Mathematics Grade 12 P1 Pretrial FS Mock 2023 Paper & Memo

Downloaded from Stanmorephysics.com education Department of Education FREE STATE PROVINCE MARKS: 150 TIME: 3 HOURS Stanmorephysics This paper consists of 12 pages.

Qownloaded from Stanmorephysics.com FS/Mock Sep 2023 Gr. 12 Mock Exam INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the question paper. 1. This Question Paper consists of 11 questions. Answer all questions. AnswenALL the questions. Cleartyshow all the calculations, diagrams, graphs, et cetera, which you have used in determining the answers. An approved scientific calculator (non-programmable and non-graphical) may be used, unless stated otherwise. If necessary, answers should be rounded off to TWO decimal places, unless stated otherwise. Number the answers correctly according to the numbering system used in this question paper. Diagrams are not necessarily drawn to scale. Full marks will not necessarily be given to answers only. It is in your own interest to write legibly and to present the work neatly.

Qownloaded from Stanmorephysics.com F5/Mock Sep 2023 Gr. 12 Mock Exam QUESTION 1 11 13 1.5 Given x* +2x=0 Solve for x: 2x* —3x-—7=0 (correct to TWO decimal places) Given: k+5 -4 1.3.1 Solve for k. 14 1.3.2 Hence, or otherwise, solve for x if Vx+5+5= : vx+5 Solve for x and y simultaneously if: x—2y—3=0 and 4x? —Sxy+y? =0 The roots of a quadratic equation is given by x = of k for which the equation will have real roots. —2 tv4—-20k 2 . Determine the value(s) Q) (3) (4) (3) (3) (7) (2) [24]

Qownloaded from Stanmorephysics.com F5/Mock Sep 2023 Gr. 12 Mock Exam QUESTION 2 2.1 The first four terms of a quadratic sequence are: 1; —5;—13; —23 .... 2.1.1 jate the general term of the sequence. (4) 2.1.2 h term has a value of -6437 G3) 2.2. 2k +1; 3k ; 5k—S5 are the first three terms of an arithmetic sequence 2.2.1 Calculate the value of k. (2) 2.2.2. Write down this sequence in sigma notation for the first 20 terms. (3) 2.3 Given the geometric series: 8x7 + 4x’ + 2x" +... 2.3.1 For what value(s) of x will the series converge? (3) 8 23.2 Calculate the values of x if S.= 3. (4) 2.4 ‘The first, third and thirteenth terms of arithmetic sequence are the first 3terms of a geometric sequence. If the first term of both sequences is 1, determine: 2.4.1 the first three terms of the geometric sequence if r > 1 2.4.2 the sum of 7 terms of the geometric sequence if the sequence Al 2; (6) (2) [27|

Qownloaded from Stanmorephysics.com FS/Mock Sep 2023 Gr. 12 Mock Exam QUESTION 3 Sketched below are the graphs of f(x) = 2% and g(x) = —(x — 1)? + q, where q isa constant. > The graphs off, intersect at C and D. C is the y-intercepfjof both fand g. D is the point of 3.1. Show that g = 2. (2) 3.2 Write down the coordinates of the turning point of g. (2) 3.3 Determine the value(s) of ¢ for g(x) = t if the roots are equal. () 3.4 Write down the equation of f~?(x) in the form y = ... (2) 3.5 Sketch the graph of f~* on the same system of axes. Indicate the x-intercept and the (3) coordinates of one other point on your graph. 3.6 Write down the equation of f if A(x) = g(x +1) -2. (2) 3.7. How can the domain of h be restricted so that h 1 will be a function? qd) [13]

Qownloaded from Stanmorephysics.com F5/Mock Sep 2023 Gr. 12 Mock Exam QUESTION 4 The ske! low shows the graph of f(x)= oe A and B are the x- and the y-intercepts eee of f. eames ae ove Z 3 41 Write the equation of fin the form f(x) = part +4q. (3) 4.2 Determine the equations of the asymptotes of f- (2) 43 Write down the coordinates of A, the x-intercept of f (2) (7]

Qownloaded from Stanmorephysics.com F5/Mock Sep 2023 Gr. 12 Mock Exam QUESTION 5 5.1 Given: f(x) = 3-* 5.1.1 Write the equation of f~1(x), the inverse of f(x), in the form (2) x) = 5.1.2 the same set of axes, sketch the graphs of | (x) and #"(x) [4] our ANSWER BOOK. Clearly indicate the intercepts with the s and one other point on the graphs and label the TWO graphs. 5.2 A linear function satisfies the following conditions: p(—3) = 10 and (4) p'(x) = —2. Determine the inverse of p in the form p~1(x) = ++ [10 QUESTION 6 6.1 On 30 June 2013 and at the end of each month thereafter, Asif deposited R2 500 into a bank account that pays interest at 6% per annum, compounded monthly. He wants to continue to deposit this amount until 31 May 2018. Calculate how much money Asif will have in this account immediately after depositing R2 500 on 31 May 2018. (3) 62 Onl February 2018, Genevieve took a loan of R82 000 from the bank to pay for her studies. She will make her first repayment of R3 200 on 1 February 2019 and continue to make payments of R3 200 on the first of each month thereafter until she settles the loan. The bank charges interest at 15% per annum, compounded monthly. 6.2.1 Calculate how much Genevieve will owe the bank on 1 January 2019. (3) 6.2.2 How many instalments of R3 200 must she pay? (5) 6.2.3 Calculate the final payment, to the nearest rand, Genevieve has to pay to (5) settle the loan. [16]

Qownloaded from Stanmorephysics.com F5/Miock Sep 2023 Gr. 12 Mock Exam QUESTION 7 7.1 If f(x) = —2x?, determine f’(x) from first principles. (5) 7.2 Determine: 7210 @ = 2e?+1 (3) vx 7.2.2 x — 2)? (3) . 1 7.3 Given: yaa Q) Prove that the gradient of the curve is negative at each point on the curve where x>0. [14] QUESTION 8 The following information about a cubic polynomial, y = f(x) , is given: 8.1 8.2 8.3 8.4 f(-1) =0 f2)=0 fQ)=—-4 f(0) =—2 f'(-1) = fd) =0 ifx <—1then f'(x) >0 ifx > 1 then f’(x) >0 Use this information to draw a neat sketch graph of f using the grid on the (5) DIAGRAM SHEET For which value(s) of x is f decreasing? Use your graph to determine the x-coordinate of the point of inflection of f For which value(s) of x is fconcave up? (2) (2) (2) (14)

Qownloaded from Stanmorephysics.com FS/Mock Sep 2023 Gr. 12 Mock Exam QUESTION 9 A window framéiwith dimensions y by / is illustrated below. The frame consjsts/of six smaller frames. +—_y—_> 91 If 12 m of material is used to make the entire frame, show that y = ‘ (12 — 3h). 9.2 Show that the area of the window is given by A = 3h— ch 9.3 dA Find — and hence the dimensions, h/ and y, of the frame so that the area of the window is a maximum. Q) (3) (5) [10]

Qownloaded from Stanmorephysics.com FS/Mock Sep 2023 Gr. 12 Mock Exam QUESTION 10 Consider the word)*CALCULATOR” 10.1 10.2 10.3 How many different word arrangements can be made from all the letters of the word (2) CALCULATOR? What is the probability of making a word arrangement that will start and end with (2) the letter L? In how many ways can all the letters be arranged if no similar letters should be close (3) to each other? 7] QUESTION 11 11.2 Given P(A) =0,5; P(B) =x and P(A or B) =0,88. Calculate the value(s) of x, if 11.1.1 Event A and B are mutually exclusive. (2) 11.1.2. Event A and B are dependent. (3) South African Women Football Team known as “Banyana Banyana” are to play a friendly football match against their Nigerian counterparts known as “Super Falcons”. There should be a winner on the day of the match. Penalties would decide the winner in case the match is drawn after regulation time. On the day the match takes place, there is a 30% chance that it could rain (R), 45% chance that it could be sunny(S) or it could be cloudy(C). Banyana Banyana has a 24% of winning ona rainy day, a 65% of winning on a sunny day or a 33% chance of wining on a cloudy day. 11.2.1 Draw a tree diagram to represent all outcomes of the abave scenario. (3) 11.2.2 What is the probability of Banyana Banyana winning the match? (3) [11]

Qownloaded from Stanmorephysics.com F5/Mock Sep 2023 Gr. 12 Mock Exam Name and Surmame: .........00...066:00cccccccceseessceeeeeeeeeneeecseessestenneesrsseness Class: DIAGRAM S QUESTION 8.

Qownloaded from Stanmorephysics.com F5/Mock Sep 2023 Gr. 12 Mock Exam INFORMATION SHEET ye DEVE —4ac 2a A= P(l+ni) = P(1—ni) A=P(\-i)" A=P(l+i)" T, =a+(n- S, =52a+(n-Nd] T, =ar"" gs <M) pel S,=—4;-ler<l . r-l l-r a i S(x+h)—f(x) = in —————— fQ)- oA d= (x, -%)' +02 -»,) w(S22222) y=nmerte y-y, =mMx-x,) m=22 m=tané x, x, (x-a) +(y-by =y* b 2_,? 2 1 InAABC: an me a’ =b'+c°-2be.cosA area AABC = = absin sin(a+ 8) =sin a.cos f+ cosa.sin B sin(@— 8)=sin a.cos#—cosa.sin B cos(a+f)= cosa@.cos #—sin a.sin § cos(a—2)=cose.cosf+sin asin § cos’ a—sin* a cos 2@ =41—2sin* a sin 2a = 2sin a.cos@ 2cos?a-1 a 2 _ x ee yo oo =f n n P(A) = 78) P(A or B) = P(A) + P(B) — P(A and B) p=atbx pone“) De-x)°

Downloaded from Stanmorephysics.com education Department of Education FREE STATE PROVINCE MARKS: 150 This marking guideline consists of 14 pages. 1

Downloaded from Stanmorephysics.com QUESTION 1 11.1] x@+2)=0 r= x=0 OR/OF x=-2 el Vx=-2 (2) vx? + 2x> 0 ¥ graph/grafiek/ critical pts v¥x>0 OR/OF —+|—+—_|+- If x > 0 only | mark Indien x > 0 slegs 1 punt -2 0 ax>0 (3) 1.2. ~(-3)+,/(-37 - a ¥ substitution into x= Aa correct formula/ (2) substitusie in 34 V65 korrekte formule *y 4 ¥ simplify —~ oe ¥2,77 x¥=2,77 OR/OF x=-1,27 v1.27 If left in surd form only 2 marks. Indien in wortelvorm, slegs 2 punte. (4) 1.3.1} 4745k-14=0 ¥ multiplying (k-2)(k +7) =0 Y factorising k=2 o k=-T7 ¥ both solutions (3) 13.2] Jx+5=2 of Vxt+5=-7 Y substitution x+5=4 or _ invalid/no solution ¥ identifying invalid x=-l Y value of x KH 2H 3H veasases 1 14 4 o Y equation (3) 4x7 —Sxyty?=0 ..... (2) From eq (1): x = 2y +3 ...(3) Subst. (3) into (2): + A(2y + 3)? - Sy(2y +3) +y? =0 « 4(4y? + 12y + 9) — 5Sy(2y +3) + y* =0 « l6y? + 48y + 36 — 10y2 —15y+y* =0 = Ty? + 33y + 36 =0 + (Ty + 12)(y + 3) =0 —12 HY= we OLY S -3 bs => orx = -3 substitution Y simplification Y standard form Y factors Y both y values Y both x-valul (7)

Downloaded from Stanmorephysics.com 1.5 For real roots : ¥4-20k>0 4—20k >0 “20k > —4 v Answer Q) w~k< 1 “te a ‘ — QUESTION 2 2.1.1 | 13-5; —-13; -23 ¥ Value of a -6 -8 -10 ¥ Value of b ea ie Y Value of c 2a=-2 3a+b=-6 at+b+c=1 a=-1 3(-1)+b=5 -1-3+c=1 b=-3 c=5 Tr=—w —3n +5 ¥ General term (3) 24.2 | -7 = 3n + 5= 643 ¥ equating w +3n — 648=0 fants (n+ 27)(n — 24)=0 n=—27 orn=24 2.Trq = -643 y answer (3) 3k —(2k + 1)=5k—5—3k ¥" Difference k-1=2k-5 v 2.2.1 eet answer (2) 2.2.2 | 912,15 v Sequence ¥ General term T, =3nt+6 20) ¥ Sigma Notation (3)

Downloaded from Stanmorephysics.com 2.3.1 ¥ value of r ¥ Condition for converging series ¥ answer (3) 2.3.2 i ¥ — Substitution into the S.7 a correct formula x i> 2 8 = 1-2)- 8x? 3 2 8 8 2 Le 3 6 16 - 8x = 48x? ¥ — Standard form 2 = 48x* + 8x -16=0 Apaches 6x" +x-2=0 (3x +2)2x -1)=0 ¥ Answers 2 1 4) x=-= or x== 2 241 a+2d=ar at+l2d=ar a+2d=ar 1+2d=1r 1+12d=1/7 ¥ a+i2d=ar _r-l Folia v values of d de 2 142 (3 ei Y — standard form Tore =r v factors P= 6s 0 Y values of r (r= Ir — 5) =0 v sequence r=lorr=5 eeSen2y OR r=1+2d ———» 1+12d=7 1+ 12d=(1 +2dP BEAL ¥ at+2d=ar 4d —8d=0 Y atl2d=ar 4d(d — 2) =0 Y¥ values of r WE OGRae Y — standard form r=14+20) r=1+2(2) Y factors | =5 v values of d Le 5: 25 ¥ sequence (6) ies

Downloaded from Stanmorephysics.com 2.42 a(r" —1) ¥ substitute in correct eae ia | formula aes S,= 16° -)) 5-1 = 19531 Sane be Q) L Tava} QUESTION 3 3.1 y=2* ¥ substitute x = 0 and y=0 x=0 into correct formula -. f(0) =2°=1 2. C(0;1) y - intercept: a(=—(x-l) +4 i oa C(0;1) 1=-(0-1)' +q l=-l+q tS ee er pr ¥ Symmetry E(4 ; 0) 3.2 g(x)=-(x-1)° +2 v answer /antwoord DQ; 2) 3.3 t=2 ¥ answer/antwoord (1) 3.4 fol ix=2” ¥ ¥ answer/antwoord “y= log 2x 3.5 Y form of the graph ¥ /vorm van die grafiek ¥ X—intercept/afsnit (0:1) phe (231) ¥ any other coordinate on —- the graph : Xx 0/150) /enige ander koordinaat op die grafiek —t iF 3.6 g(x) =—(x—1)? +2 v d 2 g(t+1)-2=-(x- 141)? +2-2 VRMr/anwoord — 2) “ A(x) = -x? 3.7 Domain: x>0 or x <0 v Answer (1)

Downloaded from Stanmorephysics.com QUESTION 4 4 4.1 fart x=l y 22143 x-1+3 x-1 J (x)= an v a oo x-l f(x)=——+1 v FG)e 2241 x-1 x-1 (3) 4.2 x=l vx=l yel v y=1(2) 4.3 o-2%* 7 y=0 x-l 0=2+x x=-2 ear A(-2:0) aS 252) [07] QUESTION 5 5.1.1 ( ) Pope ) f"(x)=log,x OF f-'(x)=—log, x : | Slegs Antwoord:Volpunte ¥ ¥antwoord @) 5.1.2 \ } (13)\f | ¥ vorm ¥ y-afsnit {7100 = logax: 3 ¥ vorm T Y x-afsnit sb (4) f'@) 5.2 p(-3)=10 en p'(x)=-2 p(x)=-2x+e P ipeg 10=-2(-3)+e Cea -e=16 “3 omruil vai p(x)=y=-2x+16 (sam apat x=-2y+16 ¥ antwoord (4) rs 1 ‘(x)=-ax+8 Pi(x)=-5 [10]

Downloaded from Stanmorephysics.com QUESTION 6 6.1 ¥ n= 60 and i= 298 0,005 Y correct substitution into correct formula v answer GB) = RI74 425,08 6.2.1 After eleven months, Genevieve will owe/ Vn=l11 Na elf maande skuld Genevieve Y correct substitution A =82.000 1+ 0,15 Mt into correct formula 12 ¥ answer =R94006,79 (3) 6.2.2 x[1-(+i)"] ¥ 94006,79 p=_—___. is -"] v substitute into correct 2am -(14918) | formula 12 94006,79 = a 12 94 006,79 0,15 0,15)" xx 2 = 1-| 14+ 3200 12 12 ¥ correct use of logs (logs to (: : a) =1-0,3672147... be defined) 12 0,15 =! —nlog| 1+ —— | =log0,6327852.. v n=36,84 12 ¥ 36 installments —n = —36,8382... (5) n= 36,84 Genevieve will have to pay 36 installments of R3 200

Downloaded from Stanmorephysics.com 6.2.3 xl-(1+i)” r P= vn = —083826912 ¥ substitute into correct eal (4 01 ee] forrniala : ¥ answer P=2652 Outstanding balance after 36 installments is R2 652 i Final payment will be: v 2652.0 z, ae A= 2652,00114.2!5) . a on Calin ¥ answer =R 2685,00 OR/OF OR/OF 015 2 0,15)" 3.200 [1 us)" -1 94006,7! 2 0,15) 12 Balance : 94006,7{ 1+ - ) - 15 015 36 12 ed 3200 (1+) -1 12 v 12 = R2 651,72 0,15 12 v¥ 2 651,72 Final payment will be: 0,15)" v , A= 26s173{1+ 248) 2osi rf | Y¥ answer =R2 685,00 (5) [16]

Downloaded from Stanmorephysics.com QUESTION 7 7A fi) = lim Lorn F@) ¥ Substitution into correct formula v : , . —2(x+h)2—(—2x2) Expansion P= 1 ¥ Factoring ' « —2atpareh—2h? +2x P@)= a ¥ Applying limits f'(x) = lim v Answer f'(x) = lim f'@) = 4x 721 | y= Bed ¥ Simplified to binomial. ct ; oO Y 3x2 vtix? 3) 1 3 e = 3x2+=-x 2 7.2.2 | Dy[9x? — 12x + 4] 9x? — 12x +4 =18x-12 v 18x v-12 (3) 73 y=x? v —2x73 WY _ oy 3 = 2 v2 dx ax" = x3 * Therefore <0 forx >0 v oe <0 forx >0 (3) [14] QUESTION 8 [|

Downloaded from Stanmorephysics.com 8 ‘ v (-1;0) ‘ v (2;0) ; Y (0:-2) v4) v shape : _ : 0) (5) 8.2 | -1<x< 1 (2) 8.3 Pt. of inflection: “halfway” between turning points(x- mate values) v¥x= 2 ig weno) 2 8.4 f (x)>0 Yf'(x)>0 ax >0 ¥x>0 (2) [11] QUESTION 9 109A. LI] donat Et 10

Downloaded from Stanmorephysics.com 9.1 There are four horizontal bars and ¥ 4y43h three vertical bars. Y 4y+3h=12 0) => 4y+3h=12 4y =12-3h y= A (12 —3h) = 4 Im 9.2 Area = lb v Aayh =yh ¥ Subst of y =1(12-3iyh Y simplification eae G) 4 9.3 Area =3h asp Y derivative roaee | dA 3 = =3-=h dh 2 : aA a ¥ equating to 0 ae 3-2 n=0 i as ~ 1 ¥ subst of h y=7(12-3@)) Y value of y 33 (5) 2) [10] 11

Downloaded from Stanmorephysics.com QUESTION10 10.1 10! _ 222 453600 yo! 21.21.21 v Answer naa . 2) 10.2 0080 ore v Answer (2) 10.3 Same letters are together in ¥ Same letters together 2.2.2.7! ways = ! Probability At = < 38 So required probability is = OR v Probability of same letters together v Answer OR ¥ Same letter together v Same letter not Same letters not together is 41380 together ways v Answer Probabilit ired i ad 8) obability required 18 45 [07] QUESTION 11 11.1.1 | P(A or B)= P(A) + P(B) subst into correct 0.88 = 0.5+x formula ee eee answer (2) 11.1.2 | P(A or B) = P(A) + P(B)— P(A).P(B) subst into correct 0.88 =0.5+x—-0.5x formula 0.38 =0.5x 0.5x x=0,76 answer (3) 12

Downloaded from Stanmorephysics.com ¥ 25% for C 11.2.1 ¥ Correct outcomes w— RW 242 to represent 30% R 76 ogee BAL win/lose 9 v Correct diagram 65% a seta I (3) 5% S$ % 35% w— sw 3 Wr ote cw 679 ib acai 3 | 11.2.2 Probability of winning v P(W) = P(R and W) or P(S and L) or P(C and W) =30% x 24% + 45% x 65% + 25% x 67% 1889 = 0.9445 OR 2000 30% x 24% + 45% x 65% + 25% x 67% v ~ answer (3) (14) 13