MATHS N4 LEARNER PACK 1

Asa!) THE EDUCATION CONNECTION 4) Northink Colege LEARNER PACK

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higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA meat sists of 5 pages and 1 formula sheet. - Downloaded from hlayiso.com Copyright reserved dhet Please turn over

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" 16030164) -2- T1080(E)\(J30)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE MATHEMATICS N4 TIME: 3 HOURS MARKS: 100 INSTRUCTIONS AND INFORMATION 1, Answer ALL the questions. © “~~” 2. Read ALL the questions carefully. any 3. Number the answers according to th lumbering system used in this c 4, Show ALL the calculations grid ss Ey c= All the graph work musi Ke’ Graph paper is NOT and, the turning point(s} MUST & néstion paper, You are NOT compelled to use the eSsarily complete. Downloaded from hlayiso.com Copyright reserved dhet Please turn over

(16030164) 3. TLOBOE)U30F * = Cn Te Rt ° QUESTION 1 1.1 Solve for x and y if: + p-8 + J\2-j7) 1-73 1.2 Given: Z=54 75 121 Find Z 1.2.2 1.2.3 13 13.1 13.2 What is thex p= 14 1.4.1 Sketch the CH 14.2 Solve for 3eand y by the use of Cramer's rule. 2.3 Given: 2 -2 1 |Dl=3 -1 2 4 6 5 - 2.3.1 Determine the value of minor of -1 2.3.2 Determine the value of the cofactor of 5 . Copyright reserved dhee +4 OWnloaded from hlayiso.com viens tr owe 6) ay @) @) 3) @) 8) qd) [20] 4) @) @ @ hs

* (46030164) -4. T1080(E)(J30)T 2.4 Given: T= aa -én | Dp Make p the subject of the formula, 4 2.5 The difference between two numbers is 2. The sum of their squares is 100. Determine the numbers. (4) . [20] QUESTION 3 3.1 Prove that: (tein? x+ tcos 2x=2~sec* x 1) “32 Solve for x if: 6cos? x=5+sinx; (6) @) (A) 3) [20] (4) 42 Expand to FOUR terms only by using the binomial theorem if: (2x? ~3)' (4) 43 Given: yor? +9x? 415x ‘ Calculate, by the use of differentiation, the maximum and the minimum turnings. Also, calculate the point of inflection. (8) Downloaded from hlayiso.com Copyright reserved dhet Please tum over

(16030164) -5- T1080(E)30)T 44 Differentiate the following in terms of x: yer - +3e™ -x? 3% (4) sin 2x 120} QUESTION 5 5.1~ > Simplify> tan x + sin 2x ee J cosx (3) 5.2 Evaluate: : ae 6,5 47° de {4 @) = % « 5.3 53.1 Sketch and indicate — ea enclosed.by the gtaphs of y=, Also, indicate the Tepresenten vestriptorbe use sogalculate the enclosed @) area. = te = >, = 5.3.2 Calculateyaisingtintegration, thes nglosed area indicated in QUESTIONS ies & 4) 7) [20] TOTAL: 1060 Copyright reserved axe DOwnloaded from hlayiso.com a

” “(46030164) T1080B)(V30)T FORMULA SHEET a =b & loga’® =logb énx=log.x (|e =7" fue atbj=ct+d = a=candb=d sin(a + 5) = sina cosb + sind cosa cos(a + b) = cosa cosb F sing sinb tana + tanb tan (@+b)= (a+) iFtana tank _ VOx)ui(x) -— u(x)v'(x) fea]? Jsin x de =-cosx +e fcosxde=sinx +e ftanx dt =fnsecxte Jsec.x dx = én (secx+ tanx) +e Downloaded from hlayiso.com Copyright reserved dhet

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MATHEMATICS N4 MARKING GUIDELINE QUESTION 1 41 — B+/5)2-J7) xtjy= 1-fs . 6—j11—j235 e+ = Ts 1 41-1 | 143 atiy= 4-j3 “* 14)3 * 414+j112+33 an are xt+jy = 7,4+j11,9 x=7Aandy = 11,0 S 1.2.2 1.2.2 Z=-5—5 r=vat +b? = ¥E5 FCs = ¥50 6 = 180° + tan-12 6 = 225° Z=reis225° 1.2.3 1.3.1 (5) (1) 8) (2) Downloaded from hlayiso.com

1.3.2 Range = {y:y €R;-11<y <0} 1.4.1 (3) (2) (3) 1.4.2 Continuous (2) [20] C Question 2 24 gexts =5. xt 22%, 23 = 5,2*.2 8.22% = 10.2* _ 10 2* _— ee “ xlag2 = logi0 — log8 . x = 0,322 (4) Bd 2.2 x= 7 1 <1 = O-D)-@) @)-)-GD =3 P 6 C y=z {| . ti = AG)-Oo @\(-D)-@@) =0 (4) _[2 1 2.3.4 M4 = lf sl = (2)(6) - @@) =6 (2) = (1343/2 72 2.3.2 c= [2 73 = (2)(-1) - @)(-2) =4 (2) Downloaded from hlayiso.com

2.4 Ts 3 (2-In§) 25 let the two numbers bex andy x-yH2 0. x=yt2...( x? 4y*7=100...(2) substitute (1) into (2) (42)? + yy? + 100 y2+4y+4+y? = 100 y* + 2y -—48 = 0 (vy +8) - 6) =0 * y=-8ory=6 andx =—60rx=8 (4) (4) [20] Question 3 3.1 (tan?x + 1)cos2x = 2 —sec?x LHS = sec*x(cos*x — sin?x) = sec*x(2cos?x — 1) 1 = Ga, (2cos*x — 1) a o=Q- cos?x = 2—sec?x = RHS C. 3.2 6cos?x =5+sinx ; 0°<x<¢ 90° 6(L — sin?x) = 5+ sinx 6—6sin’x =5 + sinx 6sin?x + sinx-1=0 (3sinx —1)(2sinx +1) =0 . 1 2 1 . . - sinx =5 or sinx =—5 (outside domain) x = 19,97° 3.3 3 + aad Fra x ¥ 3 5 cos(x + y) = cosxcosy — sinxsiny Downloaded from hlayiso.com (4) (6)

65 3.4 slaxtcosx 1 cos2x cosx—sinx _. sine +cosx a cos*x—sin?x — cosx—sinx _ sinetcosx 1 (cosx~sinx)(cosxtsine) cosx—sinx =0 3.5 cos2x = cos(x +x) = cos*x — sin?x = cos*x — (1 — cos?x) = 2cos*x-1 (3) (4) (3) {20] Question 4 4.1 ax vp? e?*£_Inx,20e?* =’ (o>)? 1 = e2#(4-2Inx) (e?*}? +-aInx rw ek 4.2 (2x? — 3)* 4.3(2x?)*(-3)? (4) 4.3.2(2x?} (~3)* = (2x7)* + 4(2x?)3(-3) + o4 = 16x8 — 96x68 + 216x* — 216x? ++ 4.3 y ax + 9x? + 15x & 2 3x? 418x415 a ay _ qa = Ox +18 ; Turning points when e =0 « 3x? +18x +15 =0 (x +5) +1) =0 .*=-5 or x=-1 y = (-5)? + 9(-5)? + 15(-5) = 25 ory = (-1)3 + 9(-1)? + 15(-1) =-7 B21 te (4) « maxTP at (—5;25) and minTP at (—1;-7) Downloaded from hlayiso.com

2 Point of inflection at = =0 ie6x+18=0 “6x = -18 axe -3 y = (-3)° + 9(-3)? +15(-3) =9 * POI is at (-3;9) (8) 4.4 Y= t3e%- x—3™ = 2cosec2x + 3e77* — x3 — 33% i ® = -2.2cosec2x.cotlx — 73.e-7 — 3x7 — 3.3 In3 +C 1 = ~4eosec2x.cot2x — 21e77* — 3x2 ~ 3.3% m3 +C (4) {20} Question 5 tanx+tsinz 5a Irae 4 tanx 4 2sinxcosx! =f (e+ seer) dx = f(secxtanx + + 2sinx)ax = secx — 2cosx (3) 5.2 Spo 4 dx 4rex 0,5 ~ [Sil _ 4-2(0.5) 4-2(0} ~ [ ed ~ [Sel = -0,09 + 0,361 = 0,271 (3) 5.3.1 y (3) Downloaded from hlayiso.com

5.3.2 yaa x dA=ydx = aidx 6A=2 Se xadx 4 3 2 -2x/ | 2 Jo =2x[:(#!) -2(o) = 10,67 units? (4) COSECAX x 2 = J (Bsecxtanx — 4sinAx — z +e 3% — 373k E x2) dx 5.4 f (3secxtanx —~—4 24 63% 3 °3% 4 vz) dx 3x = Bsecx + St Tinx —2e-3* 4 2 4 2x8 46 3 3in3 3 = Bsecx + cos4x — Tnx —he- + 2x3 46 {7) . 3 3in3 3 {20} Downloaded from hlayiso.com

* DHET . 2 1930(E)(MO8yT DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE MATHEMATICS N4 TIME: 3 HOURS . MARKS: 100 Aen -z.2% INSTRUCTIONS AND INFORMATION 1. Answer ALL the questions. 2 Answer ALL FIVE questions in full, Show ALL the calculations and intermediary steps, Simplify where possible. 3, ALL the graph work must be done in the ANSWER BOOK. Graph paper is NOT supplied. The values of intercepts with the system of axes and the turning point(s) must be shown on the graph. 4, ALL final answers must be accurately approximated to THREE decimal places. 5. Questions may be answered in any order, but subsections of questions must NOT be separated, 6. A FORMULA SHEET is attached to this question paper. You are NOT compelled to use the formulae and the list is NOT necessarily complete. 7. Write neatly and legibly. Copyright reserved Down loa ded fro m h layiso.conr: turn over

Copyright reserved DHET -3- . T1030(E}(M28)T QUESTION 1 e + x? 4. ? Ld Bel Skeich th bh of —--s-Rz etch the graph o: 72 32° @) 1.1.2 Whats the domain of the Pe h in QUESTION 1.1.1? (2) 1.1.3 Is the graph in QUESTION 1.1.1 symmetrical about both axes? ¢9) 1.1.4 Does the graph in QUESTION 1.1.1 represent a function ora relation? (i) L2 Sketch the graph of x=—./20-y? . GB) _13 Solve for x if: Xt yt yt 2ZfoBs+7 (4) 14 Given: Z, =1-j1,732 Z, =— 6,928 — 74 141 Convert Z, and Z, into polar form. The arguments may only be positive. (2) : Z, we 1.4.2 Calculate -—— and leave the answer in d+ jb form... 2 sot : - (4) (20] Downloaded from hlayiso.ééii¥'*”

~ -4- “" -p1030(8)(M28)T A - QUESTION 2 24 Given: 2x -2y-Z=3 4x4+5y-2Z=~-3 3x t4y-3Z2=-7 241 Solve for the value of Z by using Cramer's rule, 21,2 Determine the value of the cofactor of ~3. 2.2 Given: w zy t=—fog l+— g “( a Make y the subject of the formula, 2.3 Solve for x iff ye? = gfe 24 The sum of two numbers is 10, The difference of their squares is 50, Calculate the TWO numbers, A QUESTION 3 3.1 Prove that: Sin2A ~ Cos24 +1 = tan A Sin2A + Cos2a+l 3.2 Solve for 4 ift 2ceos’ A= —Cosd; 0° < A<360° 3.3 3.3.1 Derive a formula for Cos24, 3.3.2 Use the derived formula in QUESTION 3.3.1 to determine the value of Cos120° without the use of a calculator, 3.4 if sin A=0,6 and sin B=0,4; and both A and B are acute angles, determine the value of Cos(4 + B) without the use of a calculator. 3.5 Prove that: sin(l 80° + A)= —sin A (8) 2) @ @G) 3) [20] 6) @) Q) @) @) (3) [20] Copyrightreseved DOW MIOAGeA TOM MlAayISO.COPiasetum over

QUESTION 4 . 4 Determine: tim f x-2 oe x32 (x -8 @ 42 Given: ' y=e™.cos3x. Differentiate by using the product rule. ; m)) 4.3 Differentiate the following with respect to x: y=2.3* ~4cos2x — Sénx - 3x7 4 44 Given: yer? —L1y? +32x-28 Determine, using differentiation, the maximum and the minimum turning points. Also determine the point of inflection by using the second derivative. (8) [20] QUESTION 5 Sl Integrate the following in terms of x: [@sinax eo — 3x4 + 2.3 — cosec*x + 2) dx anes) z 5.2 Determine: faras Q) 1 5.3 5.3.1 Sketch and indicate the area enclosed by the graph of y=4 cosy, the x~ axis, x=Oand x=14. Also, indicate the representative strip to be used to calculate the area of the enclosed area. QB) 5.3.2 Calculate, using integration, the value of the enclosed area indicated in QUESTION 5.3.1. ; rc) 5.4 - Simplify: Seca iy cos ecy @G) [20] TOTAL: 100 Copyright reserved Downloaded from hlayiso.com

. DHET FORMULA SHEET a* =6 & loga* = logs - T403008)(M28)T £n x= logsx (JOY =r \n@ atbj=e+d) & a=candbad sin(a + 6) = sing cosh + sinb cosa cos(a + b)= cosa cosh F sina sind tana = tanb tan (a = b) = —__—_—__. iF tana tanb ¥ ay a ny aT ax nax” ka* ka tna kenx & x sin x cos X cos x -sin x tanx sec*x cot x -cosee”x sec x sec x tanx cosec x -cosec x cot x ntl ax Sax? dy = +C n+] fde=atnxte x x katdy = ka fna Ags = {va x Copyright reserved te sin’y + cosy = 1 1 + cot?y = cosec?x 1 + tan’x = seo’x J ue)» ¥G) a8 = u(xy'(a) + a'(x) vOx)- 90) 3 _ 2Qw'@) ~uGy"@) as [eeey? dy dy du dx du dv fsin x de=-cosx +c Jeosxdv=sinx +e Stans de =fnseexte fsecx de = tn (sec x + tana) +e Downloaded from hlayiso.com

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¥ PLEASE GO ACcoRmING TT. MWuet — UA Ne “TO ME Ques Tan J/ANSHEES, neo, ., N4 MATHEMATICS ASSIGNMENT 1 MAY 2013 Question 2 2 -2 -1 211 [DJ=|l4 5 -2 3.4 -3 5 -2 4 -2 45 =2 lh “3 ~©2) [3 “3 +CD I; 4 = 2[(5)(-3) — (4)(-2)] + 2[(4)(-3) — 3)(-2)] - (4) (4) - @)G)] =~14-12-1=-27 2-2 3 - ID,,;=|4 5 —3 Ci 3 4 -7 =2[f T1-Caly Tesh 4 = 2[(5)(-7) — A)-2)) + 214) (-7) - BDI + 31) - BE)! =-46~-38+3=-81 Wal = -81 pl] aa 3 (8) 24.2 Cofactor of -3 = (-1)°*9 |? ? = [(2)(5) —- 4)(-2)] = 18 Q) om. =” x 2.2 t= 7 oa(1 + 2) 9 x tg o1+2=510w Z 221001 Pil nyse (10% -1) (4) Downloaded from hlayiso.com

23 25x-2 = 53xtd « (5x -— 2)log2 = (8x + 1)log5 - 5xlog2 — 2log2 = 3xlog5 + logS ~ 5xlog2 ~ 3xlog5 = log5 + Zlog2 «. x(5log2 — 3log5) = log5 + 2lag2 _ togs+2log2 ~ Slog2-3logs = —2,198 GB) 24 x+y=10 «x=10-y x*—y? = 50 Substitute: (10 — y)? —y? = 50 +. 100 — 20y +4%—32 = 50 « 20y = 50 “y=25andx=7,5 16) [20] Question 3 _. Ssin2A—coszAti 3.1 LHS = sin2At+cos2At+1 a _ 2sinAcosA—cos*Atsin2 Atsin2AteestA * ™ 2sinAcosAt+cos?A-sintAtsinZA+cos2A __ 2sinAcosAt+2sin?A ~~ asinAcosA+2cos?A __ 2sinAfeosAt+sins} ~~ acosAtsinA+cosA} = tanA = RHS 6) 3.2 2cos?A = —cosA;0° < A < 360° + 2cos?A + cosA = 0 + cosA(2cosA + 1) = 0 Downloaded from hlayiso. com

14 4 =1-/1,732 2, = —6,928 — j4 14.1 2, = 2|300° 22 = 8[210° (2) 2, _ 2/300° 142 P= Soins = 0,25]90° = 0,25cis90° = 0,257 (4) OO (20) Question 4 . -2 0 . 44 Jims 35 = 5 (undefined) lim, p22 = i x2 “My 22 53g — M92 Get eaet4) =H 1 = NM y92 Vay oetd 4 2422) +4 2 =a (4) ( . 42 y=e3*.cos3x = uv now 2 =ukey& = ¢5*_3(~sin3x) + cos3x. 3e3* = —3e3*sin3x + 3e3*cos3x = 3¢3*(cos3x — sin3x) (4) 43 y= 2.35% —Acos2x ~ Sinx — 3x74 ay 5. 42 8 _ 3-2 il 2.5.3>* In3 —~ 4.2(—-sin2x) 5 3(-x7*) = 10.35*in3 + 8sin2x — 5x71 + 3x7? (4) Downloaded from hlayiso.com

44 y=x?—1ix? +32x—28 Bo Bx? — 22" 432 and £% = 6x — 22 For turning points: 3x? —22x4+32=0 (3x — 16)(x - 2) =0 x= 2orx=2=5,33 atx =2: “ = 6(2) — 22 =—-10 ~ maximum turning point also y = 23 ~ 11(2)? +. 32(2) — 28 =0 (3 at x = §,33:£2 = 6(5,33) ~ 22 = 32 « minimum turning point also y = (5,33)8 — 11(8,33)? + 32(5,33) — 28 = -18,5 »: The turning points are at (2;0) and (5,33; -18,5) fil oint of inflection: #50 .6x-22=0 axa = 3,67 atx = 3,67:y = (3,67)? — 11(3,67)? + 32(3,67) — 28 “y = 9,28 ¢ : Point of inflection is at (3,67; —9,28) (8) [20] Question 5 5.1 f(@sin3x —- e~** — 3x-4 + 2,35* — cosec?x + 2)dx _ Ax -3 8x = 30S 3 4 2, St cote + 2x +6 = _ 14x =3 4 2.35% = cos3x + 7e Tx +a + cotx + 2x + {7) Downloaded from hlayiso.com

MATHEMATICS N4 ASSIGNMENT 2 Question 1 11a yet 5 ase ‘ 2. UTP 32-12. qo 8 va) a Be 2 “4 . sf (3) ae . 11.2 Domain = {x:x € R;-2 2x > 2} (2) ‘1413 yes (1) 1.1.4 relation (1) 1.2 y =v20—x2 Y pos 3 TS (3) 1.3 x+ytyf+2j=8f+7 oxbytyf=7t+ 6) “y=6 andx+y=7 wXb627 axed (4) Downloaded from hlayiso.com

oa “ cosA = 0 or cosA= 3 - A = 90° or A = 270° or A = 120° or A = 240° 3.3/1 cos2A = cos(A +A) = cosAcosA — sinAsinA = cos?A — sin?A 3.3.2 cas120° = cas(60° + 60°) = cos*60 — sin?60 = -@ ele Bw 34 sinA=06= = =i « other side = 4 sinB =0,4= = = : « other side = J21 now cos(A + B) = cosAcosB — sinAsinB 3.5 LHS = sin(180° + A) = sin180°cosA + cos180°sinA = 0cosA + (-1)sinA = —sinA = RHS 4) 3) (3) GB) 3) [20] TOTAL : Downloaded from hlayiso.com 40 marks

5.2 fp 2-dx 2 75x ~ (inal 1 9-5(2) 9-501) ™ W\osine) — “sine, = —0,00028 + 0.009 = 0,00872 5.3.1 dx 90 ‘ y=Acosx 5.3.2 dA = ydx x A= 2 4cosxdx = [4sinx]2 = [4sin 3] — [4sin0] = 4 (units)* 54 ff cosecx = f sinxdx = —-cosx+€ (3) (4) (3) (20] Downloaded from hlayiso.com

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! oomenrenscacenorannsieitaansecreiiy F} education 4 Department: Education REPUBLIC OF SOUTH AFRICA T1430(E)(N23)T NOVEMBER 2005 Ci /” NATIONAL CERTIFICATE MATHEMATICS N4 (16030164) 23 November (X-Paper) 09:00 — 12:00 This question paper consists of 4 pages and a i-page formula sheet. |_____nawnloaded from hlayiso.com et

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2. T1430(B) (N23) DEPARTMENT OF EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE MATHEMATICS N4 TIME: 3 HOURS MARKS: 100 Answer ALJ the questions. INSTRUCTIONS AND INFORMATION 1, Answer ALL FIVE questions in full. Show ALL the calculations, and intermediary steps. Simplify where possible. . : All the graph work must be done in the answer book, Graph paper is NOT supplied, Values of intercepts with the system of axes and the turning point(s) MUST be shown on the graphs, 3. ALL final answers must be accurately approximated to ‘THREE decimal places, 4, Questions may be answered in any order but subsections of questions MUST not be separated. . 5. You are NOT compelled to use the formulae and the list is NOT necessarily complete. : QUESTION 1 11 Make x the subject of the formula: _ 107 +1 1077 ~1 12, Solve for x: 52x42 = 35a] ; 1,3 13.1 Sketch the graph of the inverse of y =3x-—2. 1.3.2 18 the graph in QUESTION 1.3.1 continuous or discontinuous? Downloaded from hlayisé.com

“N 3. TMas0CQNay a) 13.3 Is the graph in QUESTION 1.3.1 a relation or a fiinction? 13.4 Sketch the graph of y= log 14 A quarter of the sum of two numbers is 8 Three times the difference between the two numbers is 30. Calculate the values of the numbers, QUESTION 2 2.1. Given: Z 3-1 +74 24,1 Determine Z, . : 2.1.2 Convert Z into polar form, Show ALL the calculations, 2.1.3 Indicate Z and the caleulated valves on the Argand diagram, 2.2 Given; 24¢43b=4 ‘ a~4b=5 Use Cramer's mule fo solve for a and 4, 23 Given; S 6 11 qY. 45 7 2.3.1 - Write down and caloulate the co-factor of 6, 2.3.2 Write down and calculate the minor of 4, 24 Simplify: 733 Show ALL, the calculations, QUESTION 3 3.1 Prove that: cos x(1+ sin x) = cos? x seex + tanx 3.2 Solve for x: tan(x ~ 15°) = cot(x + 20°);0° <x <99° 3.3 Simplify the following expression as far ag possible: 1 + 1 cosecr~— cox cosecx + cotx a? . PTO € Lt Downloaded from hlayiso.com @) (Ss) (20) q) Q) Q) 6) 2) 0) G) [al (4) 4) (8) — :

4. T143002)(N23)T Determine the value of tan75° WITHOUT using a calculator, ‘34 3.5 Prove that: sec[90° - @]=cosec@ QUESTION 4 2 41 Differentiate y= —— by using a quotient rule, 2énx 4.2 Calculate, using cifferentiation, the co-ordinates of the maximum and minimum x? turning points of y= oe x? Also distinguish between the maximum and minimum furning points using the second derivative. fi 2 43 Calculate” xol YO yt] d?y -3x _ 95x 44 Determine —— > ig y= Se ~3 dx? QUESTION 5 51 5.11 Skeich and CLEARLY indicate the area enclosed by the graph y= 3,47; the x-axis, x= 0 and x= 2, Indicate the representative strip, 5.1.2 Caleulate, using integration, the magnitude of the area indicated in QUESTION 5.1.1. 5.2 Integrate: [(Ssin3x4 fx 4 5x9 43.5") cy 5,3 Evaluate; 1 f@x-Ide 2 5.4 Simplify: f seth TOTAL: *--00000-- Downloaded from hlayiso.com @) @) , [20] 4° () @) iO) £20} B) @) ©) (4) G) [20] 100

TM430@E)QNRayT MATHEMATICS Né4 FORMULA SHEET a* =b <= loga* =logd (ey =r" Ine atb=ctd 2 a=candb=d ; n8 yj sin@@ + b) = sina cosb +-sinb cosa cos(a + b) = cosa cosh ¥ sing sind tana + tanb. fan(@+)= 1¥ tana tanh y ay - & ae nal ka* ka* tna Here E mS , . ¥ sin x cos x cos x fesin x tanx sec’y cot x -cosec’y SEC x sec ¥ tan x cosec x -COSEC X COE ute ax . cy ox" dk = +C j nt) a s- ad =atnx te Ske de = ka log ete & oy = [ vdx NEW SYLLABUS fnx = loge sin’x + cos*y =] 1 + cot?x = cosec’x 1 + tan?x = gecty Y= ux) + v(x) > 2 = u(x)¥'() + u(x) v(x) =%Q) 22 de ad v(x) — Yeu (x) - (ev'(x) | [oeey? a dae du de fsinx d= -cosxte foosxde=sine +e Jtanx de = tnseoxte Ssecx dy =fn(secx + tanx) +e Downloaded from hlayiso.com

‘ og os MEMO REM DUM — Gq ¢ Novenbos door Mathematics Nq aan 1 [Vewncy 4 ~ - [~ WBZ. Cong "peed om / a entinuous . Ww U7 n33, Funchon Ut M10"= IOs m4 d) iO*(Mar) ome? 4 vx ) [Bee 10°. = my? Wet 2x = hos 7 ; Ze (8 Mb) WZ 02. 2a to gleme)] - te4(n-1) Guestien 2/Vesne Zz . ze 4 rye eo aan ae ee aie-y)=2 20 ~~ Rxr2\O,6 = (sen QA, 3y~ ‘© RAE - Rib? shy P- tL, Fh ye Be © i hi ee Ram tye IO ~~. 7° @ - 5879” |. i¢difrom hlayiso.com

Question 2] tae 2 \y} 7 b oy eX le |b 2b Bu AI4 Ww . . 5 6 Va en2 ys Cire of" : 5 L a ya WR oy 4jetB yo b= le ov = OS Ax deur |e . fe) =7F9L6y" — 2. O 2 lou, 036° wv By Ai ' t i &) = 3 Ze tr23{leg oe 4 7 @) ZEB Zon Deane 2.3.2 | £ ve — gn a $ VY yoy ! NY Co? BER kl ay 4 ~ZaUs> ’ cl pve J ad “it @) = ey eee. 2B Question 3 Neng 3 a ar Ib} = I° ®) 3 ag Sine) = cost mocll cose CL ot Stet D> LZ tes*z ~ i: ar [Df i 3 cosx (J4-siax) y-Cesx 2 oe| LB OF [PSI = -3! vA coSlX = CoS a fbal » -3f 2 28/9 ae e f bs "Daiwiildaded from hlayiso.com @) _

5 Question 3 /Verag 3B 3.2 deus (x, - +6? a bal Gp? _ oy] bein C215 2 ben, CPZ) z-1€" = 70°. exe gg? 08 f OF cL[Gor Gis). cbt Cred) JOS PH Xx = LP Zoe” — 29 se - es? ad 428° a) 13 - 1 | Sive Cose = oR/ oc t f COSFO° GED Pp 5IA90 Sia © L: COSELS = seed Sine = Sine g) Hos be Heg Ao Question 4 [nag 2. = age boy. ? oe a hae xt ye ex | ; 2 . i221 Kcosece Lote + Loses. — cot r a . (scx - ctr) Cesecse.t cote 4 = Adan oe & @Az): ~ ieee hawt | Cesecte. — coe, a = we al = 2. Cesecr @ oe o i > QevsceenV dle LE Oy Gubiio® 2+ han 79 2 tan S38) = Can gs + bei, BOO Lo been YS Cen BO tle a l- OF ae ame, 2 Vee jee Ps — 1334 o; S Cod b %) Downlo Sie irom hlfgis6,om nme

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higher education & training Department: Higher Education and Training REPUBLIC OF SOUTH AFRICA HN T1080(E)(A3)T AUGUST 2012 NATIONAL CERTIFICATE MATHEMATICS N4 (16030164) 3 August (X-Paper) 09:00 — 12:00 This question paper consists of 5 pages and a I-page formula sheet. DHET Copyright reserved Please tur over Downloaded from hlayiso.com

-2- T1080(E)(A3)T DEPARTMENT OF HIGHER EDUCATION AND TRAINING REPUBLIC OF SOUTH AFRICA NATIONAL CERTIFICATE MATHEMATICS N4 TIME: 3 HOURS MARKS: 100° INSTRUCTIONS AND INFORMATION 1, 2. Answer ALL the questions, Answer ALL FIVE questions in full, Show ALL the calculations and intermediary steps. Simplify where possible Read ALL the questions carefully, Number the answers correctly according io the numbering system used in this question paper, ALL the graph work must be done in the ANSWER BOOK. Graph paper is NOT supplied, Values of intercepts with the system of axes and the turning point(s) must be shown on the graph. ALL final answers must be accurately approximated to THREE decimal places, Questions can be answered in any order but subsections of questions must NOT be | separated. A FORMULA SHEET is attached to this question paper, You ate NOT compelled to use the formulae and the list is NOT necessarily complete. Write neatly and legibly, DEBT Copyrigh iso eee "Bownloaded from hlayiso.com +> -Please turn over

3 TLO8O(E(A3)T QUESTION 1 11 The breadth of a rectangle is 3 m less than its length. Calculate the length and the breadth of the rectangle if the area of the rectangle is 75 m?, 1,2 Make 'n' the subject of the formula if: os al” - i) rol 1.3 Solve for x ift 3oet = 52x42 14 Given: 3ypt axe Z +l 3x+2Z =8-Sy 3Z-l=x-2Z Determine the value of y by using Cramer's rule only. QUESTION 2 21 2.41 Sketch the graph of xy =-10, 2.1.2 Is the graph of xy =-10 discontinuous? x? 2 122 2.21 Sketch the graph of jet a =, x y 2.2.2 Is the graph of 6 + 7 = asymmetric about the x-axis or not? 23 Solve for x ift (e+ Ne +4)=-8 PHER Conreh Fy OWnloaded from hlayiso.commv: 4 @) @) (8) (20) G) q) G) 2) 4

Ae TiO8O(E)(A3)T 24 Given: rw (24 | 963° J" Use De Moivre's theorem to calculate for-Z. Leave the answer in a + jb form. @) 2.5 Given: 223-73 2.5.1 Find Z a) 2.5.2 Convert Z into polar form. @ must be positive. Show ALL steps. , 9 QUESTION 3 3.1 Simplify the following: -cot A+ tan A+1 (4) 3.2 Solve for x if 3sec’ x—-S=4tanx } 0°< x5 270° (5) 3.3 Prove that: Sin2A-l cos A-sin A sin A—cos A (4) 3.4 Calculate the value of sin— 75° without a calculator. ° 4) 35 Derive a formula for cos2A in terms of sin 4. , oe DHET Conmmie==~*! Downloaded from hlayiso.éoii”

“5 TIOBO(E)(A3)T QUESTION 4 4,1 Differentiate from first principles if: Se)=Vs* 42 Given; ye bx? 7" Differentiate by using the chain mule. 43 Differentiate with respect to x: 7 2 y=5x? ~ 2 - Ip + - tndy -cot2x ; 3x x 4.4 Given: yor’ +3y? 9x45 Calculate, with the aid of differentiation, the co-ordinates of the turning points, QUESTION 5 SE SAL Sketch and clearly indicate the area bounded by the graph of y=3x?-12, the xaxiss x=0 and x=2. Also, indicate the representative to be used to calculate the area shown. 5.12. Calculate, using integration, the magnitude of the area indicated in QUESTION 5.1.1, 52 Integrate: f 3.64 “tx ; wot +16 cos 4x4 cosec x cot x + 4m |de 5.3 Determine: "p cos xd M 5.4 Simplify: [lcos? 4—sin? \za TOTAL: meer cones" Ownloaded from hlayiso.com ) @) (6) (6) [20] @) (4) (7) @G) G) [20] 100

FORMULA SHEET a® =b loga* = logh Tlosocay(Aayr én x= logex (ey =r" |n0 atbjaetd o a=candb=d sin(a + 5) = sina cosh + sinb cosa cos(a + b) = cosa cosh F sina sinb tana + tanb tan (a b) = —-—--—----~ LF tana tanb y dy ad ax” nox”! ka* ka* tna kenx & x sin x cos x cos x -sin x tanx sec’x cot x -cosee’x sec x sec x tanx cosec x -cosec x cot x at ox a = e"ds = +C J ntl fe d=amxte x x Ska*ds = fa +e tna b Aoy = {, pdx DHET Copyright reserved sin’x + cos’x = I 1 + cot’x = cosec’x 1 + tan?x = sec’x Y= u(x) vQ) 2 2 = u(x)e'(x) + u'(x) v(x) = HG) v(x) ay _ vu) —u@)'@) de [x]? > Jsina de=—-cosxte fcosxdx=sinx +e Jtanx de = énseexte [secx de = én (secx + tanx) +e Downloaded from hlayiso.com ~

Mathematics N4 August 2012 Memo . Question 1.1 . The breadth of a rectangle is 3 m less than its length. Calculate the length and the breadth of the rectangle if the area of the rectangle is 75 m?, Area=LxB B=l-3 75 =Lx(L-3} =P ~3L I} -3L-75=0 Le= —btVb? ~4ac 2a Le 3443) -(4x-75) £=10,289m or L=-7,289 B =10,289-3 B= 7,289m Question 1.2 Make ‘n’ the subject of the equation. a(r"—1) ss r-1 tog = + 4 =logr” to =D, 7 =nlogr a i rma | = nat 4d OR nator, D4] logr a Downloaded from hlayiso.com

Question 1.3 Solve forxr | . “gfe ast . log pet = log gested _(5x-1)log3 = (2x + 2)log$ (5x ~1}0,477 = (x +2)0,699 2,385x - 0,477 = 1,398x +1,398 2,385x —1,398x = 1,398 + 0,477 0,987x = 1,875 x= 1,900 Question 1.4 3y42x=241 3x-+22 =8-Sy 3z-Lax-2z Determine the value of ‘y’ by using Cramer's rule only. 2x4+3y-z=1 3x45y+2z2=8. —a40t52=1 2+3-1 D= (34542 -~£+0+45 D= 25%5)-(0x2)-3[(3x5)-Cix2)]-1f(@x0)-(-1x5)] D= 2[25]-3ft7]-1[5] =-6 2+i-1 N,= B+8+2 T4145) = alfexs) —(ix2)}-affsxs) = (-12)] (61) -(-1»8)] 2f3s])-fi7]-fit] iit Ny N, N, Downloaded from hlayiso.com

Question 2.1.4 . Sketch the graph: xy = -10 yo aanls a t 7 a, <Ccs -) . Question 2.1.2 : " .. The graph is disco tinuous? . Question 2 1, ‘Sketth the graphy" ” Question 2.2.2 The ellipse is symmetric _ Downloaded from hlayiso.com

Question 2.3 (ct{r+4)=-8 x? ¢5x4i2=0 _~b4Vb? - 4ac 2a _ 55? - (a2) 2 Lo Sty—1xf23 7 2 _~ 5+ j4,796 ~ 2 x5-2,54 72,398 or x =~-2,5—/2,398 Question 2.4 Z =(2,4.286 a) (ea) 20,5x #63") Z =(1,549.243,15°) Z =1,549.00543,15 + /1,549.sin 43,15 Z 1130+ 1,059 Question 2.5.1 Z=-3- j3 zZ tL i Z=-3+ 73 Question 2.5.2 o= 4s” 6 = 180° - 45” a= 135" Z= 4182135" Downloaded from hlayiso.com

. . Question 3.1 cos A sind cosA+sinA sind tn d¢1o Sin yy cos A sin A +-cos A cos A +1 cot d+ | =. cotAtl= tan Atls= cot A+] = cosAt sind cosd tan A+1 sin A sin A+cos 4 cotd4+1 cosd tand+] sind colt] ig Question 3.2 3sec? x-5= tanx 3+ tan? x)-5 = 4tanx 343tan? x-5—4tanx <0 3tan? x~4tanx—-2=0 ~btVb? -4ac tany = 2a 4% J16-(4x3x-2) fan =—*—_ ns 6 C. 446,325 . tanx a tanx =1,721_or _tanx = ~0,388 x= tan?! 1,721 X= 59,841? = 239,841? Downloaded from hlayiso.com

Question 3.3 mn IA-1 : Sina 1 =cos.4—sind sin A-cosA . sin24~-1= (sin A—cos AXcos A—sin A) sin2A—1=sin d.cos d~sin? A~cos® A+sin A.cos A sin2A-1=2sin A.cos 4 — (sin? 4+cos? 4) ~ sin2d4~1=2sin A.cosA-1 . 8in24—l=sin2A-1 Question 3.4 sin(~7 sin{-75”) = -[sin 30” cos 45° + in 45*.cos30° Question 3.5 cos24 = cos’ A—sin? A cos24 =1—sin? A~sin? 4 cos24=1-2sin? 4 ¢ Yuestion 41 Se) =e & f@)=s S@)=# S(etal=(e4aY Petnsx® +2xh+H Downloaded from hlayiso.com

x + 2xh eh ax? t'@)= Lim + ss a hae f'Q)=2x4+0 Question 4.2 y=Q°-7) Let # =Px? -7) Question 4,3 yx? ~Fe Pp FInd cots wd, de ds dy 3 x 15x? +36x)3 -0+ 6x? ~44 2e0see"2x 7 0-+42c0see22x 2 Gx) * . 1 = ~<a); ~Sapr ae ~Fsnde— cote ® cons Question 4.4 pax 43x? -9n45 dy OF + 6x-9 ay Let ==0 ° ads 3x? +.6x-9=0 x?42x-3=0 Downloaded from hlayiso.com

(x-1Xx4+3)=0 . : . (x-1)=0 2 L, (x43)=0 xsl : : y=—3. yePs3—of}es y=(-3) 43-3) -9-3)45 y=0 “y=32 Tuming Point; = (1;0) Tuming Point, = 3:32) Question 5.1.1 - ; ¥ 4 a0 yr dx? -12 | yrs ~h yay? 4 - Let y=0 . dye: to ye xt-4=0 “ i 2 om x= t2 Let x=0 yard: Turning Point: yexr-4 x=-— =-4 x= Tuming Point = (0;-4) Question 5.1.2 Az fae A= (6x? -12)r Az [3x%dv— Frade 3x? A=——-2x 3 A=2-0-12(2)-0 =-16(urits)? Poe Downloaded from hlayiso.com

: Question 5.2 : Ie ay ff ave J Te ole i 60s Ad + f COS ecx. cot xely + Janet =3 frrar ~4 fe ax+7 fe de 4 16 foos dede+ feos ecx, col xdy + fande Je™ | 16sindx = —3y4diny ey x 7 ri —cosecx + 47n4+C 30° ; =-——4lnx~e"™ + 4sin 4x-coseex +4m+C x , fe C08 xx = 2 Poosxay d 4 z P2cosxex =2sinx 4 z ?2cos xa = 2sin™ 25% 4 2 4 Pecos xdy = 2-/2 4 a P2cos xate = 0,586 4 co Question 5.4 t flcos? 4—sin? Ayia = foos24aa f{cos? A-sin? AYA = nA +0 Downloaded from hlayiso.com

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