Mathematics Grade 12 KZN March 2019 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS COMMON TEST MARCH 2019 MARKS: 100 TIME: 2 hours This question paper consists of 9 pages and 1 information sheet. Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics 2 March 2019 Common Test NSC INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. This question paper consists of 9 questions. 2. Answers ALL questions. 3. Clearly show ALL calculations, diagrams, graphs, etc cetera that you have used in determining yours answers. 4. Answers only will not necessarily be awarded full marks. 5. An approved scientific calculator (non-programmable and non-graphical) may be used, unless stated otherwise. 6. If necessary, answers should be rounded off to TWO decimal places, unless stated otherwise. 7. Diagrams are NOT necessarily drawn to scale. 8. 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 March 2019 Common Test NSC QUESTION 1 Given the quadratic sequence: 2 ; 5 ; 10 ; 17 ; . . . 1.1 Write down the next two terms of the quadratic sequence. (2) 1.2 Calculate the nth term of the quadratic sequence. (4) [6] QUESTION 2 2.1 Given the combined constant and arithmetic sequences: 5 ; 4 ; 5 ; 7 ; 5 ; 10 ; ... Determine the position of the term 1051 in the combined sequence. (3) 2.2 The series 3 + 8 + 13 + … consists of n terms. The sum of the last three terms is 699. 2.2.1 Determine the sum to n terms in terms of n. (2) 2.2.2 If the last three terms are excluded from the series, then determine in terms of n the sum of the remaining terms. (2) 2.2.3 Hence, or otherwise, determine the value of n. (3) [10] QUESTION 3 3.1 The sixth term of a geometric sequence is 80 more than the fifth term. 80 3.1.1 Show that a = . (2) r − r4 5 3.1.2 If it is further given that sum of the fifth and sixth terms is 240, determine the value of the common ratio. (5) 3.2 Write the geometric series 9 + 3 + 1 ; … to 130 terms in sigma notation. (2) [9] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics 4 March 2019 Common Test NSC QUESTION 4    In the diagram below, BC = CE; E1 = x and D1 = D2 . B B 4.1 Name, with reasons, TWO other angles each equal to x and show that FD = FE. (4) 4.2 Prove that BF bisects CBˆ A . (4) 4.3 Hence, or otherwise, prove that Â1 = CBˆ A . (4) [12] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com 5 March 2019 Common Test NSC QUESTION 5 5.1 The angle at the point of contact between a tangent to a circle and a chord is --------------- . (1) 5.2 In the sketch below, circle centre O has a tangent KLM. Diameter NQ produced meet the tangent in K. Nˆ 1 = 32 º and Nˆ 2 = 230 . Calculate, with reasons, the size of: 5.2.1 P̂2 (1) 5.2.2 POˆ Q (2) 5.2.3 L̂2 (2) 5.2.4 NL̂Q (1) 5.2.5 L̂3 (2) 5.2.6 PLˆ K (2) [11] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics 6 March 2019 Common Test NSC QUESTION 6 AD AE In the diagram below, has Prove the theorem that states = . DB EC [7] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics 7 March 2019 Common Test NSC QUESTION 7 In the diagram below, EFGK is a cyclic quadrilateral with Fˆ = 900 . EK and FG are produced to meet at H. HJ is drawn parallel to FE. GK produced meets HJ at J. 7.1 Prove that: 7.1.1 JHˆ F = 900 (2) 7.1.2 Kˆ 2 = 900 (2) 7.1.3  HKG ⫼  JHG (3) 7.2 Calculate JG and KG if HG = 5cm and JH = 10cm. (4) [11] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics 8 March 2019 Common Test NSC QUESTION 8 (ANSWER THIS QUESTION WITHOUT THE USE OF CALCULATOR) 8.1 Show (4) 8.2 If sin 360 = m , and cos 240 = n, determine the following in terms of m and /or n: 8.2.1 cos 360 (3) 8.2.2 sin 12 0 (4) 8.3 Simplify: (5) 8.4 Calculate the value of if (4) [20] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics 9 March 2019 Common Test NSC QUESTION 9 Given f(x) = sin x + 1 and g(x) = cos 2x 9.1 Show that f(x) = g(x) can be written as (2 sin x + 1) sin x = 0. (2) 9.2 Hence or otherwise determine the general solution of sin x + 1 = cos 2x. (6) 9.3 Write down the range of g(x) – 1. (1) 9.4 Consider the following geometric series Determine the values of x for the interval for which the series will converge. (5) [14] TOTAL MARKS: [100] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics 10 March 2019 Common Test NSC INFORMATION SHEET: MATHEMATICS INLIGTINGSBLAD: WISKUNDE − b  b 2 − 4ac x= 2a A = P(1 + ni) A = P(1 − ni) A = P(1 − i) n A = P(1 + i) n Tn = a + (n − 1)d Sn = n (2a + (n − 1)d ) 2 Tn = ar n −1 Sn = ( a r n −1 ) ; r 1 S = a ; −1  r  1 r −1 1− r F=  x (1 + i ) − 1 n  P= x[1 − (1 + i)−n ] i i f ( x + h) − f ( x ) f ' ( x) = lim h→ 0 h  x + x2 y1 + y 2  d = ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2 M  1 ;   2 2  y 2 − y1 y = mx + c y − y1 = m( x − x1 ) m= m = tan  x 2 − x1 (x − a )2 + ( y − b)2 = r 2 a b c 1 In ABC: = = a 2 = b 2 + c 2 − 2bc. cos A area ABC = ab. sin C sin A sin B sin C 2 sin( +  ) = sin  . cos  + cos . sin  sin( −  ) = sin  . cos  − cos . sin  cos( +  ) = cos . cos  − sin  . sin  cos( −  ) = cos . cos  + sin  . sin  cos2  − sin 2   cos 2 = 1 − 2 sin 2  sin 2 = 2 sin  . cos 2 cos2  − 1  n 2  fx  (x − x ) i x=  = i =1 2 n n n( A ) P( A) = P(A or B) = P(A) + P(B) - P(A and B) n(S) yˆ = a + bx b=  (x − x )( y − y ) (x − x) 2 Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics 11 March 2019 Common Test NSC Copyright Reserved Please Turn Over

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINE COMMON TEST MARCH 2019 NATIONAL SENIOR CERTIFICATE GRADE 12 MARKS: 100 This marking guideline consists of 10 pages. Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics P2 2 March 2019 Common Test NSC QUESTION 1 1.1 26 ; 37 AA correct values 2 1.2 2 5 10 17 3 5 7 2 2 𝟐𝒂 = 𝟐 A a value 𝒂=𝟏 CAb value 𝟑𝒂 + 𝒃 = 𝟑 𝒃=𝟎 CAc value 𝒂+𝒃+𝒄=𝟐 𝒄=𝟏 CAgeneral term 𝑻𝒏 = 𝒏𝟐 + 𝟏 OR OR 4 2𝑎 = 2 A a value 𝑎=1 3𝑎 + 𝑏 = 3 CAb value 𝑏=0 Ac value 𝑇0 = 𝑐 = 1 𝑇𝑛 = 𝑛2 + 1 CAgeneral term OR T  T  n  1d  n  1n  2 d OR 4 n 1 1 2 2 Aformula  2  n  13  n  1n  2 2 Asubstituting first and second 2 difference values  2  3n  3  n  3n  2 2 CAsimplifying  n2  1 CAgeneral term OR OR n 1 Tn  2a  n  2d   T1 Aformula 4 2 Acorrect substitution into formula n 1  2(3)  (n  2)(2)  2 CAsimplifying 2  (n  1)3  n  2  2  (n  1)( n  1)  2 CAgeneral term  n 1 2  n 1 2 2 4 [6] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics P2 3 March 2019 Common Test NSC QUESTION 2 2.1 3𝑛 + 1 = 1051 Aequating nth term to 1051 3𝑛 = 1050 𝑛 = 350 CAvalue of n th 1051 is in the 700 position CAconclusion 3 2.2.1 a  3, d  5 A d = 5 Sn  2a  (n  1)d  n 2  2(3)  (n  1)(5) n CAsubstitution of a & d into sum 2 formula  5n  1 n 2 2 n3 A n – 3 2.2.2 S n3  5(n  3)  1 2 CAsubstituting into sum formula n3  5n  14 2 2 2.2.3 𝑆𝑛 − 𝑆𝑛−3 = 699 𝑛 𝑛−3 [5𝑛 + 1] − [5𝑛 − 14] = 699 CAforming equation 2 2 𝑛(5𝑛 + 1) − (𝑛 − 3)(5𝑛 − 14) = 1398 5𝑛2 + 𝑛 − 5𝑛2 + 29𝑛 − 42 = 1398 30𝑛 = 1440 CAsimplification 𝑛 = 48 CAanswer (n must be a natural number) 3 OR OR Tn = 5n – 2 Tn-1 = 5(n-1) – 2 Tn-2 = 5(n-2) – 2 CAforming equations 15n – 21 = 699 CAsimplification n = 45 3 CAanswer (n must be a natural number) [10] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics P2 4 March 2019 Common Test NSC QUESTION 3 3.1.1 𝑇6 = 80 + 𝑇5 𝑎𝑟 5 = 80 + 𝑎𝑟 4 Aforming equation 𝑎𝑟 5 − 𝑎𝑟 4 = 80 𝑎(𝑟 5 − 𝑟 4 ) = 80 Afactorizing 80 𝑎= 5 𝑟 − 𝑟4 2 3.1.2 𝑇5 + 𝑇6 = 240 𝑎𝑟 4 + 𝑎𝑟 5 = 240 Aforming equation 𝑎(𝑟 4 + 𝑟 5 ) = 240 240 CAfactorizing 𝑎= 4 𝑟 + 𝑟5 80 240 CAequating 5 4 = 5 𝑟 −𝑟 𝑟 + 𝑟4 [ 80𝑟 + 80𝑟 4 = 240𝑟 5 − 240𝑟 4 5 CAsimplifying 320𝑟 4 = 160𝑟 5 𝑟=2 CAr – value OR OR 5 Let the fifth term be = x and sixth term = y, then 𝑥 + 𝑦 = 240 → (1) Aforming equation (1) −𝑥 + 𝑦 = 80 → (2) (1) + (2) Aforming equation (2) 2𝑦 = 320 CA y – value 𝑦 = 160; 𝑥 = 80 160 CAx – value 𝑟= =2 80 5 CA r – value 3.2 130 Aupper and lower limit values 1 𝑘−1 ∑9( ) 3 A kth term 2 𝑘=1 [9] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics P2 5 March 2019 Common Test NSC QUESTION 4 4.1 𝐵̂2 = 𝐸̂1 = x …….. (BC = CE) A S/R 𝐷̂1 = 𝐵̂2 = x … (Ext. ∠ of cyclic quad = interior A S AR opposite angles) ̂1 = 𝐸̂1 ∴𝐷 A S ∴FD = FE (4) 4.2 ̂2 = 𝐷 𝐷 ̂1 = x …… given A S ̂2 = 𝐵̂1 = 𝑥 … subtended by arc AF 𝐷 A S A R  𝐵̂1 = 𝐵̂2 A S ∴EB bisects 𝐶𝐵̂ 𝐴 (4) 4.3 𝐴̂1 = 𝐹̂2 …… subtended by arc A S AR 𝐹̂2 = 2𝑥 ….. ext. ∠ of ΔDFE A S AR  𝐴̂1 = 2𝑥 = 𝐶𝐵̂ 𝐴 (4) [12] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics P2 6 March 2019 Common Test NSC QUESTION 5 5.1 …….. equal to the angle subtended by the AS (1) chord in the opposite circle segment. 5.2.1 𝑃̂2 = 23° …… (ON = OP; radii) A S/R (1) 5.2.2 𝑃𝑂̂ 𝑄 = 2𝑁 ̂2 = 46° …… (∠at centre) / A S AR (2) (Ext ∠of𝛥) 5.2.3 𝐿̂2 = 𝑁 ̂2 = 23°…… (subt. by arc PQ) A S AR (2) 5.2.4 𝑁𝐿̂𝑄 = 90° …… (subt. by diameter NQ) AS/R (1) 5.2.5 𝐿̂3 = 90° − 23° CA S = 67o CA answer (2) OR OR 𝑃𝑂̂ 𝑁 = 134° …..(angles of triangle) 𝑃𝑂̂ 𝑁 = 2𝐿̂ …..(angle at centre theorem) CA S = 67° CA answer (2) 5.2.6 𝑃𝐿̂𝐾 = 𝐿𝑁 ̂ 𝑃 …… (tan-chord theorem) A S/R = 32o + 23o = 55o A answer (2) [11] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics P2 7 March 2019 Common Test NSC QUESTION 6 Construction: Draw line h1⊥AC and h2 ⊥AB A construction Proof: 1 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐷𝐸 2 . 𝐴𝐷. ℎ2 A S 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐵𝐷𝐸 = 1 𝑠𝑎𝑚𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 ℎ2 A R . 𝐷𝐵.ℎ2 2 1 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐷𝐸 . 𝐴𝐸. ℎ1 A S/R and 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐶𝐸𝐷 = 21 𝑠𝑎𝑚𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 ℎ1 . 𝐸𝐶. ℎ1 2 but 𝑎𝑟𝑒𝑎 𝑜𝑓 ∆𝐵𝐷𝐸 = 𝑎𝑟𝑒𝑎 𝑜𝑓 ∆𝐶𝐸𝐷 ∵ (𝑠𝑎𝑚𝑒 𝑏𝑎𝑠𝑒 𝐷𝐸; 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 ℎ𝑒𝑖𝑔ℎ𝑡; 𝐷𝐸 ∥ 𝐵𝐶) AA S/ R 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐷𝐸 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐷𝐸 A S ∴ = 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐵𝐷𝐸 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐶𝐸𝐷 𝐴𝐷 𝐴𝐸 ∴ = 𝐷𝐵 𝐸𝐶 [7] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics P2 8 March 2019 Common Test NSC QUESTION 7 7.1 7.1.1 ̂ 𝐹 + 𝐹̂ = 180 ° …. Co-interior angles; JH // EF 𝐽𝐻 A S/R ̂ F = 90o …… 𝐹̂ = 90 ° (given) ∴ J𝐻 A S (2) 7.1.2 ̂2 = 𝐹̂ = 90° …… ext. ∠ of cyclic quad 𝐾 AA S/R (2) 7.1.3 In ΔHKG and ΔJHG ̂1 = 𝐺 𝐺 ̂1 …… common A S/R ̂2 = 𝐽𝐻 𝐾 ̂ 𝐺 = 90 ° …… proved A S/R ̂ 𝐺 = 𝐻𝐽̂𝐺 …… remaining ∠ ∴ 𝐾𝐻 ∴ ∆𝐻𝐾𝐺///∆𝐽𝐻𝐺 …… (∠∠∠) A R (3) 7.2 JG2 = HJ2 + HG2 …… Pythagoras = 102 + 52 = 125cm2 JG= √125𝑐𝑚2 = 5√5cm A 5√5 𝐾𝐺 𝐻𝐺 = 𝐻𝐺 𝐽𝐺 A ∆𝐻𝐾𝐺///∆𝐽𝐻𝐺 𝐻𝐺 2 𝐾𝐺 = 𝐽𝐺 52 = 5√5 CA substitution 5 = √5 5√5 = 5 = √5 cm CA answer in any form = 2,24 cm (4) [11] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com QUESTION 8 8.1 (cos 𝑥)(cos 𝑥)(− tan 𝑥) A− tan 𝑥 − cos 𝑥 Acos 𝑥 sin 𝑥 sin 𝑥 Acos 𝑥 (cos 𝑥)(cos 𝑥) ( = cos 𝑥) cos 𝑥 Acos 𝑥 = sin 𝑥 (4) 8.2 8.2.1 1 1 1  n2 m 36o 24o 1  m2 𝑛 A A diagrams (A) (B) cos 36o = √1 − 𝑚2 CA √1 − 𝑚2 answer (3) 8.2.2 sin 12 = sin (36 – 24 ) o o o A compound angle = sin 36o cos 24o – cos 36o sin 24o A expansion = m n – (√1 − 𝑚2 )(√1 − 𝑛2 ) CACA answer (4) OR OR 1−𝑐𝑜𝑠24° sin 12o = √ 2 A A CACA answer (4) 1−𝑛 =√ 2 8.3 2 cos 75° cos 15° Acos 75° cos[45° − 𝑥 + 𝑥] 2 sin 15° cos 15° Acos[45°] = Asin 15° cos 45° sin 30° = sin 30° cos 45° A 1 1 cos 45° = ÷ 2 √2 √2 CAanswer in any form = (5) 2 8.4 𝑠𝑖𝑛2 3𝑥 − 2 sin 3𝑥 cos 3𝑥 + 𝑐𝑜𝑠 2 3𝑥 Aexpansion = 1 − sin 2(3𝑥) A1 − sin 2(3𝑥) = 1 − sin 6𝑥 −2 A1 − ( ) −2 5 =1−( ) 5 2 =1 Aanswer in any form (4) 5 [20] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics P2 10 March 2019 Common Test NSC QUESTION 9 9.1 sin 𝑥 + 1 = 1 − 2𝑠𝑖𝑛2 𝑥 Aequating 2𝑠𝑖𝑛2 𝑥 + sin 𝑥 = 0 Astandard form sin 𝑥 (2 sin 𝑥 + 1) = 0 (2) 9.2 sin 𝑥 (2 sin 𝑥 + 1) = 0 Asin 𝑥 = 0 sin 𝑥 = 0 or 2 sin 𝑥 = −1 1 Asin 𝑥 = − 2 1 𝑥 = 0° + 𝑘. 360°; 𝑘 ∈ ℤ sin 𝑥 = − 2 A0° + 𝑘. 360° 𝑥 = 180° + 𝑘. 360°; 𝑘 ∈ ℤ 𝑥 = 210° + 𝑘. 360°; 𝑘 ∈ ℤ A 𝑘 ∈ ℤ 𝑥 = 330° + 𝑘. 360°; 𝑘 ∈ ℤ CA210° + 𝑘. 360° (6) CA330° + 𝑘. 360° 9.3 −2 ≤ 𝑦 ≤ 0 Aanswer (1) 9.4 𝑟 = 2 cos 2𝑥 A𝑟 = 2 cos 2𝑥 −1 < 2 cos 2𝑥 < 1 A−1 < 𝑟 < 1 −1 1 1 CA 2 < cos 2𝑥 < ; If cos 2x = 2 2 −1 1 ∴ 2𝑥 = 60° < cos 2𝑥 < then 𝑥 = 30° 2 2 ∴ 30° < 𝑥 < 90° CA 𝑥 = 30° CAanswer 30° < 𝑥 < 90° (5) [14] TOTAL MARKS: [100] Copyright Reserved Please Turn Over