Mathematics Grade 12 IEB Paper 1 2019 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 2019 MATHEMATICS: PAPER I MARKING GUIDELINES Time: 3 hours 150 marks These marking guidelines are prepared for use by examiners and sub-examiners, all of whom are required to attend a standardisation meeting to ensure that the guidelines are consistently interpreted and applied in the marking of candidates' scripts. The IEB will not enter into any discussions or correspondence about any marking guidelines. It is acknowledged that there may be different views about some matters of emphasis or detail in the guidelines. It is also recognised that, without the benefit of attendance at a standardisation meeting, there may be different interpretations of the application of the marking guidelines. IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 2 of 17 NOTE: • If a student answers a question more than once, only mark the FIRST attempt. • Consistent accuracy applies in all aspects of the marking memorandum. SECTION A QUESTION 1 (a)(1) 2( −2)2 + ( −2) + k =0 correct subst. of –2 8−2+k = 0 k = −6 k = −6 (a)(2) ∴ 2 x 2 + x − 6 = 0 Factors/correct subst. in ( 2x − 3 )( x + 2 ) =0 formula 3 ∴ Other root is 3 2 2 (b)(1) x − 2= 3 x + 2 Isolate surd x 2 − 4x + 4 ( x − 2) = (3 x + 2 ) 2 2 9( x + 2) x − 4 x + 4= 9( x + 2) 2 x 2 − 13 x − 14 x 2 − 13 x − 14 = 0 factors ( x − 14)( x + 1) = 0 answer with selection x = 14 or x = −1 Check: x = −1 is not valid (b)(2) x 2 − x − 6 ≤ 0 ( x − 3)( x + 2) ≤ 0 Factors/critical values Crit. values: 3 ; –2 Number line/graph x ≥ −2 + - + -2 3 x≤3 Solution: −2 ≤ x ≤ 3 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 3 of 17 QUESTION 2 (a) = A P (1 + i )n Sub. P into correct formula 1 0,123  0,123  = A 12 349  1 + 52  52  n =1 A = R12 378,21 (b) 1 − (1 + i )− n  P = x  0,123  i  52   0,123 −52n  1 −  1 +    52   12 349 = 94,75 Correct P & x in correct  0,123   52  formula   0,6917... = (1,00236...)−52 n log1,00236... 0,6917... = −52n conversion to logs n ≈ 3 years answer (c) =A P (1 − in ) P = 12349 =A 12 349 (1 − 0,2 × 2 ) 0,2 × 2 A = 7 409,40 answer (d) Balance outstanding = A – F  0,123 2×52  Use of correct formula 94,75  1 + − 1  0,123  2×52   52   n = 104 in A-F formula =12349  1 +  −  52  0,123 52 94,75 = 15 788,54384 − 11 156,97628 123 rate 5200 = R4 631,57 Depreciated amount = R7 409,40 Answer Therefore it would be sufficient. Conclusion OR x 1 − (1 + i )  −n P=   Correct Pv formula i 94,75 in P formula   0,123  −52  94,75 1 −  1 +    52   n ≈ 52 into formula P= 0,123 123 52 rate 5200 P = R4 630,90 Answer Depreciated amount = R7 409,40 Conclusion Therefore it would be sufficient. IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 4 of 17 QUESTION 3 (a)(1) f ( x + h) − f ( x ) f '( x ) = lim h →0 h −5( x + h )2 + ( x + h ) −5( x + h )2 + ( x + h ) − ( −5 x 2 + x ) f '( x ) = lim h →0 h Squaring & distributing −5( x + 2 xh + h 2 ) + x + h + 5 x 2 − x 2 f '( x ) = lim h →0 h Factorisation −5 x − 10 xh − 5h 2 + x + h + 5 x 2 − x 2 f '( x ) = lim h →0 h h( −10 x − 5h + 1) notation f '( x ) = lim h →0 h f '( x ) = lim ( −10 x − 5h + 1) sub. in 0 to get −10 x + 1 h →0 = −10 x + 1 OR f ( x + h) =−5( x + h )2 + ( x + h ) f ( x + h ) =−5( x 2 + 2 xh + h 2 ) + x + h −5( x + h )2 + ( x + h ) f ( x + h ) =−5 x 2 − 10 xh − 5h 2 + x + h f ( x + h) − f ( x ) =−5 x 2 − 10 xh − 5h 2 + x + h − ( −5 x 2 + x ) Squaring & distributing f ( x + h) − f ( x ) =−10 xh − 5h 2 + h Factorisation f ( x + h) − f ( x ) f '( x ) = lim notation h →0 h h( −10 x − 5h + 1) f '( x ) = lim h →0 h f '( x ) = lim ( −10 x − 5h + 1) sub. in 0 to get −10 x + 1 h →0 = − 10 x + 1 (a)(2) At: x = 1, f '(1) − = 10(1) + 1 f '(1) = −9 f '(1) = −9 ∴ Eq. of tangent: y = −9 x + c Calculating y-coord of –4 Substitute: (1; − 4 ) −4 =−9(1) + c Answer c =5 ∴ y =−9 x + 5 OR At: x = 1, f '(1) − = 10(1) + 1 f '(1) = −9 f '(1) = −9 Calculating y-coord of Substitute: (1; − 4 ) –4 y − ( −4) =−9( x − 1) Answer ∴ y =−9 x + 5 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 5 of 17 (b)(1) 3 x3 + x 2 y= x 3 x3 x 2 = y + x x 1 x2 = x +x y 2 2 1 dy 1 − 1 x2 = 2x + x 2 dx 2 2x dy 1 = 2x + 1 − 21 dx 2 x x 2 (b)(2)  (2 x − 3)(4 x 2 + 6 x + 9)  Dx   (2 x − 3)(4 x 2 + 6 x + 9)  (4 x 2 + 6 x + 9)  Dx ( 2 x − 3 ) Dx ( 2 x − 3 ) =2 =2 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 6 of 17 QUESTION 4 (4)(a) Tn =a + (n − 1)d Tn =5 + (n − 1)(4) d =4 T= 4n + 1 n T= n 4n + 1 = 4n + 1 100 Tn = 100 4n = 99 3 n = 24 Answer 4 24 pentagons (b)(1) T1 = 3 and Tn = 47 Sn = 300 n = Sn (a + l ) Correct formula 2 n = 300 ( 3 + 47 ) = 300 n ( 3 + 47 ) 2 2 n = 12 Answer (b)(2) Tn =a + (n − 1)d Correct formula 47= 3 + 11d 47= 3 + 11d d =4 Answer (c) 1 1 1 Series: + + + ... 4 8 16 Expansion 1 r = , converging series 1 2 r = 2 a =S∞ ; −1< r < 1 1− r Correct formula 1 S∞ = 4 1 1− 2 1 S∞ = 2 Answer (d) T5 T7 = Equating ratios T3 T5 4 1 = Correct substitution 5p + 1 4 p=3 Answer IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 7 of 17 QUESTION 5 (a) 50 50 (b) 2a = −4 3a + b =18 a+b+c =2 Determining differences a = −2 3( −2) + b =18 ( −2) + 24 + c = 2 a = −2 b = 24 c = −20 b = 24 c = −20 Tn = −2n 2 + 24n − 20 OR 2a = −4 a = −2 Determining differences Tn = −2n 2 + bn + c a = −2 b = 24 T1 : − 2 + b + c = 2 ∴b + c = 4 … eq 1 c = −20 T2 : − 8 + 2b + c= 20 ∴ 2b + c= 28 … eq 2 T2 − T1 : = b 24 Subst. in eq. 1: 24 + c =4 ∴ c =−20 Tn = −2n 2 + 24n − 20 OR (n − 1)(n − 2) Determining differences Tn= T2 (n − 1) − T1(n − 2) + × (2nd diff.) 2 a = −2 (n − 1)(n − 2) b = 24 = Tn 20(n − 1) − 2(n − 2) + × ( −4) c = −20 2 Tn = 20n − 20 − 2n + 4 − 2(n 2 − 3n + 2) Tn= 20n − 20 − 2n + 4 − 2n 2 + 6n − 4 Tn = −2n 2 + 24n − 20 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 8 of 17 (c) Tn = −2n 2 + 24n − 20 Determining n Tn − = 2(n 2 − 12n + 10) n=6 −2 ( n − 6 ) − 26  2 Tn =   Tn =−2(n − 6) + 52 2 Answer Max. passengers is 52 OR Determining n Tn ' = −4n + 24 n=6 −4n + 24 = 0 n=6 Subst. n = 6 Answer Tn − = 2(6)2 + 24(6) − 20 T6 = 52 Max. passengers is 52 (d) Let n be 12 stops T12 − = 2(12)2 + 24(12) − 20 ∴T12 = −20 Substitution Invalid due to negative answer Explanation OR −2n 2 + 24n − 20 ≥ 0 Passengers must be ≥ 0 Crit. Values: 6 ± 26 Substitution Explanation Hence 0,9 ≤ n ≤ 11,09 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 9 of 17 SECTION B QUESTION 6 (a) f ( x ) > g ( x ) for 0 < x < 3 x >0 x<3 (b) g ( x ) = loga x subst. ( 3;1) Substitution 1 = loga 3 Answer a=3 f ( x ) = kx subst. ( 3;1) Substitution 1 = 3k (1) = ( 3k ) 2 2 1 k= 3 Answer (c) 1 f :y = x Changing x and y 3 1 f −1 : x = y 3 1 y x2 = 3 =y 3 x 2 for x ≥ 0 y = 3x 2 Domain: x ≥ 0 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 10 of 17 QUESTION 7 (a) Shape of f and g Intersection: (1;9) Graph g: Horizontal asymptote (0;1) Graph f: Horizontal asymptote (0;3) (b) a 2 x = 3 x +1 x  a2  Isolating x :   a2 x =3  3  3x Conversion to logs x  a2  Answer   =3  3  log3 1 2  = x  a  3  OR x = loga2 3 3 OR log3 x= log a 2 − log3 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 11 of 17 QUESTION 8 ( 2x − 1) ≥ 0 2 (a)(1) Answer (a)(2) Shape 1  T.P.  ;5  2  y-int.: ( 0;6 ) (a)(3) Shift 5 units down Answer (a)(4) ( 2 x − 1)2 k = 4 x 2 − 4 x + (1 − k ) = 0 Sub in formula −( −4) ± ( −4) − 4(4)(1 − k ) 2 1± k x= x= 2(4) 2 4 ± 16k k ≥0 x= 8 4±4 k x= 8 1± k x= 2 Roots are real for k ≥ 0 OR ( 2x − 1) k 2 = 4 x 2 − 4 x + (1 − k ) = 0 Sub in formula −( −4) ± ( −4)2 − 4(4)(1 − k ) x= 4 ± 16k 2(4) x= 8 4 ± 16k x= 16k ≥ 0 ∴ k ≥ 0 8 Roots are real for 16k ≥ 0 ∴ k ≥ 0 OR 2 x − 1 =± k ( 2x − 1) k 2 = 1± k 2 x − 1 =± k x= 2 1± k x= k ≥0 2 Roots are real for k ≥ 0 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 12 of 17 (a)(5) y= 4 x 2 − 4 x + (1 + k ) For real, unequal and rational, ∆ > 0 and perfect square ∆ = ( −4)2 − 4(4)(1 + k ) ∆ = −16k ∆ = −16k accurate value of k 1 1 ∴ k =−1 , k = − , k = − etc. accurate value of k 4 16 accurate value of k OR y= 4 x 2 − 4 x + (1 + k ) Solve for 4 x 2 − 4 x + (1 + k ) = 0 through trial & 4 x 2 − 4 x + (1 + k ) = 0 improvement: When k = −1 roots are real, rational & unequal accurate value of k When k = −4 roots are real, rational & unequal, etc. accurate value of k accurate value of k (b) px 2 + qx + r =0 −q ± q 2 − 4 pr x= 2p −q + q 2 − 4 pr ∴P = −q + q 2 − 4 pr 2p P= 2p −q ± q 2 − 4 pr x= 2 −q + q 2 − 4 pr ∴Q = −q + q 2 − 4 pr 2 Q= 2 For: P : Q , 1  −q + q 2 − 4 pr   −q + q 2 − 4 pr    : 1  p  2   2  1 Ratio: :1 p OR Answer Ratio: 1: p IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 13 of 17 QUESTION 9 (a)(1) y =a( x − x1 )( x − x2 )( x − x3 ) formula  1 y =a( x + 3)( x + 3)  x −  Substitution of intercepts  2 Subst.: (0;9) Subst. of (0;9) a = −2 a = −2  1 y= −2( x + 3)  x −  2  2  1 −2  x −  ( x 2 + 6 x + 9 ) y=  2 Answer showing  1 1 a, b, c and d y= −2  x 3 + 6 x 2 + 9 x − x 2 − 3 x − 4   2 2  1 1 y= −2  x 3 + 5 x 2 + 6 x − 4   2 2 y= −2 x − 11x − 12 x + 9 3 2 OR formula Substitution of intercepts y =a( x + 3)( x + 3) ( 2 x − 1) Subst. of (0;9) Subst.: (0;9) a = −1 a = −1 y= −1( x + 3)( x + 3) ( 2 x − 1) y= ( ) −1 x 2 + 6 x + 9 (2 x − 1) y=−1(2 x 3 + 12 x 2 + 18 x − x 2 − 6 x − 9) Answer showing a, b, c and d y − = 1(2 x 3 + 11x 2 + 12 x − 9) y=−2 x 3 − 11x 2 − 12 x + 9 (a)(2) f (x) = −2 x 3 − 11x 2 − 12 x + 9 f '( x ) = −6 x 2 − 22 x − 12 f '( x ) = −6 x 2 − 22 x − 12 f ''( x ) − = 12 x − 22 f ''( x ) − = 12 x − 22 −12 x − 22 = 0 11 x= − 11 x= − 6 6 (b) f '( x ) = 8 f '( x ) = 8 −6 x − 22 x − 12 = 2 8 −6 x − 22 x − 20 = 2 0 x = −2 5 x= − or x = −2 y =5 3 E ( −2;5) IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 14 of 17 (c) 2 =y +q x+p p=3 2 y = + q subst. ( −2;5) x +3 x +3 2 Substitution =5 +q −2 + 3 q =3 q =3 2 ∴y= +3 x +3 (d) y= x + 6 y=x y= x + 6 OR Line goes through ( −3;3) ( −3;3) y x +c = ... substitute ( −3;3) y= x + 6 ∴y = x + 6 (e) ( −∞; −3 ) ∪ [ −2;0] ( −∞; −3 ) ( ) due to asymptote [ −2;0] (f) h( x ) − k =−2 x 3 − 11x 2 − 12 x + 9 For h: y -int. (0;9) y -int. (0;9)  11   11  For g: y -int.  0;  y -int.  0;   3  3 16 16 k<− k<− 3 3 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 15 of 17 QUESTION 10 4 3 4 3 V π r 2h + = πr V π r 2h + = πr 3 3 4 = π r 2h + π r 3 1000 3 4 Making h subject of πr = 2 h 1000 − π r 3 formula 3 1  4  =h 2  1000 − π r 3  …. Eq. 1 1  4  πr  3  =h 2  1000 − π r 3  πr  3  Total S.A.(S) = S.A. Cylinder + S.A. Sphere S= 2π rh + 4π r 2 …. Sub. Eq. 1 S= 2π rh + 4π r 2  1  4  S= 2π r  2  1000 − π r 3   + 4π r 2 π r  3  2 4  S=  1000 − π r 3  + 4π r 2 r 3  8 S= 2000r −1 − π r 2 + 4π r 2 3 4 S= 2000r −1 + π r 2 2000r −1 + 4 2 πr 3 3 dS 8 = − 2000r −2 + π r dS = − 8 2000r −2 + π r dr 3 dr 3 8π r 2000 − 2 = 0 3 r 8π r 3 − 6000 0 = 6000 r3 = 8π r ≈ 6,2 for S.A to be a minimum Answer IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 16 of 17 QUESTION 11 (a) 10! 10! 5!× 2!× 3! [ ] [ ] = 2 520 5!× 2!× 3! Answer (b)(1) = 0,3 × 0,65 P (not picked) 0,3 and 0,65 = 0,195 multiplication of above Answer (b)(2) P (made into juice) indicating 0,6 or 60% = ( 0,7 × 0,6 ) + ( 0,3 × 0,35 × 0,6 ) ( 0,7 × 0,6 ) = 0,483 ( 0,3 × 0,35 × 0,6 ) ∴ Approx. 48,3% will be made into juice 0,483 OR P (made into juice) indicating 0,6 or 60% = (1 − 0,195) × 0,6 = 0,483 (1 − ...) ∴ Approx. 48,3% will be made into juice (1 − 0,195) × 0,6 0,483 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER I – MARKING GUIDELINES Page 17 of 17 (b)(3) Total oranges exported = 120 × 172 = 20 640 20 640 P (exported) = ( 0,7 × 0,09 ) + ( 0,3 × 0,35 × 0,09 ) ( 0,7 × 0,09 ) = 0,07245 ( 0,3 × 0,35 × 0,09 ) ∴ 7,245% exported Let total number of oranges = x 20 640 × 100 =7,245 x ∴x = 284 886 oranges in total Answer OR Total oranges exported = 120 × 172 20 640 = 20 640 P (exported) (1 − 0,195) = (1 − 0,195) × 0,09 (1 − 0,195) × 0,09 = 0,07245 ∴ 7,245% exported Let total number of oranges = x 20 640 × 100 =7,245 x ∴x = 284 886 oranges in total Answer Total: 150 marks IEB Copyright © 2019