Mathematics Grade 12 Paper 2 KZN Prelim 2024 Memo

Downloaded from hlayiso.com FINAL MATHEMATICS P2 PREPARATORY EXAMINATION SEPTEMBER 2024 MARKING GUIDELINES NATIONAL SENIOR CERTIFICATE GRADE 12 MARKS: 150 These marking guidelines consist of 14 pages. Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 2 KZN September 2024 Preparatory Examinations GRADE 12 Marking Guidelines • 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 guidelines. • Assuming values/answer in order to solve a problem is unacceptable. GEOMETRY A mark for a correct statement S (A statement mark is independent of a reason.) R A mark for a correct reason (A reason mark may only be awarded if the statement is correct.) S/R Award a mark if the statement AND reason are both correct. QUESTION 1 Penalise only once for incorrect rounding in Question 1. 165500 1.1.1 Mean = ✓A 165 500 in numerator 12 Also accept: 13,79 Answer only: = R13792 thousand rand Full marks ✓CA answer If answer is given as 13,79 instead of R13 792, penalise in 1.1.1, but don’t penalise again for this mistake in 1.1.2, 1.2 and 1.5. (2) 1.1.2 Standard deviation = R4 404 ✓A answer (1) 1.2 R13 792 +R4 404 = R18 196 ✓CA R18 196 2 employees earn a salary more than one standard deviation ✓CA 2 employees above the mean. Answer only: Full marks (2) 1.3 a = 8, 45 ✓A correct a value b = 0, 45 ✓A correct b value Answer only: yˆ = 0, 45 x + 8, 45 Full marks ✓ CA answer (3) 1.4 r = 0,94 ✓A answer (1) 1.5 yˆ = 0, 45 ( 30 ) + 8, 45 ✓CA substitution yˆ = 21,95  R21 950 ✓CA answer (2) OR OR R21 804 (calculator) ✓✓ CA CA (2) Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 3 KZN September 2024 Preparatory Examinations GRADE 12 Marking Guidelines 1.6 Yes. ✓CA answer r = 0,94 implies a strong correlation between employee ✓CA justification experience and monthly salary and therefore a prediction would be reliable. (2) OR OR Yes. ✓CA answer r = 0,94 , which is close to 1, and therefore implies a strong ✓CA justification correlation between employee experience and monthly salary and therefore a prediction would be reliable. (2) [13] QUESTION 2 2.1 5500 ✓A answer (1) 2.2 Q1 = 29 (accept 28 – 29) ✓A value of Q1 Q3 = 39 (accept 38 – 39) ✓A value of Q3 IQR = 10 (accept 9 – 11) ✓CA answer (3) 2.3 650 (accept 620 – 700) ✓A numerator in range 620 5500 to 700 = 11,82% (accept 11,27% – 12,73%) ✓CA answer (2) [6] QUESTION 3 y2 − y1 3.1 mAB = x2 − x1 −2 − 4 = ✓A substitution −6 − 0 =1 ✓CA answer (2) 3.2 mCD = mAB = 1 ✓CA mCD = 1 y = mx + c Substitute (10 ; −1) and mCD = 1 : ✓CA substitution of gradient −1 = 1(10 ) + c and point c = −11 y = 1x − 11 ✓CA answer (3) Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 4 KZN September 2024 Preparatory Examinations GRADE 12 Marking Guidelines 3.3 Midpoint of AC is the same as the midpoint of BD ✓A midpoint of BD [diagonals of parm. bisect each other]  Midpoint of AC  −6 + 10 −2 − 1  = M ;   2 2   −3  Answer only: ✓CA x coordinate = M2 ;   2  Full marks ✓CA y-coordinate (3) OR OR ✓A coordinates of C C (4 ; − 7)  Midpoint of AC 0+4 4−7 = M ;   2 2   −3  ✓CA x coordinate = M2 ;  ✓CA y-coordinate  2  (3) 3.4 C (4 ; − 7) ✓CA x coordinate ✓CA y-coordinate (2) 3.5 mAB = 1 ˆ =1 tan AFG ✓CA tan ˆ AFG = 1 ˆ = 45 AFG ˆ = 45 ✓CA AFG −1 − 4 mAD = 10 − 0 1 1 =− ✓A mAD = − 2 2 ˆ =− 1 tan AHJ 2 ˆ AHJ = 153, 43 ˆ = 153, 43 ✓CA AHJ BADˆ = 153, 43 − 45 [exterior  of HAF ] = 108, 43 ✓CA ˆ BAD = 108, 43  BCDˆ = 108, 43 [opp  s of a parm.] ˆ = 108, 43 ✓CA BCD (6) Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 5 KZN September 2024 Preparatory Examinations GRADE 12 Marking Guidelines OR OR (10 − 4 ) + ( −1 + 7 ) = 6 2 2 2 CD = ✓CA length of CD ( −6 − 4 ) + ( −2 + 7 ) = 5 5 2 2 BC = ✓CA length of BC ( −6 − 10 ) + ( −2 + 1) = 2 2 BD = 257 ✓A length of BD ˆ BD 2 = BC2 + CD2 − 2.BC.CD.cosBCD ✓A use of cosine rule ( 257 ) = (5 5 ) + ( 6 2 ) − 2.(5 5 ). (6 2 ).cosBCD 2 2 2 ˆ ✓CA substitution into cosine (5 5 ) + ( 6 2 ) − ( 257 ) 2 2 2 rule ˆ  cosBCD= 2. ( 5 5 ) . ( 6 2 ) ˆ = 108, 43 BCD ✓CA answer (6) [16] QUESTION 4 r 2 = OJ 2 = 22 + ( −1) 2 4.1.1 ✓A substitution r= 5 ✓A length of OJ (2) 4.1.2 OK = OJ + JK = 5 + 2 5 = 3 5 ✓A length of OK (3 5 ) = ( a − 0 ) + ( −3 − 0 ) 2 2 2 ✓A substitution 45 = a 2 + 9 a 2 = 36 ✓A a 2 subject of formula a = −6 or a = 6 N/A (3) OR OR OJ = 5  JK = 2 5 ✓A length of JK ( 2 5 ) = ( a − 2) + ( −3 + 1) 2 2 2 ✓A substitution 20 = a 2 − 4a + 4 + 4 a 2 − 4a − 12 = 0 ✓A standard form ( a − 6 )( a + 2 ) = 0 a = 6 or a = −2 N/A (3) ( x − 6 ) + ( y + 3) = 20 ✓A ( x − 6 ) + ( y + 3) 2 2 2 2 4.1.3 ✓CA = 20 (2) Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 6 KZN September 2024 Preparatory Examinations GRADE 12 Marking Guidelines 4.1.4 Substitute (10; − 4 ) : (10 − 6 ) + ( −4 + 3) ✓CA substitution 2 2 = 17 17  20 , ✓CA 17  20  the point lies inside the circle ✓CA conclusion (3) 4.1.5 KO = 5 + 2 5 = 3 5 ✓ CA length of KO In POR and PKS : 1. Pˆ = Pˆ [common] ˆ 2. PRO = PSK ˆ [= 90 ; tangent ⊥ radius] 3. PORˆ = PKS ˆ [remaining s ] POR ||| PKS [  ] PO OR = [ ||| s ] PK KS OR 1 = = 2OR 2 1  PO = PK ✓A PO = 1 PK 2 2 PO = OK = 3 5 ( ) PK = 2 3 5 = 6 5 ✓ CA length of PK ˆ PSK = 90 [radius ⊥ tangent] PS2 = PK 2 − KS2 [Theorem of Pythagoras] ✓ CA substitution in Theorem ( ) − (2 5 ) 2 2 = 6 5 of Pythagoras = 160  PS = 160 = 4 10 ✓CA answer (5) 25 25 4.2.1 x 2 − 4 x + 4 + y 2 + 5 y + = −d + 4 + ✓A completing the square 4 4 2 ( x − 2 ) +  y +  = −d + 2 5 41  5 ✓A ( x − 2 ) +  y +  2 2  2 4  2  5 Answer only: Centre  2 ; −  Full marks ✓CA x coordinate  2 ✓CA y coordinate (4) 4.2.2 diameter = 24 units,  radius =12 units ✓A radius =12 units 41 −d + = 144 ✓CA equating 4 535 d =− ✓CA answer 4 (3) [22] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 7 KZN September 2024 Preparatory Examinations GRADE 12 Marking Guidelines QUESTION 5 5.1.1 tan 58 = n r 2 = x2 + y 2 [Theorem of Pythagoras] 32° ✓A subst. in Theorem of r =1 + n 2 2 2 Pythagoras n r = n2 + 1 ✓A r = n 2 + 1 n  sin 58 = 1 + n2 58° ✓CA answer 1 (3) 5.1.2 sin 296 = − sin 64 ✓A − sin 64 = − sin 2 ( 32 ) = −2sin 32 cos 32 ✓CA expansion = −2 cos 58 sin 58 ✓CA co-functions  1  n  = −2     1 + n  1 + n  2 2 −2n ✓CA answer = 1 + n2 (4) 5.1.3 cos 2 = cos ( 60 − 58 ) ✓A cos ( 60 − 58 ) = cos 60 cos 58 + sin 60 sin 58 ✓A expansion 1 1 3 n =  +  2 1+ n 2 2 1 + n2 ✓CA answer 1 + 3n = 2 1 + n2 (3) OR OR cos 2 = cos ( 32 − 30 ) ✓A cos ( 32 − 30 ) = cos 32 cos 30 + sin 32 sin 30 ✓A expansion n 3 1 1 =  +  1+ n 2 2 1+ n 2 2 ✓CA answer 3n + 1 = 2 1 + n2 (3) Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 8 KZN September 2024 Preparatory Examinations GRADE 12 Marking Guidelines 5.2.1 LHS 1 − (1 − 2sin 2 x ) ✓A 1 − 2sin 2 x = 2sin x cos x ✓A 2sin x cos x 2sin 2 x = 2sin x cos x sin x = cos x ✓A simplification = tan x = RHS (3) OR OR LHS = ( sin 2 x + cos 2 x − 2 cos 2 x − 1 ) ✓A 2cos 2 x − 1 2sin x cos x ✓A 2sin x cos x sin x − cos 2 x + 1 2 = 2sin x cos x sin 2 x − cos 2 x + sin 2 x + cos 2 x = 2sin x cos x 2 2sin x = 2sin x cos x sin x = cos x ✓A simplification = tan x = RHS (3) OR LHS OR = ( sin 2 x + cos 2 x − cos 2 x − sin 2 x ) 2sin x cos x ✓A cos 2 x − sin 2 x 2sin x2 ✓A 2sin x cos x = 2sin x cos x sin x = cos x ✓A simplification = tan x = RHS (3) Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 9 KZN September 2024 Preparatory Examinations GRADE 12 Marking Guidelines 5.2.2 tan15 1 − cos 2 (15 ) 1 − cos 2 (15 ) = ✓A sin 2 (15 ) sin 2 (15 ) 1 − cos 30 = sin 30 3 1− 2 ✓A substitution of special = 1 angle values 2  3 2 = 1 −   2  1 = 2− 3 ✓CA answer (3) sin x 5.3 sin ( 360 + x ) .cos ( 90 + x ) − cos ( − x ) .tan ( 360 − x ) = sin x. ( − sin x ) − sin x ✓A sin x ✓A − sin x cos x. ( − tan x ) ✓A cos x ✓A − tan x = − sin 2 x + 1 ✓CA 1 = cos x 2 ✓CA answer (6) 1 1 5.4 cos 2 x − = sin x 3 3 1 1 1 − 2sin 2 x − = sin x ✓A 1 − 2sin 2 x 3 3 3 − 6sin x − 1 = sin x 2 6sin 2 x + sin x − 2 = 0 ✓A standard form (3sin x + 2)(2sin x − 1) = 0 ✓CA factors 2 sin x = − 3 ✓CA x = 221,81 + k.360  x = 221,81 + k.360 or x = 318,19 + k.360 , k  Z or x = 318,19 + k.360 1 or sin x = 2 ✓CA x = 30 + k.360 or  x = 30 + k.360 or x = 150 + k.360 , k  Z x = 150 + k.360 ✓A k  Z (6) 5.5 sin ( 2 x + 30 ) + k = 3 sin ( 2 x + 30 ) = 3 − k ✓A sin ( 2 x + 30 ) = 3 − k sin ( 2 x + 30 )  −1 or sin ( 2 x + 30 )  1 ✓A sin ( 2 x + 30 )  −1 or sin ( 2 x + 30 )  1 3 − k  −1 or 3− k 1 ✓CA 3 − k  −1 or 3 − k  1 k 4 or k2 ✓CA k  4 ✓CA k  2 (5) [33] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 10 KZN September 2024 Preparatory Examinations GRADE 12 Marking Guidelines QUESTION 6 1 6.1 b= ✓A answer 2 (1) 6.2 period = 360 ✓A answer (1) 6.3 A ( 30;1) ✓A 30 ✓A 1 (2) 6.4 x = 160 ✓A answer (1) 6.5 −3  y  1 ✓✓ AA (2) OR OR y   −3 ; 1 ✓✓ AA (2) [7] QUESTION 7 sin y sin x 7.1 = ✓A substitution in sine rule 2b b 2b sin x 2b sin x sin y = OR b sin y = 2b sin x ✓A sin y = OR b b sin y = 2sin x b sin y = 2b sin x (2) AB AB 7.2 = tan  ✓A = tan  BC BC  AB = BC.tan ✓A AB = BC.tan D̂ = 180 − ( x + y ) ✓A D̂ = 180 − ( x + y ) ˆ BC = BD + CD − 2BD.CD cos D 2 2 2 BC2 = (2b) 2 + b 2 + 2(2b)(b) cos 180 − ( x + y )  ✓A substitution in cosine rule BC2 = (2b)2 + b2 + 2(2b)(b) cos ( x + y ) ✓A + cos ( x + y ) BC = 5b + 4b cos ( x + y ) 2 2 2 BC2 = b 2 ( 5 + 4 cos ( x + y ) ) ✓A simplification ✓A taking square root on BC = b ( 5 + 4cos ( x + y ) ) LHS and RHS  AB = b tan  (5 + 4cos ( x + y )) (7) 7.3 AB = 54,8 tan 42, 6 5 + 4 cos ( 31 + 75,84 ) ✓A substitution AB = 98, 76 metres ✓A answer (2) [11] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 11 KZN September 2024 Preparatory Examinations GRADE 12 Marking Guidelines QUESTION 8 8.1.1 ( ˆ = 1 COE A1 2 ˆ ) [  at centre = 2  at circumference] ✓R = 68 ✓A answer (2) 8.1.2 ˆ Ê1 = A [tan chord theorem] ✓R 1 = 68 ✓CA answer (2) 8.1.3 ˆ = Eˆ BCE [ alt  s; DF || CA] ✓R 1 = 68 ✓CA answer (2) 8.1.4 ˆ = 180 − BCE G ˆ [opp.  s of cyclic quad] ✓R = 112 ✓CA answer (2) 8.2 ˆ = 90 BED [radius ⊥ tangent] ✓S✓R = B̂1 [co-interior  s ; DF || CA] ✓ S/R 1  AB = AC [line from centre ⊥ to chord] ✓R 2 = 7 units ✓A answer (5) OR OR ˆ = 90 BED [radius ⊥ tangent] ✓S✓R  Ê 2 = 22 ( ˆ + Eˆ  ˆB1 = 180 − BCE 2 ) [sum of  s of  BCE] = 180 − ( 68 + 22 ) = 90 ✓ S/R 1  AB = AC [line from centre ⊥ to chord] ✓R 2 = 7 units ✓A answer (5) [13] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 12 KZN September 2024 Preparatory Examinations GRADE 12 Marking Guidelines QUESTION 9 9. In QVS: QU QT = [prop. theorem; UT||VS] OR [line || one side of  ] ✓S ✓R UV TS 5 = 2 5k = 2k 5 15 5k = 2 x ; or: x = k . And: 3x = k ✓ x i.t.o. k 2 2 In UPR: PS UV = [prop. theorem; UP||VS] OR [line || one side of  ] ✓S PR UR 2k = 15 k 2 4 = ✓S 15 PS 4  = SR 11 ✓answer (6) [6] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 13 KZN September 2024 Preparatory Examinations GRADE 12 Marking Guidelines QUESTION 10 10.1 A D 1 P Q E F B C Construct AP = DE and AQ = DF ✓construction In APQ and DEF: 1. AP = DE [from construction] 2. AQ = DF [from construction] 3. ˆ A= ˆ D [given] ✓ APQ  DEF APQ  DEF [SAS]  Pˆ = Eˆ 1 [from   s ] ✓ Pˆ1 = Eˆ But: B ˆ = Eˆ [given]  Pˆ1 = Bˆ  PQ || BC [corresponding  s are =] ✓S✓R AP AQ  = [prop. theorem; PQ || BC] ✓S/R AB AC DE DF  = [DE = AP; DF = AQ] AB AC (6) 10.2.1 In MKL and MNP: MK 40 1. = = 2,5 ✓S MN 16 KL 25 2. = = 2,5 ✓S NP 10 ML 30 3. = = 2,5 ✓S MP 12 MKL||| MNP [sides of  s in proportion] ✓R (4) Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 14 KZN September 2024 Preparatory Examinations GRADE 12 Marking Guidelines 10.2.2 ˆ = Lˆ NPM [from |||  s ] ✓S ✓R  KLNP is a cyclic quadrilateral [converse: ext.  of cyclic ✓R quadrilateral] OR [ext.  of quad = int. opp.  ] (3) OR OR ˆ =K PNM ˆ [from |||  s ] ✓S ✓R  KLNP is a cyclic quadrilateral [converse: ext.  of cyclic quadrilateral] OR ✓R [ext.  of quad = int. opp.  ] (3) 10.3.1 In BCE and ADE: ✓ selecting triangles 1. Eˆ 1 = Eˆ 3 [vertically opp.  s] ✓S 2. Cˆ =D ˆ [  s in the same segment] ✓S/R 1 2 ˆ 3. B̂ = A [sum of  s of  s ] ˆ ✓ B̂ = A BCE||| ADE [   ] OR BC AD [   ]  = [from |||  s ] CE DE ✓R AD.CE  BC = ✓S/R DE (5) 10.3.2 In ADE and BDC: ✓ selecting triangles ˆ =D 1. D ˆ [given] 2 1 2. Aˆ =B ˆ [  s in the same segment] ✓S/R ˆ 3. Ê3 = BCD [sum of  s of  s ] ADE||| BDC [   ] ✓R AD DE ✓S  = [from |||  s ] BD CD  AD.CD = DE.BD = DE. ( DE + BE ) ✓ substitute DE + BE = DE 2 + DE.BE (5) [23] TOTAL: 150 Copyright Reserved Please turn over