Mathematics Grade 12 KZN March Test 2025 Memo

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QUESTION 1 Common Test March 2025 El S,, =3(2)° -5(21) YA substitution =1218 vA answer (2) 1.12 |g. =3(22) —5(22) = 1342 vA value of S,, T, =S,,—-S,, = 1342-1218 =124 ¥CA answer _ (2) 1.13 8162 =3n? —5n YA equating 3n° -5n-8162=0 vA standard form -(-5 ts * — 4(3)(-8162 n= ( ) ( ( M ) YCA substitution 2(3) a=5300 or as _b4 3 N/A YCA answer (53 only) 53 terms have to be added (4) 1.2.1 | The even-numbered terms form an AS with a=7 and d=5. T,. of the AS =a+(n-l)d v — _ 7+(39-1)5 A substitution =197 ¥CA answer (2) 1.2.2 | Sum of the 52 odd-numbered terms = 52 x 7 = 364 VA 52x7 Sum of the 51 even-numbered terms n =5[2a+(n-1)d] 2! 2(7 51-1).5 = aL (7)+(S1-1). | ¥CA substitution = 6732 ¥CA sum of AS Sum of first 103 terms = 364+ 6732 = 7096 ¥CA answer (4) [14] Copyright Reserved Please turn over

Common Test March 2025 QUESTION 2 Dla 9; -27; 81 VA 9 VA -27; 81 - (2) BL2 | Fog vA answer Q) 2.1.3 | No, the series will not converge YCA answer (no) r<-] OR — rdoes not lie between —1 and 1. YCA motivation _ (2) 218 | woe r=-3 n=12 VA n=12 a(r" - 1) | 9x((-3)" -1) =~ 321 ¥YCA substitution =~] 195740x ¥CA answer (3) ~ VA 2 5 8 3 3 2"4 difference = 3 YA 2" difference = 3 _ 2nd difference 3 am 2 2 YCA value of a 3a+b=2 p=2-3(3 __5 YCA value of b 2 2 Tes =3(29)' ~5(29)+e=1166 YCA substitution in Fr. ~ 2B 2 c=-23 YCA value of ¢ Se 37 3% —23 2 2 T, =5 (1) -3(0)-23=-24 ¥CA answer (6) [14] Copyright Reserved Please turn over

Common Test March 2025 QUESTION 3 , 2x+3 3.11 | f= so HeHS _ 2x43 “I(e+3) VA -(x+3) _ (2x43) x3 _ —(2x+6-3) 7 wed ae va 22e+3)-3) x+ rts aH), 3 x+3 x+3 , 3 ifs a > (2) OR OR 2x+3 f(x) = —x-3 3 2 -3 YA -2+ (through the = to —x-3 3 method of long division) ie ae vA —(x+3) 3 =-24+— x+3 (2) 3.1.2 | For y-intercept, let x = 0: i ae x+3 yeaa- =-l YA y=-l For x-intercept. let y = 0: “__2=0 vA 3-0 webs x+3 3 -2 r+3 3 3 a YA x=-2 x 5 # 5 (3) Copyright Reserved Please turn over

Common Test March 2025 cm Bs) ¥CA intercepts vA asymptotes vA shape (3) 3.1.4 ya-xte vA equation of straight line Substitute (—3 ; —2): with a gradient of — 1 2=-(-3)+c c=-5 yes-x-5 ¥CA answer (2) 3.1.5 xeR, x#-3 YA xeER (a) VYCA x#-3 (2) a2 42K 2 ¥CAYCA 335-5 (b) 2 2 2) Copyright Reserved Please turn over

Common Test March 2025 3.2.1 [AtR: 3x+6=0 x=—2 R(-2:0) YA x-coordinate VA y-coordinate (2) 3.2.2. | From symmetry, the x-coordinate at S = | y =a(x+2)(x-1) YCA y=a(x+2)(x-1) y=ax +ax—2a 2 Substitute (-3:9} 9=4(-5] +a(-4)-20 ¥CA substitution 9= = 4 a=-4 YCA value of a y=4xr? 4x48 YCA answer..... OR (4) ye=a(x+p) +q OR 1y 2 y=a{x+3) +9 vA y=af x43) 49 Subst. S(1 ; 0) OR Subst. R(-2 ; 0): 2 2: o=a(1+5) +9 o=a{-2+3} +9 YCA substitution Es -9=4{2) 4 a= 5 YCA value of a ayaa) +9 ¥CA answer (4) [20] QUESTION 4 ait | ft: pavox He x= Br YA swopping x and y x =3y , . Pe ? v =—_ yeti x20 ay 3 YA x20 G3) 4.1.2 | 0<x<3 OR xe[0 ;3] YAVA answer (2) Copyright Reserved Please turn over

Common Test March 2025 4.2.1 ey-{t} OR gi y= as gl: r=(1} gi:x=3? vA swopping x and y y=log, x “. y=—log; x vA answer 2 (2) 4.2.2 y ¥CA shape YA x-intercept YCA coordinates of one more point (3) 1 x 4.2.3 -al1) 47 * (5) 12 Substitute (—2 ; 10): 10= a() +7 YA substitution 9a=3 1 a= 3 ¥CA answer (2) 4.2.4 he y= +}(3) +7 - (3 3 x+l 1 x vi, ( l y 7 From y=] — +7 t =|o] ; pi=| = Ee J (3) ~# (; 3 Translation of 1 unit to the right and 7 units downwards. YA | unit to the right YA 7 units downwards Answer only: (3) full marks [15] Copyright Reserved Please turn over

Common Test March 2025 QUESTION 5 : 4 4 S.1.1 sin f =— v i =_— g : A sing ; xtap -1° [Pythagoras] =5? 4? ». == 9 72.3 VA x-—value cos 8 = 2 ¥CA answer (3) 5.1.2 cos2f =2cos’ B-1 2 Bs} -(3)-1 25 sil 25 OR cos26 =cos’? B-sin? B 2 (3) -() = 2 25 25 gol 25 OR cos2f =1-2sin? B wail 25 YA double Z expansion YCA substitution ¥CA answer (3) OR YA _— double Z expansion ¥CA _ substitution ¥CA answer (3) OR YA _ double 2 expansion VA _ substitution ¥CA answer (3) Copyright Reserved Please turn over

Common Test March 2025 5.1.3 | sin38=sin(2B+£) = sin2fcos P+cos2fsin B ¥A compound 2 expansion = 2sin Bcos Bcos B+cos2fsin B YA double Z expansion _ f4)(-3)(-3) . (-7)(4 =2 BIS 5 a5 \l5 ¥CA substitution = 228 125 125 — ¥CA answer 125 (4) 52 sin(—180°—@) tan(180°— 0) cos(—0) cos? (90° +0)+3sin? 0 _ sin@,—tan 0.cos@ VA sin@ VA —tan@ (-sin0)' +3sin? @ YA cos8 VA —sind sin0 ind.—-——.cos9 i 7 sin Tr cos “A BaAd= sin@ 4sin’ 0 a __ sin’ @ 4sin’ 6 -l ¥CA_ answer 4 (6) 53 1 (1-cos4x) 8 1 =—[1-cos(2x2 : =i cos(2* 2x) ] VA gl! 20s(2x2x)] 7 sod 1 sa = gL (t-2sin 2x) VA gl !-(1-2sin’ 2x) = {[2sin* 2x] Ir. =< [sin* 2x] vA gin? 2x] 2 1), . 2 = 4 (2sin.xc0s.x) vA g(2sinxcos~) 1 . =5(4)(sin® x.c0s? x) vA glA)(sin? x.c0s” x) = sin’ x.cos’ x @ OR OR Copyright Reserved Please turn over

g(t cosas) Ls al! —vos{2x 2x) - L Z . eos? 2x | : i = st —cos* 2x)| Common Test March 2025 vA vA al! —cos(2x 2x) | g[t=(200s° 2x-1)| \ ; = ¢[2sin* 2x] TPs Ja. ein? = o[sin® 25] A [sin 2x] 1 , 2 ‘ YA — = (2sin xe0s.x) a. 4 Painzionaz) 1 . 2 = 5(4)(sin’ x.c0s? x) vA glA)(sin? x.COS: x) = sin” x.cos” x 6) [21] Copyright Reserved Please turn over

QUESTION 6 Common Test March 2025 6.1 cos(x—45°) =sin2x cos (x - 45°) = cos(90°— 2x) x= 45° = 90° —2x+k.360° or x-45°=360—(90°—2x) + k.360° 3x = 135° + k.360° x —45° = 270° + 2x +k.360° x= 45°+k.120°, keZ —x =315°+ k.360° YA co-ratio YA both equations YCA 45°+k.120° x =-315°—k.360° VA keZ x =45°+k.360°; keZ (4) OR OR cos(x—45°) = sin2x sin[ 90°—(x—45°) | = sin 2x YA co-ratio —x+135°=2x+k.360° or —x+135°=180°-—2x+k.360° vA both equations 3x = 135°+k.360° x =45°+ k.360° x =45°+k.120°, keZ x =45°+k.360°, keZ YCA 45°+k.120° VA keZ (4) 6.2.1 |360° vA answer Q) 6.2.2 |(-135°; -1) YA x-coordinate (a) YA y-coordinate (2) o. 1 YCA x-coordinate — (= , -5) YCA y-coordinate (2) 6.2.3 |90°<x<135° OR x € (90° ; 135°) YAY A answer (2) 1 6.2.4 —=(cosx+sinx wr ) = A cos + | sinx v2 2 | cos45°.cos.x + sin 45°sin x VA cos45°.cosx+sin 45° sin x = cos(x~ 45°) ¥A compound Z identity . To solve for x if sin2x = cos (x = 45°): x= 45° or x €[165°;180° | OR x = 45° or 165° < x <180° VA x=45° YCAYCA x [165°;180°] or 165° < x < 180° (5) [16] Copyright Reserved TOTAL: