Mathematics Grade 12 Paper 2 MP June 2023 Paper & Memo

We Stgumorephysics.com A, education } MPUMALANGA PROVINCE i REPUBLICOF SOUTH AFRICA MARKS: 150 TIME: 3 hours This question paper consists of 13 pages and 1 information sheet and an answer book is provided.

Mathemapenynloaded from Stanmorephysics.com MDE/Sune 2023 INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. The question pay nsists of 10 questions. 2. Answer ALL tions in the SPECIAL ANSWER BOOK provided. 3. Clearly show ALL calculations, diagrams, graphs, etc. which you have used in determining the answers. 4. Answers only will NOT necessarily be awarded full marks. 5. You may use an approved scientific calculator (non-programmable and non-graphical) unless stated otherwise. 6. Ifnecessary, round off answers to TWO decimal places, unless stated otherwise. 7. Diagrams are NOT necessarily drawn to scale. 8. Write neatly and legibly. Copyright reserved Please turn over

Mathemiownloaded from Stanmorephysics.com QUESTION 1 MDE/June 2023 An estate agent did a survey on the salaries of the people renting homes in a complex. She selected 12 homes where two salary earners live for her survey. The data collected is recorded in the table below. s a Salary of } 5 oq@ia100 | 5100 | 3560 | 6250 | 7400 | 4210 | 3200 } 2600 | 1000 | 4100 } 8000 person | ly eel 4500 | 8320 | 6500 | 3500 | 1500 } 4200 | 6420 | 3520 | 10500 | 11000 | 7800 | 19350 1.1 Determine the median salary for person 1 in this data. (2) 1.2 Determine the mean income for person 2 in this data. (2) 1.3 Determine the number of salaries for person 2 that are above ONE standard deviation from the mean. (3) 1.4 Determine the equation of the least squares regression line for the given data in the table. (3) 1.5 Draw a scatter plot and the least squares regression line for this data on the grid provided in the ANSWER BOOK. (4) 1.6 Explain what will happen to the least squares regression line if the income of the person receiving an income of R19350 decreases by 50%. (2) {16] Copyright reserved Please turn over

Mathemadownloaded from Stanmorephysics.com QUESTION 2 The ogive below shows the money collected by parents during a fundraising event at a school. MDE/June 2023 fan Ogive 10 + a a e : I: g 50 5 a & 40 ev : | £ a 2 30 + — “| = —- © 20 10 See SE : 0 oe aon = a (| 0 500 1000 1500 2000 2500 3000 3500 Amount collected 2.1 Determine the modal class of this data. qi) 2.2 Use the ogive to determine the total number of parents who raised R1000 or more. (3) 2.3 Use the ogive to determine the values of a, b and c in the five number summary, given in the table below. RO a b ec R3000 (4) [8] Copyright reserved Please turn over

Mathemaleeeynioaded from Stanmorephysics.com MDE/Sune 2023 SC’ QUESTION 3 In the diagram below ABCD is a parallelogram with A(—3; 0), B(L; 3), C(3; —2) and D(x; y). 3.1 3.2 3.3 3.4 3.5 3.6 A(-3 ; 0) BCI ;3) we D(x; y) Write down the coordinates of D. CB +2) (2) Calculate the gradients of AB and BC and state if ABCD is a rectangle. Give a reason for your answer. Determine the coordinates of M, the midpoint of AB. (4) (2) Find the equation of line MN passing through M, which is perpendicular to AB. (3) Calculate the size of BCD. Determine the area of ABCD. Copyright reserved (5) (5) [21] Please turn over

Mathemataewnloaded from Stanmorephysics.com MDE/June 2023 QUESTION 4 4.1 A(3; 2) is the centre of the circle through B and C. BC is a diameter of the circle. x B(S ;-1) 4.1.1 Write down the equation of the circle in the form (x — a)* + (y — b)? = r? (4) 4.1.2 determine the coordinates of C. (2) 4.1.3 Determine the equation of the tangent to the circle at C. (4) 4.1.4 Determine the equations of the tangents to the circle which gradients are zero. Give your answers in the simplest surd form. (2) 4.1.5 A new circle is drawn. A point P(x; y) on the circumference of the new circle, is such that it is always 4 units from the circumference of the original circle, and outside the original circle. Determine the equation if this new circle. (2) 4.2 — Determine whether the circles with equations A:(x+3) + y*=4 and B: x? +y?+4y+3=0 will intersect or not. Show ALL calculations. (6) [20] Copyright reserved Please turn over

Mathemaownloaded from Stanmorephysics.com MDE/June 2023 QUESTION 6 The graph of f(x) = —2 cos(x + 15°)is given. A B c 6.1 Give the amplitude of f(x). (i) 6.2 Give the period of f(x). (ld) 63 If g(x)=fx) + 2, give the range of g(x). (2) 6.4 Give the coordinates of: 6.4.1 A, anx- intercept of f (2) 6.4.2 B, the y- intercept of f (2) 64.3 C,aturning-point of f (2) 6.4.4. The x- coordinate of D, if D has coordinates (x; 0,87). Show your calculations (3) [13] Copyright reserved Please turn over

Mathemadownloaded from Stanmorephysics.com MDEZ/June 2023 QUESTION 7 AD is a vertical pole and points B and C are in the same horizontal plane as D, the foot of the tower., and DAC=x, CBD=2x, BDC=90°+x and AC=2 A 7.1 Show that BC=1. 6) 72 Show that BD = 253 @) cos x 19] Copyright reserved Please turn over

Mathemadaewnloaded from Stanmomephysics.com MDE/June 2023 NSC Give reasons for your statements and calculations in QUESTIONS 8, 9 and 10. Question 8 8.1 In the diagr: is the centre of the circle. KL // OM, NLT is a tangent to the circle at L. KPM =35° Determine, giving reasons, the sizes of the following angles: 8.1.1 Oe (2) B12 Oy 3) 8.13 Li 2) 8.14 § (2) Copyright reserved Please turn over

Mathemalaawnloaded from Stanmorephysics.com MDE/Sune 2023 Ni 8.2 Inthe circle below with centre O, DE = EF = 12, FH= 25, OG 1 FH, OE = 5 and OG =x. D F 4 H Determine, giving reasons, the length of: 8.2.1 OD. (4) 8.2.2 GH. (2) [15] Copyright reserved Please turn over

Mathemaownloaded from Stanmorephysics.com MDE/June 2023 QUESTION 9 9.1 InAKLMand ASTR, R=$, L=T, M=R. Prove that = =5. (6) K s /\ T R L M 9.2 In the sketch EG // FH and GI // HJ, AB = BE and AC = CD. BC = 6, DE=x, EF =x — 3, DG=y+5, GH=y, DI=2a+2and IJ =2a-3. Calculate the values of : 9.21 x (2) 9.2.2 y (3) 9.2.3 a (3) [8] Copyright reserved Please turn over

Mathemaawwnloaded from Stanmorephysics.com MDE/June 2023 QUESTION 10 In the sketch, QS = SR and POT = STR = 90°. P T al 4 | Q ' Ss R 10.1 Prove that APQR /// ASTR (4) 10.2 Hence, prove that PR. RT = SQ.RQ (2) [16] TOTAL: 150 Copyright reserved Please turn over

Mathemapownloaded from Stanmorephysics.com FORMULA SHEET __ ~btvb?-4ac x= 2a A=P(1+ni) = P(1— ni) Ty, =a+(n—-1) T, = ar? Fs x{(1+ ia 1] fi@= tim + Nase) d = J (2-1)? + 02-1)? yemxtce y— yy = m(x— x4) (x-a)?+(y—-b)* =r? In AABC: “W-=—2- = n ‘ Sind sinB sinC a® = b* +c? — 2be.cosA 1 area A ABC = zab.sinc sin(a@ + B) = sina.cosf + cos a@.sinB cos(a + 8) = cosa.cos 8 — sina sinB cos 2a = cos*a — sin’a =1-2sin?a = 2cos*a—1 _ _Lfx i=— n n(A) P(A) = RG) J=atbx Copyright reserved MDE/June 2023 A=P(1—i)” A=P(1+i)" n n = 5 [2a + (n—1)d] a(r™ — 1) Sn = aq ire So=Trpici<r<t 1- ;y-n paw +0) i 2A 2) a( 2 ' 2 m= 22 m = tan@ x2 — 4 sin(a — 8) = sina.cos B — cosa.sinB cos(a ~ 8) = cosa.cosB +sinasinp sin 2a = 2sinacos@ yer —x)? ot= _i=i n P(A of B) = P(A) + —P(AenB) b _r@-HO-/) Ge = #)?

Mi theTnerias 1 Gtinchedel Vi j MDE/June 2022 ° From Stanmerspnysics. com ox QUESTION/VRAAG 5 5.11 (3) 5.1.2 (2) 5.2 laYaYay (5) Copyright reserved/Kopiereg voorbehou HN LL Please turn over/Blaai om asseblief MPUMALANGA

4, education MPUMALANGA PROVINCE REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS P2 JUNE 2023 MARKING GUIDELINE MARKS: 150 marks This question paper consists of 13 pages and an information-sheet.

mahewhdoaded from StanmoreBhysics.com NOTE: Consistent a Assuming v; Pepypr LET WEL: 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, cy applies to ALL aspects of the marking memorandum. answers in order to solve a problem is unacceptable. MDE/June 2023 5. As 'n kandidaat 'n vraag TWEE keer beantwoord het, sien slegs die EERSTE poging na. 6. As 'n kandidaat 'n antwoord deurgehaal en nie oorgedoen het nie, sien die deurgehaalde antwoord na. 7. Volgehoue akkuraatheid is op ALLE aspekte van die memorandum van toepassing. 8. Dit is onaanvaarbaar om waardes/antwoorde te veronderstel om 'n probleem op te los. 9 . Write neatly and legibly. QUESTION/VRAAG 1 1.1 87110 2 R517 YR 1259,17 (2) 1.2 SD = R4579,26 ¥SD Above ONE standard deviation = mean+1SD ¥ boundary = R7259,17 + R4579, 26 = R11838,43 Only ONE household v answer 3) 1.3 | y=atbx ¥ a=6102,1123,47 as6Natt Yb =0,2792,85 b=0,27 y =6102,11+0,27x Y y=6102,11+0,27x (3) Copyright reserved Please turn over

Manowmoaded from Stanmorephysics.com MDE/June 2023 1.4 Scatter plot of income per household Income of parent 2 0 2000 4000 6000 8000 10000 Income of parent 1 ¥ Y points correctly plotted ¥ V regression line 4) 1.5 | The gradient of the line will become smaller, causing the data to be more symmetrical about the regression line, and there will be no P ¥ V answer outlier, (2) [14] Copyright reserved Please turn over

mahDewnrtoaded from Stanmorephysics. com MDE/June 2023 QUESTION/VRAAG 2 Ogive ” Toi | 60 AapE B : : § 50 "i 3 £ 4 : E 30 + < = 3 20 = | = + —— 10 + = : —— + 7 0 Sek a | 0 500 1000 1500 2000 2500 3000 3500 Amount collected 2.1 | 1000 <x < 1500 Y answer (1) 2.2 | 60-15 =45 parents ¥15 and 60 ¥ answer (3) 2.2 | a= lower quartile = 1000 ¥ method YQl b = median = 1350 ¥Q2 c= upper quartile = 1850 ¥Q3 (4) [7] Copyright reserved Please turn over

Matheietoaded from Stanmerephysics. com QUESTION 3 B(1; 3) A(-3; 9) a MDE/June 2023 AA O(a; y) 3.1 |D(-1; -5) vv (2) 3.2 ~ 023 7 Map = Ra 4 map v _ 3-(-2)_ 5 Mec Mec = 3 3 5 15 Sy22 23+ -1Vv aX 7re #-1 #-1 :. AB not perpendicular to BC, . ABCD is not a rectangle Not rectangley (4) —34+1 0+3. 3.3 [Myg= Cs S=(-1: >) vv @) 3.4 Map ==, « m,=-5 mv 3 4 y—7=-3@-CD) Subst ¥ = 4,443 y 3 * 3 2 v @) =—4y4H =—5% tz ae 3.5 tana = -: py a = —68,198 ...° + 180° = 111,80° tan =* (AB// CD) B = 36,87° v Copyright reserved Please turn over

Mathewnmoaded from Stanmorephysics.com MDE/June 2023 NSC-Marking ¢ uideline ©. =111,80° — 36,87° = 74,93° (ext Z of A) v BCD =105,07° _ (angles on straight line) v (5) 3.6 BC? = (1=3) + (3 — (-2)) = 45035 = 29 BC = cp? =)(34 (-1))* + (-2 = (-5))? =16+9=25 cD= s Area ABCD == BC.CD.sin BCD. = 5.V29. 5 . sin 105,07° = 13 units? Subst in formula ¥ BC ¥ CD ¥ Subst in sine form v iv (5) [21] Copyright reserved Please turn over

mathgieittoaded from Stanmorephysics. com “QUESTION 4 4.1 A(3; 2) 5; -1) MDE/June 2023 Copyright reserved 4.1.1 |r? =(5—3)*+(-1-2)? v =449=13 v “. (x — 3)? + (y — 2)? = 13 vv (4) 4.1.2 |} C=(1;5) (symmetry) vv (2) 4.1.3 a as v Mac = 73 <2 2 Mtangent ™ 3 v y-S= 2a -1) v y=sx-i+5 =x+4e v @) 4.1.4 |y = V13 and horizontal lines 13 from centre. y=2+ VI3 and v v 2) y=2- vi3 lar 41.5 | (-3)?+(y- 2)? = 44 vi3)? centre Or (x — 3)? + (y — 2)? = 57,84 radius (2) 4.2 x? +y*+4y+3=0 x? + y? + 4y + 2? = —-3 +2? x?+(y+2)2=1 v Please turn over

mathowmtoaded from Stanmorephysics.com MDE/June 2023 \C-Marking Guideline Centres: A: (3; 0) and B: (0; —2) v Distance between centres AB = /(3 — 0)? + (2— 0)? = Vi3 y = 3,61 Radii=1 axeh2 y Sum of fl >! fag y Sum of radii-< Distance between centres v 6) .. Circles do not intersect [20] Copyright reserved Please turn over

Mathawmtoaded from Stanmore i MDE/June 2023 merephysigs.com QUESTION 5 5.1 sk cos 51° = 7 4 39 1 1-k? 51° . k 5.1.1 y=P-kh Y Pythagoras yavl-kP tan 219° = tan (180° + 39°) = tan39° v¥—tan39° Vie answer (3) 5.1.2 | sin(-411°) =sin(-411 ° +360°) =sin(-51°) =~-sin51° Y reduction ¥ answer 2 -- ue (2) 5.1.3 | 0089? =cos(60°-51°) y compound angles = cos60°cos51° +sin 60°sin 51° oa 1 BY fi-R Y substitution = (5) (k) (2) Y substitution 3(1—K? 1k v8(-#) 2 2 4y0n0 Copyright reserved Please turn over

Matheryatigst oa ded from Stanmarethysics. com MDE/June 2023 5.2 | sin(45°+x)-sin(45° - x) = (sin 45cosx + cos 45sin x)(sin 45cos x —cos45sinx) | expansion ¥ expansion 2 2. 2 2. = Y2 oosx +2 sinx V2 osx 2 sin x V2 2 2 2 2 ‘> 1 1.) Y simplification =o sin x answer ; ©) 2 53 sinx+sin2x =tanx 1+cosx+cos2x LHS = sin x +sin 2x 1+cosx+cos2x __ sinx+2sinxcosx ¥ 2sin x08 x 1+cosx+2cos? x-1 ¥ 20s’ x1 sinx +2sinxcosx ~ Cosx+2eos* x Y simplification of denominator i 142 = sinx(1+ 2006) ¥ common factor cos x(1+2cos) ¥ common factor _ _ v sinx =tan = tanx = RHS cosx * (6) 54 | sinl5°=sin(45°-30°) ¥ sin(45°-30°) = sin 45cos30—cos45sin30 ¥ expansion (213) (ZI | ¥ substitution =| — |] — |-| > substituti 2 2 2 \2 _ vo-v2 ¥ simplification /6 4 (4) Copyright reserved Please turn over

manDewatoaded from Stanmorephysics.com sin’ x —2sinxcosx + cos’ x-1=2sin’ x-2 in? x—2sinxcosx+2=0 ~sin’ x—-sinxcosx+1=0 sin’ xcosx +sin? x +cos’ x= 0 cos’ x —sinxcosx =0 cosx{cosx—sinx) =0 MDE/June 2023 SC-Marl leline 5.5 (sinx—cosx) -1_, sin? x—-1 sin? x - sin xcosx + cos’ x1 = 2(sin® x-1) V sin? x—2 Y sin? x +cos? x ¥ standard form ¥ common factor cosx=0 or cosx —sinx =0 V x=4904360k keZ x = +90° + 360k cosx =sinx tanx=1 ¥ tanx =1 x =45°+180k,keEZ V¥ x= 45°+180K ) [31] Copyright reserved Please turn over

Mathipasintoa i MDE/June 2023 a ded from Stanmorephysics. com une: QUESTION 6 The graph of f(x) = —2 cos(x + 15°)is given. (x}0, 87) A c 6.1 |Amplitude = 2 v (1) 6.2 ‘eriod = 360° v (e5) 6.3 |Range of g(x): y € [0; 4] vv (2) 6.4.1 |... A=(—105°; 0) vv (2) 6.4.2 [B= (0, -1,93) vv (2) 6.43 |C= (165°; —2) vv ro) 6.4.4 | —2cos(x + 15°) = 0,87 cos(x + 15°) = —0,435 v x + 15° = 115,785 ...° + 360k, k EZ x = 100,79° + 360k ¥v -. D=(100,79°; 0,87) v G) {13] Copyright reserved Please turn over

Matheiwsatoaded from Stanmorephysics. com QUESTION 7 MDE/June 2023 AD is a vertical pole and points B and C are in the same horizontal plane as D, the foot of the tower. DAC = x, BDC = 2x,BDC = 90° +x and AC =2. A cos3x cosx Icos 3x =2cos2x ~ 1. cosx TA Nein = 2 v 2 v CD =2sinx ce v BC __ op Subst in Sin-rule sin(90°+x) sin 2x Cosx ¥ Bo _?sinx 2 sinx cosx ¥ cosx 2sinx cosx 2sinx. cosx Simplification ¥ BC == _ 2sinx cosx (6) =1 7.2 |BCD = 180° — (90° + x) — 2x int Zs of A = 90° -3x v In ABCD: BD _ 1 . sin(90°3x) _ sin(90°+x) bst in sine-form BD 1 ‘o-functionsY (3) [9] Copyright reserved Please turn over

Mathbausetoaded from Stanorephysics.com Question 8 \C-Mar! leline MDE/June 2023 8.1 In the diagram, O is the centre of the circle. KL // OM, NLT is a tangent to the circle at L. 8.1.1 |6, =70° Z at centre circle =2xZ atciroumf |SV RY (2) 8.1.2 |K, =70° alt.Zs, KL // MO S+RY L, =70° Zs opp = radii S+RY 2 6, = 40° int Zs of A S+RY (3) 8.1.3 |Z, =20° radius L tangent SV RY (2) 8.1.4 |S = 135° opp Z of cyclic quad Sv RY (2) Copyright reserved Please turn over

Mathewntoaded from Stanmorephysics.com ig Guideline MDE/June 2023 NSC-Mar! 8.2 D F mn 8.2.1 |ORD = 90° Centre circle midpoint chord |S“ RY OD? = 12? + 5? Theorem of Pythagoras v = 169 OD = 13 v (4) 8.2.2 |FG= 12,5 Centre circle L chord Sv RV (2) [15] Copyright reserved Please turn over

manbaangmoaded from Stanmorephysics.com MDE/June 2023 NSC-Marking Guideline QUESTION 9 9.1 InAKLM and ASTR, R = $, £ =f, M = R. Prove that = = =. (6) K s A B T L M 9.1 |Construction In AKAB and ASTR: AK = ST construction KB=SR construction R=§ given . AKAB = ASTR SAS sAL=f Bu? =7 given AL= 7? “AB // LM corresponding angles = KA _ KB i si “la = kM line // one side of A 6 ST SR = = an = kM AK = ST and KB=SR Copyright reserved Please turn over

mahbangroaded from Stanmorephysics.com MDE/Sune 2023 NSC-Marking Guideline 92. Inthe sketch EG // FH and GI// HJ. AB = BE and AC = CD. BC = 6, DE =x, EF =x - 3, DG=y+5,GH=y, DI=2a+2 and = 2a —3. 9.1 jx=12 midpt theorem SY RY (2) 9.2 = = line // to one side of A SvoRY (3) 9.3 |2at2 _ y+ line // to one side of A Sv RV 2a-3 y 2 4 2a—3—«S 3 6a+6=8a— 12 2a=18 a=9 i @) (14] WEEE] Copyright reserved Please turn over

Mahixawmintoaded from Stanmorephysics.com ing Guideline NSC-Mar! QUESTION 10 MDE/June 2023 HE 0 ; 2 e 10.1 [In APQR and ASTR Q=T 90° given S+RVY R=R common Z S+RY P=s int Zs of A S+R VY APQR /// ASTR AAA RV @) a imilari y or ar RS similarity «. PR.RT = RS.RQ but =9 RS =QS Y @) ©. PR.RT = QS.RQ [10] TOTAL: 150 Copyright reserved Please turn over

Downloaded from Stanmorephysics.com MATHEMATICS P2 JUNE 2021 GRID 1 2 3 4 TOTAL 1.1 median 2 1.2 mean 2 1.3 standard deviation 3 1.4 regressionline 3 1.5 scatter plot, 4 1.6 interprete——> 2 2.1 modal class 1 2.2 interprete ogive 3 2.3 interprete ogive 4 3.1 coordinates 2 3.2 Rectangle 4 3.3 midpoint 1 3.4 equation of line 2 3.5 size of angle 3 3.6 area triangle 3 4.1.1 equation circle 4 4.1.2 coordinates 2 4.1.3 equation tangent 4 4.1.4 horisontai tangents 2 4.1.5 new circle 2 4.2 circles intersect 6 5.1.1 reduction 3 5.1.2 reduction 2 5.1.3 compound angle 4 5.2 compound expansion 5 5.3 identities 6 5.4 identities 4 5.5 general solution 7 6.1 amplitude 1 6.2 period 1 4 6.3 range 2 I 6.4.1 x-intercept 6.4.2 y-interccept 2 6.4.3 turning point 6.4.4 x-coordinate 3 7.1 show length 5 12 7.2 show that 8.1.1 angle at centre 3 8.1.2 int angles triangle 3

Downloaded from Stanmorephysics.com 8.1.3 tan perp radius 3 8.1.4 opp angles cyclig 2 11 8.2.4 Pyth 8.2.2 centre circle perp chord 5 9.2.1 cyclic quad 4 9.2.2 congruency— 3 9.2.3 angle at Genitre 5 9.2.4 same septne n 5 22 9 pllaTayl 10.1 properties of parm I 1 10.2 ‘one line // to 3 line in triangle 10.3 area rule 2 6 11,1 similarity 3 11.2 ratio 3 11.3 ratio 5 11 TOTAL 29 55 58 14 66