Mathematics P2 Grade 11 Exemplar 2013 Eng Memo_hlayiso.com_.pdf

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 11 MATHEMATICS P2 EXEMPLAR 2013 MEMORANDUM MARKS: 150 This memorandum consists of 13 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 2 DBE/2013 NSC – Grade 11 Exemplar – Memorandum NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • If a candidate has crossed out an attempt of a question and not redone the question, mark the crossed out version. • Consistent accuracy applies in ALL aspects of the marking memorandum. • Assuming answers/values in order to solve a problem is NOT acceptable. QUESTION 1 1.1 n 408 ∑x 408 i  19 Mean = i =1 = = 21,47 answer n 19 (2) 1.2 Standard deviation = 7,81 answer (2) 1.3 The one standard deviation limits are ( x − 1σ ; x + 1σ ) = (21,47 – 7,81; 21,47 + 7,81) = (13,66 ; 29,28)  interval ∴ 13 people lie within 1 standard deviation of the mean. 13 people (2) 1.4 5 12 13 15 18 18 18 19 20 21  Q 1 = 18 21 22 23 23 26 29 33 35 37  Q 3 = 26 IQR = 26 – 18 IQR = 8 =8 (3) 1.5 box whiskers 4 5 8 12 16 18 20 21 24 26 28 32 36 37 40 (3) 1.6 There is a marked difference between the lowest value (5) and the next lowest value (12) whilst the differences between all other data reason points are within at most 3 values. 5 is an outlier (2) ∴ 5 is an outlier [14] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 3 DBE/2013 NSC – Grade 11 Exemplar – Memorandum QUESTION 2 2.1 Cumulative Class Frequency frequency 0≤m<2 7 7  first three 2≤m<4 15 22 cumulative 4≤m<6 26 48 frequencies correct 6≤m<8 29 77 remainder 8 ≤ m < 10 36 113 correct (total = 160) 10 ≤ m < 12 31 144 (2) 12 ≤ m < 14 14 158 14 ≤ m < 16 2 160 2.2 160 150 grounding at 0 140 plotting cumulative 130 frequencies at 120 upper limits  smooth shape 110 of curve (3) 100 90 Cumulative Frequency 80 70 60 50 40 30 20 10 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Number of sms messages Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 4 DBE/2013 NSC – Grade 11 Exemplar – Memorandum 2.3 The median for the data is approximately 8 messages. Median (1) 2.4 Approximately 130 learners sent 11 or fewer messages. Therefore 30 learners 30 learners sent more than 11 messages. answer 30 × 100% = 18,75% 160 (2) 2.5 Skewed to the left or negatively skewed answer (1) [9] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 5 DBE/2013 NSC – Grade 11 Exemplar – Memorandum QUESTION 3 y D 45° 18,43° A(1 ; 6) C(12 ; 3) E θ x B(3 ; 0) 3.1  3 + 12 0 + 3   substitution into E ;  midpoint  2 2   1 1 formula = 7 ;1  answer  2 2 (2) 3.2 3−0  substitution into m BC = gradient formula 12 − 3 1 = 3 answer (2) 3.3 1  tan θ = m BC tan θ = m BC = 3 1 answer θ = tan −1   = 18,43° 3 (2) 3.4 1 1 m AD = m BC = AD||BC, equal gradients m AD = 3 3 6−0 m = −3 m AB = = −3 AB 1− 3 ∴ m AD × m AB = 1 × −3 = −1 m AD ×m AB = −1 3 (3) ∴AD ⊥ AB 3.5 inclination of new line = 45° + 18,43° = 63,43° 18,43° ∴ tan 63,43° = 2 = m line 63,43° m = 2 y − 6 = 2( x − 1) ∴ subst of (1 ; 6) y = 2x + 4  equation (5) [14] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 6 DBE/2013 NSC – Grade 11 Exemplar – Memorandum QUESTION 4 4.1 mQP = mOS = 6 QP||OS, equal gradients  mQP = 6 y − 17 = 6( x + 3) subst (–3 ; 17) y = 6 x + 35 into formula equation (3) 4.2 6 x + 35 = − x  setting up 7 x = −35 equation OR x = – 5 x = −5 y = −(− 5) = 5 y = 6(– 5) + 35 = 5 y=5 ∴ Q(–5 ; 5) coordinates of Q (4) 4.3 OQ 2 = (−5 − 0) 2 + (5 − 0) 2  substitution into distance formula = 50 5 2 OQ = 50 = 5 2 units (2) 4.4 mOS = 6 80,54° ∴ inclination of OS is tan −1 (4) = 80,54° mOQ = −1 ∴ inclination of QO is 180° - tan −1 (1) = 135° 135° α = 135° − 80,54...° 54,46° = 54,46° (3) 4.5 QS = OS + OQ − 2OS .OQ. cos α 2 2 2 correct use of cosine rule = 148 + 50 − 2( 148 )( 50 . cos 54,46° substitution into QS = 9,90 units formula 9,90 (3) [15] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 7 DBE/2013 NSC – Grade 11 Exemplar – Memorandum QUESTION 5 5.1.1 5 5 cos α = − − 13 13 (1) 5.1.2 (− 5) + b = 13 2 2 2 b 2 = 169 − 25 = 144 b = 12 b = 12 tan (180° − α) – tan α = − tan α 12 = −( − )  12 5 5 12 = (3) 5 5.2.1 sin(θ − 360°) sin(90° − θ ) tan(−θ ) cos(90° + θ ) sin θ cos θ (− tan θ )  reductions = − sin θ sin θ tan θ =  sin θ  cos θ = − cos θ  −   cos θ   sin θ = sin θ (5) 5.2.2 From 5.2.1: sin θ = 0,5 sin θ = 0,5 Ref ∠ = 30° 30° ∴θ = 30° or θ = 150° 150° (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 8 DBE/2013 NSC – Grade 11 Exemplar – Memorandum 5.3.1 8 4 LHS = − sin A 1 + cos A 2 8 4  sin2A=1– cos2A = − 1 − cos A 1 + cos A 2 8 4 = − (1 − cos A)(1 + cos A) 1 + cos A factorising 8 − 4(1 − cos A)  addition = (1 − cos A)(1 + cos A) 8 − 4 + 4 cos A = (1 − cos A)(1 + cos A) simplification 4(1 + cos A) =  factorising (1 − cos A)(1 + cos A) (5) 4 = = RHS 1 − cos A 5.3.2 Identity is undefined when sin 2 A = 0 . That is when sin A = 0 or  each value cosA = ±1 ∴A = 0° or A = 180° or A = 360°. (3) 5.4 8 cos 2 x − 2 cos x − 1 = 0  factorising (4 cos+ 1)(2 cos x − 1) = 0 values of cosx 1 1  104,48° or cos x = − or cos x = 4 2 255,52° ∴ x = 104,48° + k .360°; k ∈ Z or x = 60° + k .360°; k ∈ Z  60° or 300°  + 360°.k x = 255,52° + k .360°; k ∈ Z x = 300° + k .360°; k ∈ Z kε Z (6) [26] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 9 DBE/2013 NSC – Grade 11 Exemplar – Memorandum QUESTION 6 6.1 p = – 45°  value of p q=–1  value of q (2) 6.2 B(157,5° ; – 0,38)  value of x  value of y (2) 6.3 f(x) <g(x) when  − 180° ≤ x < −22,5° − 180° ≤ x < −22,5° or 157,5° < x ≤ 180°  157,5° < x ≤ 180° (2) 6.4.1 h(x) = cos (x – 45° + 30°)  + 30° = cos (x – 15°)  simplest form (2) 6.4.2 x = –135° – 30° = –165°  –165° (1) [9] QUESTION 7 7.1 Draw BD ⊥ AC B In ∆ABD: BD  construction sin A = ∴ BD = c. sin A c c a  sin A  making BD the In ∆CBD: subject BD sin C = ∴ BD = a. sin C A D C  sin C a ∴c . sin A = a. sin C c . sin A = a. sin C sin A sin C ∴ = a c (5) 7.2.1 sin R sin P = r p sin R sin 132° =  substitution into 27,2 73,2 correct formula 27,2 × sin 132° sin R = 73,2 making sin R the subject = 0,276... Rˆ = 16,03° 16,03° (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 10 DBE/2013 NSC – Grade 11 Exemplar – Memorandum 7.2.2 Q̂ = 180° − 132° − 16,03° = 31,97°  Q̂ = 31,97° 1 area of PQR = pr. sin Q  substitution into 2 1 correct formula = (73,2)(27,2). sin 31,97° 2  527,1 = 527,10 cm 2 (3) 7.3.1 PSˆQ = 180° − (a + b)  PSˆQ = 180° − (a + b) In ∆PSQ: SQ PQ = sin P sin PSˆQ SQ = h  sin[180° − (a + b)] sin a sin[180° − (a + b)] = sin (a + b) SQ h = sin a sin( a + b) h sin a making SQ the SQ = sin( a + b) subject (3) 7.3.2 SQˆ R = 90° − b  SQˆ R = 90° − b In ∆RSQ: RS = sin SQˆ R  use sine ratio SQ correctly RS = SQ.sin(90° − b) h sin a  sin(90° − b) = = . cos b sin( a + b) cosb (3) h sin a. cos b = sin( a + b) [17] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 11 DBE/2013 NSC – Grade 11 Exemplar – Memorandum QUESTION 8 Volume of hemisphere 1 4   substitution into =  π r3 correct formula 2 3  2 = π (3) 3 3 = 18π cm 3 18 π Volume of conical hole 1 = π r 2h 3 1 8  substitution into = π (1,5) 2 ( ) 3 9 correct formula 2 2 = π cm 3  π 3 3 1 17 π volume of metal A 26 1 ∴ = 3 =  17 π volume of metal B 2 1 3 π 3 ratio 26 : 1 Ratio of volume metal A : Volume metal B = 26 : 1 (6) [6] QUESTION 9 9.1 …bisects the chord.  answer (1) 9.2.1 OE = 10 cm … O midpoint of DE  OE = 10 OC = OE – CE = 10 – 2  OC = 8 = 8 cm (2) 9.2.2 In ∆COQ: QC2 = OQ2 – OC2 … Theorem of Pythagoras Using Theorem = (10)2 – (8)2 of Pythagoras = 36 QC = 6 cm  QC = 6 ∴PQ = 2QC … line drawn from centre ⊥ to chord  PQ = 12 (S) bisects chord  reason PQ = 12 cm (4) [7] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 12 DBE/2013 NSC – Grade 11 Exemplar – Memorandum QUESTION 10 10.1 D O A E B Construction: Produce DO to E  construction Proof: In ∆OBD: OBˆ D = ODˆ B …OD = OB = r  OBˆ D = ODˆ B EOˆ B = 2 × ODˆ B …exterior angle of triangle  EOˆ B = 2 × ODˆ B (S/R) In ∆AOD: OAˆ D = ODˆ A …OA = OD = r EOˆ A = 2 × ODˆ A …exterior angle of triangle  EOˆ A = 2 × ODˆ A AOˆ B = EOˆ B + EOˆ A (S/R)  = 2 × ODˆ B + 2 × ODˆ A AOˆ B = EOˆ B + EOˆ A = 2(ODˆ B + ODˆ A) = 2 ADˆ B (5) 10.2.1(a) M̂ = 76° … ∠ at centre = 2(∠ at circumference)  76°  reason (2) 10.2.1(b) Tˆ = 38° …ext∠ of cyc quad KTAB  38° 2  reason (2) 10.2.1(c) Ĉ = 38° … ext∠ of cyclic quad or ∠s in same  38° segment  reason (2) 10.2.1(d) CAˆ N = Cˆ = 38° …NA = NC  CAˆ N = 38° (S/R) Kˆ = 38° …ext ∠ of cyclic quad CATK  Kˆ 4 = 38° 4 (2) 10.2.2 ∴ Kˆ 4 = Tˆ2  statement ∴NK = NT …base ∠ equal s  reason (2) 10.2.3 N̂ = 180° − (38° + 38°)  N̂ = 104° (S/R) …∠s of ∆KNT = 104° Nˆ + KMˆ A = 104° + 76° = 180°  Nˆ + KMˆ A = 180° ∴ AMKN is cyclic quad …opposite ∠s = 180° reason (3) [18] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 13 DBE/2013 NSC –Grade 11 Exemplar – Memorandum QUESTION 11 11.1 ..... equal to the angle subtended by the same chord in the  alternate segment alternate segment. (1) 11.2.1 Aˆ1 = Cˆ 2 = x …tangent chord theorem  Aˆ1 = Cˆ 2 = x Cˆ = Gˆ = x 2 2 …tangent chord theorem  reason ∴ Aˆ1 = Gˆ 2 = x  Cˆ 2 = Gˆ 2 = x ∴ BCG  EA …alternate ∠s =  reason  conclusion with reason (5) 11.2.2 Eˆ1 = Cˆ 3 = y …alternate ∠s; BG EA  E1 = C3 = y (S/R) ˆ ˆ Fˆ = Cˆ = y 1 3 …ext∠ of cyclic quad CDFG ∴ Eˆ1 = Fˆ1 = y  Fˆ1 = Cˆ 3 = y (S) ∴ EA is a tangent …converse tangent-chord theorem  reason  Eˆ1 = Fˆ1 = y reason (5) 11.2.3 Bˆ = CAˆ E …tangent-chord theorem  CAE = B ˆ ˆ Cˆ = CAˆ E … alternate ∠s; BG EA reason 1  Cˆ 1 = CAˆ E (S/R) Cˆ 1 = Bˆ  reason ∴ AB = AC …base ∠s = (4) [15] TOTAL: 150 Copyright reserved