Mathematics Grade 12 Mpumalanga March Test 2025 Memo

education MPUMALANGA PROVINCE REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE MATHEMATICS TERM 1 MARCH 2025 MEMO MARKS: 58 TIME: 70 minutes This marking guidelines consists of 6 pages. Please turn over

QUESTION 1 17 March 2025 Yele'ly Y substitution BES 222 Rivce ¥ answer (2) a=8)d =7:n =36 T,=at(n-ld Tg =8+(36-1)(7) Ti, =953 1.1.2. ¥ substitution a=8;d =7:n =36 Y answer (2) 36 Sy = = (28) + (36-1)(7)] Sig = 4698 £13: ¥ value of T72 T, =at(n-ld ¥ substitution in T72.m Ty =8+(72~1(7) = 505 ee (3) 1 5_,=84+(72—1 -1\(7) 15 + Tyg = 786 505484 497 —7in = 786 —7Fm = ~224 m=32 12.1 | r=x=-2 v answer (1) 1.2.2. |-1<x-2<1 ¥ substitution 1<x<3 ¥ answer (2) [10] Please turn over

17 March 2025 QUESTION 2 2.1.1 Y value of a 16 0 -12 29 ¥ value of b LMA4 Y value of ¢ ¥ answer -16 -12 -8 Waa ° 4 4 2a=4 -16=302)+6 16=2-22+¢ a=2 b= -22 T, =2n* -22n + 36 21.2 ¥ substitution 32; 28; 24; 20; .... v answer a=32 (2) d=-4 T, =a+(n-l)d T, =32+(n-I)(4) T, =-4n+36 2.1.3 | 2n? -22n +36 =—4n +36 ¥ equating Y factors 2 - 2n° —18n=0 Yn=9 n-9=0 ¥ answer (4) n(n—9)= n=0 or n=9 -.T,, and T,, are the terms 22 | =-54, 7, =162, T, =—486 Y series a=52,r=-3,n=7. ¥ substitution v¥n=7 7 v answer s, -2%4 = I ta) =-29538 [14] Please turn over

QUESTION 3 17 March 2025 341.1 | 3-9) Vx=1 Vy=-9 (2) 3.1.2 | x=l ¥ answer (1) 3.1.3 | ye[-9; 0) ¥ answer () 3.1.4 | (0;-8) v VY answer (2) 3.2.1 | The graph shifted 9 units upwards and then ¥9 units upwards shifted 1 unit to the left v1 unit to the left (2) 3.2.2 |x>0orx<0 v answer (1) 3.2.3 | x=y? Y interchange x and y ¥ equation y= tvx. x20 Y¥ restriction (3) [12] QUESTION 4 4.1 27 _ 3 ¥ substitution reel ¥ cube root 3/— =a 8 3 a 2 4.2 h: ¥ shape Y any point v x-intercept (3) 43 3Y ¥ swapping x and y x= (3) Y answer (2) Please turn over

17 March 2025 g(x) = log, x 4.4 ve(o . 33) v answer (1) 8 [8] QUESTION 5 Bish ¥ correct diagram ¥x=V1l-m’ ¥ answer (3) l-m? cos36°=V1—m? OR cos” 36°+sin* 36°=1 g ai 2260 _1_ cin2 260 correct diagram oa Ae = lait 26 Fi ¥ substitution =l-m ¥ answer (3) cos36°=V1—m? 5.2 —2sin 15° .cos 15° ¥ —sin 30° cos(45° — x) cos x — sin(45° — x)sin x ¥ compound angle : ¥ compound angle _ —sin 30 1 [cos xcos 45° + sin x cos 45°] cos x —[sin 45° cos x — sin x cos 45%| sin;x 1 aust 7 | ¥ cos 45° = sin 45 = — "| I 1 ; 1 I v2 cosx.—=+ sin x.—= |cosx—| —=cosx-—sin x.—= |sin x ¥ common factor | 2 v2 E V2 (6) lV = 2 A iggst ee coe) gs eorex v2 2 v2 v2 Please turn over

17 March 2025 _ —sin 30 cos[(45° — x) +x] 1 ¥ —sin 30° ¥ ¥ compound angle -= 1 ao 2 v=— cos45° Y cos45° _l 1 <r v2 v2 (6) _-v2 Z 5.3.1 AC? = (12)? + (20)? — 2(12)(20) cos 110° ¥ substitution AC* =708,17 answer & “AC = 26,61 5.3.2 | sin DAC _ sin ADC ¥ ar ratio = substitution ' io . wi ¥ answer (3) sin DAC _ sin 71° 7 26,61 sin DAC = 0,2487 DAC =14,4° [14] TOTAL MARKS: 58 Please turn over