Mathematics Grade 11 March Test 2019 Paper & Memo

Downloaded from hlayiso.com testpapers.co.za NATIONAL SENIOR CERTIFICATE GRADE 11 MATHEMATICS COMMON TEST MARCH 2019 MARKS: 75 TIME: 1½ hours This question paper consists of 6 pages and 2 Diagram Sheets.

Downloaded from hlayiso.com Mathematics 2 Common Test March 2019 GRADE 11-NSC 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. Clearly show ALL calculations, diagrams, graphs, etc. which you have used in determining your answers. 4. Answers only will NOT necessarily be awarded full marks. 5. You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise. 6. If necessary, round off answers correct to TWO decimal places, unless stated otherwise. Diagrams are NOT necessarily drawn to scale. 7. Write neatly and legibly. 8. Diagrams for QUESTION 4.1, QUESTION 4.2, QUESTION 5.1 and QUESTION 5.2 9. on the DIAGRAM SHEETS provided. Detach the DIAGRAM SHEETS and hand in together with your Answer Book. Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics 3 March 2019 GRADE 11-NSC QUESTION 1 1.1 Solve for x: 1.1.1 7𝑥 2 − 2𝑥 − 3 = 0 (correct to TWO decimal places) (3) 1.1.2 (𝑥 − 2)2 − 4 = 0 (3) 1.1.3 √7𝑥 + 2 + 2x = 0 (4) 1.1.4 𝑥 2 − 𝑥 − 56 < 0 (3) 1.2 Solve for x and y simultaneously: 2 x + y = 1 and 2 x 2 − xy + y 2 = 4 (6) [19] QUESTION 2 3 2.1 Solve for x without the use of a calculator : x 4 = 64 (2) 2.2 Simplify without the use of a calculator : 5− x.1251− x.252 x 2.2.1 (3) 5 2.2.2 √12 − √147 + 31,5 (3) 52006 − 52004 + 24 2.3 If = a , calculate 𝑎 without the use of a calculator. (3) 52004 + 1 [11] QUESTION 3 ANSWER QUESTION 3 WITHOUT USING A CALCULATOR. 9 3.1 Given: tan  = − and 180 <  < 360 . 40 Use a sketch to determine the value of sin  + cos  . (4) 3.2 Simplify fully: sin ( 90 −  ) .tan ( 360 −  ) .sin ( − 180 ) 1 − cos 2  (6) Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics 4 Common Test March 2019 GRADE 11-NSC Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics 5 Common Test March 2019 GRADE 11-NSC 3.3 Determine the value of the following in terms of p, if cos32 = p : 3.3.1 cos 212 (2) 3.3.2 sin ( −328) (3) [15] GIVE REASONS FOR YOUR STATEMENTS AND CALCULATIONS IN QUESTIONS 4 AND 5. QUESTION 4 4.1 In the diagram, O is the centre of the circle and R is a point on chord ST, such that OR is perpendicular to ST. Prove the theorem which states that SR = RT. (5) Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics 6 March 2019 GRADE 11-NSC 4.2 In the diagram, O is the centre of the circle and D is a point on chord AC such that AD = DC = 24 mm. OD is drawn and produced to meet the circle at B. OD = 7 mm. OA and OC are drawn. Calculate the length of BD. (5) [10] QUESTION 5 5.1 In the diagram two circles intersect at L and H. O is the centre of the circle passing through F, G, H and L. GO and LO are drawn. LHJK is a cyclic quadrilateral. FLK and GHJ are straight lines. GOLˆ = 136 5.1.1 Calculate the size of F̂ . (2) Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics 7 Common Test March 2019 GRADE 11-NSC 5.1.2 Calculate the size of K̂ . (4) Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics 8 Common Test March 2019 GRADE 11-NSC 5.2 In the diagram, O is the centre of the circle. Diameter BOA is produced to E such that EDG is a tangent to the circle at D. C is a point on the circle such that BA || CD. AD, BD and AC are drawn. F is a point of intersection of AC and BD. Let B̂ = x . 5.2.1 Write down, with reasons, four other angles each equal to x. (6) 5.2.2 Determine the size of Ê in terms of x. (4) 5.2.3 Prove that CA is a tangent to the circle passing through A, D and E. (4) [20] TOTAL: 75 Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics 9 Common Test March 2019 GRADE 11-NSC NAME & SURNAME: DIAGRAM SHEET 1 QUESTION 4.1 TEAR OFF QUESTION 4.2 Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics 10 Common Test March 2019 GRADE 11-NSC NAME & SURNAME: DIAGRAM SHEET 2 QUESTION 5.1 TEAR OFF QUESTION 5.2 Copyright Reserved Please turn over

Downloaded from hlayiso.com MATHEMATICS COMMON TEST MARCH 2019 MARKING GUIDELINES NATIONAL SENIOR CERTIFICATE GRADE 11 MARKS: 75 These marking guideline consists of 8 pages.

Downloaded from hlayiso.com Mathematics 2 Common Test March 2019 GRADE 11-Marking Guideline 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 QUESTION 1 1.1.1 7𝑥 2 − 2𝑥 − 3 = 0 −𝑏 ± √𝑏2 − 4𝑎𝑐 𝑥= 2𝑎  substituting in correct −(−2) ± √(−2)2 − 4(7)(−3) 𝑥= formula 2(7)  x-values 𝑥 = −0,53 or 𝑥 = 0,81  x-values (3) 1.1.2 (𝑥 − 2)2 − 4 = 0 (𝑥 − 2)2 = 4  isolate (𝑥 − 2)2 𝑥 − 2 = ±2  ±2 𝑥 = 4 or 𝑥 = 0  both answers OR OR (𝑥 − 2)2 − 4 = 0 𝑥 2 − 4𝑥 + 4 − 4 = 0  𝑥 2 − 4𝑥 + 4 𝑥 2 − 4𝑥 = 0 𝑥 ( 𝑥 − 4) = 0  factors 𝑥 = 4 or 𝑥 = 0  both answers (3) 1.1.3 √7𝑥 + 2 + 2x = 0 (√7𝑥 + 2)2 = (−2𝑥)2  isolate √7𝑥 + 2 7𝑥 + 2 = 4𝑥 2 4𝑥 2 − 7𝑥 − 2 = 0 standard form (4𝑥 + 1)(𝑥 − 2) = 0  factors 1 𝑥=− or 𝑥 = 2 4 1 ∴ 𝑥 = − 4 only  correct solution (4) Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics 3 Common Test March 2019 GRADE 11-Marking Guideline 1.1.4 𝑥 2 − 𝑥 − 56 < 0 (𝑥 − 8)(𝑥 + 7) < 0 correct factors CV 𝑥 = 8 𝑜𝑟 𝑥 = −7 + − + −7 8 −7 < 𝑥 < 8 correct solution (3) 1.2 2 x + y = 1 and 2 x − xy + y = 4 2 2 y = 1 − 2x  y = 1 − 2x 2 x 2 − x(1 − 2 x) + (1 − 2 x) 2 = 4 substitution 2 x2 − x + 2 x2 + 1 − 4 x + 4 x2 = 4 8x2 − 5x − 3 = 0  standard form (8x + 3)( x − 1) = 0  factors 3 x=− or x = 1  x values 8  3 y = 1 − 2  −  or y = 1 − 2 (1)  8 3 y =1 or y = −1 y values 4 (6) OR OR 2 x + y = 1 and 2 x 2 − xy + y 2 = 4 1− y 1− y x=  x= 2 2  1− y   1− y  2 2 substitution 2  − y + y = 4  2   2   1− 2 y + y2   1− y  2 2 − y + y −4 = 0  4   2  1− 2 y + y2 − y + y2 + 2 y2 − 8 = 0 4 y2 − 3 y − 7 = 0  standard form (4 y − 7)( y + 1) = 0  factors 3 y =1 or y = −1  y values 4 3 1 −1 x= 4 or x = 1 − (−1) 2 2 3 x values x=− or x = 1 8 (6) [19] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics 4 Common Test March 2019 GRADE 11-Marking Guideline QUESTION 2 2.1 3 x 4 = 64 4 4  34  3 4  raising both sides to the   x = (2 6 3 ) 3   x = 256 or 28  answer (2) 2.2.1 −x 1− x 2x 5 .125 .25 5 5− x. ( 53 ) . ( 52 ) 1− x 2x = rewriting as base 3 5 − x 3 −3 x 4 x 5 .5 .5 = 5 =5 − x +3−3 x + 4 x −1 using exponential rules = 52 =25  answer (3) 2.2.2 1,5 √12 − √147 + 3 3 = √4 × 3 − √49 × 3 + 32  simplifying surds = 2√3 − 7√3 + √9 × 3 = 2√3 − 7√3 + 3√3  3√3 = – 2√3  answer (3) 2.3 5 − 5 + 24 2006 2004 =a 52004 + 1 52004 (52 − 1) + 24 factorising =a 52004 + 1 52004 (24) + 24 =a 52004 + 1 24(52004 + 1) factorising =a 52004 + 1 a = 24 answer (3) [11] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics 5 Common Test March 2019 GRADE 11-Marking Guideline QUESTION 3 3.1 y x  correct sketch in 4th quadrant (40 ; – 9) x2 + y2 = r 2 40 2 + (− 9 ) = r 2 2 r = 41  value of r sin  + cos −9 40 = +  substitution 41 41 31 =  answer 41 (4) 3.2 ( ) ( ) ( sin 90 −  . tan 360 −  . sin  − 180    ) 1 − cos 2   cos  cos  . − tan  . − sin   − tan  =  − sin  sin 2  sin   sin 2  cos  . − . − sin  = cos  sin  sin 2   cos  =1  answer (6) 3.3.1 cos 212 = cos (180 + 32) = − cos32  − cos32 = −p  answer (2) 3.3.2 sin ( −328) = sin 32  sin 32 1 = 1 − cos2 32  correct sketch or identity p = 1 − p2  answer (3) [15] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics 6 Common Test March 2019 GRADE 11-Marking Guideline QUESTION 4 4.1 Construction: Draw OS and OT.  construction Proof: In Δ OSR and ΔOTR : 1. OS = OT [radii]  S/R 2. OR = OR [common]  S (OR is common) ˆ ˆ 3. SRO = TRO = 90 [ s on a straight line]  S/R  Δ OSR  ΔOTR [ 90 ; H; S ]  S/R  SR = RT [  s ] (5) 4.2 OD ⊥ AC [line from centre to midpoint of chord]  S/R OA = AD + OD [Pythagoras] 2 2 2  S/R = 24 + 7 2 2 = 625 OA = 25 mm  length of the radius OB = OA [radii]  S/R  BD = 25 − 7 = 18mm  answer (5) [10] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics 7 Common Test March 2019 GRADE 11-Marking Guideline QUESTION 5 1 ˆ 5.1.1 F̂ = GOL [  at centre = 2   at circumference] R 2 = 68  answer (2) 5.1.2 ˆ +H F ˆ =180 [opp.  s of cyclic quadrilateral] R 1 Ĥ1 =180 − 68 =112  size of Ĥ1 Kˆ = H ˆ 1 [ext.  of cyclic quadrilateral] R =112  answer (4) OR OR ˆ = Fˆ = 68 H [ext.  of cyclic quadrilateral] S R 2 K̂ = 112 [opp.  s of cyclic quadrilateral] S R (4) 5.2.1 Ĉ = x [  s in the same segment] S R Â1 = Ĉ = x [alt.  s; BA || CD]  S/R D̂ 2 = Â1 = x [  s in the same segment]  S/R ˆ =B D ˆ =x [tan-chord theorem] S R 4 (6) OR OR D̂ 2 = x [alt.  s; BA || CD]  S/R Â1 = D̂ 2 = x [  s in the same segment] S R Ĉ = B̂ = x [  s in the same segment]  S/R D4 = ˆB = x ˆ [tan-chord theorem] S R (6) 5.2.2 D̂3 = 90 [  in a semicircle] S R Eˆ = 180 − B ( ˆ + BDE ˆ ) [sum of  s in  ] S = 180 − ( x +90 + x ) = 90 − 2x  answer (4) OR OR D̂3 = 90 [  in a semicircle] S R Ê + CD̂E = 180  [co-interior  s; BA || CD] S Ê = 180 − CD̂E  = 180  − (x + 90  + x ) = 90 − 2 x  answer (4) Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics 8 Common Test March 2019 GRADE 11-Marking Guideline 5.2.3 ˆ = 180 − B A 2 ( ˆ +D ˆ +A 3 ˆ 1 ) [sum of  s in  ABD] S = 180 − ( x +90 + x ) = 90 − 2x  Â2 = 90 − 2x ˆ A2 = ˆE [both = 90 − 2x ] ˆ = Eˆ A 2  AE is a tangent to the circle through A, D and E [converse: tan-chord-theorem] R (4) OR OR D1 = 180 − ˆ ˆ ( D2 + ˆ D3 + ˆ D4 ) [ s on a straight line] S = 180 − ( x +90 + x ) = 90 − 2x ˆ D̂1 = A [tan-chord-theorem] 2  Â 2 = 90  − x  Â2 = 90 − 2x ˆ = Eˆ A [both = 90 − 2x ] ˆ = Eˆ 2 A 2  AE is a tangent to the circle through A, D and E [converse: tan-chord-theorem] R (4) [20] TOTAL MARKS: 75 Copyright Reserved Please turn over