MP MATHS P1 MG ENG AFR Sept 2024 hlayiso.com

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS P1/WISKUNDE V1 SEPTEMBER 2024 MARKING GUIDELINES / NASIENRIGLYNE MARKS / PUNTE: 150 The marking guidelines consist of 12 pages. Die nasienriglyne bestaan uit 12 bladsye. Copyright reserved/ Kopiereg voorbehou Please turn over / Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1 / Wiskunde V1 2 MDE/MDO/September 2024 NSC TAKE 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 guiselines. • Assuming values/answers in order to solve a problem is unacceptable. LET WEL: • As 'n kandidaat 'n vraag TWEE keer beantwoord het, sien slegs die EERSTE poging na. • As 'n kandidaat 'n antwoord deurgehaal en nie oorgedoen het nie, sien die deurgehaalde antwoord na. • Volgehoue akkuraatheid is op ALLE aspekte van die nasienriglyn van toepassing. • Dit is onaanvaarbaar om waardes/antwoorde te veronderstel om 'n probleem op te los. QUESTION 1 1.1.1 x = 2 or x = −3  x=2  x = −3 (2) 1.1.2 3x 2 − 4 x − 5 = 0  standard form − (− 4 ) ± (− 4)2 − 4(3)(− 5)  substitution x= 2(3)  x − values x = 2,12 or x = −0,79 (4) 1.1.3 5 − x =1+ x  isolating surd 5 − x = (1 + x) 2  squaring both sides 5 − x = 1 + 2 x + x2  standard form  factoring x 2 + 3x − 4 = 0 ( x + 4)( x − 1) = 0  x ≠ −4 or x = 1 x ≠ −4 or x = 1 (5) 1.1.4 0< x<5  answer (2) Copyright reserved/ Kopiereg voorbehou Please turn over / Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1 / Wiskunde V1 3 MDE/MDO/September 2024 NSC 1.2 x =−1 + 2 y  x = −1+ 2 y  substitution of x 2 2 (−1 + 2 y ) − 7 − y =− y 3y2 − 3y − 6 =0  standard form  factors y2 − y − 2 = 0  both y − values ( y − 2)( y + 1) = 0 y = 2 or y = −1  both x-values x =−1 + 2(2) or x =−1 + 2(−1) (6) = 3 = −3 OR 1+ x 1+ x  y= y= 2 2 2 1+ x  1+ x   substitution x2 − 7 −   =−    2   2  1 + 2x + x2 1+ x  x2 − 7 − =−   4  2  4 x 2 − 28 − x 2 − 2 x − 1 =−2 − 2 x x2 − 9 =0  standard form ( x − 3)( x + 3) =0  factors  both x-values x = 3 or x = −3 1+ 3 1− 3 =y = or y  both y − values 2 2 (6) = 2 = −1 1.3 ∆ = (ab) 2 − 4(a 2 )(b 2 )  substitution 2 2 = −3a 2b 2  − 3a b 2 2 − 3a 2b 2 < 0  − 3a b < 0 +conclusion (3) ∴ roots are non-real for all real values of a and b ≠ 0 [22] QUESTION 2 Copyright reserved/ Kopiereg voorbehou Please turn over / Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1 / Wiskunde V1 4 MDE/MDO/September 2024 NSC 2.1 6 ; x ; 26 ; 45 ; y x−6 26 − x 19 y − 45 1st differences 2nd differences 26 − x − ( x − 6 ) = 19 − ( 26 − x ) = y − 45 − 19 equate 2nd differences − 2 x + 32 = −7 + x − 3 x = −39 value of x x = 13 substitute 19 − (26 − x) = y − 45 − 19 value of y (6) y = 70 2.2.1 220 + 213 + 206 + … − 11 d = −7 Tn = 220+(n − 1 )( − 7 ) = −11  substitution in Tn − 7n + 227 = − 11 − 7 n = −238  equate to − 11 n = 34  value of n 34 34  substitution in Sn S 34= [2(220) + (34 − 1)(−7)] OR S34 = [220 + (−7)] 2 2  answer (5) = 3 553 = 3 553 34 2.2.2 34  ∑ ∑ (−7n + 227) n =1 n =1  ( − 7 n + 227 ) (3)  13,5  15 2.3 S n = 15 + 2   1 − 0 ,9  2 = 285 13,5  < 290 1 − 0 ,9 285 (4) OR OR 15 15 13,5  S n= + 1 − 0 ,9 1 − 0 ,9 1 − 0,9 = 285 13,5  < 290 1 − 0,9  sum  285 (4) 1 1− t  2.4.1 r   and the series converges for − 1 < r < 1 r 5 3  Copyright reserved/ Kopiereg voorbehou Please turn over / Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1 / Wiskunde V1 5 MDE/MDO/September 2024 NSC 1 1− t  −1 <   <1 5 3   substitution − 15 < 1 − t < 15 − 16 < −t < 14  answer (3) − 14 < t < 16 1  1 − 15  − 14  value of r 2.4.2 r=  = 5  3  15 ∴ 𝑆𝑆𝑛𝑛 exists  1 − 15  − 350 a = 25 =  value of a  3  3 − 350 S∞ = 3  − 14  1−    substitution into S∞  15  − 1750 = 29  answer (4) 2.5 S70 = 270 − 5 + 3 = 265 + 3  S 70 S69 = 269 − 5 + 3 = 264 + 3  S 69 ∴ T70 = S 70 − S 69  substitution in T70 64 64 = 2.2 − 2  1.264 (4) = 1.264 [29] QUESTION 3 3.1 x=0  answer (1) 3.2 x = −2 equations (2) y = −1 3.3 g ( x) = b x + c y = bx − 4 substitution of asymptote 5 = b2 − 4 9 = b2 substitution of (2 ; 5) 32 = b 2 b=3 value of b x g ( x) = 3 − 4 equation (4) Copyright reserved/ Kopiereg voorbehou Please turn over / Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1 / Wiskunde V1 6 MDE/MDO/September 2024 NSC 3.4 y ∈ ( −∞; −1) ∪ ( −1; ∞ ) critical points OR y < −1 or y > −1 notation (2) OR y ∈ R, y ≠ −1 3.5 k ( x) = −(3 x − 4) − 4  − g ( x) − 4 = −3 x + 4 − 4 answer (2) = −3 x 3.6 y = −x + c negative gradient − 1 = −(−2) + c y=−( x + 2) − 1 OR substitution −3= c =− x − 3 ∴ y = −x − 3 answer (3) 3.7 x ∈ [0; ∞ ) OR x≥0 answer (2) [16] QUESTION 4 1 4.1 x=− OR f ′( x) = −2 x + 1 = 0 2(−1) 1 x= 1 1 2 x= x= 2 2 2 1 1 y = −  + + 6 2 2 y-value 25 = 4  answer (3)  − 1 25  ∴ T .P ;   2 4  2 4.2 −x + x+6=0 x2 − x − 6 = 0  factors (x − 3)(x + 2) = 0 x = 3 or x = −2 ∴ CD 5 = units answer (2) 4.3 ( AB(x) = 3 x + 10 − − x + x + 6 2 )  subtracting  answer (2) = x2 + 2x + 4 4.4 AB′( x) = 2 x + 2 = 0  AB′ = 0 x = −1  x − value Minimum of AB = (−1) 2 + 2(−1) + 4  substitution =3 answer (4) 10 10 4.5 x<− or − 2 < x < 3 x<− 3 3  − 2 < x < 3 (3) Copyright reserved/ Kopiereg voorbehou Please turn over / Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1 / Wiskunde V1 7 MDE/MDO/September 2024 NSC 4.6 k = f ( x) + 2 = − x 2 + x + 8 2  − x2 + x + 8 1 1 TP: f ( x) + 2 = −  + + 8 2 2 33 33 y= y= 4 4 33 33 ∴ k < (3) ∴k < 4 4 [17] QUESTION 5 5.1 =A P(1 − i ) n 1 1 =P P(1 − 0,1184) n A= P 4 4 1 = (1 − 0,1184) n 4 use of logarithms 1 = log(1 − 0,1184) n log 4 answer (3) 1 ∴n = log (1− 0,1184) 4 = 11   0,098  −5×12  x 1 − 1 +    n = 60   12   5.2.1 72 000 = 0,098 0,098 i = 12 12  0,098  substitution of P 72000   x=  12    0,098  −60  1 − 1 +     12   answer (4) = R1 522,71   0,098 −1,5×12  substitution of x 1522,71 1 − 1 +     12    n = 1,5 × 12 = 18 5.2.2 OB = 0,098 12 answer (3) = R 25 393,30 Copyright reserved/ Kopiereg voorbehou Please turn over / Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1 / Wiskunde V1 8 MDE/MDO/September 2024 NSC 5.2.3 R1522,71 × (1,5 × 12) = R 27 408,78  R 27408,78 ∴ R 27 408,78 − R 25393,30 = R 2015, 48  R 27408,78 − R 25393,30  answer (3) Therefore by settling the amount he will save R2 015,48. 3  0,1025  = 5.3 A R793 749, 25 1 +  12  = R814 263,3052  R814 263,3052 New instalments will therefore be calculated as follows: x 1 − (1 + i ) − n  substitution into P P= i  n = 231   0,1025 −231  x 1 − 1 +   0,1025   12   12 R814 263,3052 =  0,1025 12 answer (5) x = R8089, 20 [18] Copyright reserved/ Kopiereg voorbehou Please turn over / Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1 / Wiskunde V1 9 MDE/MDO/September 2024 NSC QUESTION6 6.1 2 f ( x + h) = − x+h  f ( x + h) 2  2 substitution of − − −  x + h  x f ( x + h) f ' ( x) = lim h →0 h 2 2 − + = lim x + h x h →0 h simplification − 2 x + 2 x + 2h x ( x + h) = lim h →0 h 2h 2 = lim x + xh h →0 h factoring h 2 = lim 2 h→0 x + xh 2 answer (5) = 2 x 6.2.1 y ( x − 2) = x 2 − 4 factors y ( x − 2) = ( x − 2)( x + 2) y = x+2  y = x+2 dy =1 dx answer (3)   2 2  3 6.2.2 Dx  3   5 x5 x  − 3  −3  2x 5 Dx 2 x 5 6 5 −8   x −8 5 (3) 6 =− x5 5 [11] Copyright reserved/ Kopiereg voorbehou Please turn over / Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1 / Wiskunde V1 10 MDE/MDO/September 2024 NSC QUESTION7 7.1 f ′( x) = 0 − 3x 2 = 0  f ′( x) = 0 x=0 ∴ y =1  (0 ;1) (2) 7.2 f ′′( x) = 0  f ′′( x) = 0 − 750 x = 0  (0 ;1) (2) x=0 y =1 Stationary point and point of inflection in this instance occur at same point. 7.2 Shape  y - intercept  x-intercept (3) 7.3 x < 0, x ∈ R answer (1) 3 1 1 7.4 f   = 1 − 125   10   10  7 7  = 8 8 2 1 1 15 f ′  = −375  −  10   10  4 15 =− 4 substitution 7 15  1 y − = − x −  8 4 10  15 5  equation y=− x+ 4 4 (4) [12] Copyright reserved/ Kopiereg voorbehou Please turn over / Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1 / Wiskunde V1 11 MDE/MDO/September 2024 NSC QUESTION 8 V πx 8.1 = =2 h 440  equated to 440 440 440 ∴h = h = 2 (2) π x² πx 8.2 =SA 2π x 2 + 2π x h correct formula  440  SA 2π x + 2π x  2  = 2 πx  substitution of h (2) 880 = 2π x 2 + x 8.3 SA′( x) = 0  SA′( x) = 0 4π x − 880 x −2 =0 880 derivative 4π x = 2 x simplification 4π x = 880 3 ∴x = 4,12 cm answer (4) [8] Copyright reserved/ Kopiereg voorbehou Please turn over / Blaai om asseblief

Downloaded from hlayiso.com Mathematics P1 / Wiskunde V1 12 MDE/MDO/September 2024 NSC QUESTION9 9.1 P( A or B) = P( A) + P( B) − P( A and B)  P( A and B ) = 0 1 1 0,57 = P( B) + P( B) − 0  P( A) = P( B) 2 2 3  substitute into addition rule 0,57 = P( B) 2  answer (4) P( B) = 0,38 9.2.1 7 P(First client takes a loaf of white bread) = answer (1) 12 5 4 4 9.2.2 P( BB) = ×  12 11 11 20 5  multiplication = = 132 33 20  (3) 132 9.2.3 Branch 1 Branch 2 66 W  branch 1 1212 W 7 12 6 12 B  branch 2 4 BB 12 5 12 B 7 6 5 8  × and × 8 12 12 12 12 W 12  7 6   5 8  41 answer (4) P (WB) or P( BW ) =  ×  +  ×  =  12 12   12 12  72 9.3.1 3 × 5 = 15  answer (1) 9.3.2 3 × 5 × 5 × 3 = 225  3× 5× 5× 3  answer (2) 9.3.3 3 × 5 × 4 × 2 = 120  3 × 5 × 4 × 2 (2) [17] TOTAL MARKS: 150 Copyright reserved/ Kopiereg voorbehou