N2 MATHEMATICS SPOT ON R80

higher education iT = & training | {== Department: = Higher Education and Training REPUBLIC OF SOUTH AFRICA T1382(E)(N19)T NOVEMBER 2010 NATIONAL CERTIFICATE MATHEMATICS N2 (16030192) 19 November (X-Paper) 09:00 — 12:00 REQUIREMENTS: Graph paper. This question paper consists of 7 pages and a 2-page information shéet, - Downloaded from hlayiso.com

DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE MATHEMATICS N2 TIME; 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2. Read ALL the questions carefully. 3. Number the answers correctly according to the numbering system used in this question paper. 4, Write neatly ard legibly. QUESTION 1 Ll Various options are given as possiblé answers to the following questions. Write only the letter (A ~ D) next to the question number (1.1.1 — 1.1.10) in the ANSWER BOOK. tid How can % be expressed in terms of base 27 A 2? B 23 Cc 2? D 2! ) 1.1.2 The /og/00 is the same as: A 20 B 10 c i D 2 dd) Downloaded from hlayiso.como

115 1.1.7 1.1.8 ‘The trig-function, tan@ is positive in... quadrants, first and fourth second and fixst first and third all four Dom> If you factorise (a? —(a-by ) the answer will be... A (b-a)(a+d). B (a-(a-b) (a+ (a-0)). c (a?+a)(a? -2). D (a? +5? }(a? - 27). How many deprees are in a radian? A 57,3° B 60,5° C 67,5° Dp 43,6° (2") is the same as ... A ate Bab Cc 4 D -3— ais the same as ... I A= a np oo ax c 4 a p # x V8 can be expressed as ... A wW3. B 22. Cc 32. i D 3/3. Downloaded from hlayiso.com,,,, () Q) @ (1) (1) @)

1.2 L19 How can 2* = 3 be expressed in terms of a logaxithm? A 3=)og,2 x=log, 3 3=log, x 2=log,x vaw 1.4.10 Express log, 2+ log, 14 as a single logarithm: > log 24 Blog, 7 Clog, 28 D_ log, 16 Indicate whether the following statements are TRUE or FALSE, Write only ‘true’ or ‘false’ next to the question number (1.2.1 — 1.2.10) in the ANSWER BOOK. 1.2.1 log, x = y can be expressed ag x=5" , 1.2.2 A #i, 1.2.3 If 8 =1; thenis x=0, 1.2.4 The log, 0,25=-2, 1.2.5 In the figure below, the cosec@ ==. y | x 1.2.6 The expression «fx? ty is not équal to x+y. 127 Mi2sx® =25x, 12.4 There is 0,143 revolutions in 51°22', Downloaded from hlayiso.conro (1) Q) qd) (1) (1) ) (1) (1) @) (1)

1.2.9 tan @ cosec@ is not equal to sec@. 1.2.10 The log, a can be written as ee QUESTION 2 21 Factorise: 2.14 5x? +14x—3 2.1.2 a‘ 81 2.1.3 28x2 —Q1lyz ~ 2O0xy +15y? 2:2 Determine the highest common factor (HICF) oft 12(x? ~4) 30(x? 4x44) 6(x? --x-6) ° 2.3 Simplify: 2.3.1 jog, 2+ log, 10 + log, 3-log, 40 242 a? =2a-8 | 4a +12 2a +6 a4 QUESTION 3 3.1 Solve forx by using the quadratic formulae: 2x? —3x~-4=0 3.2 Solve for x in the following equations: 3.2.1 3.2.2 3.2.3 log, x=4 log, 9=2 43° =36 Downloaded from hlayiso.com;> @) Q) ) @) @) | (4) [20] 4) (2) ® ®

3.3 3.4 3.5 Use natural logarithms to determine the value of: 4[4357,45 Make 'a! the subject of the formulae: 2 Ha fash? Solve for 'x' and 'y' in the following simultancous equation: 3x~-4y=3 and 2x+5y=25 QUESTION 4 41 42 43 4.4 If the radius of a wheel is 20 mm and it turns at 32 revolutions per second, then calculate: : . 41 Angular velocity 4.12 Peripheral velocity The circumference of a circle is 155 em, Calculate the radius as well as the arc jength of a sector which is 30°. A metal structuie forms the pirinacle of a church tower built of face bricks. The brick section is 43,42 m high. The angle of elevation frorti a point on the ground to the top of the pinnacle is 40°15' and to the bottom of the pinnacle, the angle of elevation is 36°10, Calculate the height of the pinnacle. If sin@ = : and 90° < @ 180° determine the following: , 4.4.1 tan@ 442 secd Downloaded from hlayiso.com, 3) (4) 0} (2) Q) 6) (6) @) @ [20]

QUESTION 5 5.1 3.2 5.3 54 ysx* +2x-3, determine the following: 5.A1 The roots 5.1.2 The y-intercept 5.1.3 The symmetry-axis 5.1.4 ‘The turning point Draw on graph paper, and on the same set of axis, the graphs of pox? +2x-3 yerx-l. port ee a fe ~2 =3 Use the graph to determine the solutions for the following equations: yar $2x3 and y=x-1 Sketch, on graph paper, the graph of: y=sinx—0,5 for 0° $x 5180" Use the graph in QUESTION 5.3, to determine the value of x for which: 5.4.1 sitix -0,5 =0 5.4.2, sinxy=1 TOTAL: Downloaded from hlayiso.com @ @) Q) a) @) (3) Q) @) 2] 100

MATHEMATICS N2 INFORMATION SHEET This sheet must accompany the question paper. The right cone Volume = “iu Surface area = ar fh? +r? tnt? = wre +a? The right pyramid ; Volume = % (area of base) x (perpendicular height) The prism Volume = (area of base) x (perpendicular height) The cylinder Volume = n77h Surface area = 2nr? + 2arh 'Thé sphere v= Sa A= Ain? Degrees and radians 180° = at rad c Sector: ¢ = & radius ~;d= yo Auigular velocity and cireuniferential velocity Angular velocity: w = 2nn Circurnferential velocity: y= Dn n=rotation frequency (t/s = revolution per second) Mid-ordinate rule Area = (distance between ordinates) x (sum of mid-ordiniates) [A= ordinate + Last ordinate 2 Multiply by the distance between the ordinates. +Sum of other orsaates Graphs Straight line: y= mx + ¢ Parabola: y= ax’ + dx+c Axis of symmetry: x = = 2a ~ bt fb? ~ aac Roots: x = aa 2a Downloaded from hlayisq;g¢om

Trigonometry 90° < 6 < 180° sin. 0 = sin (180° ~ @) cos 0 = -cos (180° - 6 ) tan @ = -tan (180° - ¢) Segments of circles Chord length =x Height of segnient = Diameter of circle = D 2 x D=h+ + 4h Regular polygons Angle subtended at centre of circumscribed circle by one side: ____360° muumber of sides R= radius of circumscribed circle x= length of side x= 2R sin £ 2 Annulus: A = {e? -r?) Downloaded from hlayiso.com

education Department: Education REPUBLIC OF SOUTH AFRICA November 2010 MATHEMATICS N2 16030192 November 2010 WISKUNDE N2 Downloaded from hlayiso.com

DEPARTMENT OF HIGHER EDUCATION AND TRAINING T1550 (A) (AG) T NOVEMBER 2010 S2 education NATIONAL CERTIFICATE Lia Sonar MATHEMATICS N2 (16030192) NOVEMBER (%-PAPER) O9h00 : 12h00 This question paper consists of 05 pages and __ page (s) of formula sheet. Downloaded from hlayiso.com

Question 1 1.11. 1.1.2, 1.1.3. 1.1.4. 1.1.5, 1.1.6, 11.7. C 1.1.8. B 1.1.9. B 1.1.10. C QAQr Fa p> 1.2. 1.2.1. True 1.2.2. False 1.2.3. True 1.2.4. True 1.2.5, False 1.2.6. False 1.2.7. False 1.2.8. True 1.2.9, False 1.2.10.True Question 2 2.1.1. (Sx— 1) (« +3) 21.2.(2-3) (434 =(a- 3) (a+ 3) (a +9) Downloaded from hlayiso.com [10} {10} [20] [2] 3]

2.1.3. 4x (72 - 5y) - 3y (72~ Sy) =(71z - Sy) (4x - 3y} ee 2.2. Determine the highest common factor (HCF) of: 2.6 (x - 2) (x +2) 5.6 (x ~ 2) (x - 2) 6 (k+2)(x- 3) HEC =6 2.3. Simplify: 4x40 2.3.1. log 3 10 Log =~ +1 —_, (at2Ka-4) 4(a+3) 2.3.2. 2€a+3) (a-2)(a+2} 2(a—4) a-2 Question 3 3.1.1. xexnb+ vh? — 4ac 2a _ O32 VOR IOCH 2(2) _ 3449-32 - 4, 34v4i g—Vaa & 4 2,351 -0,851 Downloaded from hlayiso. {4] [4] [3] 14] 14) com

3.2.1. 3.2.2. 3.2.3. x=2" x= x=3 3*= 36 = 3% = 32 X=2 3.3, log (4357,45)" = 710g4357,45 3.5. substitute x in I 3 CS) ~4y=3 Downloaded from hlayiso.com [2] [2] [2] [3] GB)

4.1.1. 4.1.2. 4.2. 75 —1Sy -4y=3 75 —l5y - By = 6 —2By =6-75 substitute yin 3 = 25758) 2 x=5 (4] [20] Question 4 Given : 20mm N = 32REV/per second V=500N V= 3,142 x 40 x 32 4021,76 t/s 2] W = 2nN = 20n32 = 201,062 r/s 2] Given : Circumference = 15,5 cm @ = 30° = 0, 524 r/s Downloaded from hlayiso.com

a) Circumference = x D circumference D = renee cid _ 45, ti cid D= 4,933 ent = 2,467 om [2] b) a=" i=6 xr = 0,524 x 2,467 4.3. 43.42 / / fi / fi fl hi i i f i i f Wr ne nnn dn _ 31 B 5D _ 43,42 a) tan 36,167 =— a 43,42 BC =—*242_ tan 36,167 = 59,398m b) tan 40°15’= = AB =BC, tan 40°15’ = 59,398 x 0,847 = 50, 77m Downloaded from hlayiso.com

AD = AB -BD AD = 50,777 - 43,42 =7,357m 4.4.1, Question 5 514, x2+2x—3=0 (&+3)«-)D=0 x=3o0rx=1 5.1.2. y-intercept: let x = 0 y= -3 Downloaded from hlayiso.com 16] B] E] 2] (1)

5.1.3, x= S14. y=(-ll + 2C1)-3 yFl-2-3 =-4 T.P = {-1; -4} 5.2. GRAPH 5.2. 5 a4 5.3. xX |o° 30° | 60° 90° | 120° | 150° 180° Y joe 0,5° | 0,866° | 1° | 0,866° | 0,5° | 0° 0,5 | -0,5° | 0° | 0,366° | 0,5° | 0,366° | 0° | -0,59 3] Downloaded from hlayiso.com {1} ea i] 3]

b /soe 60? 90° 120° NU8O: Lo,5 -1 13} S41. sinx-0,5=0 sinx= 0,5 x= 30° or x=150 (2} 54.2. sinx =1 [3] x= sin’ 1 x= 90° (20} Total Marks {100] Downloaded from hlayiso.com

higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA T1360(E)(N18)T NOVEMBER 2011 NATIONAL CERTIFICATE MATHEMATICS N2 (16030192) 18 November (X-Paper) 09:00 - 12:00 REQUIREMENTS: TWO sheets of graph paper. This question paper consists of 8 pages and a 2-page information sheet. Downloaded from hlayiso.com

QUESTION 1 Ll Jarious options are given “Se beens eos as possible answers to the following questions. Write only the letter (A - D) next to the question number (LLL = 1.1.10) in the ANSWER BOOK. ie aw x : (=| can be written as ... y A m x Boo C le a D ml ¥ 12 The log327 is the same as ... gOWy We wo 8 7 1.1.3 fa + b is equal to ... A Ja+b B (a+b) C fa. vb Dab Downloaded from hlayiso.com PTO Q) a) @)

1.1.5 a 11380(E)(N18)T The trigonometric function cotx can be expressed as ... A D sin x cosN 1 cosecy 4a" is the same as ,.. A Cc D ae" in a? The trigonometric function cosec @ is positive in which quadrant? A B Cc D Third and fourth quadrant First and second quadrant Second and third quadrant First and fourth quadrant 3loga73 is the same as ... A B Cc D 9 18 I 27 The expression Ure’ can be written as ae B Cc log ,x loge e loge Dowhloaded from hlayiso.com PTO (1) Q) qd) QQ) ()

“5. Fou eyes 119 V8 can be expressed as .. A 4 Bo 242 C42 , p 2 ) L140 sin’! 0,5 is equal to ... A 60° B 90° Cc 30° D 45° () 1.2 Indicate whether the following statements are TRUE or FALSE. Choose the answer and write only ‘true’ or ‘false! next to the question number (1.2.1 - 1.2.10) in the ANSWER BOOK. 121 scan be expressed as x 16 Q) 1.2.2 p™ is the same as 5 a) 123 x=log, fean be expressed as p* =! re) (x! ae ; - 1.2.4 xt yl.z®}° is the same as Xyz a) 1.2.5 logs 25 can be written as 2 ql) 1.2.6 There are 61,3° in 1,07 radians (ly 1.2.7 9" is equal to 3 ran) 1.2.8 sec is the same as u cos Q) 1.2.9 afx'.y® can be written as vy qd) 1.2.10 Converting 3,5 radians to degrees, the answer will be 20,0535 (1) 220] Downloaded from hlayiso.com PTO

wu 11380(E)(N18)T QUESTION 2 2.2 Find the lowest common multiple (L.C.M) and the highest common factor (HCF) of the following: AV + ye 2 2 xo ~y x tay 23 Simplify the following: 23.1 a+a-6 _a7n4 P-a- 12 @ -16 2.3.2 27a? 34 aA? QUESTION 3 3.] Solve for x in the following equations: 3.1.1 [Z)s-2 2 3.1.2 2773" =81 3.2 Find the factors of the following: 211 2a? ~ 8b? 212 3x? + 36x + 96 3213 wy -a'x’y + acy a! Simplify WITHOUT using the calculator: 3.2. ° log327 + log,36 - = log;49 Downloaded from hlayiso.com PTO @) (3) (6) (3) (2) 20]

cas) ras fe peu Use logarithms to find the value of the following: P=(3,8)' + 0,943 Solve for x and yin the following simultaneous equations: Solve for x in the following equation with the aid of a quadratic formula: 3x? -2x-}=0 QUESTION 4 41 4.2 4.4 Calculate the surface area and volume of a sphere with a diameter of 68 mm. The ordinates of an irregular sheet of metal are as follows: 120; 127; 139; 143; 147; 163; 157; 153; 148; 136 en 118 mm. The common interval is 16mm, Use the mid-ordinate mule to calculate the area of the sheet metal. A wheel with a diameter of 80 mm rotates at 21 r/s. Calculate the following: 4.3.1 The peripheral velocity 4.3.2 The angular velocity of the wheel The angle of a sector of a circle is 55 ° and its radius is 3,978 cm. Calculate the following: 44. The arc length of the sector incm 442 The area of the sector in om’ Determine the value of the following: sin(24,8° + 173,4°) Downloaded from hlayiso.com PTO @) “) (4) (22) (4) ) (2) {18]

- HISBULE}INT8)T QUESTION 5 5. ta w te 1 io Calculate the vatue of the gradient and the y -intercept, if: aps 4dy-6 Given the functiony =x? + 2-8 Determine the following: 5.2.1 The axis of symmetry 5.2.2 The turning point 5.2.3 The x-intercept 5.2.4 The y-intercept Draw the graphs of y=cos@and y=sin@on the same system of axis for the values of @ where 0° < @ < 180°. From the graphs above determine the values of 0 where: 3.3.1 cos8 +1=0 5.3.2 sie =] 5.3.3 sin@ =cos@ 5.3.4 sin@ =0 5.3.5 cos@ =} TOTAL: Downloaded from hlayiso.com i) (2) (2)

TL360(E)(NIS)T MATHEMATICS N2 i INFORMATION SHEET This sheet must accompany the question paper. The right cone 3 Volume = '4ar7h —— Surface area = ay Vh? +r ar? = are +a? The right pyramid Volume = V% (area of base) x (perpendicular height) The prism Volume = (area of base) x (perpendicular height) The eylinder 2 Volume = wh ~ 3 Surface area = 21? + 2arh The sphere ro 425 2 Vs <a? 4 = 4dqr J Degrees and radians [80° = wt rad are radius Sector: @ = -:A=Y%Pe@ Angular velocity and circumferential velocity Angular velocity: 1 = 2m Circumferential velocity: vy =Dn '= rotation frequency (1/s = revolution per second) Mid-ordinate rule Area = (distance between ordinates) « (sum of mid-ordinates) a ordinate + Last ordinate =) ET Est ordinate: 5 + Sum of other ontnats Multiply by the distance between the ordinates, Graphs Suaight line: p= my + ¢ Parabola: y= ax? + by +¢ a -b Axis of symmetry: x = — 2a b+yb* ~4a —br — +4ac e Soot: «=~—y Swinloaded from hlayiso.com

we 11360(E)(N18)T Trigonometry : 90° < a < 180° sin @ = sin (180° - 8) COS 8 = -cos (180°- @ ) tan @ = -tan (180° - @ ) Segments of circles Chord Jength = x Height of seement = / Diameter of circle = D 2 D=h+* Regular polygons Angle subtended at centre of circumscribed circle by one side: 360° 8 number of sides R = radius of circumscribed circle x= length of side 0 x=2R sin — 5 Annulus: A = alr? -7?] Downloaded from hlayiso.com

education yer a LAs Department Oy : Education wy REPUBLIC OF SOUTH AFRICA : “NATURAL SCIENCE _ MARKING GUIDELINES a 2011 MATHEMATICS N2 16030192 November 2011 WISKUNDE N2 Downloaded from hlayiso.com

MATH MEMORANDUM EMATICS N2 WISKUNDE QUESTION/VRAAG 1 1.1.1 4.1.10 1.2.4 1.2.2 1.2.3 1.2.4 1.2.5 1.2.6 1.2.7 1.2.8 1.2.9 1.2.10 ow TRUE x FALSE TRUE ~~ FALSE“ oO TRUE “ 4 “ uw“ TRUE FALSE TRUE TRUE FALSE Downloaded from hlayiso.com

QUESTION/VRAAG 2 2.1 2.4.4 2.1.2 2.1.3 2.2 LOM: HCF: 2.3 2.3.1 2.3.2 te Sad a 2a ERAN Xo 128 32) x atx 4) (x + 8) a v vw x*y(xy ~ a*) + alxy -a?) (yt a)(xy - a*) f a ee xyt yo sy (ety) a wea apn yh, Cat ») . xe + xy =e(r ty) Vay (e+ yye- yy x+y y (a¥2y(a43) “(a+ 4)ia-4) (a+ 3)@ed * @-Dat2) va (a+4) (a+ 2) 3344, q-3-2_ pra+3 / = 3.05.5 % 2187b \/ ae QUESTIONVRAAG 3 3.1 3.1.1 3.4.2 3.2 3.2.1 33a 2 3tk — 34 asaya 4 #1 1 logs 33 + log, 6? ~ 7 loer pe 1 : 3ogs 3 + 2loge6 ~(5 x 2) log, 7 ¥” 2342-1~ =4 7 Downloaded from hlayiso.com (3) (3) (3) (6) (3)

3.2.2 log3—log5 | log 34 — log 52 log3 ~ logs SS ~ 2log3—2log5 _ (log3 logs) ~ 2 (log3 — 2log5) 4 3.3 og (3,8)* JA logP = 4 log 3,8 — log0,943 logP = 4 x 0,579 ~ (— 0,025) “ P= 192316 A ‘ P= 221,117 ¥ 3.4 2x + 2y = Devel 2X — BY = Bee @-@sy= -1 y= -0,2 Va substitude yin 1 3.6 _ b+ vb? = 4ac yo LO MEVEY? 4A 2(3) _ 2476 “16 4 244% 2-4 x= 6 or 6 ty x=lorx= -z nd Downloaded from hlayiso.com

QUESTION/VRAAG 4 4A 4.2 4.3 4.5 4.3.1 4.3.2 4.4.1 4.4.2 A= 4nr? (4) = 4n(34)? wv = 14526,724mm? “ 4 V= grr? = 573) vo = 164636,25mm3 ist ord + last ord “ 4 spore ast ore + sum of other ord| x dist between them (4) 2 120+ 118 a [+ 127 +139 4143 + 147 +163 41574 1534148 4 136! x16 ~~ = 1194131 x16“ = 1432 x16W = 22912mm? or 2,2912 x 107*m? v V = 2nn (2) = 2n x 0,04 x21 =5.277m/s a =2an (2) =2n x21” 134,946rad/s~ =r¢ (2) = 3,978 x 0,959 ~ =3814em ~ 1 A= 5 or? (2) = = x 0,959 x (3,972)2 2 =7587em? ~~ sin(24,8° + 73 4°) (2) sin(498, 2% sin a [18] Downloaded from hlayiso.com

QUESTION/VRAAG 5 54 _4 6 A eR 3 =tx-2t 3 fromy =mx+c 4 4 gradient(m) = = a 3 aw y —intercept(c) = —2 52 5214 yo 2b * 2a _ 2 ~ Oy =-1 5.2.2 rp y= (1)? + 2-1-3 =1-2-8 = 9 . TP (-1;-9) ¥ 5.23 pery=0 x? + 2x>8=0 x= -*0rx= ww 5.2.4 for y ~ intercept let x = 0 y= -8Wv §.3 y= sind jx | o® 130? [60° | gc%] 120° | 180° ) 180° y [o [os |o866 |1 |0866 |o58 fo y = cosd x |o°| 30° | 60? | 90°] 120° | 150° ee y |i /oseejos |o |o5 0,886 | 4 4 1¢ Lt J wa y= sina Q o° 30° 60° 99° 120° 150° 180° oy = cos “ Downloaded from o.com

5.3.1 5.3.2 5.3.3 5.3.4 5.3.5 6 = 180° 6= 90° 6 = 45 O= o%nd 6 = 180°" = 0° Downloaded from hlayiso.com {18}