Mathematics Grade 12 P2 Pretrial FS Mock 2023 Paper & Memo

Downloaded from Stanmorephysics.com education Department of Education FREE STATE PROVINCE MARKS: 150 DURATION: 3 HOURS This question paper consists of 15 pages. 1

Downloaded from Stanmorephysics.com INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. This question paper consists of 9 questions. 2. Answer AEE*the questions. 3. Number theyanswers correctly according to the numbering system used in this question paper. 4. Clearly show ALL calculations, diagrams, graphs, etc. which you have used in determining your answers. 5. Answers only will NOT necessarily be awarded full marks. 6. You may use an approved scientific calculator (non-programmable and non- graphical), unless otherwise stated. 7. If necessary, round off answers to TWO decimal places, unless stated otherwise

Downloaded from Stanmorephysics.com QUESTION 1 The speeds, in kilometres per hour, of cyclists that passed a point on the route of the Ironman Race were recorded and summarised in the table below: Speed (km/h) EL Frequency i} Cumulative On Frequency 0<x<10 7 10 10 10<x<20 5 30 20<x<30 © 45 30<x<40 72 40<x<50 170 1.1 Complete the above table in the ANSWER BOOK provided. (2) 1.2. Make use of the axes provided in the ANSWER BOOK to draw a cumulative frequency curve for the above data. (3) 1.3 Indicate clearly on your graph where the estimates of the lower quartile (QI ) and median (M) speeds can be read off. Write down these estimates. (2) 1.4. Draw a box and whisker diagram for the data. Use the number line in the ANSWER BOOK. (2) 1.5 Use your graph to estimate the number of cyclists that passed the point with speeds greater than 35 km/h. qd) [10] QUESTION 2 During the month of June patients visited a number of medical facilities for treatment. The table shows the number of patients treated on certain dates during the month of June Dates in the month of June 3 5 8 12 | 15 | 19 | 22 | 26 Number of patients treated 270 | 275 | 376 | 420 | 602 | 684 | 800 | 820 2.1 On DIAGRAM SHEET 2, draw a scatter plot of the given data. (3) 2.2 Determine the equation of the least squares regression line of patients treated (y) against date (x). (3) 2.3. Estimate how many patients have been treated on the 24th of June. (2) 2.4 Draw the least squares regression line on the grid on DIAGRAM SHEET 2. (3)

Downloaded from Stanmorephysics.com 2.5 Calculate the correlation coefficient of the data. Comment on the strength of the relationship between the variables. (3) 2.6 Given that the mean for patients treated on certain dates is 528,63 calculate how many patieats were within one deviation of the mean (3) [17] QUESTION 3 In the diagram below, A (-3 ; k), B(4 ; 8), C(5 ; 0) and D (-2 ; -4) are vertices of the parallelogram ABCD. Diagonals AC and BD bisect each other at P. The angles of inclination of AD and BD are a and B respectively. AD cuts the x-axis at E. F is a point in the fourth quadrant. e B(4; 8) A334 Pp as B E 0 C(5; 0) x D(-2;-4) e F 3.1 Determine the gradient of BC. (2) 3.2 If the distance between points A(-3 ; k) and B(4 ; 8) is 65, calculate the value of k. (4) 3.3 Prove, using analytical geometry methods, that BP | AC. (3) 3.4 Calculate the coordinates of F if it is given that ACFD is a parallelogram. (2) 3.5 Calculate the size of EDO (correct to ONE decimal place). (6) 3.6 Calculate the area of AADC. (4) [21]

Downloaded from Stanmorephysics.com QUESTION 4 4. A circle with centre at C passes through the origin, O, and also intersects the x-axis at F and the y-axis at E. The tangent to the circle at B ( 4 ; 6) intersects the x-axis at K and the F o c B(—4; -6) E L 4.1 Calculate the length of the radius of the circle. (3) 4.2. Determine the equation of the circle in the form (x _ ay + (y - by =r. (4) 4.3. What type of a triangle is AOBL ? Give reason for your answer. (2) 4.4 Determine the equation of the tangent KL. (4) 4.5 Determine the co-ordinates of E. (2) 4.2.6 Determine whether EF is a diameter of the circle. Show all working. (5) [20] QUESTION 5 5.1 If tan 58° = m, determine the following in terms of m without usifi 5.1.1 sin 58° (2) 5.1.2 sin 296° (3) 5.1.3 cos2° (3)

Downloaded from Stanmorephysics.com §2 If 5tan@+ 26 =0 and 0°< 6 < 270° , determine with the aid of a sketch and without using the calculator, the value of : 5.2.1 sinO 5.2.2 cos? 5.13 Afggs0+7V6sind Bs(E 240°), tan 225° QUESTION 6 cos(180° + x). tan(360° — x).sin*(90° — 9 5.1 Determine the value of os( x) - ( x).sin ( x) +sim° x sin(180° — x) 5.2.1 Prove the identity: cos(4 - B)-cos(A + B)= 2sin Asin B 5.2.2 Hence calculate, without using a calculator, the value of cos15°—cos75° 53 Find the value of tan @, if the distance between A(cos6; sin 2) and BO; 7) is 86. QUESTION 7 Consider: I (x)= cos(x — 45°) and J (x)= tan x for xE [- 180° ; 180°] 6.1 6.2 Use the grid provided to draw sketch graphs of f and g on the same set of axes for xe [-180° 3 180°]. Show clearly all the intercepts on the axes, the coordinates of the turning points and the asymptotes. Use your graphs to answer the following questions for x € [- 180° ; 1 80°] 6.2.1 Write the solutions of cos(x - 45°) =0 6.2.2 Write down the equations of asymptote(s) of g. 6.2.3 Write down the range of f. . . . ° 1 6.2.4 How many solutions exist for the equation cos(x—45 )s tan ? 6.2.5 For what value(s) of x is fi xdg(x)> 0? (2) (1) (4) [15] (6) (3) (4) (4) [17] (6) (2) (2) (1) (1) G3) [15]

Downloaded from Stanmorephysics.com QUESTION 7 In the figure below Thabo is standing at a point A on top of building AB that is 45 m high. He observes two cars at C and D respectively. The cars at C and D are in the same horizontal plane as B. The angle of elevation from{G+to A is 43° and the angle of elevation from D to A is 50° and C AD = 69° 45m (44 7.1. Calculate the lengths of AC and AD, correct to 2 decimal places. (4) 7.2 Calculate the distance between the two cars, the length of CD. (3) (7]

Downloaded from Stanmorephysics.com QUESTION 8 PROVIDE REASONS FOR ALL YOUR STATEMENTS AND CALCULATIONS IN QUESTION 8, 9 AND 10 Tn the diagram, is a cyclic quadrilateral. G is a point on AD such that BG || CD. ECF is a tangent to tl cle at C. BD is a chord of the circle. GB D =30° and DC F = 60° 8.1 Calculate, with reasons, the size of: 8.1.1 D, 812 «iB, 8.1.3 C, 814 DAB 8.2 Is BD a diameter of the circle? Motivate your answer. (1) (2) (1) (2) (2) 18]

Downloaded from Stanmorephysics.com QUESTION 9 9.1. In AABC below, D and E are points on AB and AC respectively such that AD AE DE || BC ve the theorem that states that DE = BEC (6)

4 education Department of Education FREE STATE PROVINCE 9.2 In the diagram below, DGFC is a cyclic quadrilateral and AB is a tangent to the circle at B. 2 Chords BD and BC are drawn. DG and CF produced meet at E and DC is produced to A. EA||GF. Downloaded from Stanmorephysics.com, ‘ 9.2.1 Give areason why B, = D, (1) 9.2.2. Prove that AABCII| AADB. (3) 9.2.3. Prove E, = D, (4) 9.2.4 Prove AE? = ADx AC (4) 9.2.5 Hence, deduce that AE=AB (3) [21| 10

Downloaded from Stanmorephysics.com SCHOOL NAME: Name: Grade 12 lnar 4000, - Speed (km/h) Frequency (f) Cumulative Frequency 0<x<10 10 10 10<x<20 30 20<x <30 45 30<x<40 72 40<x<50 170 11

Downloaded from Stanmorephysics.com QUESTION 1.2 Cumulative frequency 100 70 50 30 20 10 Cumulative Frequency Graph Ee 20 30 40 50 60 70 80 12

Downloaded from Stanmorephysics.com DIAGRAM SHEET 2 QUESTIONS 2.1 AND 2.4 tter plot of dates (in June) versus patients 900 ony =! It 800 =) 700 600 =] 2 Fy 500 Fao & 300 200 100 0 iS 10 15 20 Dates in June 13

Downloaded from Stanmorephysics.com QUESTION 9.1 14

Downloaded from Stanmorephysics.com INFORMATION SHEET: MATHEMATICS —b+aJb? —4ac 2a A=P( A=P(l=ni) A=P(l-i)" A=P(l+i)" n a | Sige yi-M or =a+(n-Nd 5, ="(2a+(n—1d) ial =) 2 2 cae" 5-0) pel S, =—%;-I<r<l r-l l-r pals -) pla") i i (x)= lim F(x+h)- fe) ho0 h + d= V(x, -x,)7 +(¥2 -y,)? m2 272) a 2 y=mxt+e yy, =m(x—x,) m=227* m=tan@ ¥q—% (x-—af +(y—byY =r? masse, 2 = 6 2 © qr sh? 40?-2be.cosA area ABC =| absinC sind sinB sinC 2 sin(a + £8) = sina.cos + cosa.sin # sin(a — 8) = sina.cos # — cos a.sinB cos(@ + 3)= cos a@.cos #—sina.sin 8 cos(@ — 3)= cos @.cos #+sina.sin 8 2 a] cos’ a@—sin @ cos 2a =41—2sin’ a sin2a@ = 2sina@.cosa 2cos? a —1 Ie D(z, -3) y=-—— o = n n nat P = 715) P(A or B) = P(A) + P(B) — P(A and B) peatbe p= 2G-20-9) De@-xy 15

Downloaded from Stanmorephysics.com education Department of Education FREE STATE PROVINCE MARKS: 150 This marking guideline consists of 14 pages. 1

Downloaded from Stanmorephysics.com QUESTION 1 1.1 Speed (km/h) Frequency (/) Cumulative frequency Y column 1 values O<x 10 10 ¥ column 2 values fnYaTa 10<x 20 30 (2) 20 < xjS30 45 75 30<x < 40 72 147 40<x<50 23 170 1.2 ¥ Grounding(0; 0) $98 v¥ Maximum (50; 170) & a8 - ¥ Curve G3) 13 | = YQ # 2 iS e Z: 3 Es 1.4 ¥ Q3 1 , 7 Y Box (2) 1.5 164 (Accept 163 — 165) v Answer (1) [10] I]

Downloaded from Stanmorephysics.com QUESTION 2 2.1 EE v¥ v v ALL points plotted correctly -1 for every 2 mistakes -1 if points are joined 1 t 1 t @) YY t E soo Ai 175; sa0.a75) | H A bade tt ' i Ea . + es i] a3 H t 100 + i ' ot } 1 o 5 10 41s 20 2 » Dates in June 2.2 | y=26,88x +161,24 ¥ value of a= 161,24 v value of b = 26,88 ¥ equation of the line (3) 2.3. | y=26,88(24) 4161,24 ¥ ssubstitute in formula y =806,36 y 806 people treated on 24th of June answer as Natural Number (2) 24 | seegrid ¥ straight line v graph passes through (x ; y) ¥ graph passes through y- intercept from regression line formula (3) 2.5 | r=0,98 This is a very strong positive, linear correlation. As the days | YY correct answer . . . v interpretation (3) increase the number of patients increase 2.6 | (x-o;x+o) (528,63 —210,46;528,63 + 210,46) (318,17 ; 739,09) 17 4 patients. 9,09) v answer (3)

Downloaded from Stanmorephysics.com [17] QUESTION 3 31] 92 ~*daoat * x x ada _ 8-0 cal Y correct substitution into the BC 4-5 correct formula Y answer mM, =—-8 Q) 3.2 2 z v substitute A and B into B=y(-3-4 k-8 A ( ) * ( ) distance formula 65=49+k? -16k +64 Y standard form 2 Y factors k —-16k +48 =0 v¥ k=4 (k -4\(k -12)= 0 k=4o0r k=12 k=4 4) 3.3 8- (- 4) Y My Myc My =2 v Mp XM 4c -4-9 _ 1 “-3-5 2 1 Myc XM = 2%— =I GB) AC | BD 3.4 midpoint of AC = midpoint of DC ¥ x-coordinate Z _ _ ¥ y-coordinate x+(-3)_=2+5 ) yt4_-4+0 2 2 2 2 x=6o0r y=-8 F(6;-8) 3.5 | m, =m, =-8 Map tana =-8 vY tana =— a =180° tan '(8) ¥ a =97,13° — ° a =97,13° ¥ fRa3,43 veal t = =2 Ml an B Mop al er B=63,43° PS O, = 63,43°(vert opp Z’) (6)

Downloaded from Stanmorephysics.com EDO=a-B EDO =97,13° - 63,43° =33,7° 3.6 = —(—3)/ +(Q—4Y = = Y length of AC AC = J (5-(-3)) + (0-4) = V80 = 45 Y leneth of DP DP=, I- 2- ly; == (- 4- 2y = 45 =3V5 Y correct substitution into i formula Area of AADG= pac x DP v answer (4) Area of AADC = 5 (4y5 avs) Area of AADC = 30 square units [08] QUESTION 4 4 41 = (x, 1) +(v,-y,) OB & * y +O.-y y v substitution OB =,|(-4-0) +(-6-0) Y OB=V52 OB = v52 v r=vl13 OB = 213 (3) r=v13 4.2 ("=F *) 2 2 C(-2;-3) (x+2) +(y+3) =13 4.3 | Right-angled triangle, tangent perpendicular to the radius. Y right angled A v reason (2) 44 - —~ J. Mog = ox ¥ Mop —6-0 ¥ Me, BO yp 3 4-0 Y substitution Mao = 2 v answer (4) m, =-2 ‘KL 3 Y-YN= n(x» yt6= 2 r+ 4)

Downloaded from Stanmorephysics.com __2 2% rr ae 4.5 F(0;-6) vv answer (2) 46 | (+2) +(063=13 F(-4:0) any mm evocl a 3 944 2 _-346 3 e240 2 Points F, C and E are collinear EF is a diameter v xX, Y Ve Y Mac Y Mey Y Points F, C and E are collinear (5) [20] QUESTION 5 5.1.1 tan 58° =m vVl4m =r x+y =r v answer P+m=r ; Vita =r @) sin 58° = —" l+m 1+m 5.1.2 | sin 296° =—sin 64° v —sin 64° sin 296° =—sin 2(32°) Y substitution sin 296° = —2sin 32° x cos32° ¥ answer sin 296° =-2x. yp l+m Vl+m GB) . 2m sin 296° =— —, l+m? lar 52-1 | 5tan@+2V6 =0 lonat

Downloaded from Stanmorephysics.com y ._v49 r= ; 5 —— 2vél v sind = 20 sin 9 =——— it v649 ; 5 tan 0 = -—— 0°< 6 < 270° ne 5 (2) re=yptx r =(-2V6) +(5) r =49 r=7 sind= 26 7 5.2.2 cos =— 5 i" answer dd) cos(— 240°), tan 225° 5 26 u- 3 + ra 255) v —sin 30° cos(90° + 30°) tan(180° + 45°) Y¥ tan45° —sin 30°. tan 45° u i v -=—xl -=xl Y answer (4) =-4 [15] QUESTION 6

Downloaded from Stanmorephysics.com 6.1 cos(180°+ x) .tan(360°~x).sin?(90°=2) J sa y —cosx sin(180°—x) ~ tana v cos’ x v sinx = Y cos x+sin’ x v answer (6) = cos’ x + sin’? x =] 6.2.1 | cos(A — B)—cos(A + B) v =cos A cos B+ sin A sin B — [cos A cos B —sin A sin B] v v = cos A cos B + sin A sin B—cos A cos B +sin A sinB =2sin A sinB G) 6.2.2 | ©os15°—cos 75° = cos(45° — 30°) — cos(45° + 30°) v v = 2sin 45°.sin 30° v v = axr2 yh or/of 2x-txt 2-2 apy) eae (4) 2 V2 OR cos15°— cos 75° = cos(45° — 30°) — cos(45° + 30°) = cos 45°cos30°+sin45°sin 30°— [cos 45°cos 30°— sin45°sin30°] = 2sin45° sin 30° =2 2d or/of 2 xxl 2°23 ond) V2 =— of = 5 of ii

Downloaded from Stanmorephysics.com 6.3 AB? = (cos 0-6) Hsin@-7)° 86 = cos’O — 12cos 6+ 36 +sin’0 — 14sin0 +49 86 = 1+ 36; — 12cos 0 —14sin 6 O=-1 14sin 0 14sin6 = -1 sin® _ -12 cos@ 14 tan@ =—°/-0,86 i LAKK (4)

\o cS = = a=! R= 228 22 Soh F op 9 foo g Sead feces Esaas SSssS>S v shape f eoSeeS: Downloaded from Stanmorephysics.com QUESTION 7 ' ' i ° 1 1 ' ial i xr 48 ' ' b H H H har 1" ks ar Be me Po ase : . ' by ov : ' : es ee eonorer sl psceks = H ' : eo oii im : ' ; ES ~ S H ' ' es Jy - Ss ' ; ' Pee =| So) 8 ° o--4 donenaoee banaes Wyn yy : , . n x ~~ wy : e|ig) & * ' i ‘ sss] ! 1 ‘ ' ~~ i} Vv ' A |o. Ss se Eel Reem caren ene nets eer! FSeeRenay eee Cenece eke ee —| & i ‘ H eioi=| & &® ' 4 ' 1|S\7| BY ' t H ay) fos i \ "Tllul fy 3% idea crtinerdetrsindedvaleciel stele ain’ ainid atiladarahatad ben] fe left oF = lan} iin = ai ailalaila i [ey eye ie Q) (2) Q) (1) U5] Y notation v answer QB) v Y answer v¥_v_answer v answer ¥ end points v answer 10

Downloaded from Stanmorephysics.com QUESTION 7 TA sin 43° = V sin 43° = > AC AC AC =66m v AC . 45 . 45 43° = —— ¥ sin 43° = —— sin 4c tC v AD sin 50° = = 4 AD =58,74m (4) 7.2. | CD? = AC’ + AD’ —2AC.AD cos 69° Y using cosine rule CD =(66) +(58.74) —2(66)(58.74)cos69° CD* =5027,72 CD =70.91m Y substitution v answer (3) 11 WEES: RU

Downloaded from Stanmorephysics.com QUESTION 8 8.1.1] A * ¥ S&R 1 , = B, = 30° (alter angles DC parallel GB) (1) 8.1.2] 4 4 ° vs B, =C, =60 (tan-chord theorem) YR (2) 8.1.3] 4 ¥ S&R 1 C, =90° (sum of angles in triangle) @ 8.1.4 2 n vs DAB+ C2 =180° YR (2) DAB =90° 8.2 | Yes is diameter, converse angle in a semi-circle vs VR (2) 12

Downloaded from Stanmorephysics.com QUESTION 9 9.1 Y constr vs vs YS YR Cc ¥ conclusion Construction: Connect DC and BE and draw the altitudes k and h i Area AADE _5*ADxK _ aD (6) Area ABDE 1xpDxk BD i Area AADE _7*%AEXh ag ‘Area ADEC — LxECxh ~ EC but/maar: Area ABDE = Area ADEC [DE common base and DE | | BC/ DE gemeensk basis en DE | | BC] Area AADE _ Area AADE “Area ABDE Area ADEC AD AE BD EC 9.2.1 tangent-chord theorem YR (I) 9.2.2 In AABC and AADB: Ag A =A, [common/gemeenskaplik] B, =D, [proven/bewys in 10.2.1] vs :. AABC ||| AADB[Z; Z; 2] YR OR In AABC and AADB: A, =A, [common/gemeenskaplik] “Ss a" a [proven/bewys in 10.2.1] vs BCA=B, [Zs of A=180°] mR :. AABC ||| AADB 9.2.3 E, =F [verwiss Ze/alternate Zs ; EA | | GF] vS YR F=5 5 [ext Z of cye quad DGFC/buite Z v kdvh YS YR E,=D, 13

Downloaded from Stanmorephysics.com 9.2.4 In AAEC and AADE: vs [common/gemeenskaplik] v [proven/bewys in 10.2.3] s v C ||| AADE (4; 2; 2] e vs = AD AC AAEC and AADE: [common/gemeenskaplik] vs [proven/bewys in 10.2.3] vs [4s of A= 180° OR ext 4 of cye quad DGFC/ VR buite 2 v kdvh DGFC] .. AAEC ||| AADE _ AE _ AC 2 AE? = ADx AC 9.2.5 AB LAC [AARC ||| AADR] “S AD AB AB? = ADx AC vs 2 “8 = AE~ [from 10.2.4] -. AB= AE 24) 14