Mathematics Grade 11 KZN March Test 2024 Paper & Memo

ownloaded from Stanmorephysics.com KWAZULU-NATAL PROVINCE EDUCATION ee Se REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE MARKS: 75 TIME: 1% hours This question paper consists of 6 pages.

Mathematics 2 Downloaded from StanmorephYSics. com Common Test March 2024 INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. tw This question paper consists of 4 questions. Answer ALL the questions. Number the answers correctly according to the numbering system used in this question paper. Clearly show ALL calculations, diagrams, graphs, etc. which you have used in determining your answers. Answers only will NOT necessarily be awarded full marks. You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise. If necessary, round off answers correct to TWO decimal places, unless stated otherwise. Write neatly and legibly. Copyright Reserved Please Turn Over

Mathematics j Downloaded from Stanmorepfysics.com QUESTION 1 | Solve for x: at 2? +5r=0 12 4x? +3x+6=0 (answer correct to TWO decimal places) 13 Wx-2=x 12 py x? -x- 20-205 *x-20<0. determine the sum of all the integers satisfying the 0<0. 13 1.4 The roots of a quadratic equation are x = feb 5 Calculate the value(s) of k for which the roots are equal. 1.5 Given: (x+5)° =1-p? Calculate the values of p for which the roots of the equation are non-real. (3) (3) (4) @) (2) (6) (2) (5) [28] Copyright Reserved Please Tum Over

Mathematics 4 Common Test March 2024 NSC Downloaded from Stanmorephysics.com QUESTION 2 2.1 Simplify fully without using a calculator. Your answer must be free of negative exponents and irrational denominators. 26h G3) 322 _ ort 2 ae ey V3N6 (4) 2.2 Solve for x, without using a calculator: 2.2.1 9/32 =128 (4) a eo 2.2.2 x? —3x4-10=0 4) [20] Copyright Reserved Please Tum Over

Mathematics 5 } ey Downloaded from Stanmorephysics. com QUESTION 3 DO NOT-USE A CALCULATOR WHEN ANSWERING QUESTION 3. 3.1 In the diagram below, P(-6 ; 8) is a point in the Cartesian plane, A is a point on the positive x-axis. AOP = 0. Determine, with the aid of the diagram, the following: 344 The length of OP. (2) 3.1.2 sin0+cos@ @) 32 If sin25° = 7 determine the value of tan 25° in terms of k. (3) 33 Simplify the following expression fully, without the use of a calculator: sin 698°.cos 300° tan 135°.cos(—248°) 2) [15] Copyright Reserved Please Tum Over

6 NSC Downloaded from Stanmorephysics.com QUESTION 4 Mathematics 4l Simplify the following expression fully: sin(180° + x).sin(90° +, 7 NY ld wn lal + sin. x.cos(90°— x) (7) tan(-x) 4.2 Given; —Si08 1+cos0 4.2.1 Prove that 899 __ 1-cosé (3) 1+cos0 sin? 42.2 Forwhich value of @ in the interval 8 <[ 0"; 360°] is : am? +cos 7 undefined? — (2) [12] TOTAL: 75

Downloaded from Stanmorephysics.com |} KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE GRADE 11 | These marking guidelines consist of 8 pages. Copyright Reserved Please turn over

Mathemewnloaded from Stanmorephysics.com Common Test March 2024 GRADE 11 Marking Guideline QUESTION 1 1.1.1 2x7 +5x=0 x(2x+5)=0 Y factors 5 x=0 orx= “3 Y answer Y answer GB) 1.1.2 | -4x?+3x+6=0 OR 4x’ —3x-6=0 yn Tbe vb? ~4ac ye DD END? ~4ac 2a 2a 2 = 343° ~4(-4)(6) = 34 (-3) -4(4)(-6) | v substitution 2(-4) 2(4) =—0,91 or x =1,66 Y answer Y answer QB) 1.1.3 3/ x-2=x (3 |p 2) =x? Y squaring both sides 9(x-2) =x? x? —-9x+18=0 Y standard form (x-6)(x-3)=0 Y factors x=3 or x=6 Y both answers (4) 1.2.1 x’ —x-20<0 (x+4)(x-5) <0 Y factors v Y answer (3) 12:2) Sum of integers = (-3)+(-2)+(-1)+04+14+24+3+4 =4 Y adding integers between —4 and 5 Y answer (2) Copyright Reserved Please turn over

Mathemawnloaded from Stanmorephysics.com GRADE 11 Marking Guideline Common Test March 2024 1.3 4x+y=7 V=7H4X eeeeceeeeee line 1 v making y the subject of the Bx? +.2KV = V2 eeeeceeeeeeee line 2 formula Substitute line 1 into line 2: 3x* + 2x(7-4x) =(7- 4x)’ Y substitution 3x° +14x —8x° = 49 —56x + 16x" 21x’ + 70x-49 =0 3x? -10x+7=0 Y standard form (3x-7)(x-1) =0 Y factors x -2 or x=1 Y x-values 7 yerg or y= 3 Vv y-values (6) 1.4 For equal roots: 13—2k =0 v 13-2k=0 2k =13 13 1 k= > or 85 v answer (2) 1.5 (x+5)" =1-p° x+5=4/1- p’ Y square root taken on both sides x=-5+,/1-p* Y x subject of formula For non-real roots: 1-p’ <0 v¥1-p’<0 “.p’-1>0 OR +o oO —_ > -1 1 =} 1 p<-lor p>1 OR pe(-%;~-1) or (1; ~) v ¥ answer (5) OR OR Copyright Reserved Please turn over

Mathemewnloaded from Stanmorephysics.com Common Test March 2024 GRADE 11 Marking Guideline (x+5)' =1-p? x +10x+25=1-p* x? +10x+24+ p? =0 b* —4ac = 10° — 4(1)(24+ p*) =100—96—4p” =4-4p° Y b? -4ac=4-4p* For non-real roots: 4—4p’ <0 v¥ 4-4p’ <0 p-1>0 (p-1)(p+1)>0 Y factorisation OR +o oo mm -1 1 =] 1 p<-lor p>1 OR pe(-#;~-1) or (1; ») v ¥ answer () [28] QUESTION 2 2 2.1.1 64 3 2 2 =(2°) 3 Vv (2°) =2* vo -t 16 Y answer G3) OR OR 2 64 3 2 2 =(4) 2 v (aya =4? v4? at 16 v answer (3) OR OR Copyright Reserved Please turn over

Mathemewnloaded from Stanmorephysics.com Common Test March 2024 GRADE 11 Marking Guideline ae 1 1 wIEIE! v 2 64° 643 1 1 la 3; 2 v a er _t = Co ‘ Answer only: 7 0 marks Y answer (3) gen _ gmt 21.2 | Sa gene -(¥ y" ~ BgBe genre gir Vv ge = (eas v 23") =6.3 Y factorising numerator Y simplification Y answer (S) 2 3-3 2.1.3 (3-v8) V3.6 = 9-63 +3 Y multiplying out numerator V3.6 _ 12-6y3 V3.6 _ 6(2 ~ v3) Y factorising numerator Vile S28), V6 Be “V6 _ ¥6(2-v3) V6(2-V3) ~ V3 aoe only: ve iB = ¥2(2-3) OR =2/2-J6 marks Y answer (4) Copyright Reserved Please turn over

Mathemewnloaded from Stanmorephysics.com GRADE 11 Marking Guideline Common Test March 2024 2.2.1 | ¥32=128 1 324 +128 ¥ 32 Pc (2°): £2’ ¥ 32=2° and 128=2’ E =)’ —— Y equating exponents x x= 5 v answer 7 (4) i 1 2.2.2 x? —3x4 -10=0 1 1 ¥Y one mark for each correct [« +2 -s}=0 factor 1 1 1 x? 4-2 or xt =5 v xt #-2 x=5* =625 Y answer x=-1 (4) OR OR i i x? —3x4 -10=0 1 Let k = x?: k* —3k-10=0 (k+2)(k-5)=0 ¥V one mark for each correct factor k#-2 or k=5 1 1 A ox4#-2 or xt =5 V xt #-2 x=5* y =625 answer (4) [20] Copyright Reserved Please turn over

Mathemewnloaded from Stanmorephysics.com Common Test March 2024 GRADE 11 Marking Guideline QUESTION 3 3.1.1 OP? =(-6)' +8 Y substitution in r? = x? + y” =100 “OP =10 Y answer (2) 3.1.2 | sin@+cosé _ 8 (8) vi _=8 10 10 10 10 —2 21 “0 5 Y answer (3) 3.2 x’ =r?—y? [Pythagoras] Y substitution in Pythagoras or =k2-1 k sketch x=Vk? -1 v Veo 25° ° 1 2 tan 25° = —— kl Y answer 33 sin 698°. cos 300° , tan135°.cos(—248°) v —si Oo _ _—sin 22°.cos 60° v ceo —tan 45°. — cos 68° Y —tan 45° 1 1 v —cos68° —sin 22°.— —cos 68°. — = ——_2. OR = ——_—. Y special angle values ~1.—sin 22 —1.—cos 68 Y cos68° = sin 22° or vice versa 1 “9 Y answer (7) [15] Copyright Reserved Please turn over

Mathemewnloaded from Stanmorephysics.com QUESTION 4 Common Test March 2024 GRADE 11 Marking Guideline 4.1 sin(180° + x).sin(90° + x) + sin x.cos(90" - x) v sin (180° + x) =—sin x tan(-x) ¥ sin(90° + x) =Cosx —- +sinx.sinx Y tan(—x)=-tanx —tanx Y cos (90° - x) =sinx pg SUERTE sin x.sin x sinx F ta cos x cos x sinx.cosx_cosx a = x —* + sinx.sinx 1 sinx =cos’ x + sin’ x Y simplification =1 ¥ answer (2) 4.2.1 sind 1+cosO . i= _ __ sind A cos) y m cos 0) 1+cos@ (1—cos@) (1-cos) sin @(1—cos@) Y 1-cos’ @ in denominator ¥ 1-cos’ @=sin’ 6 _ 1-cosé sin (3) 4.2.2 | Undefined when 1+ cos@ =0 cos9=-1 Y¥ cos9=-1 “0 =180° Y answer (2) [12] TOTAL: 75 Copyright Reserved