Mathematics Grade 12 Paper 2 Free State Mock 2018 Memo

Downloaded from hlayiso.com PREPARATORY EXAMINATION GRADE 12 MATHEMATICS P2 SEPTEMBER 2018 MARKS: 150 MARKING GUIDELINE This marking guideline consists of 14 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 2 FS/September 2018 Grade 12 Marking Guidelines QUESTION 1 1.1 Lower quartile : Q1 = 89 ✓ 89 Upper quartile : Q3 = 113 ✓ 113 (2) 1.2 ✓ Min = 58 and max = 145 ✓ Q1 = 89 and Q3 = 113 ✓ Q2 = 100 (3) 1.3 Symmetrical accept both ✓ Answer Marginally skewed to the right (1) 1.4 1 522 1522 Mean : x = ✓✓ 15 15 x = 101,47 (2) 1.5 Standard deviation :  = 19,07 ✓  = 19,07 (1) 1.6 (x −  ; x +  ) ✓ 82,4 = (82,4 ; 120,54) ✓ 120,54 ✓ 2 days 2 days (3) [12] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 3 FS/September 2018 Grade 12 Marking Guidelines QUESTION 2 2.1 𝑟 = −0,95 ✓ ✓ value of r (2) 2.2 There is a strong negative relationship. ✓strong negative (1) 2.3 𝑎 = 11,71323529 ✓ 𝑎 = 11,71 𝑏 = −1,11764 ✓ 𝑏 = −1,12 𝑦̂ = 11,71 − 1,12𝑥 ✓ correct equation (3) 2.4 𝑦̂ = 11,71 − 1,12(6) ✓ Substitution 𝑦̂ = 4,99 ✓ 𝑦 ≈ 5 times 𝑦̂ ≈ 5 times OR 𝑦̂ = 5,007 ✓ calculator = 5 times ✓ 𝑦̂ ≈ 5 times (2) [8] QUESTION 3 3.1.1 M EF = M FG ✓ M EF = M FG − 1 − 3 1 − (−1) = ✓ Substitution into the 0−4 t correct formula − 4 1+1 = −4 t 2 2 ✓ 1= 1= t t ✓ Answer t=2 (4) 3.1.2 M EF  M FG = −1 2 2 1  = −1 ✓ 1 = −1 t t ✓ Answer t = −2 (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 4 FS/September 2018 Grade 12 Marking Guidelines 3.2.1 𝑦 = 1 𝑥 + 8 ✓ rewrite equation of 2 PR 1 1 M QR = M PS = QR//PS ✓ M PS = 2 2 y −5 = 1 (x − 4) ✓ Substitute (4; 5) 2 1 correctly y= x−2+5 2 1 ✓ Equation y = x+3 2 (4) 3.2.2 1 0 = x+3 2 ✓ 𝑦=0 x = −6 ∴ S(−6; 0) ✓ 𝑆(−6; 0) (2) 3.2.3  2 + ( −6) 6 + 0  2 + (−6) M ;  ✓  2 2  2 6+0 = M (−2; 3) ✓ 2 (2) 3.2.4 Midpoint of QS = midpoint of PR (diagonals bisect) x+4 ✓ −2= x+4 2 −2= 2 𝑥 = −8 y+5 ✓ 3= y+5 2 3= 2 ✓ R(−8 ; 1) y =1 ∴ R(−8 ; 1) (3) OR P → Q : (x − 2; y + 1) S → R : (x − 2 : y + 1) ✓ → R : (x − 2 : y + 1)  R(− 6 − 2 : 0 + 1) ✓✓  R(− 8 : 1)  R(− 8 : 1) (3) 3.2.5 1 𝑚𝑃𝑆 = 2 PS // QR 1 1 tan O 𝑆̂ 𝑃 = 𝑚𝑃𝑆 = ✓ tan O 𝑆̂ 𝑃 = 𝑚𝑃𝑆 = 2 2 ∴ 𝑂𝑆̂𝑃 = 26,57° ✓ 𝑂𝑆̂𝑃 = 26,57° 1 1 𝑚𝑅𝑆 = − 2 ✓ 𝑚𝑅𝑆 = − 2 𝑅𝑆̂O = 180° − 26,57° ✓ 𝑅𝑆̂ O = 153,43° = 153,43° ̂ ∴ 𝑅𝑆𝑃 = 153,43° − 26,57° = 126,86° ✓ 126,86° (5) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 5 FS/September 2018 Grade 12 Marking Guidelines [22] QUESTION 4 4.1  − 2 + 4 3 + (−7)  ✓ Subst. into the M ;  correct formula  2 2  = M (1;−2) ✓ Centre of the circle ∴ centre of the circle is (1; −2) (x − 1)2 + ( y − (−2))2 = r 2 ✓ Subst. 𝑁(−2; 3) (− 2 − 1)2 + (3 + 2)2 = r 2 9 + 25 = 𝑟 2 ✓ 34 = 𝑟 2 ∴ 34 = 𝑟 2 (x − 1)2 + ( y + 2)2 = 34 ✓ Equa of the circle (5) 4.2 −2−3 ✓ Subst. into the M rad = correct formula 1 − ( −2 ) −5 −5 ✓ M rad = = 3 3 3 M tan = radius ⊥ tangent ✓ M tan = 3 5 5 y − 3 = (x − (−2) ) 3 5 ✓ Subst. 𝑁(−2; 3) 3 6 and gradient y = x + +3 5 5 3 21 = x+ 5 5 ✓ Equation (5) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 6 FS/September 2018 Grade 12 Marking Guidelines 4.3 Let the angle of inclination of line NP be α and the angle 5 ✓ tan  = − adjacent to it be θ 3 −5 tan  = 3 5  = 180 − tan   ✓ 𝛼 = 120,96°  3   = 180 − 59,036243... ✓  = 59,04 = 120,96° OR  = 59,04 𝑁𝑄̂ 𝑃 = 120,96° − 90° ^ ^ ✓ NPQ = 30,96 NPQ = 180 − (59,04 + 90) = 30,96° (4) = 30,96° (ext.< 𝑜𝑓 ∆) OR ✓ 𝑁𝑄̂ 𝑃 = 90° 𝑁𝑄̂ 𝑃 = 90° diameter subtends a right angle ✓NQ = 6 and PQ = 𝑁𝑄 = 6 and 𝑃𝑄 = 10 10 6 6 ∴ tanN 𝑃̂ 𝑄 = ✓ tanN 𝑃̂ 𝑄 = 10 10 ✓ 𝑁𝑄̂ 𝑃 = 30,96° ∴ 𝑁𝑄̂ 𝑃 = 30,96° (4) 4.4 QP = 10 units and NQ = 6 units 1 Area ∆NPQ = 2 (6)(10) ✓ = 30 square units Area ∆NPQ 30✓ ∴ Area 𝑜𝑓 𝑐𝑖𝑟𝑐𝑙𝑒 = 34𝜋 ✓ = 0,28 ✓ (4) [𝟏𝟕] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 7 FS/September 2018 Grade 12 Marking Guidelines QUESTION 5 5.1.1 Reflex BÔP = 𝜃 y 𝜃 ෡ O x B P൫𝑐; −√21 ൯ c2 + ( 21) = 5 2 2 ✓ subst into pyth c 2 = 25 − 21 c2 = 4 ✓ Answer c = 2 (2) 5.1.2 2 a) cos = 5 ✓ Answer (1) 2 2 − 21  − 21  21  21  b) = +    ✓ +    2  5  2  5  − 25 21 + 42 for corr subst = ✓ Answer 50 (2) c) 2 sin  cos ✓ Identity  − 21  2  ✓ Answer = 2    (2)  5  5  − 4 21 = 25 5.2 (− sin x). tan x. cos(360 − 30) (sin x) 2 ✓ − sin 𝑥 ✓ (sin 𝑥)2 − sin x. tan x(cos30) ✓ cos 30° = sin 2 x. −√3 = tan 𝑥 ÷ sin 𝑥 2 sin x ✓ −√3sin 𝑥 1 cos x = X 2 cos 𝑥 sin𝑥 ✓ answer −√3 = (5) 2cos𝑥 [𝟏𝟐] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 8 FS/September 2018 Grade 12 Marking Guidelines QUESTION 6 6.1 2 sin 2 x ✓ sin x LHS = sin x cos x 2 − 2 sin x cos x cos x ✓ 2𝑠𝑖𝑛 𝑥𝑐𝑜𝑠 𝑥 2 sin 2 x =  1  1 2 sin x − cos x  ✓ 2 sin 𝑥 (cos 𝑥 − cos 𝑥)  cos x  sin x = sin x 1 − cos2 x ✓ 1 − cos x 2 cos x cos x cos x = sin x  2 sin x ✓ 𝑠𝑖𝑛2 𝑥 cos x = sin x ∴ LHS = RHS (5) 6.2 This identity will be undefined when 2 + 2 cos 2 x = 0 cos 2 x = −1 ✓ cos 2 x = −1 ∴ 2𝑥 = 180° + 𝑘. 360° 𝑘𝜖ℤ 𝑥 = 90° + 𝑘. 180° ✓ 2𝑥 = 180° + 𝑘. 360°; 𝑘𝜖ℤ NB: Penalise 1 mark for leaving out 𝑘𝜖ℤ ✓ 𝑥 = 90° + ✓𝑘. 180°, ; 𝑘𝜖ℤ (4) OR 2 + 2 cos 2𝑥 = 0 1 + cos 2𝑥 = 0 ✓ Identity 1 + 2𝑐𝑜𝑠 2 𝑥 − 1 = 0 cos 𝑥 = 0 ✓ cos 𝑥 = 0 ∴ 𝑥 = 90° 𝑥 = ±90° + 360°𝑘; 𝑘 ∈ ℤ ✓ ± 90° + ✓k 360°; 𝑘 ∈ ℤ (4) [9] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 9 FS/September 2018 Grade 12 Marking Guidelines QUESTION 7 7.1 𝐷𝐸෡ 𝐹 = 180° − 𝜃 opp angle of cyclic quad ✓ Answer (1) 7.2 ✓ correct sub into 𝐷𝐹 = √𝑟 2 + 𝑟 2 − 2𝑟 2 cos(180° − 𝜃) cosine 𝑟𝑢𝑙𝑒 = √2𝑟 2 + 2𝑟 2 cos 𝜃 ✓ simplification (2) = 𝑟√2 + 2 cos 𝜃 given 7.3 𝐷𝐹̂ 𝐺 = 90° angl in semi circle 𝐷𝐹 In ∆𝐷𝐹𝐺: sin 𝜃 = 𝐷𝐺 𝑟√2 + 2 cos 𝜃 ✓sin 𝜃 = 𝑟√2 + 2 cos 𝜃 2𝑟 sin 𝜃 = 2𝑟 ✓ (2 sin 𝜃)2 = (√2 + 2 cos 𝜃)2 4 sin2 𝜃 = 2 + 2 cos 𝜃 4 sin2 𝜃 = 2 + 2 cos 𝜃 ✓2 sin2 𝜃 = 1 + cos 𝜃 (3) 2 sin2 𝜃 = 1 + cos 𝜃 7.4 ✓ 2(1 − cos 2 𝜃) = 1 + 2 2(1 − cos 𝜃) = 1 + cos 𝜃 cos 𝜃 2 − 2 cos2 𝜃 = 1 + cos 𝜃 −2 cos2 𝜃 − cos 𝜃 + 1 = 0 2 cos 2 𝜃 + cos 𝜃 − 1 = 0 ✓ standard form (2 cos 𝜃 − 1)(cos 𝜃 + 1) = 0 ✓ factors cos 𝜃 = 2 1 or cos 𝜃 ≠ −1 ✓ both equations ✓ 𝜃 = 60° ∴ 𝜃 = 60° θ ≠ 180° (5) [11] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 10 FS/September 2018 Grade 12 Marking Guidelines QUESTION 8 8.1 𝑎=1 ✓ 𝑎=1 𝑏=2 ✓𝑏=2 𝑐=2 ✓𝑐=2 𝑑=1 ✓𝑑=1 (4) 8.2 𝑃(21,44°; 0,73) ✓ correct substitution (2) 8.3.1 𝑥 = 90° ✓ 90° (1) 8.3.2 x  [45;135] ✓ 45° 𝑎𝑛𝑑 135° OR ✓ Notation 45  x  135 (2) [9] QUESTION 9 9.1.1 𝐶̂ = 𝐴̂1 = 78∘ alt <'s AP // CB ✓ R (1) 9.1.2 𝐴𝐵̂ 𝐶 = 𝐴̂1 = 78∘ tan chord theorem ✓ S✓R (2) 9.1.3 𝐴̂3 = 𝐶̂ = 78∘ tan chord theorem ✓ S (1) 9.1.4 𝐴̂1 + 𝐴̂2 + 𝐴̂3 = 180∘ straight line ✓ S 78∘ + 𝐴̂2 + 18∘ = 180∘ 𝐴̂2 = 24∘ (1) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 11 FS/September 2018 Grade 12 Marking Guidelines 9.2.1 𝑌̂1 = 𝑎 tan chord theorem ✓S✓R✓ 𝑌̂1 = 𝑌𝑋෡ 𝑃 = 𝑎 alt angle XY // QT ✓S 𝑌𝑋෡ 𝑃 = 𝑄̂ = 𝑎 = angles opp = 𝑠𝑖𝑑𝑒𝑠 ✓S 𝑌̂2 = 𝑄̂ = 𝑎 angles in same circle segment ✓S✓R✓ (6) 9.2.2 ෡2 = XY T ෡T alt angles; XY // QT ✓ S/R =෡Y1 + ෡ Y2 =෡ T1 + ෡ T1 ✓ ෡ = 2T1 (2) 1 9.2.3 𝑇̂2 + 𝑇̂1 = 90° angle in 2 ⨀ S✓R✓ 𝑇̂2 = 90° − 𝑎 (2) 9.2.4 𝑇̂2 = 90° − 𝑎 proven But YO෡ Q = 2𝑇̂2 angle at centre = 2 angl on circle ✓ S = 2(90° − 𝑎) = 180° − 2𝑎 QO෡ R= 180° − 180° + 2𝑎 = 2𝑎 ∴ SORT is a cyclic quad ext angl = opp int angle ✓ R OR ෡O = 90° XY tan ⊥ radius ෡ XOY = 90°−∝ sum of < 𝑠 of ∆ ✓S ෡ ̂ XOY = 𝑇2 ∴ SORT is a cyclic quad/converse: ext. <s of cyclic quad ✓R OR ෡ Y = 2𝑇̂1 XO < at centre = 2 < at circum. ✓S ෡ ̂ XOY = 𝑇2 ∴ SORT is a cyclic quad/converse: ext. < of cyclic quad ✓R (2) 9.2.5 𝑇̂1 + 𝑇̂2 = 90° ✓ S 𝑎 + 2𝑎 = 90° 3𝑎 = 90° ∴ 𝑎 = 30° ✓ answer (2) 9.2.6 𝑋𝑌̂𝑂 = 90° rad ⊥ tangent ✓ S 𝑅̂2 = 90° alt.angles; XY // TQ ✓ S ∴ 𝑇𝑅 = 𝑅𝑄 line from centre of circle ⊥ ✓ R chord (3) [22] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 12 FS/September 2018 Grade 12 Marking Guidelines QUESTION 10 10.1.1 Const: Join KZ & LY & draw ℎ1 from K ⊥ XL & ℎ2 ✓ const from L ⊥ XK Proof 𝐴𝑟𝑒𝑎 ∆𝑋𝐾𝐿 𝑋𝐾 1 ✓ = 𝐴𝑟𝑒𝑎 ∆𝑋𝐾𝐿 2 𝑋𝐾×ℎ1 𝑋𝐾 𝐴𝑟𝑒𝑎 ∆𝐿𝑌𝐾 𝐾𝑌 = 1 = 𝐾𝑌 𝐴𝑟𝑒𝑎 ∆𝐿𝑌𝐾 𝐾𝑌×ℎ1 2 1 𝐴𝑟𝑒𝑎 ∆𝑋𝐾𝐿 𝑋𝐿 Area ∆XKL = 2 XL×ℎ2 XL = LZ ✓ = Area ∆KLZ 1 LZ×ℎ2 𝐴𝑟𝑒𝑎 ∆𝐾𝐿𝑍 𝐿𝑍 2 ✓ R Area of ∆𝑋𝐾𝐿 = 𝐴𝑟𝑒𝑎 ∆ 𝑋𝐾𝐿 common But Area ∆ 𝐿𝐾𝑌 = 𝐴𝑟𝑒𝑎 ∆ 𝐾𝐿𝑍 same base & height; LK// YZ ✓S (5) 𝐴𝑟𝑒𝑎 ∆ 𝑋𝐾𝐿 𝐴𝑟𝑒𝑎 ∆𝑋𝐾𝐿 = 𝐴𝑟𝑒𝑎 ∆𝐿𝑌𝐾 𝐴𝑟𝑒𝑎 ∆𝐾𝐿𝑍 𝑋𝐾 𝑋𝐿 ∴ = 𝐾𝑌 𝐿𝑍 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 13 FS/September 2018 Grade 12 Marking Guidelines 10.2.1 𝐸𝐷 𝐴𝑇 2 ✓S✓R = 𝐴𝐶 = 3 line // to one side of  𝐷𝐶 2 DE = 3 × 9 =6 ✓DE = 6 ∴ 𝐷 is the midpoint of BE (3) 10.2.2 BF= FT line from midpt drawn // to 2nd side bisct ✓ S/R 3rd 1 FD= 2TE midpoint theorem ✓R ✓ 𝑇𝐸 = 4 ∴ 𝑇𝐸 = 4 (3) 10.2.3(a 𝐴𝑟𝑒𝑎 𝑜𝑓 ADC ) 𝐴𝑟𝑒𝑎 𝑜𝑓ABD 𝐷𝐶 = 𝐵𝐷 same height ✓ same height 3𝑦 = 2𝑦 3 =2 3 ✓2 (2) 10.2.3(b 1 ) 2 × 𝑇𝐶 × 𝐸𝐶 × sin 𝐶 1 1 × 𝑇𝐶 × 𝐸𝐶 × sin 𝐶 2 × 𝐴𝐶 × 𝐵𝐶 × sin 𝐶 ✓2 1 2 × 𝐴𝐶 × 𝐵𝐶 × sin 𝐶 (𝑥)(𝑦) = (𝑥)(𝑦) (3𝑥)(5𝑦) ✓ (3𝑥)(5𝑦) 1 = 15 1 ✓ 15 (3) [16] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 14 FS/September 2018 Grade 12 Marking Guidelines QUESTION 11 11.1 In TPS and  QSR 𝑃𝑆 1,5 3 𝑃𝑆 1,5 = = ✓ = 𝑄𝑆 4 8 𝑄𝑆 4 𝑇𝑃 4,5 3 𝑇𝑃 4,5 = = ✓ = 𝑆𝑅 12 8 𝑆𝑅 12 𝑇𝑆 3,6 𝑇𝑆 3,6 3 ✓ = 𝑄𝑅 9,6 = = 𝑄𝑅 9,6 8 ∴ TPS/// QSR sides of triangles in proportion ✓✓ S/R ∴ 𝑃𝑇̂S= 𝑅̂ s are equiangular ✓ 𝑃𝑇̂S= 𝑅̂ ∴ 𝑇𝑃 is a tangent converse of tan chord theorem ✓R (7) 11.2 𝑃̂ = 𝑄𝑆̂𝑅 s are equiangular ✓ S/R 𝑄𝑆|| TP corres. <s are = ✓ S/R 𝑇𝑄 = 12 1,5 proportion theorem; PT||SQ ✓✓ S/R 9,6 ✓ 𝑇𝑄 = 1,2 ∴ 𝑇𝑄 = 1,2 (5) [12] TOTAL: 150 Copyright reserved Please turn over