Mathematics Grade 12 Paper 1 Limpopo June 2023 Paper & Memo

Rawnaaded from Stanmanepbysias.com May/June 2023 LIMPOPO PROVINCIAL GOVERNMENT REPUBLIC OF SOUTH AFRICA DEPARTMENT OF EDUCATION MOGALAKWENA DISTRICT MARKS: 150 TIME: 3Hours Page | 1Copyright reserved Please turn over

Rawnleaded. from Stanmanepbysias.com May/June 2023 INSTRUCTIONS AND INFORMATION: Read the follo’ instructions carefully before answering the questions. 1. This ques' japer consists of 9 questions. 2. Answer luestions. 3. Clearly show ALL calculations, diagrams, graphs, etc, that you have used in determining your answers. 4. Answers only will NOT necessarily be awarded full marks. 5. If necessary, round off answers correct to TWO decimal places, unless stated otherwise. 6. You may use an approved scientific calculator (no programmable and no graphical). 7. Diagrams not necessarily drawn to scale. 8. An information sheet with formulae is included at the end of the question paper. Page | 2Copyright reserved Please turn over

Rawnleaded. from Stanmenepbysias.com QUESTION 1 1.1 Solve 1.10Dfx - 3)(5x + 2) = 0 1.1.2 3x?- 10x- 1 = 0 (Correct to two decimal places) 113 x(x- 4) > 21 14 2342 2 1.2. Solve for x and y simultaneously. x+2y=2 x? + 8y =8 1.2.1 Given: 2**! +2*=3)*2_}, where x and y are integers. Determine the value of x and y. 1.2.2 10 {x2 43= eee e-2 1.3. Given: , -ttye—@ 1.3.1 If p = 4, determine the nature of the roots. 1.3.2 Determine the value(s) of p for which the roots are non-real. 1.4 The equations x? +rx+m=0 and x° +mx+r=0 have real and Solve for the values of r and m if r > 0 and m>0. Page | 3Copyright reserved May/June 2023 (2) (3) (3) (4) (6) (3) (5) (1) Q) roots. (6) [37] Please turn over

Rawnleaded. from Stanmanepbysias.com May/June 2023 QUESTION 2 The sequence: —67 ; x ; y ; —28 ; —19; ... has a quadratic pattern. 2.1 Dete the value of x and y 2.2 Find pression for the nth term 2.3 Pro\ it the sequence of numbers will never contain a positive term. QUESTION 3 3.1 The fourth term of a geometric progression is 24 and the ninth term is 768. Determine the first three terms of the Progression. 3.2 IfsS,, = 207 +3n then 3.2.1 Calculate 7}, 3.2.2 Find 7, im its simplest form. 3.3 Caleulate the value ofS (2n _- 1) 3.4 40 Ina converging geometric series S,, = and 7, = > calculate the z possible value(s) the first term of the series. 3.5 Given the following geometric sequence: 24 = © 12+ 6x + 3x7+... 3.5.1 Calculate the sum to infinity of the series 3.5.2 Write down the values of x for which this sequence converges. (6) (4) (3) [13] (5) (3) (4) (3) (3) (4) (2) [24] Page | 4Copyright reserved Please turn over

4.1 4.2 4.3 44 45 4.6 4.7 Rawnleaded. from Stanmanepbysias.com May/June 2023 QUESTION 4 The grap) f(x) and g(x) are shown in the diagram below. / \ 7 The turning point of f(x) is A (2 ; 9) and the graphs f and g intersects at B (3; 8). C is a point on g(x) and is on the axis of symmetry of f. Show that the function f can be defined by the equation: (4) fx) = —x?+4xe4+5 Write down the equation of the axis of symmetry of f. (1) The graph g(x), has the equation y = a*. Determine the value of a . (2) If it is given that (—1; 0) is one root of f, write down the coordinates of the other root. (1) For which value(s) of x will f(x) < 0? (2) Determine the length of AC. (2) Discuss the nature of the roots of h(x) if, h(x) = f 9. (2) [14] Page | SCopyright reserved Please turn over

Rawnladed. from Stanmenepbysjas.com May/June 2023 QUESTION 5 In the diagram below are the sketches of: f(x) = —2x +2 and g(x) = —+ q x+p The graph cuts the y-axis at (0; 4). 5.1 Calculate the values of p and q (3) 5.2. Show that a = —2 (2) 5.3 Determine the coordinates of point A the x — intercept of g(x) (2) 5.4 The value of q is increased by 1 unit. What effect will it have on the graph of g? = (1) 5.5 Write the equations of the asymptotes. (2) [10] QUESTION 6 1 x Given: P(x) = (3) 6.1 Write down the equation of p=! inthe form y = -- (2) 62 Sketch the graphs of p and p~! on the given system of axes on the diagram sheet. () Show clearly all the intercepts with the axes as well as the line of symmetry. ©3 Determine the value(s) of x for which p71(x) > 1 (2) Page | 6Copyright reserved Please turn over

Rawnaaded from Stanmanepbysias.com May/June 2023 64 Write down the equation of the graph of h(x), if h(x) is formed by shifting the graph of p(x) 2 units to the right and then 3 units up. (2) [11] QUESTIO 71 Determine f'(x) from first principles if f(x)=5—2x* . (5) 7.2 Determine 2 if dx 72 ay ag (3) 7.2.2 xy=5 (2) 7.2.3 so [37 —7x—s [ ; J 6 [14] Page | 7Copyright reserved Please turn over

Rawnleaded. from Stanmenepbysias.com QUESTION 8 Given that = x3 — 6x? + 9x — 4 with x-intercepts at A (1 ; 0) and B, y-intercept at C(0 ; AandD turning points of f. yu f(x) A(1;0) Be C (0;-4) D 8.1 Calculate the coordinates of B. 8.2 Calculate the coordinates of the turning point D. 83 Calculate the x-coordinate of the point of inflection of f. 8.4 Write down the values of k for which f(x) = k will have only ONE root. 8.5 Determine the values of x where f '(x) > 0 8.6 Determine the gradient of the line DC. Page | 8Copyright reserved Please turn over May/June 2023 -4). (4) (5) (2) (2) (3) (1) (17

Rawnladed. from Stanmenepbysjas.com May/June 2023 QUESTION 9 Two cyclists start to cycle at the same time. One starts at point B and is heading due north to point A, whilst the other starts at point D and is heading due west to point B. The cyclist starti B cycles at 30 km/h while the cyclist starting from D cycles at 40 km/h. The between B and D is 100 km. After time ¢ (measured in hours), they reach points F and ively. A q Fe O * B Cc D 9.1 Determine the distance between F and C in terms of ¢. (4) 9.2 After how long will the two cyclists be closest to each other? (4) 9.3 What will the distance between the cyclists be at the time determined in (2) QUESTION 9.2? [10] Page | 9Copyright reserved Please turn over

Downloaded from Stanmorephysics.com LIMPOPO PROVINCIAL GOVERNMENT REPUBLIC OF SOUTH AFRICA DEPARTMENT OF EDUCATION MOGALAKWENA DISTRICT MARKS: 150 Page 1 of 18

Downloaded from Stanmorephysics.com INSTRUCTIONS: NOTE: ¢ This memorandum consists of 19 pages including the cover page. e If the’candidate answers a question TWICE, only mark the FIRST e Consistent accuracy applies in all aspects of the marking memorandum. Page 2 of 18

Downloaded from Stanmorephysics.com QUESTION 1 1.1.1 x=3: 2 v¥ ¥each root . 5 ‘al —b+Vb?-4ac 2a = ~(10)+ ¥C10)?7-4)(-1) ¥ subst 2(3) _ 104 Viz ¥ Yeach root = 4 a Pama dil (-1, if incorrect rounding) ‘ , (3) 1.1.3 xv -4x>21 ¥std form . _— e 0 0 ¥ factors (x — 7)(x + 3)> a x<-3andx>7,xeR *eritical pts ¥ ¥ conclusion (5) 114 | 23°45 = 72 (x 2) p+ gra Sy? 2.2.37 + = >*F 4.F+F=15 ¥ simplify 3° (441) =15 Y factorisation 3=3 ¥ simplify x=l ¥x=1 (4) Page 3 of 18

Downloaded from Stanmorephysics.com 1.2 XF DY =D vrrcersee (1) P+8y=8 (2) x=—2yt2...(3) ¥ substitution Subst 3 in 2. (-2y + 2)(-2y + 2) + 8y=8 4y’ -8y+4+8y-8=0 4yr-4=0 ¥std form y-1=0 Y factorisation v- Dy bs “a vy-value y=lory=-1 Subst y= 1 in3 Subst y =— 1 in3 x=-2(1)+2 x=-2(-1)+2 x=0 x=4 v¥ vx —values. Solution (0 ; 1) (43-1) (6) 1.2.1 2 +2 = pet = 2°(2' +1) =3’G?-1) Y factorise 2*(3) =3'(8) z= a) ees a 2 3 ¥ simplify equated 2 =3y" bases “x-3=0 and y-1=0 “x=3 and y=1 ¥ answers OR OR ge +2" — 37? ir 2(2' +1) =3°(3?-1) rise 2*(3) =3' (8) 72) — 2993 implified equated 2*(3)=3*(2°) bases “x=3 and y=1 v answers (3) Page 4 of 18

Downloaded from Stanmorephysics.com 1.2.2 fe—3 4300 x—-2 Let k= k+3= v k? 43k-10=0 standard form (k+5)(k-2)=0 v factors k=-5 or k=2 Y_ both answers for k Vv ; Vx-2#-5 or Jx-2=2 selection 2 2 (ve=2) =(2)° ape Y answer OR Je-24+3=—8 x-2 (Vx=2)(vx=2)+3Vx=2 =10 x-2+3V¥x-2 =10 3 v simplified both sid Ix-2=12-x 2 Y squaring both side (3Vx=2) =(12-x) 9x-18=144-24x42 x? ~33x+162=0 Y standard form (x-6)(x-27)=0 F tee x=6 or x#27 lection (5) n Page 5 of 18

Downloaded from Stanmorephysics.com ¥ equal & rational (1) ase A= 16-4p = 16-4(4)=0 +: Roots equal rational — YA<0 (2) 13.2) a=] <0 non-real vp>4 <-16 >4 1.4 x +rxt+m=0 and x°+mx+r=0 x +rxtm=0 For real and equal roots, A= 0 b? -4ac =0 (r¥ —4(@)(n) = 0 r?>-4m=0 r?>=4m r m= -- () x°+mx+r=0 b* -4ac =0 nm? —A(1I(r)=0 m’?-4r=0_... (2) Substitute (1) in (2) 2 2 (= —4r=0 4 4 7 -ar=0 16 r*—64r=0 rr —64)=0 Y substitute into A=0 Y equation for m Y equation 2 Y substitute for m Page 6 of 18

Downloaded from Stanmorephysics.com r(r—4)(r? +4r +16) =0 ¥ value of r “7=4 r n=— 4 4 =—- 4 m= ¥ value of m OR x +rx+m=0 and x*°+mx+r=0 For real and equal roots quadratic must be a perfect square. x +rxtem=0 . (x+4m) =0 (x+ Vm) 5 x +2ymx+m=0 r=2Nm Y equation for m —=m ra 0 | x +mx+r=0 (x+F) =0 y : equation 2 x 4+2Wr+r=0 m= vr m=4r +2) Y substitute for m Page 7 of 18

Downloaded from Stanmorephysics.com Substitute (1) in (2) 4 r* a | 16 r*-64r=0 r(r’ -64)=0 r(r—4)(r? +4r +16) =0 “rs4 a|%\ = 1 a|*% 3 ll os v value of r ¥ value of m (6) [37] Page 8 of 18

Downloaded from Stanmorephysics.com QUESTION 2 21| —6 Fa \ \ rll 7 Y first differences Caw vs 2x—-67 -—2y+x-28 y+3 Y second differences awe ¥ equating second mnie differences / aes vx-value x=-52 ¥ equating second y-2x-67=-2y+x-28 differences/ 3y—-3x=39 y-x=13 i vy-value y=-39 6) 2.2 -67 -52 -39 -28 15 13011 -2 -2 2a=-2 a=-1 ¥ value of a 3a+b=15 - =1 —— ¥ value of b b=18 a+b+c=-67 —1+18+c=-67 lue of c c=-84 T, =—n" +1884 ¥ answer (4) Page 9 of 18

Downloaded from Stanmorephysics.com 2.3 T,>0 v¥—n’ +18n-84>0 —n+ 84>0 n?-1 4<0 ¥ n° -18n+84<0 no Solution ¥ conclusion ~. the nce will never contain a positive term (3) [13] QUESTION 3 3.1 T, =24 T, = 768 vA - rt T, 768 = r v 18 =r 24 24 32=r° ¥ simplification 2? =r’ vr r=2 Lop or Ta 7, T, 24 _ ys 768 _ a a 24 768 a=— = 8 256 tz a=3 a=3 T, =ar"" T, =3-2"" n v T3 3 & so... OR (5) Page 10 of 18

Downloaded from Stanmorephysics.com T,=24 T, =768 T, =ar"" Ty Or? = Wa ccsscemesccesase’ pation V equations T, = CS equation2 eq e 1 ar® =768 ar’? =24 r> =32 vr =32 ao vr r=2 ar? =24 or ar’ =768 a-2?=24 a-2° =768 a=3 a=3 va 7.20 T, =3-2"" 3 6 1B ¥ T&T; 3.2.1 | Fiz = Siz — Sir “T,=S2—S, _ a ~ 278 v substitution . v answer (3) 322| T,=S,—S,., YT=5.-S,, =2n? +3n—[(n-1) +3(n-1)] Faibetitetions =2n? +3n—[2n? —4n+2+3n-3] =2n? +3n-2n?+n+1 =4n+1 ¥ simplification v answer (4) Page 11 of 18

Downloaded from Stanmorephysics.com 3.3 is Sen 1) 3454+7+... a=3 S, =F Pa+(n-1a] Yaad d= =17f.34+(17-12] 2 ¥n=17 n=17 =323 answer (3) 3.4 s, -40 7-3 3 rs a 40 —= et yt &, = T, =a BE) f=ar=7 a=a-r) r= = aC - =) “substitution 3 2a 40 100 ¥ standard form rr ar ¥ factors ¥ answer(both values of a) (3) Page 12 of 18

Downloaded from Stanmorephysics.com aes) a Y value of r x - &* 4, 3X 4 x 1 6x 2 S. = Y — substitution in correct r formula 4 S.= —*— lz Y correct numerator 48 Y correct denominator So = 2x —x’? (4) 3.5.2 | It exists when: ind «3 ie: -—l<r <1 7 Ari § ot <2 v answer ANSWER ONLY: Award full marks. (2) [24] QUESTION 4 4.1 f(x) = a(x -2)?+9 YT, =2" a(3-2)?+9=8 feat a+9=8 vb 1 » f(x) = -1(x- 2)? +9 v Answer f(x) = -1(x? -4x +449 f(x) = -x? +4x-449 f(x) = -x? +4x4+5 Page 13 of 18

Downloaded from Stanmorephysics.com 42|x=2 v Answer (1) 43 g(x) = a* ¥ Substitution a=8 a*=2 ¥ Answer a=2 (2) 4.4 | (5; 0) ¥ Answer (1) 45 |x<-lorx>5 Mark per answer v¥x<-1 ¥x>5 (2) 4.6 | g(x) = 2% g(2) =27 =4 ¥g(2)=4 AC =9-4 AC = 5 units ¥ Answer (2) 4.7 | Roots will be real and equal ¥Real ¥ Equal (2) [14] QUESTION 5 5.1 x= intercept of the line: = 2x+2=0 « x=1 ¥x int p=-l Yp : . “q y-intercept of line: y = 2 ge? @) 5.2 - v i g(x) = +2 ‘Substitute point x-1 ' a (0:4) doit 2=4 v a= -2 (2) Page 14 of 18

Downloaded from Stanmorephysics.com 5.3 ¥Subst y = 0 Sue 2 °A=(; (2:0) ¥ Answer (2) 5.4 | The graph moves up 1 unit ¥ Answer (1) [8] QUESTION 6 6.1 ¥ Interchange ¥ Answer (2) 6.2 Intercept of p ¥ Intercept of p~* v Line: y=x ¥ Shape of p and pivy (5) Page 15 of 18

Downloaded from Stanmorephysics.com 6.3 0 < 1 S*Sq Y critical points ¥ answer (2) 6.4 x-2 ws —2 h(x) +3 wr aS (2) —" [11] QUESTION 7 7.1 ve) =m Lt W-SLO) I'(x= a 2 v ituti 5-2(x+h)'-(5-2x°) substitution = lim ——_—_ a h Y expansion = lim 2 4A 2h" = 5+ 2x" = h Y simplification <lim —4xh-2h° ho h Y notation and lim (—4x—2h) ._ h(—4x-2h) hoo = lim ———. ms h v answer = lim(-4x-2h) ho =-4x (5) 7.2.1 25" =7x*+ Ba x aie ; =7x*+2x? v 2x i o. = 28x? +3x? Y 28x° v at (3) Page 16 of 18

Downloaded from Stanmorephysics.com 722 | *v¥=5 F J r= x » =S5x v y=5x" v answer [aTaYey (2) 723 wp [32-18-6 -Tx- -4| V3x-7 ¥ 6x" _ x _ 4 ¥ 3and differentiating constant = D,[3x-7-6x"' | sr =3+6x° ”) [14] QUESTION 8 8.1 x3 —6x?+9x-4=0 (x -1)0?-5x+4) = 0 @-)@-4)A-1) = w= I) X=1;x=4 (x? -Sx+4) B(4 ; 0) (x — 4)(x — 1) (4) .f(@) = -4 ¥ coordinates D: (3; -4) 8.2 | (x) = x9 -6x?+9x-4 f (Qo) = 3x2 -12x+9=0 3x2 —12x +9 x*-4x+3=0 ¥=0 -(-4) ¥ {C42 - 4D) ¥ subs. into = 2d) formula x=3;x=1 = 4 *fG)= -4 « D(3;-4) = (5) 8.3 | f(x) = 3x?-12x+9 % 12 =0 f"(x) = 6x-12=0 wee ) ND 8.4 | k <-4 orlof k>0 wk <-4 (2) “k>0 Page 17 of 18

Downloaded from Stanmorephysics.com 85 | x <1 or/of x>3 v¥x<1 Yor ¥%x >3 (3) 8.6 |0 v (1) tra Jana? [17] QUESTIO! 9.1 Afters hours: BF = 301 km and CD= 401km -Y BF =30t -.BC = 100-401 ¥ BC=100—40r Fc = (Gory + (100 — 407)" ¥ Pythagoras = V900r? + 10000 — 80007 + 160077 = 425007 — 80007 + 10000 “answer (4) 9.2 FC is a minimum when FC is a minimum. Y BF =30t ¥ BC =100—40r 2 2_ FC* = 2500r° — 80007 + 10000 Py eenisseen 2 are = 50007 — 8000 = 0 dt “answer t =—— =1,6hrs (96 minutes) (4) 5000 9.3 FC = 4250077 — 80007 + 10000 ¥ subs into equation = 42500(1.6)° — 1.6)+ 10000 - ¥ 00(1.6)° — 8000(1.6) + Vanswer They will be 60km apart. (2) [10] Page 18 of 18