Mathematics Grade 12 KZN March Test 2026 Paper & Memo

Downloaded from Stanmorephysics.com KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE MARKS: 100 TIME: 2 hours This question paper consists of 7 pages and an information sheet. ight Reserved Please Turn Over

Mate"®Swnloaded from Stanmorepkysics.com Maren 2026 INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. This question paper consists of 9 questions. 2 Answer ALL the questions. 3. Number the answers correctly according to the numbering system used in this question paper. 4. Clearly show ALL calculations, diagrams, graphs, etc. which you have used in determining your answers. 5 Answers only will NOT necessarily | il marks. 6. You may use an approved sci¢ unless stated otherwise. 8. Write neatly and legibly. Copyright Reserved Please Turn Over

Mate" Swnloaded from Stanmorephysics.com Maren 2026 QUESTION 1 Given: 26, 45 and 70 are the third, fourth and fifth terms of a quadratic number pattern. 1.1 Calculate the first term of the quadratic number pattern. qd) L.2 Determine the expression for the n" term of the pattern. (4) 1.3. Which two consecutive terms of the first differences sequence will have a product of 7735? (5) [10] QUESTION 2 2.1 The 5" term of a geometric seqt e 9" term is = where r <0. Determine the first three terms 0} ence. (5) 2.2 Given: }°2p'*, where p#0~ k=0 Calculate the value(s) of p, if itis given that S,, =12,5. (5) [10] QUESTION 3 After undergoing knee surgery, Patrick had to return to his running program gradually. His physiotherapist advised him to do it in the following way: He has to run daily. In the first week he must run 6 minutes per day. Every week thereafter he must increase the time per day by 3 minutes. 3.1 During which week of his program will he be running 60: minutes per day? (3) 3.2 Calculate the total number of minutes he would have run by the end of the 19" week. (3) [6] Copyright Reserved Please Turn Over

Matte"®Swnloaded from Stanmorepbysics.com Maren 2026 QUESTION 4 Given: fOyo2 +1 and g(x)=k*. The point (2 ; 9) lies on g. 2 4.1 Determine the value of k. qd) 4.2. Write down the equations of the asymptotes of f (2) 4.3. Determine the equation of g”', in the form y=... (2) 4.4 Draw sketch graphs of f and g™ on the same set of axes, clearly indicating all asymptotes and intercepts with the axes. (5) 4.5 For which values of x will f(x).g '(x) <0? (2) 4.6 Determine the x-coordinates of the points of i pie ection between / and its axis of symmetry that has a positive gi ysics.com (3) [15] QUESTION 5 Sketched below is the graph of f(x) = th x-intercepts of — 4 and 2. A(4;8) isapoint onthe graph, 5.1 Write down the coordinates of the image of A after reflection in the axis of symmetry of f- (2) 52, Show that a=5, b=land c=-4. (3) 5.3. Determine the values of d such that $(x+ d y +x=4-d will have two positive real roots. (3) [8] Copyright Reserved Please Turn Over

Matte" Swnloaded from Stanmorephysics.com Maren 2026 QUESTION 6 The graph of 2 Oni ce ched below. The point P(3 \ 01 6.1 Determine the equation of f in th Indicate all restrictions. (3) 6.2 Sketch the graph of f in your an ate any intercepts with the axes, as well as the coordinates of one other point, Stam orephysics.com (3) 6.3 Describe the transformation from f" to g™', if g" (x) = 12x , where x>0. (1) 17] QUESTION 7 7.1 Itis given that acos28°=b. Express the following in terms of a and b, WITHOUT the use of a calculator: 7.1.1 cos(—28°) (2) 7.1.2 sin118° (2) 7.1.3 sinS6° @) 7.1.4 cosl4° (3) 7.2 Simply to a single trigonometric ratio, WITHOUT the use of a calculator. Show ALL working details. Stanmorephysics.com [2 cos” (180° + x)— 1 ].cos 67° (6sin? x—3).tan 23° (6) 73 Prove the following identity, WITHOUT the use of a calculator: sin(45°+ 0) —cos(45° + 0) = V2 sin’ 0+ V2 sin Ocos* 0 (5) [21] Copyright Reserved Please Turn Over

“ate Swnloaded from Stanmorephysics.com Mare 2026 QUESTION 8 In the diagram t \elow, the graphs of f(x)=asinx and g(x)=2cosby are drawn for the interval 80°] . g passes through (90° ; 0), and i n ynoreph s.ccim 8.1 | Determine the values of a and b. (2) 8.2 Determine the values of x, in the interval x <[—180° ; 0°], for which f(2x)<0. (2) 8.3. P and Q are two points of intersection of f and g. 8.3.1 Calculate the x-coordinates of P and Q. (4) 8.3.2 Hence, determine the values of x in the interval x <[-180° ; 180°], for : 2 which tanx >=. x>3 @) 8.4 What will the equation of g be after the y-axis has been shifted 45° to the left? (2) [14] Copyright Reserved Please Turn Over

Mate Swnloaded from Stanmorepkysics.com Maren 2006 QUESTION ‘eel pole AB is shown in the diagram below. AC and AD are steel cables, anchoring the pole to the ground at C and D. B, Cand D ¢ rizontal plane. The length of AD is 46 m. D 9.1 Write down the size of CAD. () 9.2 Calculate the length of AC. (3) 9.3. Calculate the height of the pole AB. (2) 9.4 Calculate the area of triangle ACD. (3) 9] TOTAL MARKS: 100 Copyright Reserved Please Turn Over

March 2026 “ate Swnloaded from Stanmorepkysics.com INFORMATION SHEET: MATHEMATICS _ =btvb2 —4ac 2a A=P(ltni) A=P(1-ni) A= P(\-i)" A=P(+i)" T, =a¥(n=N)d S, =Sba+(n—nd] T, =ar"" KY _ a" -1) ;r#l S, =“ 3-ler<l n i r-l l-r pasts =) p_Ml-a+I U i . h)— fQ) ! = jf S(xt+ P= tm h = V3)" +02-)? ) y=mte yy, =m Y27 "1 m=tanO a Si (x-a) +(y-by =r? InAABC: ne sinA sinB_ sinC a? =b? +c? ~2be.cosA area AABC =sabsin€ sin(a + B) = sina.cosf + cosa.sin B sin(a ~ B)=sina.cosf —cosa.sinB cos(a + 8)=cosa.cos fh —sina.sin B cos(a - 8) = cosa.cosf +sina.sin B cos’ a—sin’ a cos2a ={1—2sin’ a sin2a = 2sina.cosa 2cos*a-1 m 2 >}. _ Lb -¥) 1 oe = n n A P(A) = aS) P(A or B) = P(A) + P(B) — P(A and B) nN x-x\y-y feotie p= Le NY-D) S@-x) Copyright Reserved

Downloaded from Stanmorephysics.com KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE TIME: 2 hours NAME OF CANDIDATE: 100 This answer book consists of 16 pages. Copyright Reserved Please Turn Over

he i . h 202 “eGwnloaded from Stanmarephysics.com Maren 20% QUESTION 1 Solution/Oplossing Marks/ Punte 1.1 Q) 1.2 (4) is (5) [10] Copyright Reserved Please Turn Over

Math i . March 2026 ““"6wnloaded from Stanmerephysics.com _ QUESTION 2 Solution/Oplossing Marks/ Punte 21 (5) 22 Copyright Reserved Please Turn Over

hemati ‘ h 202 “*"Bownloaded from Stanmarephysics.com Marenanns Solution/Oplossing Marks/ Punte jorephysies.com (5) [10] Copyright Reserved Please Turn Over

h i . h 202 “"Bownloaded from Stanmarephysics.com Marenanns QUESTION 3 Solution/Oplossing Marks/ Punte 3.1 @) 3.2 GB) [6] Copyright Reserved Please Turn Over

Mathematics March 2026 wnloaded from Stanmerephysics.com QUESTION 4 Solution/Oplossing Marks/ Punte 4.1 dQ) 4.2 (2) 4.3 (2) 44 (5) Copyright Reserved Please Turn Over

“te®Swnloaded from Stanmerephysics.com Maren 2026 4.5 @) 4.6 @) [15| Copyright Reserved Please Turn Over

“eownloaded from Stanmerephysics.com Maren 2026 UESTION 5 ——— Solution/Oplossing Marks/ Punte 5.1 dfimorenhVsics.com (2) 5:2 (3) 5.3 (3) [38] Copyright Reserved Please Turn Over

ath i . h 2026 “aneéwnloaded from Stanmerephysics.com ware Solution/Oplossing Marks/ Punte 6.1 Stafimorephysics.com Q) 6.2 Q) 6.3 Q) 17] Copyright Reserved Please Turn Over

Mathematics March 2026 wnloaded from Stanmerephysics.com QUESTION 7 Solution/Oplossing Marks/ Punte TAA Q) TA2 (2) IA3 1 jorephysies.com G) 714 GB) Copyright Reserved Please Turn Over

“ie Swnloaded from Stanmarephysics.com Maren 2026 Solution/Oplossing Marks/ Punte 7.2 (6) 73 (5) [21] Copyright Reserved Please Turn Over

Mathematics wnloaded from Stanmerephysics.com March 2026 QUESTION 8. : ‘ x —180° 180° " Mimoreph ysics.com Solution/Oplossing Marks/ Punte 8.1 (2) 8.2 (2) 8.3.1 (4) Copyright Reserved Please Turn Over

Mathematics wnloaded from Stanmerephysics.com March 2026 Copyright Reserved jorephysies.com Solution/Oplossing Marks/ Punte 8.3.2 (4) 8.4 (2) [14] Please Turn Over

Mathe: March 2026 Swnloaded from Stanmereplixsics.com QUESTION 9. dfimorephysics.com Solution/Oplossing Marks/ Punte 9.1 Q) 9.2 @) 9.3 Q) Copyright Reserved Please Turn Over

Mathematics March 2026 wnloaded from Stanmerephiysics.com Solution/Oplossing Marks/ Punte 9.4 (3) 19] Marks/ Punte Additional space/Bykomend Copyright Reserved Please Turn Over

Mathematics wnloaded from Stanmerephysics.com March 2026 Additional space/Bykomende ruimte Marks/ Punte TOTAL MARKS: 100 Copyright Reserved

Downloaded from Stanmorgphygics.com KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA MARKS: = 100 These marking guidelines consist of 11 pages. Copyright Reserved Please turn over

Mathemawnloaded from Stanmorephysics.com KZN/March 2026 GRADE 12 Marking Guidelines QUESTION. Ll T=6— VA answer > dd) 12 2a=6 nis = 3 VA value of a 3at+b=7 3(3)+b=7 b=-2 YCA_ value of b a+b+c=6 3-2+¢=6 c=5 YCA value of c T. =3n? —2n+5 ¥CA_ answer " (4) 1.3 Ist diff: 73135193... T, =6n+1 YCA expression for T, Ty =6n+7 YCA expression for T,,,, (T, a} Gun ) = 7735 (6n+1)(6n+7)=7735 ae YCA substituting 36n* +48n—7728=0 2 a= ~(48)+-y(48) -4(36)(-7728) ¥CA formula or factors 2(36) n=14 orn#-15,33 T,, and T,, ¥CA answer (5) OR. OR YCA expression for T, YCA expression for T,,_; YCA substituting ¥CA formula or factors ¥CA answer (5) [10] Copyright Reserved Please turn over

MathemMawnloaded from Stanmorephysics.com Marking Guidelines KZN/March 2026 First three terms: 22 vA ef iep OR wel oe 48 768 YCA rat 16 YCA value of r YCA value of a ¥CA answer (5) VA expanding YCA substitution into S,, formula ¥CA equating to 12,5 YCA standard form ¥CA both answers (5) [10] Copyright Reserved Please turn over

Mathemawnloaded from Stanmorephysics. com KZN/March 2026 GRADE 12 Marking Guidelines QUESTION 3, ~ il 64.9; 12 oe ae 60 > T,=at(n-l)d =6+(n-1)(3) vA expression for T, =3n+3 ee YCA equating T, to 60 n=19 Answer only: % ¥CA answer ©. during the 19" week @) 3.2 In week 1: 6x7 =42 minutes In week 2: 9x7 = 63 minutes In week 3:12 x7 =84 minutes ete. .a=42; d=21 VA values of a and d Ss, =5[2a+(n-1)xa] S.= 5 [2(42)+(19-1) 21] +60(7)] | CA substitution S,, =4 389 minutes ¥CA answer GB) OR OR Total minutes for | da: ics.com a=6; d=3 vA values of a and d x S,= n [ Maxicloal 2 E - 19 if 55 = 6+60] ¥CA substitution =627 .. Total minutes fo ¥CA answer GB) [6] Copyright Reserved Please turn over

Mathemawnloaded from Stanmorephysics. com KZN/March 2026 GRADE 12 Marking Guidelines QUESTION 4 4.1 | g(x)=k™ Subst.(2;9): 9=k? mk=3 vA answer (1) 4.2 | x=0/and y=1 YA x=0 YA y=l (2) 43 g(x): y=3* g(x): x=3" Answer only: | | “CA x=3" y=log, x Full marks ¥CA answer (2) 4.4 | For x—intercept: 241 =0 x 2 YA shape of f YCA shape of g! ¥YCA asymptotes of f YA x-intercept of g”! VA x-intercept of f (5) 4.5 | 0<x<l ¥¥CA CA answer (2) 4.6 Substitute (on): c=l yextl VA y=x4l At points of intersection: 2 x+1l=—+1 ‘ = YCA equating 2. “x x? 2 x=+y2 =+1,41 ¥CA answers GB) [15] Copyright Reserved Please turn over

Mathenewnloaded from Stanmorephysics.com KZN/March 2026 GRADE 12 Marking Guidelines QUESTION 5 51 perm | ; vA axis of symmetry (-6;8) Pall marks | | “CA answer Q) 52 Fs) =a(x+4)(x-2) “A f(s)=a(x+4)(x-2) Subst. (4 ; 8): 8=a(4+4)(4-2) VA substitution of (4 ; 8) l6a=8 1 a= — 2 F(x) =5(044)(x-2) 1 f(x)=5(x? +2x-8) vA f(2)=5(x +2x-8) P(x)=px x4 1 a= b=1 and c=-4 (3) 5.3 glrtay tx=4-d S(x+d) +x+d-4=0 f(x+d)=0 Answer only: VA f(x+d)=0 d<-4 OR de(-;-4) Full marks ¥VAA answer @) [8] Copyright Reserved Please turn over

Mathemewnloaded from Stanmorephysics.com GRADE 12 Marking Guidelines KZN/March 2026 QUESTION 6 6.1 f-\(x)i y=-V12x for x>0 £ (x): x=—J12y vA swapping the values of x and y be? =12y x ¥ CA making y the subject of the y=, x50 formula 12 ae v A restriction @) 6.2 VA shape YA ending at the origin vA any other point on the graph @) 6.3 reflection in the x-axis OR reflection about the line y =0 v A answer Q) 17] QUESTION 7 7.1.1 | cos(-28°) = ° = cos 28 YA cos28° =& ¥CA answer a (2) 7.1.2 | sin118° = cos 28° YA cos28° b =e ¥CA answer (2) Copyright Reserved Please turn over

Mathemewnloaded from Stanmorephysics.com GRADE 12 Marking Guidelines KZN/March 2026 7.1.3 | y?=a°=b? [Pythagoras] YA third side y= Va? —b* VA double angle expansion ¥YCA answer OR substitution G) OR yeva-b YA third side y= Va? —b* sin 56° = sin(28° + 28°) = sin 28°cos28° +cos 28°sin 28° YA compound angle expansion a a ¥CA answer OR substitution _ 2bya? -b? a GB) 7.1.4 | cos2(14°)=2cos*14°-1 YA double angle expansion 1+cos 28° _ 2cos* 14° 2 2 ° : 2 1+cos28° _ aeaetiye vA isolate cos* 14° 2 142 ¥CA answer 0s 14° =| —4 cos = @) cos 28° = Lj a 2cos*14°-1= a YA double angle expansion a 2cos’ 14° id ban YA isolate cos” 14° cos* 14° = —— 2a cos 14° = pe ¥CA answer 2a 3) Copyright Reserved Please turn over

Mathemewnloaded from Stanmorephysics.com KZN/March 2026 GRADE 12 Marking Guidelines 7.2 [2cos* (180°+x)-1].cos67° (6sin? x—3).tan 23° [2(-¢08 x) -1].c0s67° SR, —eORx “j =3(I =2sin? x).tan 23° (2cos* x-1).sin 23° YA sin23° ~ 30082, 518 23° YA -3co0s2x COs 2x. cos 23° YA quotient identity cos 2x =——_..cos 23° YA cos2x —3cos2x 1 =-—cos 23° ¥CA answer (6) OR OR 2 o a Oo [ 2cos (180° + x) 1].cos67 YA cos2(180°+x) (6sin? x3).tan 23° _ cos 2(180° +.x).cos 67° -3(I —2sin? x).tan 23° _ cos(360°+2x).sin 23° YA sin23° 7 yy, 8in23° YA -3cos2x —3c0s 2x. cos 23° “A quotient identity ez SF agone YA cos2x —3cos 2x 1 = <5 23° ¥CA answer (6) 73 LHS =(sin 45° cos 6 + cos 45° sin @) —(cos 45° cos 6—sin 45° sin 0) = sin 45°cos 0 +cos 45° sin 0 —cos 45° cos 0 +sin 45° sin? v2 v2 v2 v2 =— cos 0 + —— sin 0-—— cos 9 + —— sin® 2 2 2 2 _ 2v2sind 2 = V2sin0 = V2 sin (1) = 2 sin O(sin’ @+cos” 0) = 2sin’ 0+ V2 sin@cos’ 0 =RHS OR ¥AYA compound angle expansions ¥A special angle values YA V2sin0 YA V2sin 9(sin? 0+cos* 0) (5) OR Copyright Reserved Please turn over

Mathemawnloaded from Stanmorephysics.com KZN/March 2026 GRADE 12 Marking Guidelines LHS =sin(45°+0)—cos(45°+0) =(sin 45° cos 0 + cos 45° sin 8) —(cos 45° cos @—sin 45° sin 0) YAYA compound angle expansions 2 v2.) v2 v2. =e cos + 2 sino 050+ 2 sino VA special angle values <V2sin8 YVA \2sind RHS = 2 sin’ 0+ V2 sin Acos” 0 =\2 sin 0(sin’ 0+cos’ 0) YA common factor = V2sin0 «. LHS = RHS (5) [21] QUESTION 8 8.1 a=3 a=3 b=1 Ab =l @ 8.2 | -90°<x<0° OR x€[-90° ; 0°] YAY A answer (2) 8.3.1 | 3sinx =2cosx VA equating YCA tanx== YCA x-coordinate of P YCA x-coordinate of Q (4) 69° <x < 90° ¥CA VA answer ¥CA VA answer (2 marks for each interval) €(-146,31° ; —90°) or xe(33,69° ; 90°) (4) 8.4 | y=2cos(x-45°) ¥CA YCA answer Q) [14] Copyright Reserved Please turn over

Mathemewnloaded from Stanmorephysics.com KZN/March 2026 GRADE 12 Marking Guidelines QUESTION 9 € D 9.1 CAD =74° VA answer dd) 9.2 sin ACD _ sin ADC V . = A fi i ak rs use of sine rule sin 56° _ sin50° 46 AC 46sin 50° AC=——__ YA AC subject of formula sin 56 AC = 42,50 m YCA answer Q) 93 | sint2°=—A8 YCA sin12°=—2B_ 42,50 42,50 AB = 8,84 m ¥CA answer (2) 9.4 Area AACD = 3 AD.AC.sin CAD vA use of area rule = 5 (49)(42, 5)sin 74° YCA substitution = 939,63 m* “CA answer QB) 19] Copyright Reserved TOTAL: 100 MARKS