Mathematics Grade 12 Paper 1 KZN June 2025 Memo

KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA / MATHEMATICS P1 iq JUNE 2025 MARKING GUIDELINES NATIONAL SENIOR CERTIFICATE These marking guidelines consist of 15 pages. Copyright Reserved Please turn over

QUESTION I GRADE 12 Marking Guidelines KZN/June 2025 LL. (=-3 eat “ VE? ON = 5 YA answer “A answer (2) 1.1.2 | 7x7+5x~8=0 _ bs Vb" —4ac 2a ce ~(5)4 y(5) -4(7)(-8) Penalise for incorrect YA substitution into formula : 2(7) rounding only in this x=-1,48 or x=0,77 sub-question. ¥CA answer “CA answer : . (3) 113 | ~~ —s0 x -2x-8 WP —2x-8>0 YA x°-2x-8>0 (x-4)(x+2)>0 a3 ——~OR —— o—__—» + + YA critical values x<-2 or x>4 ¥YCA answer (4) 1.1.4 3°? =42-5,3" 3°32 4.5.3° =42 YA splitting exponents 3° (9+5)=42 ¥CA factorising 3" =3" g=1 ¥CA answer G3) 1.1.5 —V5x-1=5-x vA isolating the surd (-V5x -1 ) =(5 ~x) ¥CA squaring both sides 5x-1=25-10x+2° x -15x+26=0 ¥CA standard form (x-2)(x-13)=0 x#2 or x=13 YCA rejecting x=2 ¥CA answer x =13 (5) Copyright Reserved Please turn over

KZN/June 2025 GRADE 12 Marking Guidelines 1.2 x=3+y YA making x the subject of the 5 ; formula (34+y) -y(3+y)=2y° +7 YCA substitution 9+6v4¥ -3y-y?-2y°-7=0 2y? -3y-2=0 ¥CA standard form (2y+1)(y-2)=0 YCA factors 1 y = or v=2 YCA y-values fd x=3+j “4 or x=34+2 (2 on 5 or x=5 ¥CA x-values 2 (6) OR OR yox-3 YA making y the subject of the 5 2 formula x —x(x-3)=2(x-3) +7 ¥YCA substitution x? —x? +3x—2x? +12x-18-7=0 2x? —-15x425=0 ¥CA standard form (2x-5)(x-5)=0 ¥CA factors Sie 2 or x=5 ¥CA x-values 5 y=——3 or y=5-3 J 3 J 1 = “3 or y=2 YCA y-values (6) 1.3 2a° + px~p =0 A=b?—4ac =(p) ~4(2)(-p’) YA substitution into b* —4ac =p +8p? Op YCA 9p? =(3p) YCA (3p) [showing that A isa perfect square] «. The roots are rational (3) OR OR Copyright Reserved Please turn over

KZN/June 2025 GRADE 12 Marking Guidelines 20° + px—- p> =0 _ —b+yb? -4ac : 2a ~(p) +, (py) -4(2)(-r? = (P) (?) ( )( f ) YA substitution into quadratic 2(2) formula _ =p 9p? _ 2 = 4 YCA pee _—ptip _ 4 vca —P#3P a 4 “> or —p .. The roots are rational GB) coe [26] QUESTION 2 2.1.1 | T=-14 vA answer Q) 212 T, =a+(n-l)d 6+(n-1)(-5) =-239 vA substitution in 7, ve i Pa 6—5n+5=-239 CA equating T, to —239 5n=250 n=50 YCA value of n n n §,=—[a+/ OR S,==| 2a+(n-1)d = 2fasl alee] Sig =F [6+(-239)] (26) +(50- 1)( )(-5)] ¥CA substitution in S, =—5825 SS 95 [-233] = 5825 ¥CA answer (5) 221 9 z Le SE Y x+S [5—% vA Ist differences: 4; x + 5; 15—x RAI H=1S KIN ES) YCA x+5—4=15-x-(x+5) 3x=9 x=3 ¥CA answer (3) Copyright Reserved Please turn over

KZN/June 2025 GRADE 12 Marking Guidelines 2.2.2 | I“ differences: 4; 8; 12 2” difference =4=2a Rese) VA a=2 3a+b=4 3(2)+b=4 b=-2 YCA b=-2 ath+c=-9 2 2tEH-9 c=-9 YCA c=-9 T, =2n?-2n-9 VY CA T,=2n*-2n-9 (4) 2.2.3 | T,=2n’-2n-9=2(n* -n-4)-1 YCA T,=2(n?-n-4)-1 2(n? -n—4) is even An even number minus | is an odd number ¥CA conclusion (2) OR OR T, = 2n? -2n-9 =2(n* =n)-9 YCA T, =2(n?-n)-9 2(n° =n) is even; 9 is odd. An even number minus an odd number is an odd number ¥CA conclusion (2) [15] QUESTION 3 3.11 | 444.pt+4p r+... =6 4+ Pa a ws =6 Pp garde Phy pk P P So l=r . 4 . I =6 YCA substitution in S,, p 4= c(i -+) P 6 4=6 “>? YCA simplification 6 _» P p=3 ¥CA answer (4) Copyright Reserved Please turn over

KZN/June 2025 GRADE 12 Marking Guidelines 3.12 | 4:43:43? or 4; 4; = oR 4; 4; 2 ¥CA answer a 3 39 () 3.2.1 | 5*=625 vA SP Answer only: VA 625 Full marks. (2) 3.2.2 | +5 4+5* 4.45" alr” - 1} i VA r=5 i , _3(5"=1) YA substituting n=11 into S, 5-1 formula =61 035 155 ¥CA answer (3) [10] QUESTION 4 4.1 A(-4;3) vA answer Q) 4.2 For x-intercept, let y=0: = 43 vA substitute y =0 x+4 ag = x+4 x+4=3 ee | vA answer B(-1; 0) (2) 43 For y-intercept, let ¥-0: ys 2 43 YA substitute x =0 0+4 yer VA answer 4 # (2) 4.4 Translate # 4 units to the right, and 3 units down. YA 4 units to the right YA 3 units down (2) Copyright Reserved Please turn over

GRADE 12 Marking Guidelines KZN/June 2025 4.5 The points of intersection between / and the axis of caged -9 symunetry of /, ie. between y= > and y=—-x: YA points of intersection we 9 x VA —=-x rad x=3 or x=-3 y=-3 or y=3 The points are (3 ; —3) and (-3: 3} VA (3;-3) “A (-3;3) (4) OR OR Distance between origin and any point on the graph = fo-x'+(o-(2) oy si YA ,|(0-x)° +{0-(2)) >, 81 x =p" += € For a minimum distance, d(, 81 al’ *e)° d(+ 81 VA —| xX? 4+— |= 2x-162x° =0 a ( +2) 2 2x* =162 4 =81 x=3 or x=-3 Therefore, the points on j closest to the origin are: (3;-3) and (-3; 3). VA (3;-3) WA (-3;3) (4) (11) Copyright Reserved Please turn over

KZN/June 2025 Marking Guidelines QUESTION 5 5.1 Substitute [-4 : 5] in f(x)=a": 5 zig : t on! ] ats (5) YA substitution a. gh va 2 va =2 a=4 VA value of a Substitute (-3 : | in g(x)=bx: 2 . 2 o> - 1 igi > b ay YA substitution Lily 2 4 Sb=2 YA value of b (4) 5.2 y & (852) vA shape vA coordinates of any point on the (0;0) 3 graph, e.g. (0:0), (8:2), (231), (8;-2). (23-1). 7 (2) a No YA answer ‘ d) 5.4 S(x)=4 or y=4" CA x — 4° (swopping x and y) Inverse: x=4° : y=log, x YCA y=log,x domain of inverse = range of function Therefore: | 1 yolog. a we) Fal6 YCAVYCA xe zis (4) 55 ee, vA VA answer 3 (2) [13] Copyright Reserved Please turn over

KZN/June 2025 GRADE 12 Marking Guidelines QUESTION 6 [6.1 | x? 45x-6=0 (x+6)(x-1)=0 vA factors y=+6 or x=1 If A and B are not A(-6;0); B(1;0) mentioned, max. 2/3 YCA answer VCA answer 30); 3 GB) 6.2 C(0 ;-6) vA C(0;-6) 6-0 m= 0+6 =-] ¥CA m=-l y=-x-6 ¥ CA answer (3) 6.3 The x-values of the points of intersection are the roots of: g(x)=f(a) +k —x-6=x°+5x-6+k YCA -x-6=x° +5x-6+k x +6x+k=0 YCA standard form If there are no points of intersection, the equation will have no real roots, i.e.: b —4ac <0 YA b?—4ac <0 6° -4(1)(k) <0 YCA substitution —-4k <-36 wk>d ¥CA answer (5) OR OR f is translated upwards until there are no points of intersection between fand h. Just before they do not intersect any more, / will be a tangent to f- Calculating the valuc of & when /# will be a tangent to f m,=h (x) obs 2x4+5 YCA -1l=2x+5 x=-3 YCA x=-3 g(-3)=-(-3)-6— 3 YCA calculating g(—3) ">, The contact point will be: (-3;-3). /(-3)=(-3Y +5(-3)-6=-12 YCA calculating f(-3) From -12 to —3 indicates an upward translation of 9 units. -. For gand # not to intersect, it means that f has to be translated upwards by more than 9 units. “k>o ¥CA answer (5) (1 Copyright Reserved Please turn over

QUESTION 7 Tiled ] Bank B. GRADE 12 Marking Guidelines KZN/June 2025 Because interest is compounded more frequently, she will get more interest from Bank B. If the candidate calculated amounts of money from the two banks, and then conclude correctly OR If the candidate calculates effective interest rates and then conclude correctly AWARD THE MARKS YA Bank B YA more interest (2) TAQ | 1+i, -(1+) YA formula m 42 i, -(14 9) =] vA substitution i, =8,84% p.a. ¥CA answer G3) 7.2 A=P(I-i)' 230 476,05 = P(1-0,13)' YA substitution 230 476,05 poe (1-0,13) P=R 350 000 ¥CA answer Q) 4 30 24 14 73 gouis7=x(1+ 2087) (1.20%) 4 730(1+202) = o(+ 0082) 4 12 12 4 30 cn o{1e0t2) 12 ( 0,087 x} 1+—— 4 x=R6 500 OR 4 30 24 } (14°?} =8 944,97 + 1 750 14007) 12 12 a4 vA -1 750( 1+ 93?) 12 ¥CA equated to R8944,97 ¥CA answer OR (5) Copyright Reserved Please turn over

GRADE 12 Marking Guidelines KZN/June 2025 0,087 14 Amount after 3% years=a( + } =1,351528006x 6 Six months interest added: 1,35 15280061 + oe) Withdrawal of R1750: 1,414902142x—1750 (1,414902142x-1750) (1,201 172602) = 8944,97 1,699541693x—2102,05 =8944,97 x=R6 500 4 Two more years’ interest added: (1,414902142x-1 750)(1 + oe n.087) ) 24 VA -1 7s0( 1+ 93?) 12 YCA equated to R8944,97 vA a(t 0, sia NIS is} vA -{+ ¥CA answer (5) [12] QUESTION 8 Penalise once only for incorrect notation in this question. My) atimf Gt LO) 8.1.1 f (x)= lim -7(x+h) -3-(-7x° -3 = lim ( ) P ( YA substitution into the formula =lim ai — ~3-(-78"-3) YCA f(x+h)=-7x? -14xh—7h? -3 hot h —14xh-The =n h ¥CA simplification h(-l4x-7h = ee ce) ¥CA factors Ans bh =lim(-14x-7h) =—l4x ¥CA answer (5) 8.1.2 gradient of tangent at x= 5 1 -r(+3} | 1 --14(-3 YCA -14{ -— 2 2 A ¥CA answer (2) 8.2.1 | y=3x(x?-2) =3x°-6x VA 3x°-6x dy 2 7 4 YCA 9x? CA ¥-6 GB) Copyright Reserved Please turn over

KZN/June 2025 GRADE 12 Marking Guidelines 5 Be 2 =D ; 8x in, 985 a, it f ‘-s| ¥CA o,|s ‘| =D, 14 =5" ¥CA answer (3) [13] QUESTION 9 9.1 f (x)= + px” +qx +30 Sf (x)=3x +2pr+qg=0 VA derivative f (-I)=3(-1) +2p(-1)+q=0 YA f'(-1)=0 G = 2p r—Bevsssesee > |va g=2p-3 f (-1)=(-1) + p(-ly +9(-1)+30=36 | et hell En — (2) VA p-q=7 Substitute (1) into (2) p-(2p-3)=7 YA solving simultaneously pow q=2(-4)-3 g=-ll (5) OR OR f (2)=(2) + p(2) +9(2)+30=0 YA substitute in f(2)=0 et > (1) YA p-q=7 f (-1)=(-1) + p(-1) +9(-1)+30=36 YA substitute in f (-1)=36 2p+q=-19 VA 2p+q=-19 G=—2 p19... — (2) substitute (2) into (1): p-(-2p-19)=7 vA solving simultaneously p=-4 =-l1 4 (5) Copyright Reserved Please turn over

KZN/June 2025 GRADE 12 Marking Guidelines 92 | f(x)=3x-8x-11=0 YA f (x)=3x°-8x-11=0 (3x-11)(x+1)=0 YCA factors ] wel or x#-1l ¥CA im only 3 3 (ly (uy “ary (4) 400 11) 400 {=| —] -4)—] -11] — |+30=-—— =-57,14 | VCA — |e (3) (5) \z) 3 27 (3) 27 o1(H-P) -[ 2 ,-s14] 3 27 3 (4) 9.3 For x-intercepts, let 1 = 0: x3 -4x? -11x+30=0 (x-2)(x? -2x-15)=0 VA (x-2) WA (x?-2x-15) (x-2)(x-5)(x+3)=0 x=2 or x=5 ory=-3 YCA x-values Length of AC =8 units ¥CA answer (4) 94 | f'(x)=6x-8=0 VA f'(x)=6x-8=0 4 Xp => CA answer 3 (2) OR OR ll afte AS 3 YA using midpoint formula es ¥CA answer 3 Q) 9.5.1 | x<-l or x>— ¥AYCA answer OR OR xe(-0; -1) or xe u ‘ ~| VAY CA answer (2) 9.5.2 | 0<x<= VAYCA answer OR OR xe( 0-4] ¥AYCA answer 4 (2) [19] Copyright Reserved Please turn over

KZN/June 2025 GRADE 12 Marking Guidelines QUESTION 10 t eC 10.1 afrs9-£] YAt VA 9-— (2) 10.2 | Arca=_bh rq digg 2 2 1 r 1 r A(¢j=—4| 9-— YA —t| 9-— (AS = ) A us ; 9 # =—t-— 2 18 (2) 10.3. | Fora maximum: A’(1)=9 vA A (t)=0 Bd pai YA al()=2-le 2 6 2 6 7 = 27 t=5,2 units YCA value of t 3 A(5,2)= 3(5 2)- (5.2) YCA substitution 18 =15,59 ¥CA answer (5) [9] QUESTION 11 11.1.1 Combined Outcomes: yellow; yellow . YA first branch = _ yellow 12 . YA second branch 1 yellow; green 2 green; yellow vA probability values on diagram green £ green green; green 11.1.2 5 7 7 5 =—x— + —x— 12 11 12 iI - 0 35955 132 66 Probability of 1 green and | yellow ball in any order Hon Sul Fen, 42d 12 11 12 11 ¥CA answer Copyright Reserved Please turn over

KZN/June 2025 Marking Guidelines 11.2.1 No, because P(A and B)#0 VA No VA P(AandB)#0 (2) 11.2.2. | P(AorB)=P(A)+P(B)—P(A and B) =0,714+0,83-—0,58 YA 0,714+0,83-—0,58 = 0,96 YA 0,96 Probability that smoke is not detected a P[ not (A or B) | =I-P(AorB) =1-0,96 = 0,04 ¥CA answer (3) OR OR B VA Venn diagram with probability values IF P(A or B) = 0,13+0,58+0,25 VA P(A orB) = 0.96 -.P[not(A or B}}=1-0,96 leah ¥CA answer GB) [11] TOTAL: 150 Copyright Reserved