Mathematics Grade 12 Paper 1 NSC June 2013 Memo

Downloaded from hlayiso.com basic education Department: Basic Education REPUBLIC OF SOUTH AFRICA SENIOR CERTIFICATE EXAMINATION MATHEMATICS Pi HIGHER GRADE 2013 POSSIBLE ANSWERS MARKS: 200 This memorandum consists of 21 pages. Copyright reserved

Downloaded from hlayiso.com Mathematics/HG/P] 2 DBE/2013 SCE — Memorandum QUESTION 1 il 7a 3x(2—x)=7x-2 —3x? +6x-7x+2=0 3x* +x-2=0 Y standard form Gx-2\x+1)=0 Y factors /substitution 25 or x=—1] (ca) OR ¥ both values 3x(2—x)= 7x-2 —3x? + 6x-7x+2=0 pe (y+? 4-3) _-I+v@?-4GX2) , ORx a=) a 2G) _-1tv25 _-145 6 6 =tor-1 Vv @) 112 | x43 .¥x-5 =3 -—5)= Wrong factors, RE) Coat aks ® 2 ¥ square both sides x? -2x-15=9 breakdown: 5 x? -2x-24=0 Y standard form (x-6Xx+4)=0 7 factors 60 on fe 4 NIA both values “.x=6 (since x> 5) (5) v answer 1.13 3-x? =2x+1 x? +2x-2=0 —b+ Vb? ~4ac Y standard form oe ) ¥ substitution in formula _—@)+ Vey? - 4-2) 2() [p_>y2 v Vi2 -2)tyC2)° -4-DQ) OR x= 2(-1) YY values a= 242 -1 incorrect rounding off 2 =0,73 or —2,73 0 Wrong formula: 5 Check the division line if used correctly in subsequent steps G) GEFARTMENT OF BASIC EDUCATION Copyright reserved Please turn over 2013 -06- 20 PRIVATE BAG K 110 PUBLIC EXAMINATIONS

Downloaded from hlayiso.com Mathematics/HG/P 1 3 SCE — Memorandum DBE/2013 OR 3x? =2x41 x? 42x=2 2 2 2 ie 1 ¥ complete square of ye (23) = +(24) Y write as perfect square 2 on LHS (x+1) =3 x+1=4V3 Y find square roots x=-14 3 sq =0,73 or —2,73 ¥Y values 7 -1 incorrect rounding off 4 114} x-35— x . multiply by x (since it is positive) to get v oa by a non-neg 2 factors Nae V~15x54 (e+1)x-4)< 0 Gee -ls<x<4 and x>0 “ O<xs4 4 (4) OR x-3s- x x? -3x—4 vB a3x-4 ——_——-s0 5 x ; (&+1)-4)<0 since x >0 ain -l<x<4 and x>0 OF *s 1 O<x<4 dee Wrong inequality sign or using equal sign: -1 4 : marks OR x° —3x-4 <0 x DEPARTMENT GE Basia EDUCATION Copyright reserved Please turn over 2013 -06- 20 PRIVATE BAG X 110 PRETORIA 0001 PUBLIC EXAMINATIONS

Downloaded from hlayiso.com Mathematics/HG/P 1 4 DBE/2013 SCE - Memorandum ne |-2-x|>5 v¥ removing | | sign —2-x<-5 or —2-x>5 x>-24+5 or x<-2-5 V¥x<-7 x<-7 or x>3 VY x>3 OR the solution set C;-7)UG;%) OR ; (-2- Oe >25Vv Incorrect squaring: v square both sides 4+4x+x°>25 1 marks ¥ factors x +4x-21>0 4 Yx<-7 (x—3)xt 7) >0VY Yx>3 x<-7 Yor x>3 ¥ OR |2+x|>5 v v —~<¢ > 7 2 sj x<-7 Yor x>3 Vv (4) 12 2 =a @ Y=V20— x7 eee (2) From (1); 2” =227-6 ¥ 2% =2?*6 _ y=2x-6 ... 8) ¥ equation (3) Equating (2 and (3) or substituting (3) in (2): 2x-6 = V20-x? Y substitution 4x? -24x +36 =20-x? Y standard form 2 - 5x" —24x+16=0 ¥ factors/subst in formula Gx-4Yx-4)=0 Y both values of x x= 4 orx=4 Vv y= 2(CA) y =-2 or y=2 Y y=-2 (CA) x=4 and y=2 (since y> 0) (9) v answer | MEPARTHENT OF Basic 7 EDUCATION 2013 -06- 20 i PRIVATE BAG X 11 i PRETORIA 0001. : |_PUBLIC EXAMINATIONS Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/HG/P | J DBE/2013 SCE — Memorandum OR = 92x-6 12 | From GQ) 2” =2?*6 ee até (3) ¥ equation (3) 2 Substitute (3) in (2): i +6)" ye y-(225} Y substitution (CA) 2 y? =20 Yes v 4y? =80—y? -12y-36 standard form Sy? +12y-44=0 (5y +22Xy—2)=0 Y factors/subst in formula ) n Y both values of y Very 2 Substituting y Mee in (3) gives: Yx= 4 (CA) x=forx=4 Vv x= 4 (CA) x=4 and y=2 ¥ answer 13 x +mxt+1=0 y yaxrtmetl a t+ 1) 5 +H x fe) (4) *. Roots are non-real. “ OR Roots are equal, therefore, m’—4 = 0, m’=4v Forx?+mx+6=0,Y A=m?-24=-20V .. roots are non-real / imaginary . ¥ OR Y substitution into formula x +mxt+1=0 Equal roots: -. A=b? —4ac =0 Y both values of m m? — 4(11) =0 (m+2\m-2)=0 YA=-20 m=-2 or 2 Y conclusion (non-real) When m=-2: A=(- 2) —4(1X6)=-20 the roots of x” + mx + 6 = 0 are non-real When m= 2: A=(2) -4(1)6)=-20 the roots of x” + mx + 6 = 0 are non-real P"BEPARTMENT OF BASIC } EDUCATION Copyright reserved | 20 13 -06- 2 0 Please turn over PRIVATE BAG X 110 PRETORIA GOO1 { PUBLIC EXAMINATIONS

Downloaded from hlayiso.com Mathematies/HG/P 1 6 DBE/2013 SCE - Memorandum 14 | x?_(k45)e4+2k=0 A=(ck~5)? —4()(2k) Y substitution in A =k? +10k+25—-8k =k? 4 2k4+25 Y simplify to # + 2k +25 . ; v completing the square: ee aah R42k+1424 =(k+1P +24 v (k+ly +24 A>O forALL keR Y conclusion ~. Roots are real for allk eR (5) OR From +2k+25 dac—b? VV +2k+25 Minimum value of A = ———— 4a > ¥ calculate the min value _ 40)(25)-(2) _ 96 _ a 4() _ ra! Y value of A y : <A>0 forallkeR mate OR If y= +2k+25, VY the discriminant is 4 — 100 =-96 ¥ So the graph of y does not cut the k-axis. V Since a=1>0, +2k+25 >0 for all values of k. ¥ (39} ' DEPARTMENT OF BASIC b EDUCATION | 2013 -06- 20 | PRIVATE BAG X 110 i PRETORIA 0001 i PUBLIC EXAMINATIONS Copyright reserved Please turn over

Downloaded from hlayiso.com 1 PUBLIC EXAMINATIONS Mathematics/HG/P 1 7 DBE/2013 SCE — Memorandum QUESTION 2 2.1 | Given: f(x)=-x?-2x+3 and g(x)=|x+3| 211 Ro f'(x)=0 -2x-2=0 x=-] FCI=-C1P -26-1)+3 gr y= sagt (-1)(3 =-1+2+3 Al =4 =4 Turning point is c 1 34) y-intercept: (0 ; 3) x-intercepts: —(x + 3x—-1)=0 x=-30rx=1 . (330) or (1; 0) For g: (0; 3) and F3:9) For ft k vv turning point . vv x-intercepts (- 1 4) v shape 0; 3) For g: g Y 330) & (0; 3) arm Y other arm 7 x If graph is not drawn: ~ 3530) (150) (033) ¥ (330) v C4) vv If not in coordinate form: Maximum of 2 marks () 2.1.2/@) | e(x)>3 .. graph of g is above the horizontal line ¥ statement/reasoning y=3. y =3 intersect y = x+3 at (0; vix>0 3)and—x—3 at (-6;3) v x<-6 (CA 5-0 of 7-6 (3) | applies from graph) ) | #&)> e&) vv answer (CA -3<x<0 (2) | applies from graph) © -x? -2x+k=0 ie. —¥? -2x+3=3-k has non-real roots when the horizontal line y=3~-k and the parabola ye =-x* —2x+43 do not intersect. That happens. when 3-4 > 4 ie. k<41 DEPARTMENT OF BASIQ3) | vvv answer EDUCATION Copyright reserved r Please turn over t \ 2013 -66- 20 PRIVATE BAG X 110 PRETORIA G001

Mathematics/HG/P1 Downloaded from hlayiso.com 8 SCE — Memorandum DBE/2013 OR No roots when the graph is shifted vertically downwards through more than 4 units. Y The y-intercept is then <—1 ¥ ie k<-1V¥ IF - (9% a For non-real roots: (—2) - 4( it) < a Maximum: 1 mark < k<-1 (d) | For -3 <x <0; g(x)=x+3 Hx) = f()- g(x) v identify equation of g =x? -2x+3-x-3 =x? —3x Y equation -{21a03}e3 ha)=-x? ~3x 3/7 9 ~ completing a square : {x + 3) Be 2 4 : Pop ee Y answer Therefore maximum value is: — 4 4 (4) OR x)= f(x)- g(x) Y identify equation of g =-y?_ oo _ i Sit ae Y equation a 3 A(x) =-x? -3x Max value = —— au 4a —~1)(0) — (~3)? = 490) = (3) substitution 4-1) Therefore maximum value is: 2 4 v answer OR A(x) = F(&)—a() noe ; = oy? 2x 43—-x-3 Y identify equation of g =-x? -3x ¥ equation 2 Maximum when A'(x)=0 AG)=—x" —3x That is: T 2x3 = 0 Y h(x)=0 x=-> 2 ~3)\.(-3) _of_3).9 2) \ 2) V2) 4 Y answer P DEPARTMENT OF BASIC EDUCATION Copyright reserved | 2013 -06- 20 Please turn over PRETORIA G091 PRIVATE BAG X 119 | PUBLIC EXAMINATIONS

Downloaded from hlayiso.com Mathematics/HG/P 1 9 DBE/2013 SCE - Memorandum OR Roots of ~ x” —3x =—x{x+3)=04% are Oand—3. “ 0+(-3 3 Maximum value is at “3 ) “3 v ~37_343)29v Maximum value = 2 2 4 22 (x)= b* and a(-1d] a_l 22.1 | ¢-N=b ar; 1 Y substitute a(-13] = ay 4 b=4 Q) 2.2.2 | t(x)=-q(x)=-4" vv answer OR #x)= —5F Q) =a 2.2.3 q , v interchanging x and y oy = loge x v answer q2'@) = loge @) 2.2.4 J Y shape of q Y shape of t q Y shape of g™ “1 Y point e.g. (1;-1 ; | ¥ point e.g. (t;-1) (43) Y point e.g. (1;-2) x -1 G) CEPARTMENT OF BASIC EDUCATION -06- 20 Copyright reserved 2013 06 2 Please turn over PRIVATE BAG X 110 PRETORIA 0004 PUBLIC EXAMINATIONS

Downloaded from hlayiso.com Mathematics/HG/P 1 10 SCE — Memorandum DBE/2013 2.3 B(2;9) 2.3.1 | Substitute into _k=(29)= 18 (1) _| ¥ answer 2 24 23.2 yotvr’—-x 7? = (2)? +9 =85 ¥ formula (positive) —— v 85 a y=4v85—x? v answer OR Y formula (7 +7 =r’) v 85 x’+y=85 and y>0 3) |“ y>0 C032 233|°C? ¥ CQ ;2) -c-Yo Moc = xc —Xo 2-0 m= 9-0 2 _— Y gradi 9 Q) gradient [35] Copyright reserved PARTMENT OF BASIC EBUCATION 2013 -06- 20 PRIVATE BAG X 110 PRETORIA 0007 PUBLIC EXAMINATIONS Please turn over

Downloaded from hlayiso.com Mathematics/HG/P 1 i DBE/2013 SCE — Memorandum QUESTION 3 3.1 | Ifa polynomial f(x) is divided by (ax +5) until the remainder Y state theorem Proof: Any polynomial can be expressed as S(e)=(@% +b).O(%) +R where Q(x) is a polynomial in x and R is independent of x. contains no more x terms, then the remainder is given by /| t 4) F(x) =agx" +. ax") +ayx"? +..44, x +a,, a;€R and neN v express f(x) as product of factor & quotient +R F)=(+8). +R ¥ calle 2 fh 2)=(0.-8+b).0C-2)+R s4) =0+R=R oy | teers 3.2 | f(x) = 4x? + ax? +x -3 Since f(x) is divisible by 4x? -1 fG¢)= (4x? -1)x43) vv -. other factor is (x +3) ¥ and f(x)=4x° +12x? -x-3V a=12 Y and b=-1v © OR 4x? —1 = (2x—1f2x+1) SF (x)= 4x3 +ax? +bx -3, v 1 1 remainder/factor 3] =0 and f- 4 =0 theorem y as : +2)-3 _ Y substitution 2 2 2 Ltt e-3=0 4 FRG) (1) f- i] =0 2 Y substitution 3 2 2 2 2 1414-15320 24° 2 Py) a (2) ()+@Q): 2a=24 ¥ value of a ; = 12 v value of 5 and b=-1 f()=4x3 +12x? —x-3 = (2x41)\2x-1)x+3) oa «. other factor is x +3 {10} LEFARTMENT OF BASIC Copyright reserved EDUCATION Please turn over PRIVATE BAG X 110 PRETORIA 0001 PUBLIC EXAMINATIONS | | 2013 -08- 20 | i

Downloaded from hlayiso.com Mathematics/HG/P 1 QUESTION 4 12 SCE — Memorandum DBE/20}3 ay | ovenx_ = x* —yyx” vxly- vx) “ahs 1 vv G) Y factorising numerator v factorising denominator Vv dividing Y factorising numerator Y factorising denominator Y dividing 17 vY ee PU DEPARTMENT OF BASIC i EDUCATION t nian \ ra ‘ORIA 0001 Please turn over PUBLIC EXAMINATIONS

Downloaded from hlayiso.com Please turn over Mathematics/HG/P] 13 DBE/2013 SCE ~ Memorandum 4.2 p+q pitq 11 =(p+q)+|—+— Pq Y positive exponents (b+2) + =(p+4) (22) Pq =(p+q)x ( = } Y simplification pt+q i'l G)__ |“ answer 2 1 43 143.1 x3 5x3 +4=0 1 1 [? = ‘| = J =0 Y factorising 1 i : PO ee ee (1) ~ both values of x? x=4 or x= aN x=64 or x=1 (4) | ~~ answer (each value) OR 1 Let x3 =k k? -5k+4=0 (e-4Xk-1)=0 k=4 or k=lv 1 J w=4 o x=] x=(4? or x=() — i=. 43.2] log,(x+7)+2log,5=4 log, (x +7)+log, 25=4 logs 25(x+7)=4 Y write as single log 25x +175 =3' =81 ¥ write without logs 25x =-94 94 jae v PC RERARTMENT Or BASIC i EDUCATION Z | \ 2013 -06- 20 H 3 RIVATE BAG X 110 | PRETORIA 6004 Copyright reserved En renga rary Te E 5 P rn necerennceenrne OULIC EXAMINATIONS

Downloaded from hlayiso.com Mathematics/HG/P 1 14 SCE — Memorandum DBE/2013 OR log; (x+7)+2log, 5=4 log; (x +7)+log; 57 = 4log; 3 log, 25(x+7)=log,3* v 2 25(c+7)=3* v 25x =-94 94 v x=-S 25 4.3.3 logi9(2x—5)< logo, (x-3) x>3 and 2x>5 so x>3 logio(2x—5)< —log;o(x—3) log19(2x—5)+ logy9(x—3)<0 log g(2x~-5x-3)< 0 2x? 1x +151 2. 2x? -11x+14<0 . x—7x-2)<0 x>3 and 2exst ae oe 2 OR logio(2x-5)< logo, (x-3) x>3 and 2x>5V sox>3Vv logyq(2x—5) < —log;o(x-3) ¥ 1 logi9(2x—5) < logyg —- x-3 gx-s)s tv x-3 2x? -11x+15<1 (ince x—3>0)% 2x* -11x +1450 v. (2x-7)x-2)< 0% 2. x>3 and asxsiv sexs vv 2 (9) ¥x>3 and 2x>5 vv x>3 V —logio(x—3) v inequality (1) V 2x? -1x+15<1, Y factors Y solutions (2) Y end points v inequalities IF x>3 is missing and answer is given as 3 <x < s award max 5/9 marks Copyright reserved j DEPARTMENT GF BASIC EQUCATION 2013 -06- 20 PRETORIA O01 PRIVATE BAG X 110 PUBLIC EXAMINATIONS Please turn over

Downloaded from hlayiso.com Mathematics/HG/P 1 15 SCE — Memorandum DBE/2013 44 Let log,x=M «.a° and log, y=N -.a ¥Y writing in exponential form coxypsa” aN =qtt¥ Y exp. law - log, xy =M+N Y log, y=M+N =log, x+log, y (4) [26] QUESTION 5 5.1 |1343;7;10;... a=1 and d=3 5.1.1 Tr=a + (n-1)d v value of a Tos = 1 + (66-1)3 Y value of d = 196 v answer (3)_| Answer only: 2/3 marks 5.1.2 | Even sequence: 4; 10; 16;22;... ¥n=33 a= 4; d= 6 and n=33 Y value of new a Sy ="ha +(n-1)d] Y value of new d 2 3 Y formula 53 =F P@)+ 63-6] (6) | V substitution = 3 (8 +192) =3300 v answer OR Tas =4+32(6)=196 VV %4 n S,=—la+T,] ¥ . 2 7 Y formula 33 S33 =—|4+196} Vv (33 and 196 Dery @ Aco) ¥ “substitution of 33 &196 =3300 Vv ¥ answer © 2 D4... : 5.2 {5.2.1 x53 *=15+5+34+3+ Iino ¥ expansion . a=15; r=t expansion, ' : give 1 mark v value of r Se = 2 for a=15 I-r 15 ¥ formula & substitution fe 1 3 a 22,5 v answer 2 @) : DEPARTMENT OF BASIC : EDUCATION Copyright reserved { Please turn over 2013 -06- 28 PRIVATE BAG X 110 PRETORIA 0061 PUBLIC EXAMINATIONS

Downloaded from hlayiso.com Mathematics/HG/P | 16 DBE/2013 SCE — Memorandum 5.2.2 55.327 =15+54+54+84---45.9?" = zeae r=l 81 S, = _ali-r”) Y formula l-r fF] 1820 ¥ substitution LHS 1- ao “8, substitution RHS n 1-(3)- 1820 2 GF 0,998628.., we -(tY' = 0,998028... (LV 28 3 729 Wee 728 | =| =1-——=— =0,001371... (3) 729 729 v Qf =.= 0001371... n 6 (2) - 4 Aine 1og0,001371.. 3 : log 3 v answer n=6 Answer only: 2/6 (6) 5.3.1 | Salary: a = R108 000, d = R8 640 Tr=a + (@-—l)d Y formula Ts = 108 000 + (8-—1)8 640) V substitution = R168 480 v R168 480 -. gross monthly salary = nae =R14 040 (4) | ¥ answer OR Present monthly salary R9000 ¥ Monthly increase R720 “ Gross monthly salary = 9 000 + (8 — 1)720 ¥ =R14 040 ¥ 5.3.2 | Expenditure: a= R72 000, r -(1 +3] =11 . ¥ value of r Ss —a-r Y formula 0 1-r ae 72 ooo(t-f1,1F) Y substitution into formula = 1-11 ° % = R823 383,94 ~ R823 384 (4) | ~ nswer DEPARTMENT OF BASIC I EDUCATION i 2013 -06- 20 Copyright reserved i PRIVATE BAG X 110 Please turn over i PRETORIA 6001 : ™BURLIC EXAMINATIONS

Mathematics/HG/P I Downloaded from hlayiso.com 17 SCE - Memorandum DBE/2013 OR Expenditure: a= R72 000, r -(\ +3) =11 v value ofr S, _a”-1) Y formula r-l 7 72 ooo Lik - 1) Y substitution into formula a oAd= I y = R823 383,94 = R823 384 neil {26} QUESTION 6 61 f= lim ZEt4-20) Y formula h0 4 = limi (c+? - Ql -x? Y substitution into formula rev) h 2 = lim accu eLe -1 mark if Y simplification moh lim = L&e#-/@) a(- 2x- h) noo h = lim ———— hoo h Answer only, 7 _ - lim (-2x- nh) no mark ¥ factorising & dividing by h =-2x (5) | ¥ answer OR f(@)=1 -x? S(x+h)=1-(x+AP ¥ 1 (eh? =1-x? -2xh-7? Sle+h)— f(e)=~2xh 1? v ~2xh-#? Oy) = fim LEta-£) f @) a) : Y formula = lim h(—2x—h) h>0 h gy an = lim(-2x-A) factorising & dividing by h hv =-2x ¥ answer 2 62 }o21| y=*-3 2 1» 3 Y exponential form 3x7? ype-x* -3x Pr ; vv derivative (each term) oe ol ara : , ee) 3{-32*) - ifNO 2 or y =x+9x7 @G) CEPARTMENT OF BASIS EDUCATION Copyright reserved Please turn over 2013 -06- 20 PRETORIA 0004 PRIVATE BAG X 110 "Senate Arter asaeennerare tar PUBLIC EXAMINATIONS

Downloaded from hlayiso.com Mathematics/HG/P I 18 DBE/2013 SCE — Memorandum 3 3 6.2.2 x)=| x+—= fx- = an Aeron x? ~— ‘pansion ve Y ial fc ay exponential form =x? 9x3 ie g(x) =2x+ 6x73 (4) YY derivative (each term) 63 Wx) = 2x3 + 7x? +4x-4 At point of contact h'(x)= 4 ¥ e)=4 (xe) = 6x? +14x4+4=4 v derivative 6x? +14x=0 Y factors 2x(3x+7)=0 =-2 x=0 or x=-} Vxs-F . x= (x<0) AL 2)= 2-2) +7-2f +4{-2)-4 ( ) i. ( iy ( 3) ¥ substitution of x=-2 SOT Y y-coordinate i i Tid Therefore point of contact is C $-# VF AFC) ceesreee (1) . yu a at Ire Y substitution into eq. (1) c= 2817 3 27 . = 235 =8,70 (8) Y value of c [20] PARTE) Fs rat EDUCATION.” 2013 -05- 20 PRIVATE BAG xX PRETORIA gooT~ PUBLIC EXAMINATIONS Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/HG/P 1 19 DBE/2013 SCE ~ Memorandum QUESTION 7 qa F(x)=a3 +bx and g(x)=x. Turning point (—1 ; 2) TA SED =aC1)P +6C1)=2 —A-b=2 eee i a-b=2 a ¥ Eq (1) f'd= Bax? +b Fi a ¥ derivative 7) — -1 marki Y f'-l)=0 fCy=0 derivative not rey) equal to zero fC) =3a(-1? +5 =0 V Fa. 2) 3atb=0 was. (2) Q)+@: 242 oe) ee “ a=i Dia) Substitute in (1) or (2): =-3 (6) a2 qi2 F(x) =x3 -3x I “(x) =3x?-3 "(x)= 0 J ®) Y derivative =0 3x°-3=0 3(x—-1Xx +1)=0 x=1 oF ee Y both values of x AtC, f(l)= (1 -30)=-2 Y y-value (C) -. C(1; -2) (3) | Answer only: full marks OR 9 fEx)=al - x3)+a(-x) = ax? + bx) =-f(&) .. f is symmetric about the origin (~ 1 2) is a turning point ¥ -.(1_; 2) is the other turning point ¥ 713 At A fx) =e) @-3x=x ¥ equating in terms of x x -4x=0 x(x —2)(x + 2) =0 Y factors x =Oorx=2o0rx=-2 At A, x=-2 Y values of x oye? v y-value at A =y-value atC .. A & C have the same y-coordinates Y ya=Ye .. AC is parallel to the x-axis (y = 0) or mac=0 OR eee ae 2) 29 a 274 TT DEPARTMENT OF BASIC :. AC is parallel tothe x-axis EnucaTion | (6) 2013 -05- 20 Copyright reserved . Please turn over PRIVATE BAG X 110 PRETORIA 0001 i s_ PUBLIC EXAMINATIONS

Downloaded from hlayiso.com Mathematics/HG/P I 20 DBE/2013 SCE ~ Memorandum 7.1.4 | x-intercept of f f(x)=x7-3x=0 v f(e)=x° -3x=0 x(x- V3\x+ v3)= 0 x=0 or x=V3 or x=-V3 1 & BL43;0) BL3;0) and _D{v3;0) 3) _|% 2b3;0) TAS -~V3<x<0 VV -V3<x<0 or Yor “ x> 3 x>v3 (4)__| (CA from 7.1.4) 7.1.6 [(0; -3) (2) _| “each coordinate 72 7.21 VU =12—2x Y length SV =xtan60°= 3x v breadth Area = length x breadth - Y substitution in formula A= (12-2x)03 x) = V3 x(12—2x) square units (3) 722 A=12W3x-23 x? oo. a) iad Area will be maximum when: a =0 aa ay IK dx Y derivative ie. 12V3-4 V3x=0 4J3x=12V3 y 4 value of x b OR from Eq. (1): 2 oF v v answer = 13, y 2-23 =3V A= ¥3.3(12-2.3) A=18y3 or 3118 sq.units | (4) (30} POP Sarees - a MENT OF BASIC ~~ - Pe fos - e ann EUCATION | i F | 2013 -05- 20 i PRIVATE Bag x | ipso OMG X 44 fomopremee PRETORIA D004 Lead SBLIC EXAMINATIONS Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/HG/P 1 21 DBE/2013 SCE — Memorandum QUESTION 8 ¥ x>1 8.1 v y<s SEE GRAPH BELOW VY 8y <60—-4x VV 2x<18-y ¥¥ shading of feasible (8) __| region 8.2 | T=600x+400y (2)_| “answer 3 if oo Y correct gradient 83 | 8.3.1 oe mm (dotted line: SEE GRAPH) Y line at optimum position (2) _| (CA from 8.2) 8.3.2 | 7 Comfort beds and 4 Deluxe beds (1) __| ¥ answer {13] TOTAL: | 200 20 7 15 n a kl i} 3 D 8 wa wa f 10 o a be fe o io} 7 Fa = : a n, = 3 zs Z 3 a) (050) 10 15 NUMBER OF co a BEDS | SEPARTMENT GF 4 i. - JP B, 2013 -06- 20 Copyright reserved i PRIVAT Bac i PR ORG ue L_ PUBLIC E (AMINATIONS