Mathematics Grade 11 March Test 2020 Paper & Memo

Downloaded from hlayiso.com testpapers.co.za

Mathematics Downloaded from hlayiso.co NSC m 2020 Common Test INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. This question paper consists of 5 questions. Answer ALL the questions. Number the answers correctly according to the numbering system used in this question paper. Clearly show ALL calculations, diagrams, graphs, etc. which you have used in determining your answers. Answers only will NOT necessarily be awarded full marks. ‘You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise. If necessary, round off answers correct to TWO decimal places, unless stated otherwise. Diagrams are NOT necessarily drawn to scale. TWO DIAGRAM SHEETS for QUESTION 4.1, QUESTION 4.2, QUESTION 5.1 and QUESTION 5.2 are attached at the end of this question paper. Detach the DIAGRAM SHEETS and hand in together with your ANSWER BOOK. Write neatly and legibly. Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics March 2020 Common Test NSC QUESTION 1 ll Solve for x: Lid 2s(-4)=0 (2) 2) 1.1.2 4x?+11x=2 (answer correct to TWO decimal places) (4) 1.1.30 x-y8-2x =0 (5) 1.1.4 x(x~7)+12<0 (4) 1.2. Solve simultaneously for x and y: x=3y-5 and (2x-y-2)(x+y)=0 (4) [19] QUESTION 2 2) Simplify WITHOUT the use of a calculator: VI8 +2V2 + z Write your answer with a rational denominator. (3) 2.2 Solve for x without the use of a calculator: 2.2.1 16 =128 (4) geet 321 222 9 24 (4) (1) QUESTION 3 3.1 If cosO= -2 and 0eé [180° : 360°], determine the value of tan@ by using a diagram, and without the use of a calculator. (3) cos115°.cos214° 3.2 Evaluate, without using of a calculator; ——_———_ 5 valuate, without using of a calculator’ 70865° sin 236° (5) 3.3. Simplify to a single term, without the use of a calculator: cos(90° + x).tan(540° + x).cos(180° — x) + sin(-90°) q) [15) Copyright Reserved Please Turn Over

Downloaded from hlayiso.caor NSC Mathematics rch 2020 Common Test GIVE REASONS FOR YOUR STATEMENTS AND CALCULATIONS IN QUESTIONS 4 AND 5. QUESTION 4 4.1 In the diagram, AB is a chord of the circle having centre O. C is a point on AB such that ACO = 90°. OC =5 cm and AO = 10cm. ‘ Calculate, with reasons, the length of AB. (4) 4.2 In the diagram, O is the centre of the larger circle, which passes through P, R and S. R is a point on chord PV of the smaller circle PVSO. SV is produced to T. RVT = 96°. 4.2.1 Calculate, with reasons, the size of R, . (7) 4.2.2 Prove that ARVS is isosceles. (3) [14] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com NSC Mathematics March 2020 Common Test QUESTION 5 5.1 In the figure below, JLH is a tangent to the circle having centre O at L. K andM are points on the circle. K M o° J L H Prove the theorem which states that MLH =K. (5) 5.2 AE is a diameter of the circle in the diagram below. AE is produced to D. DB is a tangent to the circle at B. DC is perpendicular to AD. CBA is a straight line. BE and EC are drawn. Let A=x. E D A x 1A\3 1 2 : 3 1 B 1 Cc 5.2.1 Prove that BCDE is a cyclic quadrilateral. (3) 5.2.2. Prove that c =X. (4) 5.2.3 Prove that E, =£,. (2) 5.2.4 Is CE a tangent to the circle through A, B and E? Give a reason for your answer. (2) [16] TOTAL: 75 Copyright Reserved

Mathematics Downloaded from hlayiso.com March 2020 Common Test NSC NAME & SURNAME: DIAGRAM SHEET 1 QUESTION 4.1 A II. II. a: <C w QUESTION 4.2 I- Copyright Reserved Please Tum Over

Mathematics Down loaded from h layiso . com 2020 Common Test NSC t NAME & SURNAME: DIAGRAM SHEET 2 QUESTION 5.1 TEAR OFF QUESTION 5.2 Copyright Reserved

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 11 MATHEMATICS MARKING GUIDELINE COMMON TEST MARCH 2020 MARKS: 75 This marking guideline consists of 7 pages. GEOMETRY  MEETKUNDE A mark for a correct statement (A statement mark is independent of a reason) S 'n Punt vir 'n korrekte bewering ('n Punt vir 'n bewering is onafhanklik van die rede) A mark for the correct reason (A reason mark may only be awarded if the statement is correct) R 'n Punt vir 'n korrekte rede ('n Punt word slegs vir die rede toegeken as die bewering korrek is) Award a mark if statement AND reason are both correct S/R Ken 'n punt toe as die bewering EN rede beide korrek is Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics-(Grade 11) 2 March 2020 Common Test NSC – Marking Guideline QUESTION 1 1.1.1  1 2 x x    0  2 1 1 x  0 or x   x0  x 2 2 (2) 1.1.2 4 x 2  11x  2 4 x 2  11x  2  0  standard form  b  b 2  4ac x 2a  11  (11) 2  4(4)(2)  substitution into the x correct formula 2(4) x  0,17 or x  2,92  x-value  x-value (4) 1.1.3 x  8  2x  0 x  8  2x  isolate 8  2x x 2  8  2x squaring both sides x 2  2x  8  0 ( x  4)( x  2)  0  factors x  4 or x  2  rejecting x  4  x  2 only  correct solution (5) 1.1.4 x( x  7)  12  0 x 2  7 x  12  0 standard form ( x  3)( x  4)  0  method + + OR 3 4 3 4 correct solution 3  x  4 OR x  (3 ; 4) (4) 1.2 x  3 y  5 ……….(1) (2 x  y  2)( x  y)  0 ………….(2) Substitute (1) into (2): (2(3 y  5)  y  2)(3 y  5  y)  0 correct substitution (6 y  10  y  2)(4 y  5)  0 (5 y  12)(4 y  5)  0  factors 12 5 y or y  values of y 5 4 11 5 x or x   values of x 5 4 (4) [19] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics-(Grade 11) 3 March 2020 Common Test NSC – Marking Guideline QUESTION 2 2.1 1 18  2 2  2 1  for 3 2 = 3 22 2 2 1 2 = 5 2  rationalizing denominator 2 2 2 = 5 2 2 11 2  answer 2 (3) 2.2.1 x 16  128 4 4 2 2x 7  2  27 x 4 7  equating exponents x 4 x answer 7 (4) 2.2.2 32 x 1  32 x 1  24 3x 32 x (31  31 )  factorising  24 3x  2 3 x  2   24  3 simplification 3 9 x 3 x  32 like bases x2  answer (4) [11] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics-(Grade 11) 4 March 2020 Common Test NSC – Marking Guideline QUESTION 3 3.1 y2  r 2  x2  5 2   2 2   correct sketch in correct  21 –2 quadrant y   21 y 5  21 tan    value of y 2 21  answer  (3) 2 3.2 cos115. cos 214 cos 65.sin 236  cos 65.  cos 34    cos 65  cos 65.  sin 56    cos 34  cos 34    sin 56    sin 56  cos 34  cos 56   OR    cos 34  OR sin 56  cos 34  cos 56   1  1 (5) 3.3       cos 90  x . tan 540  x . cos 180  x  sin  90      ( sin x). tan 180  x ( cos x)  sin 90   sin x  tan x  ( sin x)(tan x)( cos x)  1   cos x  sin x   ( sin x) ( cos x)  1  1  cos x  sin x  sin 2 x  1  cos x    1 sin 2 x   sin 2 x  1   cos 2 x  answer (7) [15] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics-(Grade 11) 5 March 2020 Common Test NSC – Marking Guideline QUESTION 4 4.1 AC2  AO2  OC2 [Pythagoras]  S/ R  10 2  5 2  75 AC  75  8,66 cm  value of AC AB  2  AC [line from centre  to chord] R  17,32 cm  answer (4) 4.2 TV̂R  Ô1 [ext.  of cyclic quad.] S R  96  Ô 2  360   Ô1 [  s around a point]  360   96   size of Ô 2  264  SR 1 R̂ 1  Ô 2 [  at centre = 2   at circumf.] 2  size of R̂ 1  1 2 264   132   answer R̂ 2  180   R̂ 1 [  s on a straight line]  180  132  48 (7) 4.3 Ŝ1  RV̂T  R̂ 2 [ext.  of RVS ] S/R  96  48   48  Ŝ1  R̂ 2 [both  48 ] S  VS  VR [sides opp. to =  s] S/R And RVS is isosceles. (3) [14] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics-(Grade 11) 6 March 2020 Common Test NSC – Marking Guideline QUESTION 5 5.1 Construction: Draw diameter LOP and join P and M..  construction P K M O 2 1 J L H Proof:  S /R L̂1  L̂ 2  90  [tangent  radius] PM̂L  90 [  in a semi-circle] P̂  L̂ 2  90  [sum of  s of PML]  S/R  P̂  L̂1 But: P̂  K̂ [  s in the same segment]  S R  L̂1  K̂ (5) 5.2.1 B̂3  90  [  in a semi-circle] S R  B̂3  CD̂E [both  90  ]  BCDE is a cyclic quadrilateral R [converse: ext.  of a cyclic quadrilateral] (3) 5.2.2 B̂ 2  Â = x [tan-chord-theorem] S R B̂ 2  Ĉ1 = x [  s in the same segment] S R  Ĉ1  Â  x (4) Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics-(Grade 11) 7 March 2020 Common Test NSC – Marking Guideline 5.2.3  Ê1  180   Â  B̂3  [sum of  s of ABE ]  180  x  90      90  x   Eˆ1  90   x  Ê 3  180   Ĉ1  CD̂E    180  x  90    [sum of  s of CDE ]  90  x  Eˆ 3  90   x  Ê1  Ê 3 (2) 5.2.4  Ê 2  180   Ê1  Ê 3  [  s on a straight line]   180  90  x  90  x      2x  Ê 2  2 x  Ê 2  Â   Ê 2  Â  CE is not a tangent to the circle through A, B and E. [converse of tan-chord-theorem does not apply] (2) [16] TOTAL: 75 Copyright Reserved Please turn over