Limpopo-Maths-Grade-11-June-2023-P1-and-Memo_hlayiso.com_.pdf

Downloaded from hlayiso.com WVowhloaded Tro anmorepnysics.co LIMPOPO PROVINCIAL GOVERNMENT REPUBLIC OF SOUTH AFRICA Seseeeeest @uzegeeeezeseens 2a2xe2eeeeneesess MARKS : 100 DURATION : 2 hours This question paper consists of 7 pages including 1 diagram sheet. Copyright reserved Please turn over

Downloaded from Stanmorephysics.com | Downloaded from hlayiso.com impopo Province/June 2023 Grade 11 INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. 2. 3 This question paper consists of 6 questions and | diagram sheet. Answer e questions. Clearly ALL calculations, diagrams, graphs, et cetera that you have used in determi’ ur answers. Answers only will NOT necessarily be awarded full marks. You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise. If necessary, round off answers to TWO decimal places, unless stated otherwise. 7. Diagrams are NOT necessarily drawn to scale. Number the answers correctly according to the numbering system used in this question paper. Write neatly and legibly. Copyright reserved Please turn over

Rewnloaded from Stanmorephysics.com Downloaded from hlayiso.com Grade 11 QUESTION 1 1 Solve for x in each of the following: Limpopo Province/June 2023 La 2x )=0 (2) 11.2 3 x =4 (correct to TWO decimal places) (4) i =xt+1 (5) 1.1.4 2x? +5x < 3 (4) 2 1.1.5 5x5 =20 G3) 12 116.4 -ZH4 (5) 1.2 Solve for x and y simultaneously: Sy-x=2 and x*-3xy+4y=4 (6) [29] QUESTION 2 2.1 Simplify the following, without using a calculator: bE 8\3 211 (=) 3) 2.1.2 (VIZ +2)(v3-1) @) g2xti . 152*-3 2.1.3 Zyxt. gx, gare (4) WITHOUT usi alcul how th 2S implifi 2+2. 22 using a calculator, show that "> + > simplifies to / (4) 2,3 Determine the value of m if x? — mx + (m+ 3) = 0 has equal roots. (5) [19] Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com Limpopo Province/June 2023 Grade 11 QUESTION 3: iven: poy pee Given: f(x) = = 2 3.1 Write dow, equation(s) of the asymptote(s) of f. qd) 3.2 Determin -intercept of f. G3) 3.3 Sketch t h of f on the attached diagram sheet. Clearly show ALL the intercepts with the axes and the asymptote(s). (3) 3.4 Determine the equation of the axis of symmetry of f having a positive gradient. (2) 3.5 The graph of f is transformed to obtain the graph of h(x) = =. Describe the transformation from f to h. 2) 3.6 Write down the domain of h. qd) {12} QUESTION 4 Given: f (x) = —x? + 6x +7 4.1 Determine the coordinates of the turning point of f. (3) 4.2 Write down the equation of the axis of symmetry of f. qd) 4.3. The graph of f is shifted 4 units to the left and reflected in the x-axis to form h. Write down the equation of h in the form h(x) = a(x + p)* + q. (2) [6] QUESTION 5 , ane 5.1 Given: g(x) = (3) -4 5.1.1 Write down the equation of asymptote of g. (1) 5.1.2 Write down the range of g. qd) 5.1.3 Determine the coordinates of the x- intercept of g. (3) 5.1.4 Hence, write down the values of x for which g (x) < 0. qd) Copyright reserved Please turn over

Downloaded from hlayiso.com Rewnloaded from Stanmorephysics.com Grade 11 Limpopo Province/June 2023 5.2 The point A(3 ; 54) lies on the graph of f(x) = 3**? — 27. 5.2.1 Determine the value of p. (3) i ql) (3) [13] QUESTION 6: The diagram shows the graphs of f(x) = —x* + 2x +15 and g(x) =—-3x+k. © Graph f cuts the x-axis at A(—3 ; 0) and B(5 ; 0), the y-axis at C and has a turning point at D. e Graph g cuts the x-axis at B and the y-axis at C. e Eis a point on g such that DE is parallel to the y-axis. P op c A(—3:0), 6.1 Show that k = 15. a) 6.2 Determine the coordinates of D, the turning point f. (3) 6.3 Determine the values of x for which f is increasing. dd) 6.4 Calculate the average gradient between points A and D. (2) 6.5 Calculate the length of DE. (2) 6.6 If h(x) = f(x — 1) —2, determine the equation of h in the form: h(x) = a(x + p)? + q. (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Bewnloaded from Stanmorephysics.com Limpopo Province/June 2023 Grade 11 6.7 For which values of x is: 6.7.1 f(x) 20 qd) 6.7.2 f¢ g(x). (2) 68 Determi aximum value of p(x) = 3f0-22, (3) 6.9 Determii values of k for which f(x) + k = 0 will have two distinct real fools: (3) [21] TOTAL: 100 Copyright reserved Please turn over

Downloaded from hlayiso.com Rewnloaged from Stanmorephysics.com Grade 11 Limpopo Province/June 2023 DIAGRAM SHEET Name of learner: Question 3.3. Se aa nm wo > oa co2) ™~ oe $7654 3 24 o —_. nN oo oS on on CO ' ery hi & © Nob & oo (4) Copyright reserved

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com LIMPOPO PROVINCIAL GOVERNMENT REPUBLIC OF SOUTH AFRICA EDUCATION MARKS : 100 DURATION : 2 hours These marking guidelines consist of 10 pages including the cover page Copyright reserved Please turn over

Downloaded from Stanmorephysics.com Downloaded from hlayiso.com Grade 11-Marking Guidelines Limpopo Province/June 2023 QUESTION 1 1.1.1 | 2x(x-— 3) =0 ¥ x=0 “x=O0orx=3 v x=3 (2) 1.1.2 | 3x? —2an= 4 3x? S292 4=0 Y standard form _ AGL (-2)?=4(3)(-4) Y — substitution into correct * = 23) formula x = —0,87 or x = 1,54 v V answers (4) 113) /5—x =x4+1 v . . (Vo—x y = (x +1)? squaring both sides Sox =x74+2x41 x? +3x-4=0 Y standard form @+)Q-)=0 v factors x#—4or x= 1 Y _ both solutions to x axel v rejecting x = 4 (5) 1.1.4] 2x7+5x <3 2 2x +5x—- 3.50 Y — standard form (x+3)(2x-1) <0 1 Y critical values CV: x = —30r x=5 ALTERNATIVE METHOD: No Yes No +4 = + 3 i 2 —_ 1 — —3SxS5 vv answerlll (4) Copyright reserved Please turn over

Downloaded from Stanmorephysics.com Downloaded from hlayiso.com Grade 11-Marking Guidelines Limpopo Province/June 2023 1.1.5 5x5 = 20 e=4 Y divide both sides by 5 22 os vY _ apply exponent law (x5) @’y: Ipply exp din x= RO ¥ x=32 ALTERNATIVE METHOD: ALTERNATIVE: 2 xs=4 2y* 2)5 (x5) = (27) Y aise both sides to power 5 x2 = 210 x = ¥210 Y apply exponent law x=25 x = 32 x = 32 (G3) 1.1.6] 2% Ba _ LCD 2x (2*)? —12 = -4.2* v (2*)?-4.2* -12 =0 (2*)? +4.2* -12 =0 v (2* + 6)(2* — 2) <0 factors 2* #-6 or 2%= v 2X 4-6 “x= vo x=1 ALTERNATIVE METHOD: ALTERNATIVE: Let k = 2* 12 k =-4 k v k-method k? —12 = —4k v 2 — = k?+4k—-12=0 Keak 12 = 0 v fact (k + 6)(k—2) =0 fevers 22° #-6 or 2%=2 v axa v (5) Copyright reserved Please turn over

Downloaded from hlayiso.com Dowrleaded from Stanmorephysics.com Limpopo Province/June 2023 Grade 11-Marking Guidelines 12 ]Sy-x=2 () x? -—3xy+4y =4 (2) v generate 3 equation x=S5y-2 (3) Sub (3)/ant (2): x? “= 4y=4 (5y+ aye 3y(S5y — 2) +4y-4=0 Y — substitution 25y? —20y +4-15y? + 6y+4y-4=0 10y? — 10y = 0 v — standard form 10y(y—1) =0 v factors y=0 o y=1 v both solutions to y v . x=5(0)-2=-2 or x=5(1)-2=3 both solutions to x 6 [29] QUESTION 2 2 21d (2) _ (2) 27) ~ \Xaz. 2 = (iE Y apply exponent law = () v method 3 —+ v answer “9 ALTERNATIVE METHOD: ALTERNATIVE: 2 2 (2) = (5) v prime factor base = (2) Y~ method 3 =‘ Y answer (3) 9 2.1.2 | (V12 + 2)(V3 — 1) = (2v3 + 2)(v3 - 1) Yo 2 naTaN = 2.3-2V3 +2V3 -2 Y¥ 23- Wa WT -2 =6-2 aay =4 v answoe—l Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com Grade 11-Marking Guidelines Limpopo Province/June 2023 ALTERNATIVE METHOD: ALTERNATIVE: 12 +2)(vV3 —1) =v36 —2v3 +2V3 -—2 (VIZ +2)(V3-1) = V36-2WF+2VF-2 |) ee =6—2v3 +2V3 -2 Y 6-2V3 +2V3 -2 =4 Y answer (3) 21.3 32xtd msax-3 27x-1 Hicaln B2x-4 = oe) v prime bases (3) )* 1 3%, 52x: 4 32xt1 | 52X-3 32x-3 =e ee Y apply exponent law = 32x4142x-3-Bx43—x GB 2x-3-2x+4 v adding and subtracting =3.5 exponents -15 Y answer (4) 2.2 vz 7 v2+1 0 y¥2 _ (x2 JL ve 4. v2+1 . = (Sax et |) Y- Multiply by LCD: _ FAT, ars V2 (v2 +1) ~ 242 24+ ¥2 ; : Y — simplifi _ (V2) + v2 +22 sumpiny ~ 2+Vv2 _ (ve+2)° v factors 24+V2, =24+V2 Y answer ALTERNATIVE METHOD: ALTERNATIVE: Veo 4 vat Va Y Multiply by LCD: _ (x2 ve 4. y241 = (Gaxe)+ ea) VE(VE +1) =—2_4 4v2 +4 242 " 242 vgn _64nE ) 2-vE simplify “242 ~ 2-VE — e+ 2v2 8 ¥Y multiply-with the conjugate 4-2 2-V2 —4+2N2 Pad 2 _ 2(2+y2) “28 Y — factorise (IICT:2) =2+V2 v answer (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Bownleaded from Stanmorephysics.com Limpopo Province/June 2023 Grade 11-Marking Guidelines 2.3 |x? -—mx+(m+3)=0 A= b* —4ac For equal roots: A = 0 VA=0 a 4(m + 3)() = 0 Y substitution 2 = m w=0 ¥ simplify (m —6)=0 Vv factorise “m=-2 or m=6 Y both answers (5) [19] QUESTION 3 3.1 x =3andy=-2 v x=3 and y=-2 ql) 3.2 x-intercept: y = 0 1 0= aT 2 v y=0 24(x-3)=1 2x-G6G=1 v simplify 2x =7 =? vY xa eo 2 (3) 3.2 % Y asymptotes if ¥ — shape 0 3 EX at = v x-and y- intercepts a oN (3) 3.3 y=xte = _ at —-2=3+¢c v - substit of (3; —2) c=-5 2 y=x—5 Y y= (2) 3.4 Translate f 3 units to the left and 2 units up. Y¥ 3 units to the left v2 units up (2) Copyright reserved Please turn over

Downloaded from Stanmorephysics.com Downloaded from hlayiso.com Limpopo Province/June 2023 Grade 11-Marking Guidelines 3.5 x € (—00; +00) ;x +0 v answer ALTERNATIVE ANSWER: x€ R;x #0 (1) [12] z cae UESTION-+- Q Nan 4.1 At tpre= —2 = -_§ = 3 ¥ method 2a 2(-1) Vv x. . ny =—(3)2 +6(3) +7 = 16 x-coordinate Vv . + (3516) y coordinate ALTERNATIVE METHOD: 2 ALTERNATIVE: f(x) =x? +6x+7 2 . v method = —(x%* — 6x — 7) = —[(x? — 6x +9) -9-7] =-— —3)- [@— 3)" ~ 16] v x-coordinate — _fy — 2)2 =—@~ 3)" + 16 Y y-coordinate + (3; 16) ALTERNATIVE METHOD: ALTERNATIVE: x- intercepts at 7 and —1: V method _ 7H(-1) Xrp =~ = 3 Y x-coordinate sy =—(3)? + 6(3)+7=16 v y-coordinate + (3316) (3) 4.2 x=3 v answer () 4.3 | f(x) =x? +6x+7 =-(x-3)?+16 ¥ a=1and q=16 h(x) = (x -— 3 +4)? + 16 = (x+1)? +16 v p= 1 loot (2) INNAT Copyright reserved Please turn over

Downloaded from Stanmorephysics.com Downloaded from hlayiso.com Limpopo Province/June 2023 Grade 11-Marking Guidelines QUESTION 5 5.1.1 | y=-4 Y answer () 5.1.2 | y>-4 v answer (1) 5.13 logy g@) ms) -4 im ) a-inteept y=0 = 3) -4 v substitute y = 0 1 x 4=() 22? =2-* : v mathematical procedure “x= —2 + (-2;30) v coordinate of x-intercept (3) 5.1.4 | x > —-2 v answer (1) 3.2.1 | y = 3**? — 27 54 = 33+? —27 v substitution (3; 54) 81 = 334? 34 = 33+? Y equating indices ~4=3+p v 5 apel answer (3) 5.2.2 | y > —-27 ory € (—27; ©) v answer qd) 5.23 | g@ =-f (x) ¥ g@~)=-f(*) g(x) = -@**1 — 27) = —3**1 427 Y new equation y = —3°1 427 = 24 +. y-intercept at (0 ; 24) ¥ answer (3) [13] QUESTION 6 — {Oy 6.1 |y=—-3x+k ¥ substitutéwith (5 ; 0) 0=-3(5) +k ma ok =15 — (1) Copyright reserved Please turn over

Downloaded from Stanmorephysics.com Downloaded from hlayiso.com Limpopo Province/June 2023 Grade 11-Marking Guidelines 62 |x=-ha eat Vv x=l = ~(4)? +2(1) +15 = 16 Y — substitution Y y=16 ALTERNATIVE METHOD: ALTERNATIVE: BEL, v x=l #y = —()? +20) +15 = 16 Y — substitution +. D(1;16) v yH16 6 6.3 x<1 ¥ answer ALTERNATIVE ANSWER: x € (—«;1) dd) 6.4 A(—3;0) and D(1;16) v subst. into gradient formula 6.5 | D(1;16) and E(1;12) Y E(1; 12) « DE = 4 units Y answer (2) 6.6 | h(x) =f(x-1)—-2 = —-(x-1)? +2(-1) + 15-2 Y -(x-1)?+2%-D+ =x? +2x—-14+2x—-2413 15-2 =—x? + 4x +10 v x? +44x+10 = —(x? — 4x — 10) = —(x2 — 4x | 4-410) =-(x—-2)? +14 Vv h(x) =-(x- 2)? +14 ALTERNATIVE METHOD: ALTERNATIVE: D(1;16) and a=—1 v¥ — substitute D (1; 16) anda = -f@=-@-1? +16 1 h(x) = ft-1)-2 vh@) = -(¢ = 1-1)? + 16— =—-(x-1-1)?+16-2 2 =-(@-2)? +14 Yh) nee 2)* +14 G) 6.7.1 |-3<x<5 v awe qd) Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com Grade 11-Marking Guidelines Limpopo Province/June 2023 6.7.2 |x <Oandx>5 vy x<0 vv x>5 (2) 6.8 Max value of f(x) is 16 v Max value of f(x) is 16 » f(deal2 = 16-12 =4 VY f(x)-12=4 * ma e of p(x) = 3* =81 v answer (3) 6.9 | —x?+2x+15+k=0 b? —4ac >0 vy A>O (2)? — 4(-1)(15+k) > 0 Y correct substitution 4k > —64 k>-—16 Y answer (3) [21] Copyright reserved TOTAL: 100