Mathematics Grade 12 KZN March 2018 Memo

Education KwaZulu-Natal Department of Education I REPUBLIC OF SOUTH AFRICA SENIOR CERTIF GRADE 12 go FE SRS SE SSSR ESS SSSR Es SR ESS SEES | i MATHEMATICS ' * COMMON TEST * * MARCH 2018 : See ee eee i MARKS: 75 TIME: 1% hours N.B. This question paper consists of 5 pages and an information sheet. Copyright Reserved = Please Turn Over |

Downloaded from hlayiso.com Mathematics 2 March 2018 Common Test NSC INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. ls This question paper consists of 6 questions. 2s Answer ALL questions. 3. Clearly show ALL calculations. diagrams, graphs, et cetera that you have used in determining your answers. 4. Answers only will not necessarily be awarded full marks. 5. An approved scientific calculator (non-programmable and non-graphical) may be used, unless stated otherwise. 6. If necessary, answers should be rounded off to TWO decimal places, unless stated otherwise. 7. Diagrams are NOT necessarily drawn to scale. 8. Number the answers correctly according to the numbering system used in this question paper. Write neatly and legibly. Copyright Reserved Please Turn Over

Downloaded from hlayiso.com NSC Mathematics QUESTION 1 5; 12; 21 ; 32 ;... isa Quadratic Sequence. 1.1 Write down the next term of the sequence. "term of the sequence. 1.2. Determine the n 1.3 Which term of the above sequence is 1152? L.4 Prove that none of the terms in the sequence are perfect squares. QUESTION 2 2;5;8;5 115... isan Arithmetic Sequence. 2.1 Determine the first term that will be greater than 2012. 2.2 Calculate the sum of the first 671 terms of the sequence. QUESTION 3 ° z in ot I-r" 3.1 Prove that the sum to n terms of a Geometric series is given by S,= ar) : -F 3.2 The first term of a geometric series is 12, the last term is Bee and the sum of the -. 6141 . . . . series is a6 Determine the common ratio and the number of terms of the series. March 2018 Common Test a) (4) (4) GB) [12] (3) (G3) [6] (4) (6) [10] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics 4 March 2018 Common Test NSC QUESTION 4 I 5 ‘ The graphs below are f(x) =—x° and g(x) =— | ‘i +3. A(a; b) and B(e: d) are points(s) 5 + = x of intersection f and g. ax 4.1 Write down the equations of the asymptotes of g. (2) 4.2 If y = x+c isa line of symmetry to the graph of g, calculate the value(s) of c. (2) 4.3. Write down the range of fi qd) 4.4 Calculate the value(s) of x for which g(x) > 0. (4) 45 Write down the x — value(s) for which f(x) = g(x), x 2-1. (2) 4.6 Determine the value(s) of x if f(x) < g(x), for the interval x 2-1. (2) 47 If A(x) =—f (x + 2) +1, then write down the new equation of / in the form A(x) = a(x + py +q. (2) 48 Use your graph to determine the maximum value of gids (2) 4.9 Determine the value(s) of k for which re +5) = khas one root equal to 0. (2) . [19] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics March 2018 Common Test NSC QUESTION 5 5.1 Sketch the graph of /(x)=3" ona set of axes in your answer book. (2) 5.2 Write down the equation of the inverse of f in the form aren (2) 5.3. Sketch f'' on the same set of axes indicating the intercepts with the axes and the line of symmetry with the graph of f. (3) 5.4 Write down the equation of g(x) if g(x) = f'Cx). (2) 5.5 Determine the value(s) of x for which f~'(x) >—-1. (4) [13] QUESTION 6 6.1 How long will it take for a motor vehicle to depreciate to half its original value if the rate of depreciation is 12,35% p.a., on the diminishing balance method ? (Give your answer in years and months). (3) 6.2 On the 1 July 2017, a businessman wanted to buy a new house that costs R900 000. He took out a loan to the value of the house with a bank at an interest rate of 10.25% p.a. compounded monthly for 20 years. The bank allowed the businessman to start repayment towards the loan in three months time and indicated that his first instalment will be deducted on the 1 October 2017. 6.2.1 Calculate the value of the loan on 1 September 2017. () 6.2.2 Determine the monthly repayments if he now makes 238 monthly payments. (3) 6.2.3 Ifhe does not make the 100", 101%, and 102" payments, calculate the new instalment that he will have to make to settle the loan in the fixed time period. (7) [15] TOTAL MARKS: 75 Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Mathematics 6 March 2018 Common Test NSC INFORMATION SHEET: MATHEMATICS _p+vn? —4ac 2a A=P(+ni) A= P(=ni) A=P(-i)" A= P(1+i)" T, =at(n—-ld S, = 52a) T, =ar" Ss _ alr’ 1) ; r#l &, a4 ler n a lr pas +i" -1 p—al=0+i) "J i i fi) = lim F(xt+h)— f(x) h>0 h d= (x, =x)? HO, -¥,)* H(i gt 324) a 2 v2) ponte yoy, =m(x-x)) m= m=tan@ eK (x-ay +( s-by =r? f é £ 2 =? 4.6? —2he.cos A area AABC = asin C = = — a sn A sm Bo smc In AABC: sin(a+ B) = sin a.cos 8 + cosa.sin £ sin(@ — B) = sin a.cos f — cosa@.sin # cos(a + £)= cosa@.cos f -sin asin B cos(a — 2)=cosa.cos # +sin asin 8 2 es cos’ a@—sin” @ cos2@ =41—2sin° a sin 2a = 2sin a.cosa@ 2cos’a-l s (x, = x) ves n # P(A) = ae P(A or B) = P(A) + P(B) - P(A and B) Me Dsatbx p= KID) Copyright Reserved

Downloaded from hlayiso.com Education KwaZulu-Natal Department of Education REPUBLIC OF SOUTH AFRICA " MATHEMATICS " COMMON TEST " MARCH 2018 « MEMORANDUM NATIONAL | SENIOR CERTIFICATE GRADE 12 MARKS: 75 TIME: L.5 hours This memorandum consists of 9 pages. Copyright Reserved Please Turn Over |

Downloaded from hlayiso.com Mathematics 2 March Common Test 2018 QUESTION 1 [il 145 }Av (1) 1.2 5 \ A “32 Oo ; 1D 7 \ / \ / 2D 2 2 2a=2 val AY avalue Batb=7 ..b=4 CAM value atb+c=5 ..c=0 CAY c value T= w+4n CAM answer OR (4) OR T, =T, +(n—Vd, _ hada 2), - “ =) d, AY formula =54(n—-I(1)+ (n a =2) (2) AY substitution into correct formula =547n-T+H -3n+2 CAY simplifying ata (4) . owen “ CAY answer 13 } n+4n=1152 ~ i /M¥ equating to 1152 n’ +4n—-1152=0 my CA” standard form (n+36)n-32)=0 we lias | CAY factors n=-36 or n=32 nla CAV n values and rejecting (4) 14 Suppose i 4An= 1? ho MY equating 1" term to n? n=0 ’ i 4 AY value of 1 Bute N Kas | : Ves | Therefore no such 1 exists and no terms are, <j fh perfect squares in the sequence. '| AYfeason (3) [12]

Downloaded from hlayiso.com Mathematics 8 QUESTION 2 March Common Test 2018 \ 5, = a(l=r") l-r = 2.1 T, =3n-1 AY n term=3n—1 3n-1> 2012 or 3n-1= 2012 AY setting up inequality /equation 3n > 2013 3a = 2013 n> 67 n=671 CAY answer The 672™ term will be the first term to be greater than 2012. (3) 2.2 = AY Sq formule 8, == Pat (n-Dd] onmre 2 AY substitution of a, n and d into 671 a - Sen = -—[2(2) + 670(3)] correct formula 2 _ CAY answer = 675697 G) OR OR T, =a+(n—l)d Tey, = 2+ (671-3) = 2012 n AY Sn formula 8, = 5k +7] AY substitution of a, n and 7, into 671 correct formula Son = se +2012] CAY answer = 675697 (3) [6] UESTION 3 3.1 S,satartar +..+ ar tar > AY setting up equation | nS, =artar tar +....+ar"| tar" > (2) | AY setting up equation 2 Fd () -(2):S, -1S, =a-ar" AY subtraction LHS and RHS f a \ Sir) = aly AV factorizing LHS and RHS 4)

Downloaded from hlayiso.com Mathematics 4 : March Common Test 2018 3.2 a=12 3 ar” a 2r't=— >() . . 256 256 AY setting up equation S = I2d-r') = 6141 > (2) AY setting up equation " l-r 256 iil From (1): lar" = =—_ Opt = 3r >) AY making 12r the subject r 256 Substituting (3) into (2) 3r 56 _ 6141 I-r 256 CAY substitution of 127” into (2) 3r 6141614 Ir 256-256-256 3072 3r _ 614i G14Ir 256 256 256 ©6256 61387 = 3069 CAY r value CAY nr value (6) OR

Downloaded from hlayiso.com Mathematics March Common Test 2018 a=12 3 ar"! =—— [2r"'= 2. >) 256 256 120—r" 14] 5, - A) l-r 256 From (\): 12" 3 t 2% = Sr"= r r 256 1024 Substituting (3) into (2) ae 1024) 6141 l-r 256 it» 1024. _ O14) l-r 3072 1 6141 6141 1-—_r= ———7 1024 3072 3072 6141 | 1 6141 , _ 3072 1024" 3072 6141 3 6141 3072 p= r= aa 3072 3072 »=—- 3072-3072 6138 _ 3069 3072 3072 I pet 2 1y\" 1V1 1 2) ae ha}™ a n=11 AY setting up equation AY setting up equation AY making 7” the subject CAY substitution of ” into (2) CAY r value CAY n value (6) (10) |

\ Downloaded from hlayiso.com J \ | Mathematics : 6 9 QUESTION 4 March Common Test 2018 4.1) x=-l , 5} p AY vertical asymptote Doe cA y=3 ‘ ry AY horizontal asymptote (2) 42 yoxte AY substituting (-1 ; 3 ) 8 elie CAY c— value (2) c=4 43 y20. or yel[0;0) AY answer “(D 4.4 —— sh a 0 Ifa candidate atl . AY determining LCD and =14+3%43 50 cross multiplies by HUERATOR ; i (xt) only — 0/4 3X + wen . pe . 20 Tt CAY simplification x+1 (4) 2 CAYCA VYanswers x<-lL oor x2- = 3 OR OR j I 41 +3=0/, AY setting up equation to calculate | xX — intercept 3= x+1 3x4+3=1 3x S2 2 x= a CAYx — value 2 x<-l oor V2 op CAY CAY answers 3 (4) X=a or X=C Av AY answers (2) If Candidates solved simultaneously - i ; have irrational roots - full marks 45 GSxSC AAY Veritical values, inequality (2) ( If Candidates solved simultaneously .- f- “A 416 have irrational foots - full:marks Ay

Downloaded from hlayiso.com Mathematics _ March Common Test 2018 4.7 1 2 AQ) = ~z(«+2) +1 AAY Yanswer (2) 4.8 Maximum value of A(x) =1 CAY 1 Therefore maximum value of 3!" CAY 81 =31=81 (2)

Downloaded from hlayiso.com Mathematics March Common Test 2018 49 Ji, a Be +5) =k Answer only lo +5 =k full marks AY substituting #= 0 3 25 i AY answer (2) —=k 3 : [19] QUESTION 5 5.1 : yo Ff AY shape i AY y= intercept = 1 1 * 4 y Ko , ’ 4 F y 4 ‘a ’ @) 5.2 / 1 yo log, x =—log, x = log, ~ AAYY answer 3 ~ (2) 5.3 see graph in 5.1 to allocate marks f':CAY shape AY x—intercept at 1/(1;0) AY y=x line G3) 5.4 g(x) =log,(-x) ;.x<0 OR 3 g(x) =—log,(—x); x<0 OR g(axy= toe (—} x<0 me, ¥VCACA answer [any form] (2)

Downloaded from hlayiso.com Mathematics 9 March Common Test 2018 5.5 f'@)2-1 log, x2—-1 or log, aes al CAV setting up incquality/equation 3 3 x<3 x=3 CAV x-— value =3 (4) O<xs3 CACAY V critical values and inequality : [13] QUESTION 6 6.1 A=P(l-i)" oP = P(1—0,1235)" If P=tPU — 01235)" AY substitution into correct formula 1 Breakdown ~ 0/3 — = (0,8765)” 2 N = 1085 9765 > MY use of logs n = 5,258342252 years Time period will be Syears and 4 months CAY answer (3) OR OR A=P(l-iy” 3? = P(1—0,1235)" AY substitution into correct formula _ (0,8765)" 5 , logs = log(0,8765)" MY use of logs , = 10808765 7 1 log— 1 = §,258342252 years Time period will be Syears and 3 months CAY answer (3) 6.21 | 4=P(+iy" _ -_ =900 0001 + 0,1025 } AY substitution into correct formula 712 = R915440,66 CAY answer Q)

Downloaded from hlayiso.com Mathematics «ds ee March Common Test 2018 6.2.2 pe a= (4+i9") t AY substitution of i and value [ ( aes)" into correct formula xj 1-] 1+ 12 CAY substitution of P into correct 15440,66 = . aus 0,1025 formula 12 y (3) x = R9009,43 CAY answer 6.2.3 | Balance outstanding after the 99" payment : — afl-(L+"] ‘ 8 AY substitution of 1 =139 into 9009,43] 1—| 1+ 0,1025 : correct formula ; 12 CA” substitution of x into correct = 0,1025 formula 12 CAY R731384,7351 = R731384,7351 Value of the loan afier 3 months of non - payment : 0.1025\ CAY substitution into correct A =73138,735 (1 + | = R750287,0104 formula . CAY R750287,0104 The new monthly instalment to settle the loan : 7 136 0,1025 750287,0104 = ———_______= AY substitution 1 =136 into correct 0,1025 — : formula 12 x = R9349,11 We , CAY answer (2) [15] Total Marks : 75

Downloaded from hlayiso.com 623. Out-Bal = A - fy = pAyT — x Canard ——__- GY = 4iSy 40, bh (i+ id, 2k) 100443 [C4 9a) i| ig 2% [a = 2124513, %0s — 1393 428,418 =R7 31384 585 46 oad Value of Loan fue Zmuth = TBI3EY 5 y eu (1+ lyase) of Non payment = 78026 7, 16ds Sa New mom Unk payment x = TSO287, 166 X ib,25% 12 if | - + Ions me] = 93-49 108827 R43451 Oo

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