Mathematical Literacy Grade 11 P2 2018 Memo

Dewnleadedfromhlayiso.com education Department: Education PROVINCE OF KWAZULU-NATAL NATIONAL SENIOR CERTIFICATE GRADE 11 THE EMATICS P P2 COMMON TEST JUNE 2018 MARKS: — 100 TIME: 2 hours This question paper consists of 6 pages. Please turn over Copyright reserved Sa ee ee =e

Downloaded from hlayiso.com Mathematics Paper 2 June 2018 Common Test GRADE i INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions: 1. This question paper consists of 5 questions. 2. Answer ALL the questions. 3. Number the answers correctly according to the numbering system used in this question paper. 4. Clearly show ALL calculations, diagrams, graphs, et cetera, which you have used in determining the answers. 5. 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. 9. Write neatly and legibly. Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics Paper 2 June 2018 Common Test GRADE 11 QUESTION 1 A(O;w), BO; 3), C(4 ; 0) and D(10 : 8) are the vertices of a quadrilateral in the Cartesian plane. AD is parallel to BC. A(O;1v) D(10;8) B(0:3) 4 C(4:0) “ LL. 1.1.1 Calculate the gradient of CD. (2) 1.1.2. Hence, determine the angle of inclination of CD. (2) 1.2 Prove that BCD = 90°. (3) 1.3 1.3.1 Write down the gradient of AD. (1) 1.3.2 Hence, or otherwise, calculate the value of iv. (3) 1.4 Tfitis given that w= 5a, calculate the length of AD. (3) 1.5 Calculate the area of quadrilateral ABCD. (6) (20] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics Paper 2 June 2018 Common Test GRADE 11 QUESTION 2 A, B(1;2), C(-2;p) and D(S5 ; 0) are the vertices ofa KITE. M is the point of intersection of the diagonals of the kite. ADB=45”. . D(5:0) &. C(-2:p) 2.1 Determine the coordinates of M. (4) 2.2. Calculate the value of p. (6) 2.3. If p=~—9, determine the equation of AC. (5) 2.4 Determine the angle of inclination of AD. (5) 25 Determine the coordinates of A. (6) [26] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics Paper 2 5 June 2018 Common Test GRADE 11 QUESTION 3 - If tan@ = ; and OE [90":360"] , determine the value of 2sin6.cos@ without the we use ofa calculator. (4) ve io Simplify without the use of a calculator: sin (300° - x) + cos (90° + x] 3.2.1 sin (180° — 1) + tan 540° (5) . 1 3.2.2 cos 330°.tan (-120") + sin 73°. | (6) cos 197°, [15] QUESTION 4 (sin y=cos 4.1 The identity = =2tanx is given. sin’ x-1 4.1.1 Prove the identity (5) 4.1.2. For which values of » in the interval 0° s x <360° will the identity in 4.1.1 not be defined? (3) 4.2 Solve for x if tan(3v+40°) =—] and x] -90°;90"] . (5) 43. Determine the general solution of 2sinx = V3+3cosy. (8) [21] Please turn over Copyright Reserved

Mathematics Paper 2 Downloaded from hlayiso.com June 2018 Common Test GRADE 11 QUESTION 5 wn Copyright Reserved wn wa aay wn The €[- i) Nv sketch represents the graphs of 7 (x) = acos(x +b) and g(x) =csinx for 180°; 1 80"| 5 bo -180° 135° ~90° 45° Write down the values of a, b and ce. [fthe points of intersection of f and g are (14.64%) and (i:-0,51), write down the values of k and m7. For which values of x in the interval [-! 80°:0"] will (a) f(x)-g(x) <0? (b) f(x).g(x) =O? Determine the minimum value of A(x) if A(x) =x) + 2. Draw a sketch graph of m(x) = tan 2x for vE[0", 90") . Describe (). the graph m has to be transformed to form the graph 1 where n( x)=tan(2 450"). TOTAL MARKS: (4) (2) [18] 100

Downloaded from _hlavisocom education 5 Department: Education PROVINCE OF KWAZULU-NATAL f MATHEMATICS | “ : COMMON TEST " JUNE 2018 " MARKING GUIDELINE : | - NATIONAL SENIOR CERTIFICATE [ GRADE 11 | MARKS: 100 | This marking guideline consists of 8 pages. | Copyright reserved Please turn over

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Mathematics Paper 2 GRADE 11 Marking Guideline Downloaded from hlayiso.com Common Test June 2018 BC= (0-4) +(3-0) =5 CD = (10—4)’ +(8-0)° =10 Area of ABCD = area of trapezium = (sum of parallel sides) x height OR > BC= (0-4) +(3-0) =5 CD = (10-4) +(8-0)" = 10 Area of ABCD = area of rectangle + area of triangle =(/xb)+(45h) =(5x10)+|4x10x (124-5) | = 874+ square units OR BC=,/(0—4) +(3-0)' =5 > CD = y(10-4)' +(8-0) =10 Area of ABCD = area of ABCD + area of AABD = (£x5«10)+[4x(154-3)x10] =25 + 624 = 874 square units Y length of BC ¥ length of CD Y formula Y substitution of(5+124) ¥ substitution of 10 v answer (6) OR Y length of BC Y length of CD Y substitution of (5x10) ¥¥ [4x 10x(124—5) | v answer (6) OR Y length of BC Y length of CD v (4x5x10) v v [4x(154-3)x10] v answer (6) [18] Copyright Reserved Please turn over

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Mathematics Paper 2 GRADE 11 Marking Guideline Downloaded from hlayiso.com Common Test June 2018 My, = tan 108, 43° figrutomat Ade y=-3xte Substitute (5;0): 0=-3(5)+e e=15 y=-3x+15 Solve simultaneous equations for AC and AD: 2xn-5=-3x415 | ¥ gradient of AD Y substituting (5 ; 0) Y equation of AD Y solving simultaneously 5x =20 e4 Y value of x y=3 v value of y A(4;3) (6) OR OR DM = AM = BM [sides opposite equal angles] ¥ DM=AM=BM= V5 AB=AD = J10 [Theorem of Pythagoras] AB = AD W(e-1) +(v-2)' = (x5) +(y-0) ¥ yx) +(y-2y xv -2xtlty?-4y44=x?-10x42545° = (x-5) +(y-0) 8x—4y = 20 ye QN—Soecccccee line 1 Y simplification to y=2x-5 AD=¥10 (e-3) 0-97 =Vi0 ¥ Io-5) +0) = si Xx? -10K4 25497 =10 ee line 2 x -10x+25+(2x-5)' =10 x =10x +2544" —20x+25=10 5x° ~30x+40=0 x -6x+8=0 (x-4)(x-2)=0 x=2orx=4 vx=4 N/A y=3 vy=3 (6) [26] Please turn over Copyright Reserved

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Mathematics Paper 2 QUESTION 4 GRADE ILL Marking Guideline Downloaded from hlayiso.com Common Test June 2018 tan (3x-+40°)=-1 reference angle: 45° 3x+40° =180° -45° +7.360° or 3x+40° = 360° —45° + 1.360° 3x = 275° +1.360° x =91,67 +7.120° x =—28, 33° 3x = 95° +.360° x = 31,67" +7.120° x =—88,33° or 31,67° where n <Z OR 3x+40° = 180° —45° +.180° 3x = 95° +n.180° x = 31,67 +7.60° x = —88,33° or -28,33° or 31,67° where n € Z 4.1.1 (sin. —cos sin” x—-1 _ sin’ x~2sinxcos.x+ cos’ x1 Y multiplying out sin’ x-1 _ —2sin xeosx+sin’ x+cos’ x1 : sit xl ¥ applying identity —2sinxcosx+1-1 i 2 = es sin’ x+cos’ x =1 -(1-sin? x) v ~(1=sin’ x) _ 72sin.xcos x Y applying identity = cos" x 1-sin’ x=cos’ x 2sin x ~ cosx ¥ simplification =2tanx - __ (5) 4.1.2 | sin?x—-1=0 Y sin’ x-1=0 sin? x=1 siny=-l or sinx=l x=270° or aco? ¥ 270° v 90° (3) 4.2 ¥ 3x+40° =180° —45° +1.360° v 3x+40° = 360° —45° +.360° v -88,33° ¥ 31,67 v¥ -28,33° (5) OR v 180° —45° ¥ +n.180° v¥ -88,33° v¥ 31,67 v¥ 28,33 (5) | Copyright Reserved Please turn over

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