Mathematical Literacy Grade 11 NSC Prelim 2017 Paper & Memo

Education | KwaZulu-Natal Department of Education REPUBLIC OF SOUTH AFRICA Pa A an a Gn ae ¢ ‘e i MATHEMATICS | i COMMON TEST : SEPTEMBER 2017 | NATIONAL SENIOR CERTIFICATE [ GRADE 11 | | “{ARKS: 75 TIME: 1% hours This question paper consists of 6 pages.

Mathematics 2 September 2017 Common Test NSC INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions: 1. 2. This question paper consists of FIVE questions. Answer ALL the questions. Clearly show ALL calculations, diagrams, graphs, et cetera, 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 to TWO decimal places, unless stated otherwise. Diagrams are NOT necessarily drawn to scale. Number your answers correctly according to the numbering system used in this question paper. Write neatly and legibly. Copyright Reserved Please turn over ()

Mathematics 3 September 2017 Common Test NSC QUESTION 1 1.1. Simplify without the use of a calculator: Ld sin? 300? + tan 315" 3) 7 C 8).cos(90° + 9} 6) a tan{180° —6) 1.2 If sin29° = Jt , determine the following, in terms of 4, without the use of a calculator. \ 1.2,1 cos 61° qd) 1.2.2 tan 209° (3) 1.2.3 sin(-601) (2) cos’ x 1.3 Prove the identity: ]-———— =sinx (3) l+sinx 1.4 Solve the equation tan x = — 7 for the interval 0° < x <360°, without the use of a calculator. (2) [19] (-, QUESTION 2 2.1 On the same set of axes, sketch the graphs of (x)= sin(x ~ 60°) and glx) =cos2x for the interval ~180° <x <180°. Indicate clearly the intercepts with the axes and the turning points. (6) 2.2 Write down the range of qd) 2.3. Write down the period of g. (1) 2.4 Determine the general solution of sin(x ~ 60° )= cos 2x. (5) 2.5 Hence, for which values of x in the interval ~180° <x <0° will f(x)= g(x)? (6) 2.6 If p(x) = cos(2x +20° } describe fully, in words, the transformation from & top. G3) [19] Copyright Reserved Please turn over

p Mathematics 4 September 2017 Common Test NSC QUESTION 3 3.1 Redraw the sketch below in your answer book and use it to prove that sinB sinc ; . Cc A L™ (A) 3.2 AIK has K =124°, HK =8,6cm and HJ =13cm. H Calculate the size of angle H. (4) 3.3 Inthe figure below RT=RQ=x, POR = a and PQS isa tangent to the circle. P Q r s 33,1. Prove that TQ = x,/2(1+cos2a). (4) 3.3.2 If a= 63° and x=17cm, calculate the area of ARQT. (2) [14] Copyright Reserved Please turn over

| Mathematics 5 NSC QUESTION 4 September 2017 Common Test The sketch below shows a wooden building block made in the shape of a pyramid mounted on a cube. The height of the pyramid is equal to the length of a side of the cube, namely 8 cm. Volume = sarth Volume = area of base x perpendicular height Volume = area of base x perpendicular height Volume = jer ! / U < 7 7 7 a / A- a ome somo I i) | t i i Pon 8om 4.1 Calculate the volume of the building block. 4.2 Calculate the total surface area of the building block. Copyright Reserved (3) (4) 1 Please turn over

Mathematics ” 6 September 2017 Common Test. NSC QUESTION 5 5.1 5.2 5.3 A coin is tossed and a dice is rolled. 5.1.1 Draw a tree diagram and indicate all the possible outcomes. 5.1.2 Determine the probability that atail and a factor of 6 will be obtained. Given: P(A) =0,55 P(B)=0,35 P(AorB)=0,8 Determine whether events A and B are independent or not. Show all relevant calculations used in determining the answer. In a particular class at Nquiu High School all learners take Mathematics and Physical Sciences. The probability that a randomly selected learner from that class will pass... * Mathematics is 0,6; * Physical Sciences is 0,7; * both of these subjects is x; and * neither of these subjects is 0,2. 5.3.1 Draw a Venn diagram to illustrate the above information. 5.3.2 Calculate x. 5.3.3. Determine the probability that a randomly selected learner from that class (a) will pass at least one of the two subjects. (b) will pass exactly one of the two subjects. TOTAL MARKS: Copyright Reserved (3) (2) (4) (2) (2) ql) Q) 6) 5

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NSC Nw a NSC Mathematics 3 September Test 2017 Mathematics 4 September Test 2017 13 24] sin(x—60°)=sin(00" 2x) IA for changing cos2s to LHS = y- 2 sin(90" - 2x) lysing 2-60" $90 2x +360 ke Z 1A for , are 1S +E360" x= 60" = 90° —2x+k.360" = 1+sinx—cos* x 1 A for mmerator/ denominator a 1A for Lesinx x= 50" +4120" x=50° +K.120" or : ' _ sinx + sin?x LA for applying 1-cos?x=sin’ x60" =180" —(90" ~2x}+-4.360"»,ke Z ; = —. 4 —cos* x= Ltsinx Sppiying Seow x60" = 90" +2x-4k360" sinx(0-+ sins) =x=150' +4360" ! 1A for -150° + £360" = sinx(l + sinx) 1A for factorisation pene . H 1A for keZ 1+ sinx x=~150" +4360 © j oR = sinx = RHS ©) OR 15 fmx = -L cesfoo~ (e60')}= cos 2x 1A for changing 3 sin(x—60") to = 30° 1A for 150° cos(150° ~x)= cos 2x . rol 2: 30 1A for 330° cos(t50°—x) x = 150%or 330° iS} 150" —x=2x + k360", ke Z 1A for 191 * , 150 —2=22+8.360" =3x=-150' +360" . a 1A for QUESTION 2 #= 50" —B.120 x= 50° —k.120° or 21 | 150° —x=360° -2x+k360',ke Z 1A for 210° +4360" x=210' +4360" lAfor keZ ‘ U, 25 | Subst. k=] in x=50° +4120" yields x =—70" ICA for -70° & fle) = sin(e-60) and Subst. K=0 in x=-150° +4360" (or, k==Lin x=210° +4360") | 104 for -150° a(x) = cos: yields x =-150". 2A shape ‘The solution therefore is: —150° <x <-70° i it 2A correct intercepts . 1A for inequality 6 Sean ae a 7 Bo dea” | 2A correct turning points ; 7 2.6 | The graph of g has shifted horizontally to the left by 10° and is reflected | 1“ for horizontal shift to about the x axis. the I 1A for by 10° 1A for reflected about the x-axis @) f0:.0;8% 19 (6) 22| E11] or TA for correct answer ~I<y<l aM 2.3 | 180° 1A for correct answer @) : Copyright Reserved Please Turn Over Copyright Reserved Please Tarn Over

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NSC > Matheraties 7 Septeniber Test 2017 an 5.12 P(tail and factor of 6) = = wie OR P(tail and factor of 6) = P(tail) x P(factor of 6) NES It elm fa ei 2A for correct answer OR 2A for correct answer e2)} 52 | P(AorB)=P(A) + P(B)— P(A and B) 0,8 = 0,55 + 0,35 — P(A and B) 1A correct substitution P(A and B)= 0,1 1A correct answer for P(AandB) P(A)xP(B)= 0,55%0,35 1A calculating P(A)xP(B) =019 - P(AandB)#P(A)x P(B) 1A concluding And A and B are not independent io) 5.3.1 Maths Physics C= ( "3 02 2.A for correct diagram. 2) 5.3.2 | 0,6—x+x+0,7-x+0,2=1 1A correct statement -x=05 1A correct answer “x= 0,5 Q) 5.3.3 | P(passes at least one of two subjects) = 0,8 1A correct answer (a) QQ) 5.3.3 | P(passes exactly one of the two subjects) (b) = 01+ 0,2 = 03 2AA correct answer Q) [16] Copyright Reserved TOTAL MARKS: 75