Mathematics Grade 11 Paper 1 KZN November 2024 Paper & Memo

Downloaded from Stanmorephysics.com KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE MLA’ Siahnorephysics.corn MARKS: TIME: 3 hours This question paper consists of 8 pages and 1 information sheet. Copyright reserved Please turn over

Mathemp\ic hloaded from Stan Re ygigs.com KZN/November 2024 INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. This question paper consists of 10 questions. Answer ALL the questions. Number the answers correctly according to the numbering system used in this question paper. Clearly show ALL calculations, diagrams, graphs, etc. which you have used in determining your 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 correct to TWO decimal places, unless stated otherwise. Diagrams are NOT necessarily drawn to scale. An information sheet with formulae is included at the end of the question paper. Write neatly and legibly. Copyright reserved Please turn over

MathempjicsPhloaded from Stan vgigs.com KZN/November 2024 QUESTION 1 11 Solve for x: Orie) x?—7x=-1 (answers correct to TWO decimal places) (4) 1.12 =x? +2x+24<0 (4) 113 (2*-1)(2x-1)=0 (3) (4) (5) xo +m? +4m 2 : Calculate the smallest integral value of m for which the roots are non-real. (4) 1.4 Solve for x if mjrleadeclicdree (4) [28] QUESTION 2 2.1 Simplify fully, WITHOUT using a calculator. (4) (3) 2 4 2:2 Solve for x , WITHOUT using a calculator, x* —5x* +6=0 (4) 2.3 Solve for x and y, WITHOUT using a calculator, shy = ap (6) [17] Copyright reserved Please turn over

Mae Swhloaded from Stanmerephysigs.com ecniammeianal QUESTION 3 Consider the linear number pattern: 12 ; 9 ; 6;... 3.1 32 Determine the n“ term (general term) of the linear number pattern. of the linear number pattern. given linear number pattern? Justify your answer with an e linear number pattern are the sequence of first differences ce: 4x+y 5 —2x-1; y+3 3... x and y. QUESTION 4 Consider the quadratic number pattern: -13 ; —1 ; 73.0... 3; -2101. 4.1 4.2 43 44 4.5 Write down the FOURTH and FIFTH terms of the quadratic number pattern. Show that the general term of the quadratic number pattern is T, =-2n? +18n-29. How many terms are in the above quadratic number pattern? Determine the maximum value of Q, if it is given that Q, = 2.7, Prove that all terms in the sequence of first differences are even numbers. Copyright reserved Please turn over (2) (2) (2) (5) [11] (2) (3) (4) (3) (2) [14]

Mae 6whloaded from Stanmerephysigs.com ecniammeianal QUESTION 5 In the diagram, the graph of f(x) = 2 +q_ is drawn. D(4;3) isa point on f and Bis the y- xX+p intercept of f. The asymptotes of f intersect at (251). D(4 ; 3) f 7) Sel Write down the equations of the asymptotes of /. (2) 52 Show that the equation of f is f(x) = oH (2) x- 5.3 Calculate the coordinates of B. (2) 5.4 Determine the equation of the axis of symmetry which has a positive gradient. (2) 5:5 Determine the values of x for which f(x) <0 (4) 1 -t 5.6 The graph of / is transformed to obtain the graph of A(x) = (3 x) ‘ Describe the transformation, in words, from f to A. G3) [15] Copyright reserved Please turn over

Mae 6whloaded from Stanmerephysigs.com ecniammeianal QUESTION 6 The graphs of A(x) =ax° +bx+e and k(x) =—x+6 are drawn below. P and Q(6;0) are the x-intercepts of A and S is the turning point of h. Graphs h and k have a common y-intercept, RM lies on h and N lies on k such that MN is parallel to the y-axis. x 6.1 Write down the coordinates of R. () 6.2 It is further given that PQ is 8 units. Write down the x-coordinate of P. qd) 6.3 Prove that a= -5 , b=2 and c=6 (4) 6.4 Calculate the coordinates of S. (3) 6.5 Write down the range of h. qd) 6.6 Determine the values of x for which = >0 (3) ~*. 6.7 If g(x) =h(x)+d , determine the values of d for which g(x) will always intersect the x-axis. (2) 6.8 Determine the maximum length of line MN when h(x) > k(x) (4) [19] Copyright reserved Please turn over

Mater SWwhloaded from Stanmerephysigs.com ecniammeianal QUESTION 7 is defined by the following properties: 7A e “and g on same set of axes. Clearly show all intercepts with the axes, and asymptotes, if any. (5) 7.2 The coordinates of the point of intersection of f and g are (-1; 4). 721 If f and gare reflected along the line x =1, write down the coordinates of the point of intersection of the images of f and g under this transformation. Q) 122 Determine the values of x for which 0< g(x)<4 (2) 72.3 Determine the values of k for which the roots of —2.f(x)=k are negative. (2) [11] QUESTION 8 8.1 Miss Gwala invested R30 000 at an interest rate of 12% p.a., simple interest. After n years, the accumulated value of her investment was R51 600. Calculate the value of n. (2) 8.2 The total number of bees in an apiary are declining each year at the rate of 2,5% of the total number of the previous year. There are currently 20 416 bees in the apiary. How many bees in total were in the apiary 8 years ago? (Round off your answer to the nearest whole number). (3) 8.3 Mrs Howard invested R10 050 on 1 January 2020 in an account that paid interest at the rate of x % p.a., compounded quarterly. The accumulated amount in the account on | January 2024 was R13 867. Calculate the value of x. (4) 8.4 Mr Mngayi invested R50 000 in a savings account for 7 years. The investment earned interest at the rate of 14% p.a., compounded monthly for the first three years. The rate of interest changed to 13% p.a., compounded half-yearly, for the remaining four years. Mr Mngayi withdrew R6 032 from the account 5 years after making the initial deposit. 8.4.1 Calculate the accumulated amount in the account three years after making the initial deposit. (3) 8.4.2 Calculate the accumulated amount in the savings account at the end of the seven-year period. (4) [16] Copyright reserved Please turn over

Mater®SWwhloaded from Stanmerephysigs.com mune QUESTION 9 9.1 A and B are events such that P(A) = 0,38 , P(not B)=0,77 and P(A or B) = 0,425. Calculate: oat P(B) 2) 92) P(A and B) (2) 9:2) certain high school. The principal observed sertain day. The results are summarised in the rived late Absent Total c 30 e m h y ’ and ‘arrived late’ mutually exclusive? Give (2) 9.2.2 The events ‘being a boy’ and ‘arrived on time’ are independent. It is also given that P(being a boy and arrived on time) = =. Calculate the value of y. (4) [10] QUESTION 10 10.1 A survey was conducted among 165 Grade 12 learners. These learners were asked which car brand they preferred among BMW, Audi and VW. The results of the survey are summarised below. e 42 learners preferred a BMW. 85 learners preferred an Audi. 106 learners preferred a VW. 18 learners preferred a BMW and an Audi, but not a VW. 40 learners preferred a VW and an Audi, but not a BMW. x learners preferred a BMW and a VW. 13 learners preferred all three car brands. 20 learners did not prefer any of these three car brands. Draw a Venn diagram to represent the above information. (4) 10.2 x collected the mathematics workbook from each of the 25 al i Some of the learners in her class are boys and the remainder are girl eC randomly selected the first book from the pile and marked it. The te agai ndomly selected the second book from the pile and marked it. The probability that the first two books selected belonged to boys is x Calculate the number of girls in the class. (5) 19] TOTALMARK [150] Copyright Reserved

MaeSWwhloaded from Stanmerephysigs.com mune INFORMATION SHEET _ -b+vb2 dae 2a A= P(L+ni) A=P(\=ni) A= P(-i)" A=P(+i)" T, =a+(n—1)d S, =F Ra+(—-pa] T.=ar"! s, te) sr#l S,=—*-3 =ler<l —_ pati -1 pals" i i rey ye F@th-S@) i a h 2 xX, +x + d=y(%)-*%) +0.-»)° M Ata Me 2 2 y=mx+e y-y, =m(x-x,) m=22— m= tan 0 Fy Hy (x-ay +(y—by =r? invdBe =9§4_= 2b =e sind sinB sinC a =b? +c? —2be.cosA area AABC => absin€ sin(a+ £)=sina.cos B +cosa.sin B sin(a@— £)=sina.cos B—cosa.sin B cos(a+ 8) =cosa@.cos f —sina.sin B cos(a—f)=cosa.cos B +sina.sin B cos’ a—sin’ a cos2a =41-2sin’a@ sin2a@ = 2sina.cosa 2cos’a-l x= i=l n P(Ay= 2) n(S) D=at+bx Copyright Reserved

Downloaded from Stanmorephysics.com KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE 150 These marking guidelines consist of 16 pages. Copyright reserved Please turn over

MathemPawnloaded from Stanmorephysics .com KZN/November 2024 G ll Marking Guidelines QUESTION 1 1.1.1 xl x=-1 x? hf i Fil= 0 VA standard form is —b+ Vb? —4ac 2a = =) + YET! ~ 400) YCA substitution 2(1) x=0,15 or x=6,85 incorrect rounding Penalise 1 mark for ¥CA answer ¥CA answer (4) x<—4 or x26 OR x€(—0;—4]U[6;0) OR —x? +2x4+24<0 x? -2x-2420 (x-6)(x+4)2>0 cevs:x=6 or x=-4 x<—4 or x26 — OR xe(—-0;—4]U[6 VA factors YCA critical values ( accept if shown on a sketch or on a number line) ¥CA answer (4) YCA_ answer OR vA factors YCA critical values ( accept if shown on a sketch or on a number line) YCA answer YCA answer (4) 1.1.3 | (2*-1)(2x-1)=0 2*=2° or 2x=1 1 x=t 2 x=0 or CA P= VA answer YA answer (3) Copyright Reserved Please turn over

Mathemawnloaded from Stanmorephysics.com GRADE 11 Marking Guidelines KZN/November 2024 1.1.4 | J2x45=x43 (v2x+5 \ =(x+3) QWxtS HN + 6x49 ¥A_ squaring on both sides x e4x44=0 YCA_ standard form (¥#2)(¥+ 2) =0 ¥CA factors x=-2 ¥CA answer (4) ie) Substitute equation | into equation 2: y=2+5x and 2xy-7 =-x* 2x(24+5x)-7=-x° ¥A_ substitution 4x+10x? -7+x? =0 11x? +4x-7=0 ¥CA standard form Y¥CA factors YCA_ x-values YCA_ y-values (5) OR OR Substitute equation | into equation 2: 2xy-7=-x? -2 —2y (2?) y-7 - (=) YA. substitution 2 2 95] a7) ed HI os 5 25 5(2y° -4y)-175=-(y?-4y +4) 0=-y? -10y* +4y+20y—4+4175 O=-lly?+24y+171 O=Ily?-24y-171 VCA. standard form 0=(1ly—57)(y +3) YCA factors y a or y=-3 ¥CA_ y-values Gal 5— |-2 see td or valri}=2 soll or x=-l YCA_ x-values 1 (5) Copyright Reserved Please turn over

MathemPewnloaded from Stanmorephysics.com KZN/November 2024 GRADE 11 Marking Guidelines 13 A=m> +4m A<0 m+4m<0 VA m+4m<0 m(m+4)<0 ¥CA factors cvs: m=0 or m=-4 <0 m must be || vCA_ interval solution for m eeeeeenioreprveies com an integer YCA_ integer answer : (4) 1.4 x= deirfert yx+ \x+ Vxtvkx.... x ae ae ee a YA. squaring on both sides x =x4x VA x =x4+x x = 2x x’? -2x=0 x(x-2)=0 ¥CA factors x=0 or x=2 ¥CA_ both answers (4) [28] Copyright Reserved Please turn over

Mathemawnloaded from Stanmorephysics.com GRADE 11 QUESTION 2 Marking Guidelines KZN/November 2024 10**7.25*.8* Ii (5.297? .5?*.2?* i 2A 2 536+) xt Ax+2 52x 73% = | PH2+3.x—4x-5 vA _ splitting to prime bases YCA_ using the exponential laws ¥CA simplification ¥CA answer (4) 2.1.2 3ie2 gio _ 3303 VA 33 ~ 37133 -6) YA correct factors in denominator - 21 =? YCA_ answer 7 (3) OR OR glcargiol 324 6.3 3003 = Pg A 3203 3° (3'-6.3°) YA correct factors in denominator 1 3 3 =a ¥CA_ answer GB) Copyright Reserved Please turn over

MathemPewnloaded from Stanmorephysics.com KZN/November 2024 GRADE 11 Marking Guidelines 2 f 2.2 x3 =5x*+6=0 1 1 [er3}-3)-0 YA factors i us eater x? =3 ¥CA both equations x =? or x3) =3° YCA raising both sides to power 3 x=8 or x27 ¥ CA both answers (4) OR OR 2 uf x3 —5x3 +6=0 pit let =k k? -5k+6=0 (k-2)(k-3)=0 YA factors k=2 or k=3 1 1 x=2 or 8 =3 ¥ CA both equations 1 1 yf ab? on tt ¥CA raising both sides to power 3 x=8 or x=27 ¥CA both answers (4) 23 xalVy _f fp 2 2V2 3 _ p 3 vA = 33 72 ¥ A rationalising the denominator 2-2 °2 ie. multiplying by oH _ fov2 4 _vova V4 _3a2 2 ally _ 32 2 2 “x=3 and y=2 YCA x—value & YCA y—value (6) [17] Copyright Reserved Please turn over

Mathemawnloaded from Stanmorephysics.com GRADE 11 Marking Guidelines KZN/November 2024 QUESTION 3 31. | L=—30tk 12=-3(l)+k vA substitution of constant rae} difference and any given (” ; Tn) Ty=A5=3n YCA_ answer (2) 32 Ty =15=3(10) YA substitution of n=10 = ¥CA_ answer Q) 33 T, =15—3n YCA T,=-316 —316=15—3n —331=-3n 110,333 =n .-316 is NOT a termina pattern ¥CA conclusion (2) 3.4. 4x+y;—-2x-1; y+3 5....... 12 9 —2x-1—(4x+ y) =12 -2x-1-4x-y=12 -6x-y=13 FB — TS iirccssasncaanl (1) YA first equation yt3—-(-2x-l=9 yt+34+2x4+1=9 YA_ second equation YCA substituting (1) into (2) YCA value of x y=-6(-4,5)-13 y=14 YCA value of y (5) 11] Copyright Reserved Please turn over

Mathemewnloaded from Stanmoreph sics.com KZN/November 2024 GRADE 11 Marking Guidelines QUESTION 4 4.1. 118.7 vv AA answers (2) 4.2 HVA TIE Tf vsscsvccssccsz 3 —2101 T2854 Since Q4.2 is a proof, mAl; —4 all marks are accuracy. 2a=-4 3(-2)+b=12 —24+18+c=-13 VA 2a=-4 VA 3(-2)+b=12 “.a=-2 b=18 c=-29 YA —24+18+c=-13 G3) OR OR 2a=-4 VA 2a=—4 a=-2 T, =-2n? +bn+e -1=-2(2) +b(2) +e YA -1=-2(2) +b(2)+e 7=2b+c Fa Ee), (1) T, =-2n? +bnt+c 7 =-2(3) +b(3)+e YA 7=-2(3) +b(3)+c C= 25 —BDrecssssssssssssssssssssnssesseesesssseees (2) “.7-2b =25-3b b=18 ce=7-2(18) c=-29 (3) 43 T, =-2n’ +18n-29 -2101=~-2n? +18 -29 VA equating T, to —2101 0 =-2n? +18n-29+ 2101 0 =-2n? +18n + 2072 0=n? —9n-1036 ¥CA standard form 0 = (n+ 28)(n-37) ¥CA factors n#-28 or n=37 There are 37 terms in a pattern ¥CA answer with rejection (4) 44 The value of a is negative. ». T, has a maximum value. Ta= Ts'= 11. YCA T4=Ts=11 .. the maximum value of T, is 11. YCA maximum value of T, is 11. The maximum value of Q, is 2(11) = 22 YCA answer Answer only, full marks (3) 4.5 T, =—4n+16 YCA general term of first =2(-2n+8) differences 2(-2n +8) is a multiple of 2. ¥CA common factor of 2 :. the sequence of first differences are even numbers. (2) Copyright Reserved Please turn over

MathemPewnloaded from Stanmoreph sics.com Marking Guidelines KZN/November 2024 [141 | QUESTION 5 Si x=2 and y=1 YA answer & “A answer Q) 5.2 : a Sx) = +q x+p a IQ)= at 1 “A substituting asymptotes(2;1) 3-2 43 “A substituting point (4;3) 4-2 2=2 2 4=a 4 Since the 5.2 is a proof, “ fQ)= = i all marks are accuracy. 2) x 53 4 0)= +1 YA let x=0 FO) 0-2 f0)=-1 B(O;—1) vA answer (2) 5.4 m=1 m=1 vA value of m y=lxt+e y-yY, =m(x-x) 1=12)+e OR y-1=1(x-2) “jm yslx-2+1 yalx-l “yelx-1 YA answer 2) 55 | g2-4 41 VA let y=0 x-2 y= 40 x-2 —2=x YCA x—value asa? ¥VCA answer (4) 1 -I 5.6 A(x) =| — (x) { i+] h(x) = (4x) h(x) =4).x1 h(x) = as VA_ simplified equation of h. x translation of 2 units to the left , and 1 unit down vA translation of 2 units to the left YA lL unit down @) Copyright Reserved Please turn over

MathemPewnloaded from Stanmorephysics.com KZN/November 2024 GRADE 11 Marking Guidelines [15] | QUESTION 6 6.1 R(O;6) VA coordinates of R Q) 6.2 xp = 6-8 YA answer Q) Since the 6.3 is a proof, all WK Sabstof —2.ands marks are accuracy. vA substitution of (0;6) n ya-Z (x4 2\0-6) v A substitution of 5 1 a aes (x -4x-12) YA simplification ayaoge? +2x +6 @) OR OR YA substitution of (0;6) YA. setting up 2 equations YA solving equations simultaneously YA simplification Copyright Reserved Please turn over

MathemPewnloaded from Stanmorephysics.com KZN/November 2024 GRADE 11 Marking Guidelines paa(x+ p)+q ySa(x-2)P +q 6=a(0—2)’+q x= —2 0=a(-2-2/ +q O0=1l6a+q —1l6a = q... see(2) -.6—4a=-l6a 12a =-6 l2a_-6 12) «12 1 a=-— 6.4 (4) y=-3(2) +2(2)+6 =8 S(2;8) Answer only, full marks ¥A_ substitution ¥CA value of x “CA value of y (3) 6.5 yell,y<8 OR ye(-~;8] ¥CA answer () 6.6 x<-2 or 0<x<6 OR xe(-—«;-2) or xe[0;6) ¥vvAA answer (3) 6.7 For roots of / to be real , h cannot be shifted downwards by more than 8 units. 2d2-8 OR ¥vCA answer OR (2) Copyright Reserved Please turn over

MathemPawnloaded from Stanmorephysics. com KZN/November 2024 GRADE 11 Marking Guidelines y= ris +2x+6+d A=b?—4ac AQ) 74{-F}(6+a) A=16+2d ¥VCA answer (2) 16+2d =0 (A 20 for intersection with x-axis) 2d >-16 “d2-8 68 | MN(@)=A(x)—k) = (-$# + 2x+6)-(6-x) vA subtracting two equations = -5 x? 43x YCA_ expression of MN b s=-— 2a __ 2 -3) 2 =3 YCA_ value of x MN(x) = -50) +3(3) ) 1 =— or 4— or 4,5 ¥CA answer 2 2 (4) [19] QUESTION 7 dole for g: YA x-intercept YA y-intercept for f: YA y-intercept YA decreasing shape YA asymptote (5) 7.2.1 | (334) vA value of x “A value of y (2) 7.2.2 | -3<xs<-l vv AA answer Copyright Reserved Please turn over

Mathemewnloaded from Stanmoreph sics.com KZN/November 2024 GRADE 11 Marking Guidelines OR OR x e[-3;-1] vv AA answer (2) 723 | k<-6 ¥¥AA answer Q) OR OR ke(-2 ;-6) vv AA answer (2) [11] QUESTION 8 8.1. A=P(1+i.n) 51600 = 30000(1+0,12.7) YA _ subst. to correct formula 51600 _ 30000 0,12 n must bea cn =6 years positive integer ¥YCA answer (2) 8.2 A=P(1-i)" vA reducing balance formula 20416 = P(1—-0,025)* YA. substitution 20416 _ (1-0,025)* Penalise a mark if NOT Px 25000 bees rounded to whole number || /CA answer GB) Copyright Reserved Please turn over

Mathemawnloaded from Stanmorephysics.com GRADE 11 Marking Guidelines KZN/November 2024 8.3 A=P(1-+i)" 139867= 100s0{1 F “y YA subst. into correct formula & YA : &16 13867 _\\4_ toos0 | YCA simplification +. X¥=0,0813000046.....x 100% x =8,13% ¥CA answer (4) os 8.4.1 aul _ Yo ¥ Ys Ys % OY % Y, i Y A , J 14% p.a. monthly 13% p.a. half—yearly A=P(1+i)” 0.14) vA _ subst into correct formula A=s0o00{ 14214) du 1. Ga 12 A=R75913,30 YA answer @) gao | AqPati" 0,13 4 A, -75913,30{ 1+ 23) A, = R97 659, 906—R6032 =R91627,906 4 A,= R91627,906{ 1+ “8) answer A, =R117876,22 (4) OF OR 0,13)" ; ‘ YCA rsoiga0(1+%2 A,= 75913,30(1+%2) 6032 (1-28) : : 0,13)' YA-R6032 VA (1-23) —_— Y CA answer (4) [16] Copyright Reserved Please turn over

MathemPewnloaded from Stanmoreph sics.com Marking Guidelines 11 KZN/November 2024 QUESTION 9 9.1.1. | P(B)=1—Pimot B) “i-RiPl ¥A_ substitution to correct rule 0/23 YA answer (2) 9.1.2 P(A or B) = P(A) + P(B)—P(A and B) 0,425 = 0,38 + 0,23 —P(A and B) P(A and B) = 0,38 + 0,23 —0,425 P(A and B) = 0,185 P(A and B) = 0,19 Y A. substitution to correct rule YCA_ answer (2) Copyright Reserved Please turn over

MathemPewnloaded from Stanmoreph sics.com KZN/November 2024 GRADE 11 Marking Guidelines 9.2.1 NOT Mutually Exclusive, YA NO P(girl and arriving late) # 0. vA reason (2) 9.2.2. | P(boy and arriving on time) =P( boy )xP (arriving on time) vA tule of independent event 30 40 Bayne asl VR Phe = 16-—yl) y y y 3y° =19200 3y? _ 19200 5 3 y? = 6400 y= 6400 y=80 y must be a positive whole number ¥CA answer (4) [10] QUESTION 10 Copyright Reserved Please turn over

Mathenawnloaded from Stanmoreph Sics.com KZN/November 2024 GRADE 11 Marking Guidelines 10.1.1 Ss YA 20 YA 18 and 40 VA x-13 YA 24~-xand 66-x (4) 10.2 Tx xol vA x va 20 25 24 25 24 7 xx vA picasa < 20 600 20 4200 = 20x — 20x x?-x-210=0 (x-15)(x +14) =0 x=15 or x#-14 YA answer (boys) total number of girls = 25-15 =10 vA answer (girls) (5) [9] TOTAL: 150 Copyright Reserved