Mathematics Grade 12 Paper 2 KZN June 2023 Paper & Memo

Downloaded from Stanmorephysics.com KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE a " re " s " a n " = " . Seesesseesseees Seeessseeeeest MARKS: 150 TIME: 3 hours per consists ooh information sheet, ‘and 1 diagram she

wae owNloaded from Stanmorephysics.com June 2023 Common Test INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. This question paper consists of 9 questions. 2; Answer ALB the questions. 3: Number the answers correctly according to the numbering system used in this question paper. 4. Clearly show ALL calculations, diagrams, graphs, etc. that you have used in determining your answers. 3: Answers only will NOT necessarily be awarded full marks. 6. You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise. 7. If necessary, round off answers to TWO decimal places, unless stated otherwise. 8. Diagrams are NOT necessarily drawn to scale. 2 An information sheet with formulae is included at the end of the question paper. 10. Write neatly and legibly. Copyright Reserved Please Turn Over

MahemP>ewnloaded from Stanmorephysics.com June 2023 Common Test QUESTION 1 Seven scores for a test out of 50 marks were recorded. However, when recording the data the statistician had some difficulty with his computer. He listed the following conditions that he had remembered, [Scores [ a fot | ¢ | d@ | e | f J g | > Maximum Valueis 42 > Range is 35 > Median is 23 > The difference between the median and the upper quartile is 14 > The Inter — Quartile Range is 22 > e=2¢ >» The mean is 25. Determine the values of the scores. (Show all working) (10) [10] QUESTION 2 The class teacher of a grade 10 class collected the following data for her learners. She recorded the time taken by the learners to complete different tasks assigned. No. of learners Time in minutes o0<t<10 5 10<t<20 8 20<t<30 18 30<t< 40 40<t<50 2.1 Calculate the estimated mean time. (3) 2.2 Draw a frequency polygon (Ogive) curve on the system of axes provided, (5) 2.3. Calculate how many learners took more than 60 % of the time to complete the tasks assigned to them. (3) (1 Copyright Reserved Please Turn Over

mavendawnloaded from Stanmorephysics.com June 2023 Common Test QUESTION 3 In the sketch below, A(-8;6), B(8 ; 10), C and D(—2;0) are the vertices of a trapezium having AD || BC. T is the midpoint of DB. From the vertex B, BM is drawn perpendicular to the x—axis and intersects DC in F (8 ;3). »4 B(8;10) A(-8;6) T Cc i (8:33) 2: fe) D(-2;0) M 3.1 Calculate the gradient of AD. (2) 32) Determine the equation of BC in the form y = mx +c. (3) 3.3 Prove that BD 1 AD. GB) 3.4 Calculate the size of BDM. (2) 3.5 If TC || DM and points T and C are symmetrical about the line BM. Calculate the (3) coordinates of C. 3.6 Calculate the area of ABDF. (5) [18] Copyright Reserved Please Turn Over

Math ics P2 jownloaded from Stanmorepikysics.com QUESTION 4 In the diagram below, the circle centred at E (3 ; 1) passes through the point, P (5 ; —5). 4.1 4.2 43 e E(3;1) P(5;-5) Determine the equation of : 4.1.1 The circle in the form x? + y? + Ax + By +C=0. 4.1.2 The tangent to the circle at P (5 ; -5) in the form y= mx + c. A smaller circle is drawn inside the circle. Line EP is the diameter of the small circle. Determine the: 4.2.1 Coordinates of the centre of the smaller circle. 4.2.2 Length of the radius. Hence, or otherwise, determine whether the point C(9 ; 3) lies inside, o} or on the circle centre, E. Copyright Reserved June 2023 Common Test (4) (5) (4) (2) (4) [19] Please Turn Over

“em 3wnloaded from Stanmoraphysics.com QUESTION 5 5.1 June 2023 Common Test If tan 12° = q , then determine the value of the following, without using a calculator: 5.1.1 cos1 5.1.2 cos 5.1.3 1—2sin? 6° (3) (3) (2) (7) (3) (2) (4) [24] (6) np 3 Evaluate the following trigonometric expression without using a calculator: 2 sin? (x — 180°) cos(180° — x) cos(90° + x) sin x — cos (x — 90) sin (720° — x) 53 cos A 1 Given: (1 —tan A) Fas mpom 6 {= 2A} cosA +Sin A 5.3.1 Prove the identity. 5.3.2 Write down the maximum value of ——— cosA +Sin A 5.4 Calculate the value of 52° >dcos* A 38° QUESTION 6 6.1 Sketch the graphs of f(x) = sin>x and g(x) = cos(x + 60°) for x € [—180°; 180°] on the grid given at the back. 6.2 Use your graphs to answer the following questions: 6.2.1 Write down the range of the graph of f- 6.2.2 State the period of g. 6.2.3 Determine the values of x for which f(x) = g(x). 6.2.4 If h(x) = g(x + 30’), write down the equation of A. (2) Q) (4) (3) [16] Please Turn Over

June 2023 Common Test ““Bownloaded from Stanmorepkysics.com QUESTION 7 In the diagram below, MP is a vertical tower. MQ and MR are wire ropes used to stabilize MP and are each 19 metres in length. R, P and Q are in the same horizontal plane. The angle of elevatioméf M from R is y. RPQ = x and MQ M 19 19 P x : Q 71 Prove that the Area of A PQR = 24si"e0s'y (6) 12 Show that, RQ = 38 cos B (6) {11] Please Turn Over Copyright Reserved

Mathemati®Gwnloaded from Stanmorephysics.com ae QUESTION 8 8.1 8.2 O is the centre of the circle, radius r, and chord AB = ¥128 cm. OCD 1 AB and OC: CD=3;2. ABD = 27° A al D 8.1.1 Calculate the length of the radius of the circle, without using a calculator. (5) 8.1.2 Calculate, with reason and AOD. (2) Complete the following statement: (2) 8.2.1 The exterior angle of a cyclic quadrilateral is ..........:.:.cceseeeeeeeeeee seen 8.2.2 TQisachord of the circle PQRST. QAT 1 PAS. Q, = 30° and P = §,. P Q R a) Name 3 angles each equal to 60°. (3) b) Calculate the size of QRS. (2) c) Prove that PS || QR. (2) d) Prove that TR is a diameter of the circle . (2) [18] Copyright Reserved Please Turn Over

Math ics P2 ° 23 Ce ™"fSownloaded from Stanmorepinysics.com ee QUESTION 9 9.1 InA ABC, DE || BC. Prove : AD _ AE DB EC re D E B c (6) 9.2 In the diagram, O is the centre of the circle ACDE. EC is produced to B. TAB isa tangent to the circle at A. EC and AD intersect at H. F is a point on BA. ED || FH || TAB. EOHCB is a straight line. AH : HD = 5: 7. ieee 4 T >> rn B 9.2.1 Prove AABC ||| AEBA (4) 9.2.2 Calculate r, the radius of the circle, if AB = 5 metres and BC etres. (4) 9.2.3 Write down, with reason the value of AF : FE. (2) 9.2.4 . Area of AAFH 4 Determine Area of AAED © ” 9.2.5 Calculate the length of OH. (3) [23] TOTAL MARKS: 150 Copyright Reserved Please Turn Over

Mater wnloaded from Stanmoraphysics.com ee INFORMATION SHEET: MATHEMATICS P(1=ni) A=P(\-i)" A=P(l+i)" S,= 322+ (Dd) al alr" -1 a T, =ar' Be A r#1 S,=——3 -l<r<l ee FI l-r Pe (eo) pe -1] poil=d+)") i ry . f(xth)- f(x) ' = lm Pe ete SE me re) 40 h d= (x, -x,)? +(, -y,)? (222% 2 y=mr+e y-y, =m(x-x,) = 22221 m= tan 0 X2 7x (x-a) +(y-sf =r? a b c sinB sinC In AABC: sin(a + #) = sin a.cos # + cosa.sin 8 cos(a + 8)=cosa.cos # -sin a.sin 8 cos? a@—sin* a cos2a@ =41-2sin? a 2cos* a1 Dix n p=atbx x= Copyright Reserved a2 =b +c? -2be.cosA area AABC =~ ab.sin Gg sin(@ — £) = sin a.cos f — cosa.sin B cos(a — 8) = cosa@.cos # +sin a.sin 8 sin 2@ = 2sin @.cosa@ SG) oo =. P(A or B) = P(A) + PB) — B) p= X09) d@-x) Please Tum Over

Mathematjes P2 1 . Jownloaded from Stanmorephysics.com June 2023 Common Test DIAGRAM SHEET QUESTION 2 — Ti minutes No. of learners < 10 5 10<t<20 8 20<t<30 18 30<t<40 40<t<50 2 QUESTION 6.1 Y LS HH i n°) + r +o re) > & * fe it , 0s : pas t Copyright Reserved

Downloaded from Stanmorephysics.com FINAL KWAZULU-NATAL PROVINCE S=EDUCATION ApAREPUBLIC OF SOUTH AFRICA i, NATION SENIOR CERTIFICATE MARKS: 150 N.B. This marking guideline consists of 16 pages.

Downloaded from Stanmorephysics.com QUESTION 1 Ll | g=42 AY Value of g Minimum : .@= 42 — 35 =7 AY Value of a d=23 AY Value of f f-23 = 14.Fherefore f= 37 AY Value of d f—b = 22 jay Therefore b = 15 CAY Value of b 7 +15 +)C#23 + 2c+374+42 _ CAY Setting up equation ~ CAY Simplifying 3c +124 25 numerator 7 ~ CAY equation inc 3c +124 =175 3c =51 c=17 CAY Value of c Therefore e = 34 CAY Value of e (10) [10] QUESTION 2 Time (in minutes) Number of Learners Midpoint of fx Cf Interval(x) 0<t<10 5 5 25 5 10<t<20 8 15 120 13 20<t<30 18 25 450 31 30<t<40 7 35 245 38 40<t<50 2 45 90 40 Total 40 930

Downloaded from Stanmorephysics.com 2.1 Estimated Mean = 22 = 23,25 Av930 Av40 40 CAV Answer (only if (3) denominator is 40) (Answer only full marks) 2.2 DO NO RK THIS QUESTION 2.3 60 % of 50 minutes = 30 minutes AY Calculation 9 learners Av 30 minutes AY Answer (AO full marks) (3)

Downloaded from Stanmorephysics.com QUESTION 3 yf B(8;10) A(-8:6) F (8 3 ;) "3 « A 3 > D(-2;0) M x v 3.1 6-0 6 AY subst. of points A and D mAD = —8+2 —-6 ol CAY Answer (2) 3.2. | mAD = mBC = --1 .....(AD || BC) CAY gradient of BC y=mxt+c 10 = —-1(8) +c CAY subst. of point C and 18=c gradient y=-x +18 CAY Answer (3) 3.3 BD = 10-0 10 1 AY subst. of points A and D ee B42 10 Av Gradient of BD Since the product of the gradients of lines AY Justification AD and BD =-1 Therefore BD 1 AD (3) 3.4 |tan@=1 CAvtan@ =1 BDM = 45° CAV Answer (2) 3.5 | T(3;5) ... midpoint A Y coor es C(13 ; 5) CA CA ordinates of C (3) i

Downloaded from Stanmorephysics.com 3.6 2 .. Area of A BDF = 10V2 x = 33,3 units? 3 1 2 20 = —8)2 —3-) =6-=— BF= |(8-8) +(10 35) 65=5 Area of ABDF = 5 (10) (F) == 33: square units. OR Area of A = Area of A DBM ~ area of A DFM 1 1 10 = — (10) (10) — — (10)| — +o) 00) 4 «0)(2) = 100 333 units ? 33 OR 1 mpr = — A 1 tan FDM = — 3 FDM = 18.43° BD = 10V2 pr — Lovio 3 10v10 sin 26.57° AY Substitution AY BF value CAY Subst. into formula CAY Simplification CAY Answer (5) [18]

Downloaded from Stanmorephysics.com QUESTION 4 4.1 y * E(3;1) fe) x P(5;-5) 4.1.1] gpa re) —5)?+(1+5)2 AY Substitution EP = /(—2)? + (6)? CAY EP value EP = V40 (x - 3)? + (y— 1)? = 40 CAY Substitution x —6x4+9+4+y? —2y +1=40 x? +y? - 6x —2y -30=0 CAY Answer (4) 4.1.2 . 1+5 AY Gradient of radius Radius: m = >—= = -3 3-5 r _ 1 CAY Gradient of tangent angent: m = 3 (provided it is positive) y=mxt+c 1 -5=3(5) +e -15 =5 + 3c CAY Subst. Of point and -20 = 3c gradient. o- 62 3 4 2 y= 3° - 65 CAM c, jue CAY er (5)

Downloaded from Stanmorephysics.com 4.2.1 | Coordinates of Centre of smaller circle: AY midpoint formula (= =) = (4;-2) AY Substitution 2’ 2 , CA CAY V Answer (AO full marks) (4) i—) 4.2.2 | Radius of | circle: 2V10 Avi (2vi0) Radius of er circle: r = 5 (2V 10) = V10 units 2 CA V Answer OR Radius = / (4-3) +(-2-1) = Vi0 OR Raduis = /(4—5) + (-2-5) - Jo @) 43 Jae = (9-3) +1) AY Substitution = {6 +2? = /36+4 = V40 CAV EC = V40 _ + EC= radius CA Y equating to radius ..C lies on the circle, centre E CA VY Conclusion OR xr -6x+9+y—2vt+1=40 x+y’ 6x—2y- 30=0 OR (x —3) +(y-1) =40 LHS = (9-3) + (3-1)? =40 RHS = 40 -. LHS = RHS «. It lies on the circle (4) aaa [19]

Downloaded from Stanmorephysics.com QUESTION 5 5.1 Cos 24° = | —2sin? 12° 2 = 1 -2| 4 l+q° 2 core =| u }-[ l+¢ -J! ¢ l+q 1+¢ -le@¢ l+q? 1 5.1.1 | cos 192 . AY Ji+@ =-—cos 12 . 1 Av —cos 12 te CAY Answer (G3) 5.1.2 | cos 24° = 2cos?12°-1 AV Double angle 2 -2 ( 1 ) 4 Expansion y 1+q? CAV Substitution "ire" CAY Answer OR (3)

Downloaded from Stanmorephysics.com 5.1.3 | 1 — 2sin? 6° =cos12° AY Double angle 1 = CAV Answer Vit @ 4 : : (2) 5.2 2 sin?(x — 180°) cos(180° — x) cos(90° + x — cos (x — 90) sin (720° — x) 2( )?.(— cos x) AY (-sinx)? ~ =sinx]sin — (sin x. —sin x) Av’— cosx . AvY—sinx —2 sin? x cos x Av'sinx = 7 Av-sinx —2sin? x cos x ; CAY——_— = undefined CAY Answer (7) 53.1 cos A LHS: (1 — tanA i ( an [ss ‘) avs” A . cos A _ ;- sn A cos A AV cos? A—sin2 A cos A }\ cos? A-sin? A = (cos Asin A __£0s A . A Y simplification cos A (cos A —sin A) (cos A + sin A) _ 1 cos A+sin A =RHS (3) 5.3.2 | DO NOT MARK THIS QUESTION 5.4__| DO NOT MARK THIS QUESTION [18]

Downloaded from Stanmorephysics.com QUESTION 6 6.1 a 50 -120 -90 -60 -30 -0.5: AS 30 60 90 120 150 Graph of f: A 1 mark for x — intercepts A 1| mark for shape Graph of g: A 1 mark for end points A | mark for x — intercepts A 1 marks for minimum and maximum points A | mark for y — intercept (6) 62.1] ye€[-1;1] CAY [-1; 1] (2) A V notation 6.2.2 | 360° AY Answer (dl) 1 ° 6.2.3 sin5x = cos(x + 60°) 1 ° °. YCo - i sin5x = sin [90° — (x + 60°)] Av Co - Ratio sin5x = sin [30° — x] AY Equation in sine only 1 1 ° 5x = 30 —x AYSXx = 30—— x 3 ° 3% = 30 ° Av Ans =2 x= 20 CA( AG-full: marks) (4) 6.2.4 | h(x) = g(x + 30) h(x) = cos(x + 30 + 60) A ¥ substitution h(x) = cos(x + 90) AY cos(x + 90) h(x) = —sinx AY Answer (AO full marks) (3) [14]

Downloaded from Stanmorephysics.com QUESTION 7 M 19 19 P x Q R TA PR —=cosy 19 AY Trig ratio PR= 19 cosy AV Length of PR Now PR = PQ .....(AMPR = AMPQ ...90deg HS) Av Congruent triangles 1 . Area of A PQR = 5 (19 cos y)(19 cos y) sinx AY Reason SAS _ 361 sinx cos? y AY Substitution into Area Formula ; ) 7.2. | RQ? = 19? + 19? — 2(19)(19) cos(180° — 28) AY Size of angle RMQ RQ? = 361 + 361 — 722(—cos 28) Av Subst. Cosine rule RQ? = 722 +722 cos 2B AY Reduction RQ? = 722 + 722(2cos? B — 1) AY Simplifying RQ? = 1444 cos? B AY Double angle expansion a RQ = 38cosB OR MR? = RQ? + MQ? —2RQ. MQ cos 192 = RQ + 192 — 2(RQ) (19) cosB 0=RQ-38RQ. cosB 38 RQcos B _ RQ? RQ RQ RQ = 38 cosf AY 1444 cos? B [1]

Downloaded from Stanmorephysics.com QUESTION 8 8.1 8.1.1 oc 3 3r AY OC and CD in CD 2 2r terms of r. AC = CB = 4y2 cm... (line from centre | chord) AV SIR In AAOC: 2_ (3,)7 2 re= ( r) + (4v2) ... Pythagoras AY Pythagoras (S/R) 16 , =32 25’ CAY Simplifying r? =2xS5 50 cm? CAY Answer r=5V2 cm x (5) 8.1.2 | AOD = 54° (angle at the centre 2 x angle at the circum) AVS AYR (2)

Downloaded from Stanmorephysics.com 8.2.1 | (Exterior angle of cyclic quad = int. opp. A) Svv (2) 8.2.2 a) |P Sv QTs Sv S, sv (3) b) | QRS = 120’...[Opposite angles of cyclic quad. ] Sv RV (2) c) | QRS +$, = 120° + 60° = 180° SY PS || QR (Co-Interior angles are supplementary) RY (2) d) | Q, = 90’ .....(Corresponding Angles ; PS || QR) S/RY TR is a diameter of the circle. (Conv. Angle in the semi-circle) | RY (2) OR §, = 60° ... given Q, = 30° ... given Q, + Q, + S, =180° opp 2's of a cyclic quad “.Q, =90° -. TR is a diameter (Z's subt by 90°) Conv Z's on semi circle) nn ! [18]

Downloaded from Stanmorephysics.com QUESTION 9 9.1 Proof: Area A ADE Area A DBE 5pB x hy 1 Area A AED _ a AE xh, Area A DBE 5 EC h Area A DBE = Area A ECD Area A ADE _ Area A AED Area ADBE Area A ECD AD _ AE “DB EC = — (Common vertex E, same DB height h2) AE (Common vertex D, same ~ EC height /) (Common base DE, same height, DE // BC) Area of A ADE is common and Area A DBE = Area A ECD SY Construction YSYR S/RY YSVYR (No Marks if no construction indicated in drawing or words) (6)

Downloaded from Stanmorephysics.com 9.2 T 9.2.1 | In A’s ABC and EBA 1) B=B.....[Common] S/RY 2) A, =E£,... [Tangent — Chord Theorem] SYRY 3) C, = EAB.... [Remaining Angles of A’s] AABC ||| AEBA...[AAA] RY (4) 9.2.2 | AABC ||| AEBA AB BC LAC tar tri SRY AB EA similar triangles 2r 5 3 5 SY Substitution N ol rt+ = (2 +2) =25 3 T3)> 4rt ios 3° 9 16r? _ os 7 = 7? = 25x 2 16 S iswer = {25x a 15 t r= 16 = 4 metres (4) SY Multiplication 9.2.3 | Let AH = 5a and HD = 7a AF AH _Sa_5 a oremPFH II ED YSYR (2) FE AD 7 Ja 7 7 [Prop eorem; FH || ED]

Downloaded from Stanmorephysics.com 9.2.4 5 AAFH = J SAHE .... (Common Vertex; Equal Heights] 57/5 AAFH = 1 AAED) .... [Common Vertex; Equal Heights] AAFH _ AAED — OR Area A AFH _ } AF. AHSin A, AreaA AED 1 AE. AD Sin A, (AE) si AF AH “.| —= | since — =— AE FE AD 12a 25 144 YSYR vS/IR vs (4) 9.2.5 | Let OH =x Then HC = r- x = = — x metres AF _ BH FE HE 15 5 4a7* 7 15... atx s($+)=7(B-) a °*)= 4 x (F+5x)=(+ 7x) aX)" \qa ... +. [Prop. Theorem; FH || AB ¥S/R vs (3) [22] THIS PAPER SHOULD BE MARKED OUT OF 139 TOTAL MARKS: 150