Mathematics Grade 11 EC March Test 2026 Paper & Memo

Downloaded from Stanmorephysics.com MATHEMATICS MARKS : 60 TIME : 1 HOUR 10 MIN This question paper consists of 4 pages, including formulae that might be needed

Grade 1Downloaded from Stanmavephysées.com 17 March 2026 INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. 1. This question paper consists of TWO questions. Answer ALL the questions. 2: Clearly show ALL calculations that you have used in determining your answer. 3. You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise. 4. Answers only will not necessarily be awarded full marks. 5. If necessary, round off answers to TWO decimal places, unless stated otherwise. 6. | Number the answers correctly according umbering system used in this question paper. 7 Write neatly and legibly.

Grade Downloaded from Stanmasephysies.com 17 March 2026 QUESTION 1 GB) G) (4) (5) G) (6) GB) gk.6k-3,91-k 1.4 Simplify: ok-ag-k (4) Btahimorephysics.com 31] QUESTION 2 21 Given tana=— = and 180° < @ < 360°, determine without using a calculator and by the use of a diagram the value(s) of: 21.1 sine (2) 2.1.2 cosa qd) 2.1.3 sin?a+cos*a (2) 2.2 (6) 2.3. Simplify witho (5)

Grade 1Downloaded from Stanmavephysées.com 17 March 2026 2.4 Prove that: 2.4.1 —“(42¢0s?x) (tanx + os) = tanx (5) sinx. 2.4.2. For which value(s) of x in the interval [0; 360] is the identity in (3) QUESTION 3.4.1 undefined? 2.5 Determine the general solution of 2sinAcosA — cosA = 0 (5) [29] TOTAL MARKS 60 FORMULAE —b+vb2 —4ac x = 2a cos*@ + sin*@ =1

Downloaded from Stanmorephysics.com _ F Province ot the sued EASTERN CAPE ~~ NATIONAL SENIOR CERTIFICATE -GRADEF 11 MARCH 2026 MATHEMATICS TERM 1 CONTROLLED TEST MARKING GUIDELINE MARKS: — 75 This marking guideline consists of 7 pages.

JG/MARCH_ 2026 Downloaded Ninane stares oregnh ysis KRSM QUESTION 1 1.1 | Solve for x: 1.1.1 //2x@- x) =0 v x=0 x=0 or x =3 ¥Y x=3 (2) 1.1.2 | ( #1)(% -— 3) = 2x x*=3x+x—-3= 2x x?-—4x-3=0 Y Standard form —btvb?2—4ac = 2a Y Substitution ~(-4)tV (-4)?-4()(-3) z= 2(4) v x= 4,65 x=4,65 or x =—0,65 ¥ x=-0,65 (4) 1.1.3 | —x? +24 <—-2x x?—2x-24>0 Y Standard form (x -—6)(x +4) 20 cv Y Critical values x=6o0rx=-4 imorephysies.com 7 e<—4 x<-4 or x26 ¥ x26 (4) OR OR —x?4+24 <-2x —x? +2x+24<0 Y Standard form Y Critical values ¥ x<-4 ¥ x26 (4) 1.1.4 Y Isolation ¥ Squaring both 7+ 3x = 4x? sides 4x? —3x-7=0 (4x —7)(x+ 1) =0 v Standard form 7 x=Q orx=—1 Y Both x values ax=-l Y Conclusion (5)

Downloaded NismmESkaresoreyah ysies eRsN JG/MARCH_ 2026 m=3 12 Y Dividing by 3 Y Adding the half square on both sides Y Introducing root on both sides ¥ x=30rx=1 (4) 13 za (a)|x+y=50 ¥ x+y=50 x-5=7(y7-5) ¥ x-S=7(y—5) (2) b))x+ty=50 eevee (1) K=—SE TVS) cccvewss From (1) YSH=50—K —s-swaxaesed Sub (3) into (2) v Equation...3 x—5=7((50—x) —-5) Y Substitution x—-5=7(50-—x—-5) { x—-5=7(45 —x) E@fimorephysics.com ¥ Simplification x—-5 = 315 —7x 7%+% =315+5 8x = 320 x = 40 Y Value of x y =50-—40 y=10 Y Value of y (5) 1.4.1 i m+tvVm2-9 2 A= (5)? -9 A= 16 ¥v A= 16 -. The roots are real, rational and unequal. v Real Q) 1.4.2} A<0 m-9<0 ¥ m’?-9<0or (m — 3)(m A<0 C.v v Factors VV -3<x<3 (4) [32]

Downloaded Ninanes OrERN 'SKOG ‘KES JG/MARCH_ 2026 QUESTION 2 2.1.1 |aoyRer | g0n+2 WUT }faxsy"*3 srt Prime bases Yay (2x52)nt2 fo gnt3ygnt3 5n-3 Addition and = Gnt2 ants subtractions of = 2nt3—n-2 y gn+3+n-1-2n-4 exponents for = 21% 5-2 same bases 2i ~ 52 _ 2 Answer ~ 35 G3) 2-1,2:| 3 *8= 5.g*F1 3xt1 __ 3%.33-5.3%.31 Rewriting with ~ 3%,31 prime bases __ 3%(33-5.31) A = 3x31 fahimorephysics.com Factorising _ 33-5.31 =" _ 27-15 Simplification ~ 3 _2 Answer i (4) 2.1.3 | V¥45-V20 3V5 _ 9x5—-V4x5 = jeu Ve Breaking surds Simplification Answer (3)

Downloaded NinamExscaresoreynh ysios ‘RSM JG/MARCH_ 2026 2.2__| Solve for x: 2.2.1 | A357 a v Applying laws of exponents v¥ Equating the exponents I v Answer GB) 2 = 2.22 | 3 —5x3+6=0 1 1 v Factors oe — = 1. (x 2) (x. 3) 9 ¥Y x3=2 and x3=2 or x3=3 erin | 3 13 xP2=2"9 or x72 = 31% £=2° or x= 3" v¥ x=8 x=8 or x=27 Vi 4 = 27 ; (4) 1 x3—5x3+6=0 Ok 1 1 pe (x3)? —5x3+6=0 Eafimorephysies.com 3 Y Fact Let x3=k actors ¥ k=2ork=3 ¥Y x=8 ¥ x=27 (4)

Sownloaded stares SMOG ‘KRSM JG/MARCH 2026 QUESTION 3 3.1 | Ssing+3=0 5sinB = —3 sing-= -2 , cosB <0 4 Vv sing =—= x Cp r y v Diagram y=-3 r=5 re=x2+y? (5)? =x? + (—3)2 x= 16 va? = VI6 # apa x=i4 v 3 x=-4 3) (-4 tanB.cosB __ ee , a 2 ~~ 2 Zahmorephysies.com 5 3 v Answer * to (6) 3.2 3.2.1 | cos42° = p v y=y1-p? a On. 1-p? 3sin42° = 3 (: 7 7 meee =3/1—-p2 (3) 3.2.2 | cos132° = cos(180° — 48°) = —cos48° —_ (=) Y —cos48° 7 v —J1-p? aii ha (2)

Downloaded MAGIOR JG/MARCH_ 2026 3.2.3 | tan318° = tan(360° — 42°) = —tan 42° ¥ —tan42° P y _ vie? = vip? D P 2) 33 3.3.1 | 2cos* x + 2sinx.cosx.tanx = 2cos?x +2sinx.cosx. ——~ vu cosx cosx = 2cos*x + 2sin? x = 2(cos* x + sin? x) Y Factorising = 2(1) =2 v Answer G3) 3.3.2 | cos(360°—x).tan(—x) sin? (x-180°) __ cos x.-tanx Y cosx (-sin x)? v —tanx cos x.tanx A = -——_ v —sinx sin’ x sinx cosx sin Vio oa __ PcoaaE cosx sin?x =-— v Answer sinx (5) 3.4 cos225°.sin(—135°) — sin330° Ti tan225° ~ cos225°.sin(—135°) — sin330° LHS = tan225° __ €0S225°.—sin 135° — sin330° Y —cos45° v¥ —sin 135° or —sin 45° ¥ —sin30° v tan45° Y Special angles (5) [26] TOTAL MARKS: 75