Mathematics Grade 11 April Test 2021 Paper & Memo

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Downloaded from hlayiso.com NSC-Grade 11 INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions . es 10. Copyright Reserved 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 4.3 AND QUESTION 5 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. Please turn over

Downloaded from hlayiso.cgm Mathematics ommon Test April 2021 NSC-Grade 11 QUESTION 1 il Solve for x: T11 (2x+3\6—x)=0 (2) 1.1.2 5x? +x-—7=0 (correct to two decimals) QB) 113 x74+8x+15>0 (3) 1 Solve for x and y: 4x—y=3 and y’—2xy+1=0 (6) 13° Given: kx*-5x-1=0, with k #0. 1.3.1 For which value(s) of k will kx? —5x—1=0 have real roots? (3) 1.3.2. Determine two integral values of k for which the roots of kx” —5x-1=0 will be rational. (2) [19] QUESTION 2 2.1 Without using a calculator, rationalise the denominator and simplify: v3 —V2 V3 +2 Q) 22 Solve for x: x-2Vx-8=0 (5) 23 Simplify: 1 3x 412.377 fe 7.9% (4) [12] Copyright Reserved Please turn over

Downloaded from hlayiso.com Common Test April 2021 Mathematics QUESTION 3 371 Simplify, without the use of a calculator: sin{- 20° cos 430° (4) 3.2 Without “nl a calculator, simplify to a single trigonometric ratio: i 1+sin x. tan{i80° — x).cos(~ x) (5) 3.3 If 3tan@+1=0 and cos@ > 0, calculate without the use of a calculator, and with the aid of a diagram, the value of: 30sin? 6+/40 cos 0. (6) [15] GIVE REASONS FOR YOUR STATEMENTS AND CALCULATIONS IN QUESTIONS 4 and 5. QUESTION 4 41 In the diagram, O is the centre of the circle. A, B and D are points on the circle. OA, OB, OD and AB are drawn. AB and OD intersect at E. AE=EB=6cm. DE=2 cm. If the radius of the circle is x cm, calculate the numerical value of x. A Nye 1) Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics Common Test April 2021 NSC-Grade 11 4.2 In the diagram, O is the centre of the circle. F, G and H are points on the circle. FH, FO and GO are drawn. Prove the theorem which states that FOG =2FHG. (5) 4.3 Qis the centre of circle ABCDE. BE || CD and O, = 130°. 4.3.1 Determine, with reasons, the size of the following angles: (a) B (2) (b) A (2) (c) D (2) 43.2 Prove that B; =B3. (4) [22} i Copyright Reserved Please turn over

Downloaded from hlayiso.com Common Test Anril 7071 Mathematics wtoururaue ot QUESTION 5 In the diagram, M is the centre of the circle and diameter KL is produced to N. MP is drawn perpendicular to KN such that PRN is a tangent to the circle at R. MP and chord KR intersect at S. MR and LR are drawn. Let R 4=Xx. M K N [3N af 2 Uo Sp \3 y R P 5.1 Write down with reasons two other angles each equal to x. @) 5.2 Prove that KM is a tangent at M to the circle passing through M, P and R. (4) 7] TOTAL: 75 Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics Common Test April 2021 NSC-Grade 11 NAME & SURNAME: supeuneeeeusnseneunenane: : DIAGRAM SHEET 1 QUESTION 4.1 SY TEAR OFF i QUESTION 4.2 Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics Common Test Anril 2091 NAME & SURNAME: DIAGRAM SHEET 2 QUESTION 4.3 A B (\ QUESTION 5 TEAR OFF Copyright Reserved

‘Downloaded from hlayiso.com MATAS Ge wh 4 )0° ——=————— ——— Mathematics 2 Common Test April 202! Grade 1 1-Marking Guideline KWAZULU-NATAL PROVINCE | EDUCATION . a GEOMETRY * MEETKUNDE REPUBLIC OF SOUTH AFRICA A mark for a correct statement (A statement mark is independent of a reason) s ; " ‘n Punt vir'n korrekte bewering NATIONAL ('n Punt vir 'n bewering is onafhanklik van die rede) SENIOR CERTIFI A mark for the correct reason : (A reason mark may only be awarded if the statement is correct) '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 sm peace ccsscssssseeeces : : ; 2 Ken 'n punt toe as die bewering EN rede beide korrek is " MATHEMATICS " a COMMON TEST a QUESTION 1 APRIL 2021 a ; 7 : = ye-> or x=6 V answer “ answer MARKING GUIDELINE é @ eee eErreerr Preceerrrer eerTEee 4 ee —b+ Vb? ~ 4ac 2a : a eal eyF -46)C7) ¥ substitution 2(5) MARKS: 75 _ nie vial es =1,09 or -1.29 v answer ¥ answer i : : (3) ° T13 | x? 48r415>0 (x+5Xx+3)>0 This marking guideline consists of 9 pages. Y critical values v ¥ answer Q) Please turn over Please tum over Copyright Reserved Copyright Reserved _————

Downloaded from hlayiso.com Mathematics Common Test April 2021 Mathematics Common Test April 2021 Grade I I-Marking Guideline Grade I I-Marking Guideline 1.2 y=4x-3 ✓ makingy the subject of QUESTION 2 r- ✓3 -✓2 Substitute in y1 -2.:\J' +I = 0: the formula 2.1 {4x-3)2 -2x(4x-3)+1 =0 ✓ substitution ✓3 +✓2 ✓2) x (../3 +✓2 ) 3✓ -✓2J 16x 2 -2 4x+9-8x2 +6x+I =0 JJi- 2 ✓3 -✓ r-Jzl 8x2 -18x+l0=0 ✓ x = 3-2.Ji✓J +2 ✓ standard fonn ✓ 4x2 -9x+5 =0 2 (4x-5Xx-1)=0 ✓ factorisation ✓ multiplying out =5-2✓6 3-2 denominator correctly x=1or x=I ✓ values of x ✓ answer (3) y=2 or r= I ✓values of y 2.2 x-2-h-8 = 0 (6) 1.3.1 h 2 -4ac�0 ✓ condition for real roots x-2x2 -8=0 ✓ exponential form (-5} -4k(-1)�0 2 ✓ substitution () -4)() +2)=0 ✓ factorisation 25+4k � O 25 I I k�- x2 =4 or x=i =-'.! .!. I ✓ x2 =4 or =-2 4 x=l6 no solution _y 2 ✓ answer: ✓ x=l6 k�-6.!_ 25 I k� - ✓ no solution 4 or k�-6 _!_ OR (5) 4 4 OR (3) 1.3.2 E.g.: k=-4:-6:6:14 ... x-2-h-8=0 ✓ isolate surd ✓ any correct value of k ✓ another correct value of k 1-h =x-8 ✓ squaring both sides Any two values of k that will result in b 2 -4ac 4x=x! -16x+64 ✓ standard form being a perfect square. (2) x 1 -20x+64=0 1191 (x-16Xx-4)=0 x=16 or x=4 ✓ x=l6 or x=4 NIA ✓ rejecting x=4 (5) Cop:,,·right Reserved Please tum over Copyright Reserved Please tum over

- Downloaded from hlayiso.com Mathematics s Common Test April 2021 Mathematics 6 Common Test April 2021 Grade 11-Marking Guideline Grade 11-Marking Guideline 23 i QUESTION 3 371 412.381 )e 195 4 : tu 3.1 _/3*B+12.37)]* Y factorising numerator ee en Y ~sin20” A Y cos 70" _[ 73°} 73s Y co-function i Y answer =b-F ¥ simplification (4) =3! OR 1 *3 v answer (4) ¥ -sin20° OR OR ¥ cos 70° Y¥ co-function ¥ answer (4) ¥ factorising numerator 3.2 T+sin x. tan{{80° —x).cos(- x) aoe = I+sinx.—tanx.cosx v—tanx ¥ cosx i sinx =1-sinx 2 cosx Ce i COs x cos 32) Y I-sin? x = - Y answer 73 . (5) | = Y simplification eu Y answer ° (4) [2] Copynght Reserved Please turn over Copyright Reserved Please tum over

Mathematics fe Grade 11-Marking Guideline Common Test April 2021 3.3 tang=—4 cal. vio reJray = Ae +P =4i0 30sin? 6+ /40 cos 8 =3+6 Y standard form ¥ 4° quadrant v value of r Y substitution Y simplification Y answer (6) {15} Copyright Reserved Please tun over - Downloaded from hlayiso.com Mathematics g Common Test April 2021 Grade 1 |-Marking Guideline QUESTION 4 41 | OELAB [line from centre to midpoint of YSYR chord] ¥ OE=x-2 OE=x-2 YR OA? =OE? +AE* [Pythagoras] xax?-4x44436 ¥ substitution ¥ simplification Y answer : (ea) 42 | Contruction: Draw line HOJ Y construction H iG F [radii Zs opp. = sides] v SIR ext. Z of AHFO] v SIR [radii: Zs opp. = sides] [ext. Z of AHGO] “Ss vs (5) 4.3.1 [Z atcentre= 2x Z at circumference} |v S YR (a) (2) 43.1 | A=65° [ Zs in the same segment] YSYR (b) (2) OR OR A=65° [ext. Z of AAOE: Zs opp.=radii] |VS VR (2) 43.1 | D=1is® [opp. Zs of acyclic quadrilateral] YSVYR (c) (2) Copyright Reserved Please turn over

Mathematics 9 Grade 11-Marking Guideline Common Test April 2021 ~~ Downloaded from hlayiso.com Copyright Reserved 4.32 YSVYR [4s opp. = radii] ae (4) | [22) {tan-chord-theorem] YSYR [radii Zs opp. = sides] vsiR @) 52 [radius L tangent) YSYR My [both = 90°} “S -. KM isa tangent at M to the circle through M. P and R [converse: tan-chord-theorem] VR (4) OR OR MRP =90° [radius L tangent) YSYR M, =2x [ext Z of AMKR] My =180°-(0°+2x) [Zs ona straight line] =90" -2x 6 =180" - (90° +90" - 2x) [sum of Zs of AMPR] vs =2y M, = ©. KM isa tangent at M to the circle through M. P and R YR (i: [converse: tan-chord-theorem] (4) = 17) TOTAL MARKS: 75