Mathematics Grade 12 KZN April 2021 Memo

Downloaded from hlayiso.com ROAWACULAS =! EDUCATION REPUBLIC OF SOUTH AFRICA SENIOR CERTIFICATE MATHEMATICS COMMON TEST APRIL 2021 MARKS: 100 TIME: 2 hours N.B. This question paper consists of 6 pages, an answer sheet, 1 diagram sheet and an information sheet. Copyright Reserved Please Turn Over ——I

Downloaded from hlayiso.com Mathematics 2 April 2021 Common Test NSC INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. i This question paper consists of 7 questions. Answer ALL questions. Clearly show ALL calculations, diagrams, graphs, et cetera that you have used in determining your answers. Answers only will not necessarily be awarded full marks. An approved scientific calculator (non-programmable and non-graphical) may be used, unless stated otherwise. If necessary, answers should be rounded off to TWO decimal places, unless stated otherwise. Diagrams are NOT necessarily drawn to scale. Number the answers correctly according to the numbering system used in this question paper. Write neatly and legibly. Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics 3 April 2021 Common Test NSC QUESTION 1 Given the quadratic sequence: 44; 52; 64; 80; ... 1.1 Write down the next two terms of the sequence. (2) 1.2 Determine the n" term of the quadratic sequence. (4) 1.3 Calculate the 30" term of the sequence. (2) 1.4 Prove that the quadratic sequence will always have even terms. (3) [11] QUESTION 2 The 8" term of an arithmetic sequence is 31 and the sum of the first 30 terms is 1830. Determine the first three terms of the sequence. {7] QUESTION 3 3.1 The second term of a geometric sequence = and the ninth term is 5. Determine the value of the common ratio. (5) 3.2. Calculate the value of m if m 255 - ko ee > -8)-(05) = k=1 (4) 3.3. Given: a+ 12+ 6x + 3x7 4+ sx #0. 3.3.1 Determine the value of x for which the series converges. (3) 3.3.2 Write down the value of x for which the series is increasing. (2) [14] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics 4 April 2021 Common Test NSC QUESTION 4 4.1. Given AABC and ADEF with A = D,B = Band¢ =F. AB _ AC Prove that ne OF (7) 4.2 Inthe figure AD, DC and BE are tangents to the circle at A, C and B respectively. O is the centre of the circle. DE and AF intersect at H. AH produced meets BE produced in F. AJ, BC and BJ are chords. D Prove that: 4.2.1 ADAH ||| AOCH. (4) 422 AO.DH OH = — (6) 4.2.3. If BAisdrawn, then BF? = JF .AF (6) {23] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics ns} April 2021 Common Test NSC QUESTION 5 In the figure AF = 2CG and FE || GB. s = : Determine (with reasons): AF 5.1 = 2 FG (2) CH 5.2 = 4 HE (4) Ee) 1 ceihcollt latches (4) Area of AAFE {10} QUESTION 6 6.1 Given cos 26° = : Without using a calculator, calculate the value of the following in terms of p. 6.1.1 cos52” (4) 6.1.2 sin71° (4) 6.2 Simplify without using into a single trigonometric ratio. cos(-180’).tan @.cos 690 .sin (9-180") 6) cos?(@—90°) 6.3 Show that cos0° +cos1° + cos2” + ++»+cos178 +cos179° + cos 180° + 6sin90° = 6 (4) {17] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics 6 April 2021 Common Test NSC QUESTION 7 71 Prove the following identity: 1 — sin 2x : ———_—— = sin x - cos x sin x — cosx 7.2 Determine the general solution of: tan 3x. -1=0 tan 24° (3) (5) 7.3. Determine the maximum value of 3 sin x + cos.x, without the use of a calculator. (4) 74 Given: f(x) = 2cos(x — 30°) 7.3.1 Sketch the graph of f for the domain x € [-90°; 270°] on the axes provided. (2) 7.3.2 Use the letters P and Q to indicate on the graph the solution of the equation cos(x — 30°) = 0,5 and the x — coordinates of P and Q. Copyright Reserved (4) [18] Please Turn Over

Downloaded from hlayiso.com i i ato | Please Turn Over April 202{ Common Test GRADE: 1x0 f iho | 130 | | a. NSC oo | | ES i I lo 3 7 & bn ats a ve : a fa an ~ 6 3 2 a g 4 = a a | z £ g | S z = 1° 5 dOvd 4dO-UVAL

Downloaded from hlayiso.com April 2021 Common Test 8 NSC Mathematics GRADE: DIAGRAM SHEET QUESTION 4.1 QUESTION 4.2 QUESTION 5 Please Turn Over Copyright Reserved

Downloaded from hlayiso.com Mathematics 9 April 2021 Common Test NSC INFORMATION SHEET: MATHEMATICS INLIGTING BLADSY _-bt1b? —4ac 2a A=P(l+ni) A= P(I-ni) A=P(-i)" A=P(i+i)" T, =a+(n-l)d S, = 5 (2a+(n-1)d) T, =ar"" s el r#l Se or a pol+iy 1 powl-d+i] i i f'(x)= lim L(x+h)- f(*) h>0 h d=\(x, =x)? +0,-9,)? M2572 21522) 2 2 y2 741 y=mx+e yry, =m(x-x,) m= m=tan@ x4 Xx, (x-a) +(y—bY =r? a b c In AABC: — : sinA sinB sinC sin(a + 8) = sina.cos f+ cos a.sin B cos(@ + 8) = cos a.cos B—sina.sin B cos’ asin? a cos 2a@ = 41-2sin’? a 2cos?a-1 p=atbhx Copyright Reserved a? =b* +¢7 -2be.cos A area KMABC = x ab.sin C sin(a- B) = sina.cos # —cosa.sin B cos(a ~ £2) = cos a.cos 7 +sina.sin B sin 2a@ = 2sina.cosa P(A or B)= P(A) + P(B)— P(A and B) Ae YG - xKy a) Sa@-xy7

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS MARKING GUIDELINE COMMON TEST APRIL 2021 MARKS: 100 This memorandum consists of 9 pages. Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics 2 April 2021 Common Test NSC – Marking Guideline QUESTION 1 1.1 100 ; 124 AA✓✓answers (2) 1.2 44 52 64 80 1D 8 12 16 2D 4 4 𝟐𝒂 = 𝟒 ∴ 𝒂 = 𝟐 A✓a value 𝟑𝒂 + 𝒃 = 𝟖 ∴ 𝒃 = 𝟐 CA✓b value 𝒂 + 𝒃 + 𝒄 = 𝟒𝟒 ∴ 𝒄 = 𝟒𝟎 CA✓c value CA✓nth term 𝑻𝒏 = 𝟐𝒏𝟐 = 𝟐𝒏 + 𝟒𝟎 OR OR A✓a value 𝟐𝒂 = 𝟒 ∴ 𝒂 = 𝟐 CA✓b value 𝟑𝒂 + 𝒃 = 𝟖 ∴ 𝒃 = 𝟐 CA✓c value ∴ 𝒄 = 𝑻𝟎 = 𝟒𝟎 CA✓nth term (4) 𝑻𝒏 = 𝟐𝒏𝟐 + 𝟐𝒏 + 𝟒𝟎 OR OR 𝑻𝒏 = 𝑻𝟏 + (𝒏 − 𝟏)𝒅𝟏 + (𝒏 − 𝟏)(𝒏 − 𝟐)𝒅𝟐 OR OR (𝒏 − 𝟏) (4) 𝑻𝒏 = [𝟐𝒂 + (𝒏 − 𝟐)𝒅] + 𝑻𝟏 𝟐 1.3 𝑻𝟑𝟎 = 𝟐(𝟑𝟎)𝟐 + 𝟐(𝟑𝟎) + 𝟒𝟎 = 𝟏𝟗𝟎𝟎 CA✓substitution CA✓answer (2) 1.4 𝑻𝒏 = 𝟐𝒏𝟐 + 𝟐𝒏 + 𝟒𝟎 𝑻𝒏 = 𝟐(𝒏𝟐 + 𝒏 + 𝟐𝟎) A✓Taking out common factor of 2 𝟐(𝒏𝟐 + 𝒏 + 𝟐𝟎) is even for all 𝒏 ∈ 𝒁 A✓Rewriting nth term (3) A✓ is even for all 𝒏 ∈ 𝒁 Note: Mark CA provided 𝑻𝒏 (from1.2) is a factor of 2 [11] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics 3 April 2021 Common Test NSC – Marking Guideline QUESTION 2 2. 𝒂 + 𝟕𝒅 = 𝟑𝟏 → (𝟏) A✓equation (1) 𝟏𝟓(𝟐𝒂 + 𝟐𝟗𝒅) = 𝟏𝟖𝟑𝟎 𝟐𝒂 + 𝟐𝟗𝒅 = 𝟏𝟐𝟐 → (𝟐) A✓equation (2) 𝒂 = 𝟑𝟏 − 𝟕𝒅 → (𝟑) CA✓making a the subject 𝟐(𝟑𝟏 − 𝟕𝒅) + 𝟐𝟗𝒅 = 𝟏𝟐𝟐 CA✓correct substitution of a 𝟔𝟐 − 𝟏𝟒𝒅 + 𝟐𝟗𝒅 = 𝟏𝟐𝟐 𝟏𝟓𝒅 = 𝟔𝟎 𝒅=𝟒 CA✓d value 𝒂=𝟑 CA✓a value 𝟑 ; 𝟕 ; 𝟏𝟏 ; … CA✓sequence [7] QUESTION 3 3.1 𝟓 A✓equation (1) 𝒂𝒓 = → (𝟏) 𝟏𝟐𝟖 𝒂𝒓𝟖 = 𝟓 → (𝟐) A✓equation (2) 𝒓𝟕 = 𝟏𝟐𝟖 CA✓ 𝒓𝟕 = 𝟏𝟐𝟖 𝒓𝟕 = 𝟐 𝟕 CA✓exponential form 𝒓=𝟐 CA✓answer (5) 3.2 (−𝟖) + (−𝟖)(𝟎. 𝟓) + (−𝟖)(𝟎. 𝟓)𝟐 + ⋯ A✓generating series −𝟖(𝟎. 𝟓𝒎 − 𝟏) 𝟐𝟓𝟓 CA✓correct substitution into =− correct formula 𝟎. 𝟓 − 𝟏 𝟏𝟔 𝟐𝟓𝟓 𝟎. 𝟓𝒎 − 𝟏 = − 𝟐𝟓𝟔 𝟏 CA✓writing in exponential 𝟎. 𝟓𝒎 = = 𝟎. 𝟓𝟖 form or using logs 𝟐𝟓𝟔 (4) 𝒎=𝟖 CA✓answer 3.3.1 −𝟏 < 𝒓 < 𝟏 A✓condition for convergence 𝒙 −𝟏 < < 𝟏 A✓r value 𝟐 −𝟐 < 𝒙 < 𝟐 CA✓answer (3) 3.3.2 𝒙 < −𝟐 𝒐𝒓 𝒙 > 𝟐 CACA✓✓answer (2) [14] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics 4 April 2021 Common Test NSC – Marking Guideline QUESTION 4 A D P Q E F B C 4.1 Draw AP = DE and AQ = DF ✓S Construction (or could be In ∆𝐀𝐁𝐂 𝐚𝐧𝐝 ∆𝐃𝐄𝐅 shown on diagram) 1. AP = DE (Construction) 2. AQ = DF (Construction) ̂ 3. 𝐀 = 𝐃 ̂ (Given) ∴ ∆𝐀𝐏𝐐 ≡ ∆𝐃𝐄𝐅 (SAS) Now 𝐀𝐏 ̂𝐐 = 𝐃𝐄̂𝐅 ✓S/R But 𝐃𝐄̂𝐅 = 𝐁 ̂ (Given) ✓S ∴ 𝐀𝐏̂𝐐 = 𝐁 ̂ ✓S PQ || BC (Corresponding angles =) ✓S/R 𝐀𝐁 𝐀𝐂 = (Prop. Thm. PQ||BC) ✓S/R 𝐀𝐏 𝐀𝐘 𝐀𝐁 𝐀𝐂 = (Construction AP = DE ✓R 𝐃𝐄 𝐃𝐅 and AQ = DF) (7) 4.2.1 In ∆𝐃𝐀𝐇 𝐚𝐧𝐝 ∆𝐎𝐂𝐇 1. D𝐀̂ H = O𝐂̂H =90° (Radius ⊥Tangent) ✓S✓R ̂ 2 is common 2. 𝐇 ✓S 3. 𝐀𝐃̂ 𝐇 = 𝐂𝐎̂H (Remaining angles) (4) ∴ ∆𝐃𝐀𝐇 ||| ∆𝐎𝐂𝐇 (AAA) ✓R(AAA) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics 5 April 2021 Common Test NSC – Marking Guideline 4.2.2 𝐃𝐀 𝐃𝐇 𝐀𝐇 = 𝐎𝐇 = 𝐂𝐇 (∆𝐃𝐀𝐇 ||| ∆𝐎𝐂𝐇 ) ✓S/R 𝐎𝐂 𝐃𝐇 × 𝐎𝐂 𝐎𝐇 = ✓S 𝐃𝐀 DA = DC (Tangents drawn from common point equal) ✓S✓R AO = OC (Radii of a circle) ✓S✓R Therefore 𝐀𝐎. 𝐃𝐇 𝐎𝐇 = 𝐃𝐂 (6) 4.2.3 In ∆𝐀𝐁𝐅 𝐚𝐧𝐝 ∆𝐁𝐉𝐅 ̂ F = J𝐁 1. 𝐁𝐀 ̂F (Tangent – Chord Theorem) ✓S✓R 2. 𝐅̂ is common) ✓S ̂ 𝐅 = 𝐁𝐉̂𝐅 3. 𝐀𝐁 (Remaining angles) ∴ ∆𝐀𝐁𝐅 ||| ∆𝐁𝐉𝐅 (AAA) ✓S✓R 𝐀𝐁 𝐁𝐅 𝐀𝐅 ∴ 𝐁𝐉 = 𝐉𝐅 = 𝐁𝐅 (∆𝐀𝐁𝐅 ||| ∆𝐁𝐉𝐅) ✓S 𝐁𝐅 𝟐 = 𝐉𝐅 . 𝐀𝐅 (6) [23] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics 6 April 2021 Common Test NSC – Marking Guideline QUESTION 5 2b 2a 3b b 3a 5.1 Let AE = 2a therefore EB = 3a ✓S ✓R (2) 𝐀𝐅 𝟐 =𝟑 (Prop. Thm.; FE//GB) or (Line // one side of ∆) 𝐅𝐆 5.2 Let AF = 2b and FG = 3b Then CG = b (Given AF = 2CG) ✓S 𝐂𝐇 𝐂𝐆 = 𝐆𝐅 = 𝟑𝒃 𝒃 (Prop. Thm.; GH//FE) or (Line // one side of ∆) ✓S ✓R 𝐇𝐄 𝐂𝐇 𝟏 ∴ = ✓S (4) 𝐇𝐄 𝟑 5.3 𝐀𝐄 𝐀𝐅 𝐅𝐄 𝟐 = 𝐀𝐆 = 𝐆𝐁 = 𝟓 …. (Prop. Thm; FE || GB) or (Line // one side of ∆) 𝐀𝐁 ̂𝐁 = 𝐆𝐅̂𝐄 ….. (Corresp Angles ; FE || GB) 𝐂𝐆 ✓S/R Let FE = 2x and GB = 5x Then 𝟏 ̂𝐁 𝐀𝐫𝐞𝐚 𝐨𝐟 ∆ 𝐁𝐂𝐆 𝟐 (𝒃)(𝟓𝒙) 𝐬𝐢𝐧 𝐂𝐆 2b ✓S = 𝐀𝐫𝐞𝐚 𝐨𝐟 ∆ 𝐀𝐅𝐄 𝟏 (𝟐𝒃)(𝟐𝒙) 𝐬𝐢𝐧 𝐀𝐅̂𝐄 𝟐 2x ✓S 𝟏 ̂𝐁 (𝒃)(𝟓𝒙) 𝐬𝐢𝐧 𝐂𝐆 = 𝟐 𝟏 ̂𝐁) (𝟐𝒃)(𝟐𝒙) 𝐬𝐢𝐧 (𝟏𝟖𝟎° − 𝐂𝐆 𝟐 ✓S b 5x 𝟓 = (4) 𝟒 OR OR 𝟏 ✓S/R 𝐀𝐫𝐞𝐚 𝐨𝐟 ∆ 𝐁𝐂𝐆 = 𝟔 𝐀𝐫𝐞𝐚 𝐨𝐟 ∆𝐀𝐁𝐂 …(Equal Heights) 𝟏 𝐀𝐫𝐞𝐚 𝐨𝐟 ∆ 𝐀𝐅𝐄 = 𝟑 𝐀𝐫𝐞𝐚 𝐨𝐟 ∆𝐀𝐄𝐂 …(Equal Heights) ✓S/R 𝟐 𝐀𝐫𝐞𝐚 𝐨𝐟 ∆ 𝐀𝐄𝐂 = 𝟓 𝐀𝐫𝐞𝐚 𝐨𝐟 ∆𝐀𝐁𝐂 …(Equal Heights) ✓S 𝟐 𝐀𝐫𝐞𝐚 𝐨𝐟 ∆ 𝐀𝐅𝐄 = 𝐀𝐫𝐞𝐚 𝐨𝐟 ∆𝐀𝐁𝐂 𝟏𝟓 ✓S 𝐀𝐫𝐞𝐚 𝐨𝐟 ∆ 𝐁𝐂𝐆 𝟏𝟓 𝟓 = = (4) 𝐀𝐫𝐞𝐚 𝐨𝐟 ∆ 𝐀𝐅𝐄 𝟏𝟐 𝟒 [10] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics 7 April 2021 Common Test NSC – Marking Guideline QUESTION 6 6.1.1 128° 32° 64° A✓diagram p √𝑝2 − 1 52° 26° 1 13° 𝐜𝐨𝐬 𝟓𝟐° = 𝐜𝐨𝐬[𝟐(𝟐𝟔°)] = 𝟐𝐜𝐨𝐬 𝟐 𝟐𝟔° − 𝟏 A✓writing as double angle 𝟏 A✓expansion = 𝟐(𝑷)𝟐 −𝟏 CA✓answer (4) 6.1.2 𝐬𝐢𝐧 𝟕𝟏° = 𝐬𝐢𝐧(𝟒𝟓° + 𝟐𝟔° ) A✓𝐬𝐢𝐧(𝟒𝟓° + 𝟐𝟔° ) = 𝐬𝐢𝐧 𝟒𝟓° 𝐜𝐨𝐬 𝟐𝟔° + 𝐜𝐨𝐬 𝟒𝟓° 𝐬𝐢𝐧 𝟐𝟔° A✓compound angle √𝟐 𝟏 √𝟐 √𝒑𝟐 − 𝟏 expansion = . + . CA CA ✓✓each term 𝟐 𝒑 𝟐 𝒑 (4) ° ). ° °) 𝒔𝒊𝒏 𝜽 6.2 𝐜𝐨𝐬(−𝟏𝟖𝟎 𝐭𝐚𝐧 𝜽. 𝐜𝐨𝐬 𝟔𝟗𝟎 . 𝐬𝐢 𝐧(𝜽 − 𝟏𝟖𝟎 A✓𝒄𝒐𝒔 𝜽 𝐜𝐨𝐬 𝟐 (𝜽 − 𝟗𝟎° ) A✓𝐜𝐨𝐬 𝟑𝟎° . 𝒔𝒊𝒏 𝜽 A✓− 𝐬𝐢𝐧 𝜽 𝐜𝐨𝐬(𝟏𝟖𝟎° ) × 𝒄𝒐𝒔 𝜽 . 𝐜𝐨𝐬 𝟑𝟎° . (− 𝐬𝐢𝐧 𝜽) = 𝐬𝐢𝐧𝟐 𝜽 √𝟑 𝐬𝐢𝐧 𝜽 √𝟑 CA✓ 𝟐 or 0,866 (−𝟏) × 𝐜𝐨𝐬 𝜽 . ( 𝟐 )(− 𝐬𝐢𝐧 𝜽) = 𝐬𝐢𝐧𝟐 𝜽 √𝟑 𝐬𝐢𝐧𝟐 𝜽 = 𝟐 𝐜𝐨𝐬 𝜽 𝐬𝐢𝐧𝟐 𝜽 √𝟑 CA✓answer (5) = 𝟐 𝐜𝐨𝐬 𝜽 6.3 𝐋𝐇𝐒 = 𝐜𝐨𝐬 𝟎° + 𝐜𝐨𝐬 𝟏° + 𝐜𝐨𝐬 𝟐° + ⋯ + 𝐜𝐨𝐬 𝟏𝟕𝟖° + A✓−𝒄𝒐𝒔 𝟐° 𝐜𝐨𝐬 𝟏𝟕𝟗° + 𝒄𝒐𝒔 𝟏𝟖𝟎° + 𝟔𝐬𝐢𝐧 𝟗𝟎° A✓−𝒄𝒐𝒔 𝟏° = 𝐜𝐨𝐬 𝟎° + 𝐜𝐨𝐬 𝟏° + 𝐜𝐨𝐬 𝟐° + … − 𝐜𝐨𝐬 𝟐° − 𝒄𝒐𝒔 𝟏° A✓− 𝒄𝒐𝒔 𝟎° − 𝒄𝒐𝒔 𝟎° + 𝟔𝐬𝐢𝐧 𝟗𝟎° A✓All terms cancel except 6 (4) = 𝟔 = 𝐑𝐇𝐒 OR 𝐋𝐇𝐒 = ( 𝐜𝐨𝐬 𝟎° + 𝐜𝐨𝐬 𝟏𝟖𝟎° ) + (𝐜𝐨𝐬 𝟏° + 𝐜𝐨𝐬 𝟏𝟕𝟗° ) + A✓( 𝐜𝐨𝐬 𝟎° + 𝐜𝐨𝐬 𝟏𝟖𝟎°) (𝐜𝐨𝐬 𝟐° + 𝐜𝐨𝐬 𝟏𝟕𝟖° ) … … … + 𝟔𝐬𝐢𝐧 𝟗𝟎° A✓(𝐜𝐨𝐬 𝟏° + 𝐜𝐨𝐬 𝟏𝟕𝟗° ) 𝐋𝐇𝐒 = (𝟎) + (𝟎) + (𝟎) … … … + 𝟔𝐬𝐢𝐧 𝟗𝟎° A✓(𝐜𝐨𝐬 𝟐° + 𝐜𝐨𝐬 𝟏𝟕𝟖° LHS = 6 A✓All terms cancel except 6 [17] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics 8 April 2021 Common Test NSC – Marking Guideline QUESTION 7 7.1 𝟏 − 𝒔𝒊𝒏 𝟐 𝒙 𝒔𝒊𝒏 𝒙 − 𝒄𝒐𝒔 𝒙 𝒔𝒊𝒏𝟐 𝒙−𝟐 𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙+𝒄𝒐𝒔𝟐 𝒙 = A✓ sin2 x + cos2x = 1 𝒔𝒊𝒏 𝒙−𝒄𝒐𝒔 𝒙 A✓ 2 sin x cos x (𝒔𝒊𝒏 𝒙−𝒄𝒐𝒔 𝒙)𝟐 = A✓ (sin x – cos x)2 𝒔𝒊𝒏 𝒙−𝒄𝒐𝒔 𝒙 = 𝒔𝒊𝒏 𝒙 − 𝒄𝒐𝒔 𝒙 = RHS (3) 7.2 𝟏 𝐭𝐚𝐧 𝟑𝒙 . −𝟏=𝟎 𝐭𝐚𝐧 𝟐𝟒° A✓transposing and forming 𝐭𝐚𝐧 𝟑𝒙 = 𝐭𝐚𝐧 𝟐𝟒° equation 𝟑𝒙 = 𝟐𝟒° + 𝟏𝟖𝟎𝒌 ; 𝒌 ∈ 𝒁 A✓3𝑥 = 24° A✓180k (5) A✓ 𝑘 ∈ 𝑍 𝒙 = 𝟖° + 𝟔𝟎𝒌 ; 𝒌 ∈ 𝒁 CA✓ 𝑥 = 8° + 60𝑘 Note: If calculator is used maximum marks may be allocated 7.3 √𝟑 𝟏  3 1  𝟐 [ 𝐬𝐢𝐧 𝒙 + 𝐜𝐨𝐬 𝒙] A✓ 2 sin x + cos x  𝟐 𝟐  2 2  = 𝟐[𝐬𝐢𝐧 𝟔𝟎° 𝐬𝐢𝐧 𝒙 + 𝐜𝐨𝐬 𝟔𝟎° 𝐜𝐨𝐬 𝒙] A✓ [sin 60o sin x + cos 60o cos x] = 𝟐[𝐜𝐨𝐬(𝒙 − 𝟔𝟎°)] A✓ cos (x − 60o)  Maximum value is 2, since maximum value of CA✓ answer cos (x − 60o) = 1 (4) OR OR √𝟑 𝟏  3  𝟐 [ 𝐬𝐢𝐧 𝒙 + 𝐜𝐨𝐬 𝒙] 1 𝟐 𝟐 A✓ 2 sin x + cos x   2 2  = 𝟐[𝐜𝐨𝐬 𝟑𝟎° 𝐬𝐢𝐧 𝒙 + 𝐬𝐢𝐧 𝟑𝟎° 𝐜𝐨𝐬 𝒙] A✓ 𝟐[𝐜𝐨𝐬 𝟑𝟎° 𝐬𝐢𝐧 𝒙 + = 𝟐[𝐬𝐢𝐧(𝒙 + 𝟑𝟎°)] 𝐬𝐢𝐧 𝟑𝟎° 𝐜𝐨𝐬 𝒙]  Maximum value is 2, since maximum value of A✓ sin (x + 30o) (4) sin (x + 30o) = 1 CA✓ answer Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics 9 April 2021 Common Test NSC – Marking Guideline 7.4.1 y 2 1.5 P 1 Q 0.5 x -90 -60 -30 O 30 60 90 120 150 180 210 240 270 -0.5 -1 -1.5 -2 A✓for both x – intercepts A✓for both turning points (2) 7.4.2 𝐜𝐨𝐬(𝒙 − 𝟑𝟎° ) = 𝟎, 𝟓 A✓𝒙 − 𝟑𝟎° = 𝟔𝟎° 𝟐𝐜𝐨𝐬(𝒙 − 𝟑𝟎° ) = 𝟏 and 𝒙 − 𝟑𝟎° = −𝟔𝟎° 𝒙 − 𝟑𝟎° = 𝟔𝟎° 𝒐𝒓 𝒙 − 𝟑𝟎° = −𝟔𝟎° CA✓ 𝟗𝟎° 𝐚𝐧𝐝 −𝟑𝟎° 𝒙 = 𝟗𝟎° 𝒂𝒕 𝑸 𝒐𝒓 𝒙 = −𝟑𝟎° 𝒂𝒕 𝑷 CACA✓✓for P and Q (4) [18] Copyright Reserved Please Turn Over