Mathematics Grade 12 LP March Test 2026 Paper & Memo

Downloaded from Stanmorephysics.com NSC REPUBLIC OF ROUTH AFRICA PERE: I_IM POPC} PROVINCIAL GOVERNMENT DOPPARTMENT Or EDUCATION MARKS: 100 DURATION: 2 HOURS This Question papers consists of 8 pages and 1 information sheet. Copyright reserved Please turn over

Dewnlsaded from Stanmorephysics.com Limpopo/March 2026 NSC INSTRUCTIONS AND INFORMATION 1. This question paper consists of EIGHT questions. 2. Clearly show ALL calculations, diagrams, et cetera which you have used in determining the answers. 3. Number the answers correctly accord un g system used in the question paper. 5. You may use an approved scientific n- grammable and non- graphical), unless stated otherwise. 6. Ifnecessary, round off answers to TWO decimal places, unless stated otherwise. 7. Diagrams are NOT necessarily drawn to scale. 8. Write neatly and legibly. Copyright reserved Please turn over

Dewnlsaded from Stanmorephysics.com Limpopo/March 2026 NSC (3) (4) [7] QUESTION 2 2.1 Given the finite arithmetic sequi 3-87 2.1.1 Write down the fourth te m (dd) 2.1.2. Calculate the number of t (3) 2.1.3. Calculate the sum of all the negative numbers in the sequence. (3) 2.2 The tenth and the seventeenth terms of an arithmetic sequence are 21 and 49 respectively 2.2.1 Determine the common difference of the sequence. (3) (3) 23 (1) (4) [18] Copyright reserved Please turn over

Dewntsaded from Stanmoreph ics.com Limpopo/March 2026 NSC quadratic number pattern. two terms of the number pattern. erm of the number pattern in the 3.2.1 Determine the n‘” term 3.2.2 Calculate the sum of the fit 3.3 The first and second term of a geom eries is ‘tai re) 1OS,CG. and (4x? — 16) respectively. De eiinenaneealiets the series will converge. en as (2x — 4) of x for which Copyright reserved (1) (4) (1) (2) (4) [12] Please turn over

Dewntsaded from Stanmorephysics.com Limpopo/March 2026 NSC QUESTION 4 Sketched below are the graphs of f(x) = —2x? + 4x +16 and g(x) = 2x + 4. A and B are the x-intercepts of f C is the turning point of f 4.1 Calculate the coordinates of A and B. (3) 4.2 Determine the coordinates of C, the turning points of /- (2) 43 Write down the range of f. () 44 The graph of h(x) = f(x + p) + q has a maximum value of 15 at x = 2. Determine the values of p and qg (3) 4.5 Determine the equation of g~', the inverse of g, in the form y =... (2) 4.6 For which value(s) of x will g7().g() = 0? (2) [13] Copyright reserved Please turn over

Dewntsaded from Stanmorephysics.com Limpopo/March 2026 NSC QUESTION 5 Given the exponential function: f(x) = 3* sit Write down the range of f- (dd) 5.2 Determine the equation of f~! in the form y=... (2) 5:3 The point M(a; 2) lies on f~?. Calculate the value of a. (2) 5.4 Sketch, in your answer book, the g f,}, showing clearly ALL intercepts with axes (include t [ ach graph). Draw and name the line of symmetry be (5) 5:5 Is f~! a function? Justify your ans (2) [12] QUESTION 6 6.1 Ifcos@ = -< where 180° < 6 < 360°, determine, without using a calculator, the value of: 6.1.1 sin? 6 (3) 6.1.2 cos (6 — 135) (4) 6.2 Prove that 2cos?(45° + 1) = a nee (4) 6.3 Consider the expressi 6.3.1 Prove that sin(A - (2) 6.3.2 Simplify the followi ingl 1: sin4dx —sinl0x. (2) 6.3.3. Hence, determine the. solutioi Ox = sin3x for x € (0°; 30°] ae (5) [20] Copyright reserved Please turn over

Bewntsaded from Stanmorephysics.com Limpopo/March 2026 NSC QUESTION 7 7.1 Write down: 7.1.1 The period of f dd) 7.1.2 The amplitude of g (1) 7.2. Without using a calculator, determine the value of x for which cos2x = —sinx in the interval x € [—180° ; 180°]. (6) 7.3 Use the graphs above to determine how many degrees apart are points A and B from each other? (2) [10] Copyright reserved Please turn over

Dewntsaded from Stanmoreph ics.com Limpopo/March 2026 NS! QUESTION 8 , A, B and C are points in the same horizontal plane. D is a point directly that is DC 1 AC and DC 1 BC. It is given that ACB = 100°, CAD = 30°, 10 units and BC = 8 units. 8.1 Calculate the length of: 8.1.1 AC (2) 8.1.2 AB (3) 8.2 If it is further given that ABD = 73,4°, calculate the size of ADB. (3) [8] TOTAL: [100] Copyright reserved Please turn over

Dewnlsaded from Stanmorephysics.com Limpopo/March 2026 NSC INFORMATION SHEET Pr. =btlb? —4ac 2a A= P(l+ni) A=P(I-ni) A=(I-i! A=P(I+i)’ T= a4(n-l)d S,=2[2a+(n-Dd] T, =ar™" S, ae), r#l r= petleton! soe tim LF d=, -x) +(¥, -yy ysme+e Y-y, = m(x-%) (x-a)? +(y-bY =r? In AABC: Be nO sinA sinB sinc a? =b? +c? —2be.cosA area AABC = Fabsinc sin(a+ f) =sinacos B+cosasinB cos(a + 8) =cosa cos S—sinasin 8 cos? a@—sin® @ cos2a= 1-2sin? a 2cos? a@—1 gue n n(A) P(A) =—— 8) poathk sin(a— 8)= sina oos B—cosasinB cos(a—£8)=cosacos §+sinasin B sin2@=2sinacosa Yo,-3F 214 n o P(A or B) = P(A) + P(B)—P(A and B) pe LE DO-P) De-xy

Downloaded from Stanmorephysics.com PROVINCIAL GOVERNMENT REPUBLIC OF ROUTH AFRICA PEPARTMENT Or EDUCATION MARKS: 100 These marking guidelines consists of 11 pages

Bawnloaded from Starmonephysiasicom Limpopo/March 2026 NOTE: e Ifa candidate answers a question TWICE, only mark the FIRST attempt. e Ifacandidate crossed out an attempt of a question and not redone the question, mark the crossed-out version. © Consistent accuracy applies in ALL aspects of the marking guidelines. Stop marking at the second calculation error. «~~ Assuming answers/values to solve a problem is NOT acceptable. Copyright reserved 2 Please turn over

Bewnmloaded from Starmonephysiasicom Limpopo/March 2026 Question 1 L.A} x?-— 2x +1>0 @=D(x-1) >0 Y factors Critical value: x =1 x<1 or x>1 ¥ ¥ correct notation Sy2 = -990 (Accuracy) OR sxERx#l1 ¥VxERxF1 (Accuracy) (3) 1.1.2 | 2x --1 =V4— 5x , 2 , . (2x — 1)? = (V4—5x) squaring both sides 4x? —4x4+1+5x-4=0 Y standard form 4x? +x—-3=0 \. / _4+./(@)2 2464) 3) (4x -3)(x+1)=0 or x= Y factors = cs or x#-—-1 ¥ answers with selection ‘ (4) [7] Question 2 P v7 ‘ d) ¥ a=5 andd=-4 ¥ -87 =5+(n—1)(-4) v¥n=24 OR v¥—-4n+ 9 “eens ¥ —4n+9=-87 an 96 Vn =A n= (3) 2.1.3 | -3 5-75... 5-87 n Sn = 51a tT] Vn=2 22 faz3 S22 = = [-3 -— 87 a v answer Copyright reserved Please turn over

Bawnloaded from Starmonephysiasicom Limpopo/March 2026 OR = 5 [2a + (n—1)d] ae geen Yn=22 22 = Soo =F [2(-3) + (22 - 1)-4)] Y answer = —990 OR Va=-3 All negative terms can be writte get | Vv Vanswer the answer of -990 OR vi 28 Sum = Sy4- (5+1) 2 bea 24 v6 =7[5-87]-6 = -—990 v answer (3) 2.2.1|a+9d=21 Yat+9d=21 a+16d = 49 Yat+16d =49 “« —7d = —28 d=4 Y value of d (3) 2.2.2} a+9(4) =21 a=-15 Ya=-15 Ya=-15 Y answer (3) 2:34 Y all three terms Q) 2:3:2 Y substitution and = 1189 re Y standard form Y Factors (2m + 41)(m — 29) =0 41 «m= 29 or m# “> Y answer (4) [18] Copyright reserved 4 Please turn over

Baunloaded from Starmonephysiasicom Limpopo/March 2026 Question 3 3.11 =58; —48 Y Both answers d) 3,2 Ty = an? + bn+c 2a=2; 3a+b=2; at+b+c=-78 a=1; 3(1) +b » 1-1+c=-78 b=-1 c=-78 “Ty =n? —n—78 Yva=1 V¥b=-1 “c=-78 Y answer (4) 3:2.1 || 5: 10:20%....%; =a? Tn = (5)(2)""* Y answer dd) 322.2 a(r™ — 1) rns a 5[(2)8 - 1] Y substitution into the Sie=— Jy correct formula Sig = 1310715 v answer (2. 4) (2x — 4) a 3.3 «i ” ae r= a 4 Vr=2x+4 Converge: -1<r<1 OLS SY 1 <i 2a 1 Y substitution —5 <2x < —3 5. 22 Y answer ge" "9 (4) [12] Question 4 4.1 | 2x? +4++16=0 x —-2x-8=0 (x-4)(x+2)=0 Y factors Vx=-2Vx=4 (3) v1 v 18 (2) vy< 8 OR ¥ ye(—0;18] () Copyright reserved 5 Please turn over

Bawnloaded from Stawmonephysicsicom Limpopo/March 2026 44 |TP(1; 18) forf TP(2 ; 15) forh ¥ TP (2:15) forh 6] ms v p=-l “p=-l q=33 P ¥q=-3 (3) AS |y=2x+4 ¥ swop x andy x=2y+4 . 1 =-x-— “y=sx-2 wy 2* 2 2 (2) 4.6 | g(x) =0 or g(x) =0 x=4 or x=-2 (product 0 at x vx=4 ¥x=2 (2) [13] Question 5 5.1 |y>0 v answer OR OR ye(0;%) v answer Q) 5.2 Agar pix = 3? vx=3 = log3x y Bs Y equation (2) 5.3 y= log; x Y correct subst into correct 2= logs a formula (a ; 2) Y answer a=3*=9 (2) Copyright reserved 6 Please turn over

Bewnmloaded from Starmonephysiasicom Limpopo/March 2026 Forf v shape v y-intercept & a point Y axis of symmetry/ y=x For f-! v shape Y x-intercept & a point (5) Yes. The vertical line test cuts f~ = Y yes (2) Y valid reason OR OR Yes. For every x — value there is a unique y — value v yes Y valid reason (2) OR OR Yes. fis a one-to-one function v yes ¥ valid reason (2) [12] Question 6 6.11] y2 = /@3%= C5? [Pythagoras] yo y =-12 y=-l2 Q 3 CS;y¥) sin*@ Y substitution _(_12)? 7 (- za) v answer (3) _ 144 ~ 169 OR sin?@ = 1-—cos?6 Y square identity Copyright reserved 7 Please turn over

Bawnloaded from Starmonephysiasicom Limpopo/March 2026 5\? Y substitution sin?e = 1-(-=) 13 a} 144 v answer sin*@ = Teo (3) 6.1.2 | cos(@ — 135°) = cos0cos135° + sin@sin135° ¥ expansion = cos@(—cos45°) + sinO(sin45°) =Cal(-Z)+C9) Y reduction Y substitution _ Ww 26 Y answer OR -a-a+Ca@) 13/\" ya. 13/ \2 7: 13V2 (4) 6.2 | LHS = 2cos?(45° + x) = 2cos?(45°+x)+1-1 v+1-1 = cos[2(45°+x)]+1 Y double angle = cos(90° + 2x) +1 = (-sin2x) +1 Y simplify =1-sin2x Y reduction RHS (4) Y compound expansion Y subst special angle Y simplification v square identity (4) 6.3.1 LHS = sin(A — B) — sin(A + B) = sinAcosB — cosAsinB — (sinAcosB + cosAsinB) = sinAcosB — cosAsinB — sinAcosB — cosAsinB v sinAcosB — cosAsinB v¥—sinAcosB — cosAsinB = —2cosAsinB = RHS (2) 6.3.2 | sin4dx — sin10x = sin(7x — 3x) — sin(7x + 3x) ¥ 7x — 3x & 7x + 3x = —2cos7xsin3x v answer (2) Copyright reserved 8 Please turn over

Bewnloaded from Stawmonephysiasicom Limpopo/March 2026 6.3.3.| sin4x — sin 10x = sin3x =2 cos 7x sin 3x = sin3x Y substitution 2cos 7x sin 3x + sin3x =0 sin 3x (2cos7x +1) =0 Y factorisation x = 210° + k.360° Or x = 330° + k.360° x= -150° or x=-30° x = 90° + k.360° or x=90° cos2x = —sinx cos2x = cos (90° + x) 2x = 90°+x+k.360° or 2x = 360° — (90° + x) + k. 360° x =90°+k.360° or 3x = 270° +k.360° x = 90° + k.120° or x=90° x=-150° or x=-—30° 1 sin3x=0 or cos7x= 3 Y both equations 3x = 0° 7x =120° or 7x = 240° v answer x= 0° x=17,14° x = 34,29°(N/A) v answer (5) [20] Question 7 7.1.1 | 180° v answer d) Tl2,| 1 Y answer Q) 7.2 |1-2sin*x =—sinx 2sin?x — sinx -1=0 Y identity (2sinx + 1)(sinx — 1) =0 Y factors sinx = —3 or sinx =1 Y sinx = —+ Ref 2 = 30° ref 2 = 90° Y sinx =1 ¥—150° and —30° (6) vy 90° Y co-function v 2x in quadrant 2 Y 2x in quadrant 3 Y both general solutions ¥—150° and —30° ¥ 90° (6) Y co-function ¥ 2x in quadrant | ¥ 2x in quadrant 4 Y both general solutions ¥—150° and —30° ¥ 90° (6) Copyright reserved 9 Please turn over

Baunloaded from Starmonephysiasicom Limpopo/March 2026 OR cos2x = —sinx sin (90° — 2x) = —sinx 90° — 2x = 180° + x + k.360° or 90° — 2x = 360° —x + k.360° Y co-function ¥ 90° — 2x in quadrant 3 ¥ 90° — 2x in quadrant 4 Y both general solutions ¥—150° and —30° Copyright reserved = —30°+k.120° or x = —270° + k.360° =-150° or x= -30° or x=90° v 90° (6) 7.3 | A(—150°;0,5) B(—30°; 0.5) AB = —30° — (—150°) VY AB = —30° — (150°) AB = 120° v answer (2) [10] Question 8 B11 [AC 30° 20. °°s Y trig ratio AC = 20 cos 30° AC = 10V3 = 17,32 units ¥ answer (2) OR AC 20 Y trig ratio sin60° sin 90° «. AC = 20sin 60° = 17,32 units Y answer (2) 8.1.2 | AB? = AC? + BC? — 2AC. BCcos ACB Y cosine formula AB2 = (10v3)° + (8) — 2(10V3)(8)Ccos 100° Y substitution into cosine . formula AB = 20,30 units v answer (3) 8.2 |sinADB sin ABD AB. AD Y sine formula in AABD sinADB_ sin73,4° Y substitution in the sine —__ ra formula v answer (3) [8] TOTA L: 100 10 Please turn over