Mathematical Literacy Grade 11 P1 2018 Memo

Downloaded fromhlaviso.com education Department: Education PROVINCE OF KWAZULU-NATAL NATIONAL _ SENIOR CERTIFICATE GRADE 11 AAAS. AEBS S888 228 Cee ie : MATHEMATICS P1 COMMON TEST JUNE 2018 MARKS: — 100 Thyuk: 2 hours This question paper consists of 8 pages. (Copyright reserved Please turn over

Mathematies PT Downloaded from hlayiso.com 2 June 2018 Common Test Grade 11 - NSC INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions: 1. to we an 6. 9. This question paper consists of 6 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, et cetera, which you have used in determining the 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 to TWO decimal places, unless stated otherwise. Diagrams are NOT necessarily drawn to scale. Write neatly and legibly. Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P| June 2018 Common Test Grade 11 - NSC QUESTION I 1.1 Solve forx in each of the following: LLL (Qv4D(v-)=0 Q) 1.1.2 2y° +411=3x+421 (correct to TWO decimal places) (4) I4 ii 02e=1e8e-— “sy fae] 1.2. Solve simultaneously for. andy: ye2v4l and 3x? -Sxy44y° =24 (5) 1.3. Consider the equation ky* + kv +2=0 1.3.1 Solve for x in terms of k. (2) 1.3.2. For which values of & will the equation Av* + kv +2 = 0 have non-real roots? (3) QUESTION 2 i 1 62 x12"! x27 2 2.1 Evaluate, WITHOUT using a calculator: ; (4) 322) 2.2. Calculate the value of ————————. WITHOUT using a calculator. 102" _ Voz" Show ALL your calculations. (4) [8] Copyright Reserved Please turn over

June 2018 Common Test Downloaded from hlayiso.com Mathematics P| Grade Tb - NSC QUESTION 3 Jabu manutactures wall clock faces in the shape of a regular octagon (cight-sided shape having sides of equal length). He cuts off identical isosceles triangles from the corners of a square sheet of glass having sides 16 cm long. Ifthe length ofa side of the clock face is x, calculate the value of v correct to TWO decimal places. (7 QUESTION 4 4.1 Consider the quadratic number pattern: 0;~-9:-16; -21:-24 4.1.1 Determine then term (T,,) of the pattern. (4) 4.1.2. Calculate the 30” term of the pattern. 2) 4.1.3. Determine which term of the pattern will be equal to 200. (4) Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematies P1 June 2018 Common Test Grade 1] - NSC 4.2 The first three figures in a pattern of grey and white squares are shown below. Figure | Figure 2 Figure 3 4.2.1 Considering Figure 4 in the pattern, determine: (a) The number of grey squares in this figure. (1) (b) The number of white squares in this figure C1) (c) The number of dots in this figure. (1) Considering the n’’ figure, determine an expression in terms of #7 for: (a) The number of grey squares. (1) (b) The number of white squares. (1) (c) The number of dots. (2) 4.2.3. There are 320 dots in one of the figures. Determine the number of grey squares in this figure. Q) Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P| June 2018 Common Test Grade | 1 - NSC QUESTION 5 5] Given: f(x) = +2. xt3 aay .L.1 Draw a neat sketch graph of f, indicating the asymptote(s) and intercept(s) with the axes. Show all your calculations. (6) n ~) Anaxis of symmetry of f has the equation + = mx + 5. Write down the value of 77. (1) 5.2 In the diagram below, g represents the function g(x) = a*,a>0. The graph / is symmetrical to g about the y-axis. The point B [? ; 2; lies on the curve of g and C is the y-intercept of both gand fr. cay h 5.2.1 Write down the range of g. (1) 5.2.2 Calculate the value of a. (2) 5.2.3. Determine the equation of / in the form y= 0b". 2) 5.2.4 B’ is the reflection of B in the y-axis. Calculate the average gradient between B’ and C. (4) Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics PI June 2018 Common Test Grade 11 - NSC 53 The diagram below shows the graph of a parabola /, with turning point A(2:—3)and y— intercept B(O ;-S). ay . > oO A(2 ;-3) B(O 5-5) in bo Show that the equation of the parabola can be written as ibe (5) 5.3.2 Use the graph to determine the value(s) of & for which the equation 1» - . a ae 2xy-—5=k will have real and unequal roots. (2) 5.3.3. The parabola / is shifted vertically until the new y — intercept is at the origin. Determine the new turning point of the parabola. (2) [25] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics PI June 2018 Common Test Grade 1] - NSC QUESTION 6 i 9 P $3h-& and g(x) =x +c. Phe diagram below represents the functions: /(.) = = e A,B and C are the points at which f intersects the axes. ¢ Dis the turning point of /. eg passes through A and B. e [isapointon g. uy o ' : x > 6.1 Determine the coordinates of A,B and C. (5) 6.2 Determine the coordinates of D. GQ) 6.3 Use the graphs to determine the values of yv for which f(x) = g(x). (2) 64 Determine the equation of g in the form g(x) = mx +e. (3) 6.5 Determine the length of DE, if DE is perpendicular to the y-axis. (4) [17] TOTAL MARKS: 100 Copyright Reserved

4OA0 LUM esBdIqj Ponioesoy 1UBuxdog, AA LOG A 1+) z= - 6 " ¢=d 10 ie (oz) 6 1=x Jo —=x 0=(-*)07 +6) 0=02-XIT+ .%6 PER FXOL+ OTF XS X01 - pte bTS(L FX t+ Xb)p +X - NOL xe Suomsue YOGA suojoey_» WLIO} psepuris > uonnysqns FO AL XZ) + XZ)KS — LE Zl (s) ~ Jomsur pa a4 ive 1 L-#I-XUP np F=I-XZV uoNNyos ON Z=I-XTP 10 [= |= xz Z=¥ W L-=¥ O=(@- L+H) {b) JSMSUB pA JOMSUL » ajnunioy onwapenb out uonmpsqns Downloaded from hlayiso.com OF bI-¥S+ 7 nS" Y= 1-2U OT L-xzp =O+ 1% eT gry IQ |-=* 40 [Pe=x We - OM@r- COP FR)= | Ta Dp Qh Fq- 0=01-*E- x7 IZ+XE=LT4 XZ CVI WSO] WONT gLgZ auNG O=U=*)+20) | PUT ENOLLSaN0 Sune avao 2 49A0 UN 9S29} qf ‘saded 6 JO sjsisuo9 auyjapind Suyjavw siyy 0OL FSMUVAL . ANITACIND ONDRIVIA 810z ANOS S&L NOWWOD Id SOLLVWAUHLVI [ Ii qavuas } GLVOMILAAD YOINAS 'IVWNOILLVN TWIYN-NINZWAAY JO ADNIAOUd uoneonp juewedeg uonesnpe ere ore ee pasioson yStXdo3

3 GRADE IL Marking Guideline sc 2018 Common Test Mathematics P 1 June 2018 Common Test 13.1 | be? hv +2=0 eek 7 _ apy | each tk = 42) Y substitution into quadratic | 2k formula | _aktvk? -8k “= 2k Y answer _ ~ Q) 13.2 [kh -8K <0 Y bP -8k <0 K(k -8) <0 Y factors Y answer >) [21] Downloaded from hlayiso.com 3x2)" (3x2? Y writing as prime bases Y simplification using laws Y simplification using laws Y answer NOTE: Ifa calculator is used, then NO marks will be awarded. (4) Copyright Reserved Please turn over 2.2 i ¥ determining square roots -10! 10" * Tg (10-1) ¥ factorising denominator __10 10°-1 210 99 Y answer 3) OR OR vl0" hom® — Ji0% —_ Vio" 10> |¥ writing numerator as, LO" CLO" = 1) ¥ factorizing denominator = Y simplification 10° -1 =12 oe ono ¥ answer 99 (4) QUESTION 3 The length of the side of the clock is x « 16 16-2 the length of each corner cut out is = v i x (16— } CS } - —"|+ Bx 2 2 256-32x+x° > 4 4 512-64x42x7 = 4x? 2x? +64x-512=0 x? +32x-256 =0 —btvb -Aac 2a —32+.4(32)° — 4(1(-256) 21) X= 6,63 “Fx # 38,63 de of the clock is 6,63 em [Theorem of Pythagoras] x= the length of the ¥ equation Y simplification Y standard form ¥ substitution into quadratic formula Y both values of x Y¥ answer Copyright Reserved Please turn over

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Downloaded from hlayiso.com Mathematics PI z GRADE II Marking Guideline QUESTION 5 +. 2018 Common Test 3 3.11 For x-intercept, substitute p= 0 For y-intercept, substitute x= 0: ~6 043 OR | The graph therefore goes through the origin; (0 ; 0) x substitution Y answer ¥ horizontal asymptote: y = 2 Y vertical asymptote: x = -3 | Y passing through the origin ¥ shape (6) fa ¥ answer dd) RIE ¥ answer | OR Y answer a) Y substitution Y answer Q) Y change sign of x ¥ answer Q) w iS & Y coordinates of Bi Y¥ coordinates of C Y substitution into average gradient formula v answer (4) Copyr Please turn over Mathematics PL 8 GRADE II Marking Guideline June 2018 Common Test 33.1. | ysatet py tg y=atx—2)-3 -5=a(0-2)'-3 -2=4a Y substitution of turning point, A Y substitution of point B Y value of a Y substitution of @ Y simplification 1 5 y= 3a +2x-5 (5) 5.3.2 | k<-3 vY answer (2) 533 [(2;2) Y x-coordinate v y-coordinate @) [25] Copyright Reserved Please turn over

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