Mathematics Grade 12 KZN March 2022 Paper & Memo

KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL S ENIOR CERTIFICATE [ GRADE 12 ] �----------------------------------r.··········� --- ♦ � � � MATHEMATICS � II II iii II II C ONTEST II I� II � II iii II ��CH2022 ,� •1 �!! !! !!!! !!!!!!!! - !!_;! !!!! !!!! !!! !!!! !! !!!! !!!! !!-'!! !!! !!!! !! !!!! !!! !! !!! !!!!---!!!! !!! !! !!� TIME: 2 hours N .B: Th isquest ion paper _consists�f 6 pages, 2 d 1agram sh ee tsan d an m f orma t ,on sh ee.t Copyright Reserved Please Turn Over

Math i : March 2022 Ci T “reBéwnloaded from Stanmorerbysics.com aren 2022 Common Test INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. This question paper consists of 8 questions. 2. Answer ALL questions. 3. Clearly show ALL calculations, diagrams, graphs, et cetera that you have used in determining your answers. & 4. Answers only will not necessarily be awarded full marks. G 5. An approved scientific calculator (non-programmable and non-real may be used, unless stated otherwise. 6. If necessary, answers should be rounded off to TWO sce unless stated otherwise. (e) 7. Diagrams are NOT necessarily drawn to scale. s 8. Number the answers correctly accordin' tothe numbering system used in this question paper. Write neatly and laity. Copyright Reserved Please Turn Over

Math i : March 2022 Ci T “reBéwnloaded from Stanmorerhysics.com aren 2022 Common Test QUESTION 1 Given the quadratic sequence: 5; x; y; 29; ... and its second constant difference is equal to 4. 1.1. Calculate the values of x and y. = 17, determine the n' term of the quadratic sequence. 50" term of the sequence. (4) (4) (2) [10] RS QUESTION 2 o: The 2™ term of an arithmetic sequence is 8 and the 7" term is eleven ieee? value of the first term. Determine the first three terms of the sequence. XN & & QUESTION 3 3.1 Given: 3.1.1 Write down the values opt three terms of the series in terms of p. 3.1.2 Determine the v efoto for which the series is converging. 3.2. The sum of the terms of a sequence is given by 4 — 4 (:) . can three terms. Copyright Reserved 7] (1) (4) (5) [10] Please Turn Over

Math i : March 2022 Ci T “*eBéwnloaded from Stanmorerhysics.com aren 2022 Common Test QUESTION 4 4.1 Given AABC and ADEF with A=D,B =EandC =F. AB Prove the theorem which states that — = — (7) DE DF A D E F B c 4.2 In the figure, AABC has HG || AC and ED || BC. ED and HG intersect at K. AP =3 and BG =26C . AB=15 units, Dc 2 A H E K D B * G Cc Determine with reasons the value of: 4.2.1 AE (3) 4.2.2 AH (3) 4.2.3 GK KH (3) 4.2.4 Area of AHEK (3) Area of AHBG [19] Copyright Reserved Please Turn Over

Math i : March 2022 Ci T “reBéwnloaded from Stanmorerbysics.com are ommon Test QUESTION 5 In the figure: PQ is the diameter of the circle. SP is the tangent to the circle at P. QT intersects the circle at W and with T on the line SP. QS intersects the circle atR. P, =x. P 1 2 x W<3 Tr ? Ss 5.1 Give areason why PRQ = 90°. (1) 5.2 Prove S=x. (3) 5.3. Prove that SRWT is a cyclic quadrilateral. (3) 5.4 Prove that AQWR ||| AQST. (3) 5.5 If QW =5 cm, TW =1cm, QR = 4cm and WR = 2 cn, calculate the lengths of: 5.5.1 TS (3) 5.5.2 SR (3) [16] Copyright Reserved Please Turn Over

Math i : March 2022 Ci T “reBéwnloaded from Stanmorerbysics.com aren 2022 Common Test QUESTION 6 6.1 Given cos20° =p andsin14° =q Without using a calculator, calculate the value of the following in terms of p, q or p and q. 6.1.1 sin20° (2) 6.1.2 cos6 (6) 6.2 Simplify into a single trigonometric ratio. 1. zsin 2x 6) ° 1 _ 2 tan(540 +\(Sa x tan x) [14] QUESTION 7 7.1.1 Prove the following identity: cos 4x = 8cos*x —8 cos?x +1 (4) 7.1.2 Hence, determine, without the use of a calculator, the general solution of 16 cos* x —16cos?x+2=1 (5) 7.1.3 Write down the minimum value of the expression 16 cos* x — 16 cos? x + 2. (2) 7.2 Calculate, without the use of a calculator, the value of: 2sin? 22.5°-1 4sin 225° cos 225° (5) sin 22.5 cos22.5 [16] QUESTION 8 8.1 Sketch the graphs of f(x) = sin(2x) and g(x) = 2.cosx for the domain x e[—90°; 180°]. (Use the axes provided) (6) 8.2 Use your graphs to determine the solution “ 21. (2) [8] Total: 100 marks Copyright Reserved Please Turn Over

Math i : March 2022 Ci T “reBéwnloaded from Stanmorerhysics.com aren 2022 Common Test INFORMATION SHEET: MATHEMATICS INLIGTING BLADSY —b+vb2 —4ac 2a A=P(1+ni) A= P(1-ni) A=P(1-i)" A=P(1+i)" T, =a+(n-1)d S, =F (2a+(n-2)d) T, =ar"* S _ alr’ -1) ; r41 8, = Sj -1<r<t . r-1 - - xi)’ -1] p x= G+i)"] (x) = lim f (x+h)— f(x) hoo h d= (X, —X,)? + (Yo - Yn)? M A*% Vat Yo 2 2 y=mx+c y—y; =m(x-x,) m= 227 m=tand Xp —% (x-a) +(y—b) =r? Inaapc; —2_=-—P___¢_ a*=b2+c%-2becosA area AABC = + ab.sinC sinA sinB sinC 2 sin(a + 8) =sin a.cos B+ cosa.sin B sin(a — f) = sin a.cos B —cosa.sin B cos(a + 2)=cosa.cos # —sin a.sin B cos(a — £)=cosa.cos # +sina.sin 2 cos’ a-sin? a cos2a =41-2sin? a sin 2a = 2sina.cosa 2cos* a@-1 d(«-») gad tn ou : P(A) = ae) P(A or B) = P(A) + P(B) — P(A and B) _ — YX &-xy-y) y=a+bx yo ae Copyright Reserved Please Turn Over

% 3 EF 5 o c E 5 5 u. 5 (6) S 8 = s a Fi Ww oO oO oO = a ro} 8 a a <3 ° ® iG : « ipa I. e < “ +e] rr} x = ~ x l, wn > - - a be = & . S | a N s ng vt re) 8 = S2 2 z Fd & zo 9 fe} & DQ wi cr Ee Ee z = s Of ry ry = g§ @ £5 5 5 2 = 2 ao o o [s} 39Vd 45O0-uV3L

B <t3 fad a 5 — 5 oO So nN a x Ss a $ os . = 48 18 <x a fo} 8 € So oO : o 2 1g 2 ® 3 8 S c wy ” > = Wi a fo) + Go fe} x ~ 3 18 Ss S 3 eo rr z S 8 = fe} re g os cL a 2 wi = 4 2 a s 2 e 8 8 BOVd 45O-UVSL

Downloaded from Stanmore tics. com KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE MARKS: 100 This memorandum consists of 9 pages. Copyright Reserved Please Turn Over

March 2022 Common Test “atteSwnloaded from Stanworephysigsacom QUESTION 1 1.1 5 x y 29 Nv IN N £ 1“ Diff x5 yrx 29-y \ VA \ LZ NE we 2™ Diff y— 2x45 29-2y+x y-2x+5=4 V sett; . sta: y-2x=-1.. (1) AY setting up equation 1” line 29-—2y+x=4 . . sty -2y+x=-25.. (2) AY setting up equation 1“ line (1) x2: 2y—4x =-2... (3) 3) + (2): Oe ‘ 7 . x=9 CAY x — value syad7 CAY y-—value (4) 1.2 2a=4 .a=2 AY a-—value 3a+b=4 -«b=-2 CAWb — value a+b+c=5 «c=5 CAV c — value Ty = 2n? -2n+5 CAYn™ term (4) OR OR 2a=4 .a=2 Ava -— value 3a+b=5 «b=-2 CAYb — value “2cC=Ty=5 CAYc — value Ty, = 2n*—2n+5 CA“ n™ term (4) OR OR n-1)(n-2 OR OR n-1 7, =! 2 20+ (n—2)d) +7, 1.3 Ts59 = 2(50)? — 2(50) + 5 = 4905 CAY substitution (from 1.2) CAY answer (2) [10] Copyright Reserved Please Turn Over

Mathenpy ics ownloaded from Stanworephysias.com March 2022 Common Test QUESTION 2 2 a+d=8 > (1) AY equation (1) a+6d=11a > (2) AVY equation (2) d=8-a > (3) CAY making d/a the subject a+6(8—a)=11a a+48—-6a=11la CAY correct substitution of d/a 48 = 16a a=3 CAV a/d value d=5 CAV d/a value 3;8; 13;.. CAY sequence 7] QUESTION 3 3.1.1 1.2.4 AV All three terms qd) "p’ p? 3.1.2 -1<r<i1 AY condition for convergence -1<-<1 CAV r value in terms of p p<-lorp>1 CACAY V answers (4) 3.2 1)" S,=4 4(;5) 1)" AY first t 1 -4-4(_) = irst term value T, =5S,=4 4(;5) 2 12 T,+Tz=S2,=4- 4(;5) =3 AY Sum of first two terms 1y> 1 Ty +1, +73 =5S;=4-4(5) =35 Av Sum of first 3 terms “T,=1 CA“ second term value 1 613 = 2 CA” third term value (5) [10] Copyright Reserved Please Turn Over

“atte Swnloaded from Stanworephysigsacom QUESTION 4 March 2022 Common Test 4.1 On AB mark off a point P such that AP = DE and on AC mark off a point Q such that AQ = DF. Join PQ. In AABC and ADEF 1. AP=DE (Construction) 2. AQ=DF (Construction) 3. A=D (Given) -. AAPQ = ADEF (SAS) Now APQ = DEF But DEF = B (Given) : APQ=B PQ || BC (Corresponding angles =) AB _ AC P. Thm. PQ||BC AP = AQ (Prop. Thm. PQ||BC) AB AC Cc ion AP = DE DE DF (Construction = and AQ = DF) YS Construction vS/IR vS vs vS/IR vS/R YR (7] Copyright Reserved Please Turn Over

“atte Swnloaded from Stanworephysigsacom March 2022 Common Test 4.2 15 6 3c 2a > B 26 G c 4.2.1 AE _ AD _ 3 15 ac 57 Prop. Thm. ED || BC) VSYR AE = 9 units Y Answer GB) 422 | AWG i @ = 30" (Prop. Thm. HG || AC) VSYR AH = 5 units Y Answer (G3) 4.2.3 HE = 4 units and BH = 6 units CAVHE = 4 units or = = = $5... (Prop. Thm. EK || BG) CAY SAYR (3) 4.2.4 | AHEK _ 2 : ; YS/R ante = 57 (As with equal altitudes) G) auBe 1075 .. (As with equal altitudes) YSIR Area of AHEK — AHEK | AHEG _ 2 2 4 Area of AHBG AHEG * HBG” 5* 5 25 v Answer OR OR 1 . og Y Area of numerator AHEK _ 2 .(4)(2c) sin EHK V Area of SHBG 5 (10)(5c) sin EAK Denominator AHEK 4 v Answer (3) AHBG 25 OR OR 2 . . ¥S/IR AHEK = 5 (AHEG) ... (As with equal altitudes) = = (2 AHBG) ... (As with equal altitudes) vSIR AreaofAHEK 4 V ———— ——— =— Answer Areaof AHBG 25 (3) [19} Copyright Reserved Please Turn Over

March 2022 Common Test “atteSwnloaded from Stanworephysigsacom QUESTION 5 5.1 Angle subtended by the diameter on a circle/ VR (1) Angle in a semi-circle 5.2 PQR = 90° — x ... (Sum of angles of A) vs (3) SPQ=90° _... (Diameter perpendicular to tangent) YSYR S=x ... (Sum of angles of A) 5.3 QWR=x ... (Angles in the same segment are equal) | ¥S/R QWR=S=x YQWR=S=x SRWT is acyclic quad. ... (Converse of exterior angle of | VR cyclic quad equal to interior (3) opposite angle) 54 In A’s QWR and QST 1) QWR=S=x ........ (Proven) vs 2) Q is common YS 3) Ty = WRQ........0...... (Remaining angles of A’s) (G3) AQWR |\i AQST .... (AAA) yR 5.5.1 QW _ WR_ QR vS/R OS Tor = gr AQWR || AQST) 24 ¥ substitution ST 6 CAY Answer (3) ST=3 cm 5.5.2 = =2 ___ (AQWR || AQST) CAY SYR QS = 7,5 cm CAY Answer (3) [16] Copyright Reserved Please Turn Over

Mathenpy ics Ownloaded from Stanworephysigsncom March 2022 Common Test QUESTION 6 6.1.1 35° 140 70 1 1-p? 20 40 P 10° 38° 152 76 1 q X 28° 14 v1-q@ 7 sin20° = 1-p? MY AY Answer (2) 6.1.2 cos 6 = cos(20° — 14°) AY Writing as difference = cos 20 cos 14 + sin 20 sin 14 =p.j1-q*+/1-p?.q AY Expansion A Y (cos 20°) value CAV J1— GCA J1 — p? AY (sin 14’) value (6) 6.2 dsin 2x tan(540° + x) ( — tan? x) cos? x 5(2 sin x cos x) 1— sin?x tan x.(——,— cos?x 5 (2 sin x cos x) sinx cos?x Av expansion of sin 2x Av reduction to tan x Av1—sin?x = cos?x sinx cos X ‘cos?x Avtanx = - Cosx cos x = _|sinxcosx x — sinx CAY simplifying Vcos? x = ycos?x =cosx CAY answer (6) [14] Copyright Reserved Please Turn Over

“atte Swnloaded from Stanworephysigsacom QUESTION 7 March 2022 Common Test 711 cos 4x = cos(2.2x) = 2cos*2x-1 = 2(2cos?x—1)?-1 = 2(4cos* x —4cos?x+1)-1 =8 cos*x—8cos?x+1 AY cos 4x = cos (2.2x) Av expansion of cos 4x AY expansion of cos 2x AY squaring bracket (4) 7.1.2 | 2cos4x=1 cos 4x = ; AY setting up equation 4x = 60 + 360k or 4x = 300° /-60' + 360k | CA”both solutions ° ° ° Av 90k x=15 +90k or x=75 /-15 +90k;keZ AVkEZ (5) CAY x =15°,75 /-15" 7.1.3 | 16 cost x—16cos*x+2 =2cos 4x AY :. Minimum Value = —2 AY (2) 7.2 2sin? 22.5°-1 4sin 22.5’ cos 22.5° —(1— 2sin? 22.5") Av re-writing numerator and “2.2 sin 22.5 cos 22.5" denominator —cos45° AY Numerator = >sin45" AY Denominator 1 AV -1 Wh Special angle values/ co - ratios =, v2 21 v2 1 1 AY Answer (5) =--=x1=-— 2 2 [16] Copyright Reserved Please Turn Over

“atteSwnloaded from Stanworephysigsacom QUESTION 8 | y March 2022 Common Test 30 60 9 +1 t at 8.1 f: x-int., Maximum and Minimum and Shape AAAVVVY gi y-int., Maximum and Minimum and Shape AAAY VY (6) 8.2 | g(*) >4 0) > FG) g(x) = f(x); f(x) >0 x €(0°;180") ;x #90 AAV Answer (2) [8] Total Marks: 100 Copyright Reserved