 0 , the gradient of g is negative. NOTE: Any other valid explanation that indicates understanding (2) 9.2 The gradient of the tangent to h at any value of x is: h′( x) = 12 x 2 + 5  derivative x 2 ≥ 0 for all x ∈  ∴12 x 2 ≥ 0 for all x ∈   explanation ∴12 x 2 + 5 > 0 for all x ∈  (the gradient is ≥ 5 ) ∴ The gradient of the tangent is always positive. It is impossible to draw a tangent to h which has a negative gradient. OR A negative gradient requires: 12 x 2 + 5 < 0 ∴12 x 2 < −5 5 ∴ x2 < − 12 This is impossible, a square is always ≥ 0  value of x 2  explanation (2) [12] 19

Downloaded from hlayiso.com 10611/24 QUESTION 10 10.1 FE = FA = y − x  expression for FE and AB + BC + CD + DE = 2 x + 2 y FA in terms of x and y ∴ 2 x + 2 y = 30 ∴ 2 y = 30 − 2 x ∴ y = 15 − x  expression for y 2 A(x) = y − ( y − x) 2  substitute into area = y 2 − y 2 + 2 xy − x 2 formula = 2 xy − x 2 = 2 x(15 − x) − x 2  substitute for y = 30 x − 2 x − x2 2 = 30 x − 3 x 2  expression for area For maximum: A '( x) = 0 ∴ 30 − 6 x = 0  derivative = 0 ∴x = 5m  value of x ∴ y = 15 − 5 ∴ y = 10 m  value of y [8] QUESTION 11 11.1 For mutually exclusive events: P( A or B) = P(A) + P(B) 0,7 = 0,4 + k  substitution ∴ k = 0,3  answer NOTE: Answer only, full marks. 0 If a candidate writes down: 1 − 0,7 = 0,3 ; award 2 (2) 11.2 For independent events: P(A and B) = P(A) × P(B) = 0,4k  P( A and B) = 0,4k P(A or B) = P( A) + P( B ) − P( A and B) 0,7 = 0,4 + k − 0,4k  correct substitution for 0,3 = 0,6k P(A or B)  answer ∴ k = 0,5 1 NOTE: Answer only, 3 0 Wrong formulas, 3 (3) [5] 20

Downloaded from hlayiso.com 10611/24 QUESTION 12 12.1 12.1.1 Recovered OUTCOMES 24% or 25 6 within 48 hrs (S; RW48 hrs 4 16% of 25 Syndicates Recovered (S; RA 48 hrs) after 48 hrs 4 80% or 5 3 60% or 5  first level Never recovered (S; NR)  second level, (syndicates Vehicles stolen branch, labels must be on.) Recovered (I; RW48 hrs)  second level. within 48 hrs 19 38% or 50 (individuals branch labels must be on.) (3) 29 58% or 50 Individuals Recovered (I; RW48 hrs) after 48 hrs 1 4% or 25 Never recovered (I; NR) NOTE: S (Syndicates) I (Individuals) RW (Recovered within 48 hours) RA (Recovered after 48 hours) NR (Never recovered) Penalty of 1 mark for a mistake in the 2 nd branch level. 2 No values but correct tree diagram, award 3 Outcomes need NOT be shown. 12.1.2 80 24  probability P(S ; RW48hrs) = × 100 100 1920 = 10000 = 0,192  answer = 19,2% 21

Downloaded from hlayiso.com 10611/24 OR 4 6 OR P(S ; RW48hrs) = × 5 25  probability 24 = 125  answer NOTE: Answer Only, award FULL marks if 12.1.2 is accurately drawn. (2) Penalty of 1 mark for writing to 1 decimal place. Accept: 0,19 and 0,192 12.1.3  80 60   20 4  P(stolen and not recovered) =  × + ×   100 100   100 100  = 0,488 = 48,8%  P(Stolen)  P(Not OR recovered)  answer  4 3 1 1  P(stolen and not recovered) =  ×  +  ×   5 5   5 25  12 1 = + 25 125 61 = 125 (3) NOTE: Accept 0,49 12.2 9× 9× 9× 5× 4  combination (A) = 14580  answer (3) [11] TOTAL: 150 22