Mathematics Grade 11 Paper 1 KZN November 2015 Paper & Memo

GR 11 MATHS ~P1 NOVEMBER EXAMINATION 2015 GSS Page 1 of 7 basic education Department: Basic Education REPUBLIC OF SOUTH AFRICA 7 _ GREENBURY SECONDARY to eo SCHOOL” i) MARKS: 150 EXAMINER: R.DEMRUGARAM TIME: 3 Hours MODERATOR: S. MUNILAL This question paper consists of 7 pages and 10 questions.

GR 11 MATHS P1 - NOVEMBER EXAMINATION 2015 ass : . Page 2 of 7 INSTRUCTIONS AND INFORMATION: Read the following instructions carefully before answering the questions: 1 2. . This question paper consists of 10 questions and 7 pages. Answer all the questions. Clearly show all calculations, diagrams, graphs, ete 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 your answers to TWO decimal places, unless stated otherwise. . Number your answers correctly according to the numbering system used in this question paper. Diagrams are not necessarily drawn to scale. It is in your own interest to write neatly and legibly. p.t.o, Question One...

Mo GR 11 MATHS P1 - NOVEMBER EXAMINATION 2015, GSS Page3 of 7 QUESTION ONE: 1.1. Solve for x: L.11. (2x-4) (3*-27)=0 (2) 1.1.2. 3x?~4x=5 (correct to 2 decimal places.) (5) 2 2 i 1.1.3. x3-—x3=6 (Hint: let xt=k) (5) 1.2. Determine the sum of the integers of the solution to x+x-650 (4) [16] QUESTION TWO: 2.1. Solve for x and y simultaneously: y- x+3=0 x -3=yt2x (6) 2.2, Simplify the following without the use of a calculator: 10* + 3 5x 1 2.2.1. a (5) 2+ 3 3. gen 1 2.3. Ifa+b= 1, determine the maximum value of ab (4) [19] QUESTION THREE: 3.1. Without solving the equation determine the nature of the roots of: x +3 =3x (4) 3.2. Prove that the equation: 3mx’ ~(2m+3n)x+2n=0 has rational roots for all rational values of m and n. (6) [10] p.t.o. Question Four...

11 MATHS Pi - NOVEMBER EXAMINATION 2015 GSS : Page 4 of 7 QUESTION FOUR: 4.1, Consider the sequence: 8311; 143... 4.1.1. Extend the sequence by two terms. (2) . 4.1.2. Is this sequence linear or quadratic? Give a reason for your answer. (2) 4.1.3. Find the general term. (2) 4.1.4. Find the tenth term. (2) 4.2, Consider.the sequence: ( ) “152375 143.. 4.2.1. Extend the sequence by two terms. , (2) 4.2.2. Is this sequence linear or quadratic? Give a reason for your answer. (2) 4.2.3, Find the general term. (2) 4.2.4, Which term of the sequence is equal to 623? (2) 4.3. IfTn=(-2)""! 4.3.1. Write down the first 4 terms of the sequence. (4) 4.3.2, What pattern do you notice? (2) 4.3.3. What is the sum of the first 4 terms? (2) 4.3.4. Which term will be equal to 64? (2) [26] ( QUESTION FIVE: 5.1. Calculate the effective annual rate of interest if the nominal rate is 7% per annum compounded monthly. (4) 5.2, A laptop costing R12 000 is purchased. Determine the book value of the laptop 3 years later if depreciation is calculated at 15 % p.a. according to the straight line method. (3) p.t.o. Question 5.3...

GR 11 MATHS Pl ~- NOVEMBER EXAMINATION 2015 GSS Page 5 of 7 5.3. R200 000 is invested for a period of 4 years at an interest rate of 6.5 % p.a. compounded monthly for the first 2 years. After the second year R10 000 is added to the account and interest changes to 8 % p.a. compounded quarterly for the next two years. Calculate the value of the investment at the end of 4 years. (5) 5.4. Karan bought a fridge for R6 000. He paid R7 125 in total over 3 years. Determine the compounded annually rate of interest that was charged. (5) [17] QUESTION SIX: 6.1. Ax) = 2x’—4x-6 g(x) =- 2x +6 ; Mis the turning point of (x). y A f(x) D E A\ {0 ”x g(x) Cc M 6.1.1, Calculate the coordinates of the intercepts A, B and C. (5) 6.1.2. Calculate the coordinates of the turning point M. (3) 6.1.3. If both graphs intersect at B and D, calculate the coordinates of D. GB) 6.1.4. Use the graph to solve the inequality f(x) . g(x) <0 (3) 6.1.5. For which value(s) of x is A(x) strictly increasing? (2) 6.2. With reference to the above graph calculate the average gradient between the points B andD of £. (3) [19] p.t.o. Question Seven...

GR {1 MATHS P1 - NOVEMBER EXAMINATION 2015 Gss Page 6 of 7 QUESTION SEVEN: The graph of f(x) = ~ =i pt2 7.1. Determine the y — intercept of f (2) 7.2. Write down the equations of the asymptotes of f (2) 7.3. For what value(s) of x is f(x) = 0? (3) 7.4, Draw a neat sketch of f(x). (5) 7.5, Write down one equation of the line of symmetry of f(x). (2) [14] QUESTION EIGHT: p(s) =t.3*+1; (t isa constant ) passes through the point (0 ; 0) 8.1. Show that t=~l1. (2) . 8.2. If p is shifted 4 units to the left to give a new function h, write down the equation of-h in the form y = ......... (3) [5] p.t.o... Question Nine...

GR 11 MATHS P1 ~ NOVEMBER EXAMINATION 2015 GSS Page 7 of 7 QUESTION NINE: 9.1. Given the following probabilities: P(A)=0,35 ; P(B)=0,2 ; P(AorB)=0,48 9.1.1. Determine P ( A and B ) if events A and B are not mutually exclusive. (3) 9.1.2. Are events A and B mutually exclusive? Give a reason for your answer, (2) 9.1.3. Are events A and B independent? Give a reason for your answer. (3) 9.2. A blue bag and a green bag each contain marbles. The blue bag has 5 yellow and 10 red marbles while the green bag has 6 yellow and 4 red marbles. If each bag has an equal chance of being selected, determine the probability of selecting a yellow marble from the blue bag or a yellow marble from the green bag with the aid ofa tree diagram. Show all outcomes. (5) [13] QUESTION TEN; 250 people have been interviewed at a local flea market and asked their drink preference: » 145 drink coffee >» 130 drink tea > 90 drink milo > 27 people indicated that they do not drink any of these » 25 people drink all three » 60 drink coffee and tea > 35 drink milo and tea Let x be the number of people that drink coffee and milo but not tea 10.1. Draw a Venn diagram representing this information and show that x = 47, (6) 10.2, Determine the probability that a person: ; 10.2.1. drinks only tea. (2) 10.2.2. drinks milo or tea. (3) [11]

2015 FINAL EXAMINATION GRADE (1 MATHEMATICS PAPER | MARKING MemoRANDUM Question One ! helt. 2a = 4 ov a%=3° zou 1a. Bx™-49-5 =O -bt ver bt be 4am xe Je 4)"~ 40)-s) va 2(aD x 2 413. 7%- d x? py Wick = 6 20 (Kt2(k- 3) = 0 kKz-2 OO oe Ke a? 4 MP2 -2 ow x3 =3 re) x 2-97" oy 2279 12. x 4% -640 a (x +3) (2-2) £0 va “3-4 6 2" 84m = -B-2-14041 42 < =37en . vn aa 2, Ov KF -0,19 {oF Question Two + 21, Y-x +B =O wa 2 7 %-3 xa = jth Km ® = 2S <3 42x x7 -3_ 20°" x (x-3) 20 x20 ow x 23U4 veh ua gre yF0 (03-3); (3)0) 224. 10 2 Boxee a ee | * (265 22 (yee 3 - 243, 5*t . 5* v7) s+ 6? zur 5 eth ch Kt3-U-2 gutete ate <4 an” a Ww NN wn k re) ya (3) ~ 260 (4) a zi-byv ab = (1-b)(b) = b-b2 = ~b the = z pened ’ d ~[o-b+4-4J aA be (Ep ed . , s max value iS ¢ (4) [9]

Question THree * eto =O b*- dag vA (-3)-4(1)(-3) at vA Bele 2732-3 ra) v ” v aus ‘ Roors are Real ineatuncaua (47 32. 3m - (2m +3n)% +2h 20 B= (2m -3n)*=4 (3m) (2) “" Am*4 Lunn $91? - 24g * n Am = lamin #In® VA (2m) = By)” “4 uw Lio} APA vA W5725 45, -85+-> “A, terms double wutl, 4.3.J. 432. ante Sign alternating guy 2-5 0% (-2)9 1 = 644% (-2)"" = (-2)8 n-l 2h aA 7 432. ABA. vow n 1. 1th tevm 15 64 (a . (2) @ 2) fe) uh O'S A PERFECT SQuARE way “ROOTS ARE RATIONAL FOR ALL % AND 1) (6) Question Four ’ Atle ITS 20% (2) v Aw2. Lineag; d= 3%" (2) 4AlZ. The at(u-~)d Ty 7 B+ (n-1) (3) T 78 t3y -3 T = aa eg (2) CA Al4. Tho = B(10) +5 = 3S vn (2) vs vA 421. 235 Bh A (2) v 42.2. Quapeanie . SECOND piFF = 2 (2) A2.2. azl be 0,¢*-2 4 Tas 4?- rz (2) ALA. N72 2 625VA nn? 2625 ne ag “. 26th tevm is 623. (2) Question Five + St- (1 + oa (14 bo 7,23 lo par™ P(i- ty) ws 2000 (1 - 18,3) Réboo “4 PC tijp?® 5 1200 7 227 685,19 a “8 Aa *227685,79 ( +i =Kz66 Te" pCi ei)? 3A = b000(i1+t) “ (141)? “) us wea }+ta t+ Ue ul i n $2. A 4 A 5.3. Ay u AY au = 200 000 (14 54. A= 2S cA 18 “so 6000 ve ay fics 6060 . _ 3 [12S t . 6000 t 5 -jue? 2X12 ‘A ry 5 190 of, PA v (» 4X2, (5) (. (© C7]

. Question Sin: . SLL. 22c?- 4-6 FO UA x*- 2% -3 70 (20-300 +1) 20 2%: Z0v xa -j7* cA wa 4 (-150)5 8 (350) © (05 Oe (5) 612. 28 eu wy 2-CA) "of 2 ro) (answer ily ) 2 2A m(2,-67" (3) GNB. 2xe%~ 4-6 2-2, +64 7 Lat =A + 2x -6-6 20 | 2% -le -12 20 2(x* =x - 6) =0 (741 IGc-3) F0 2 e-| ov x23 D(-158)4 (2 644. %7 3 Anio 4 7 43° ae, SoU XD -I | xu #3 (® 64.5. 222A (2) 62. (330); (046) 6-04 0-3“ AG > Fz 42 St (» (9) Question Seven : MW. Ys Dh 42 on; 3 97 23 (0, 23)“4 (2) 72. x 2BuA gr2~* (2) -1 Vos x-3 2 13 2(% 3)“ I = Dee = 7 8 Bx »% = 3iv% (3) { TA. Y v l 3 { a \7 ~-T-}4- ----f----2 — 74 Ly ce 7 4 L, lf, - - “A 2 t 2 Oo, 3 if 4 A | 4 yo te 5) y ( v 175. Yawn] nae Ya-xts (2) C4] Question Eieur + B41. Poco) = €.3% +1 A O’= £,3° + | O zt 4+) t aoa @) B.2+ hl) 2 TOBE pn (3 {s)

‘Question Nuwe » gL. P(A amd B) = P(x) +P(a) -P(A cv BY 2 0135 + 0,2 - 0,48” 2 0,07 v U (3) 9.42. NO” P(A AND B) = 0,07 +0 (2) “ 943. YES P(A AND B) = O07” P(A) * P(e) 2 0,35%0,2 = 0,07” ©) 2- 4 Ten + NaBs Question TEN any one o (1+ y(s) = 250 ee a(s)= 250 eUi4s) liso) diageas cU45) 130) i} 27 (90) BG-x #20 129435 $ho tlo +95-xX 429 -% = -47 x 2 47° SA ; 60 2b 2g, 10.2. P lowly bea) * 360 ~ 25 ~ 62H a pot sim pli fied ov give # ered t® = 250 (a CSL) (90) 27 aa 10.3. PCM oy ) Pim) + P(t) - P(M amd 0) 20 4 BO - BS wm 250 250 260 188 | (3) CJ u u (6) (2) u