NATIONAL SENI CERTIFICATE GRADE MATHEMATICS P EXEMPLAR 04 MEMANDUM MARKS: 50 This memorandum consists of pages.
Mathematics/P DBE/04 NSC Grade 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. As the number of days that an athlete trained increased, the time taken to run the 00m event decreased. The fewer number of days an athlete trained, the longer the time he took to complete the 00m sprint. The greater number of days an athlete trained, the shorter the time he ran the 00m sprint. explanation. (60 ; 8,). a 7,89464 b 0,07068558... ŷ 0,07x + 7,8.4 ŷ 0,07(45) + 7,8 4,67 seconds.5 r 0,74 ( 0,74077594...) r a b equation substitution.6 There is a moderately strong relationship between the variables. moderately strong () () () () () []
Mathematics/P DBE/04 NSC Grade Exemplar Memorandum QUESTION. grounding at 0 70 plotting at upper 60 limits smooth shape 50 of curve Cumulative Frequency 40 0 0 0 00 90 80 70 60 50 40 0 0 0. 40 t < 60 class. (96 ; 64) 7 64 8 learners 0 0 0 0 0 40 50 60 70 80 90 00 0 0 Time (hours).4 Frequency: 5; 44; 60; 8; 9; 6 5 0 + 44 0 + 60 50 + 8 70 + 9 90 + 6 0 Mean 7 8000 7 46,5 hours 64 8 frequency midpoints 8000 7 () () () [0]
Mathematics/P 4 DBE/04 NSC Grade Exemplar Memorandum QUESTION. K(7 ; 0)...4.5 x + 7 M and M( 5 ; ) y M + m PM 7 tan PŜK m PM PŜK tan 8,4 θ 80 90 8,4 7,57 cos 7,57 PS cos 7,57 9,49 units PK PS PS x y substitution answer () () () tan PŜK m PM P ŜK θ correct ratio PS as subject () () sin 8,4 PS sin8,4 9,49 units PK PS PS.6 N(x ; x + 7) m m TN PM x + 7 5 x ( ) 6x + 6 x + 7x 5 x 5 y + 7 7 N(5 ; 7) (TN PM) correct ratio PS as subject () N in terms of x equal gradients substitution x -value y -value
Mathematics/P 5 DBE/04 NSC Grade Exemplar Memorandum m ( TN PM) TM equation of TM: y y ( x x ) y 5 ( x ( )) y 5 x + y x + 5 x + 7 x + 5 m TM equation of TM equating x x -value x 5 y -value y + 7 7 N(5 ; 7).7. y 5 equation.7. A(a ; 5) 45 T y y x + c 5 ( ) + c 5 c y x + 5 A(a ; 5) 45 () Q( ; ) 45 5 x gradient of AQ tan 45 or tan 5 or 5 m AQ ± a a 4 or 4 a 5 or m or AQ m AQ substitution into gradient formula x-value y-value []
Mathematics/P 6 DBE/04 NSC Grade Exemplar Memorandum QUESTION 4 4. M( ; ) 4. m NT 4 m (radius tangent) AT y ( x y x 4) 4. MR AB (line from centre to midpt of chord) MB MR + RB (Theorem of Pythagoras) 0 9 ( ) RB RB + RB AB 6 units 4.4 MN ( ) + ( ) 6 + 9 5 MN 5 units 4.5 r 5 units (x ) + (y ) 4 x + y 6x 4y + 9 0 () m NT m AT substitution of m and (4 ; ) equation MR AB MB substitution into Theorem of Pythagoras AB in surd form substitution into distance formula () r substitution into circle equation equation () [5]
Mathematics/P 7 DBE/04 NSC Grade Exemplar Memorandum QUESTION 5 5.. sin α 4 4 ( ) 5 5 5.. ( 4) + b 5 b 5 6 9 b cosα 5 5.. sin (α 45 ) sin α cos 45 cos α sin 45 4. ( ). 5 5 5 5.. sin (α 45 ) sin α cos 45 cos α sin 45 4. 5 0 ( ). 5 8sin x.cos x LHS sin x cos x 4(sin x.cos x) sin x cos x 5 ( ; 4) α reduction b answer () () expansion in simplest form () expansion in simplest form sin x cos x cos x () (cos 4sin x x sin x) 4 sin x factorise 4sin x cos x 4 tan x 5.. Undefined when cos x 0 or tan x : x 45 and x 5 cos x 45 5 (6) ()
Mathematics/P 8 DBE/04 NSC Grade Exemplar Memorandum 5. sin θ + 4sin θ 5sin θ 4 0 sin θ 5sin θ 0 (sin θ + )(sin θ ) 0 sin θ or sin θ (no solution) θ 0 + 60 k or θ 0 + 60 k ; k Z θ 0 + 60 k of θ 0 + 60 k ; k Z sin θ standard form factors no solution 0 0 + 60 k ; k Z (7) [] QUESTION 6 6. b 6. A(0 ; ) 0 value of b 6. x 60 x 60 () () 6.4 h(x) cos (x 0 ) + y [ ; ] y critical values notation () () [6]
Mathematics/P 9 DBE/04 NSC Grade Exemplar Memorandum QUESTION 7 7. Draw CD AB In ACD: CD sin A CD b. sin A b b C a construction sin A making CD the subject In CBD: CD sin B a CD a. sin B A D B sin B b. sin A a. sin B sin A sin B a b 7.. S Pˆ Q 80 x (opp s of cyclic quad ) PŜQ + PQˆ S x (sum of s in ) P ŜQ PQˆ S x ( s opp equal sides) 7.. 7.. sinspˆq sinpŝq SQ PQ sin(80 x) sin x SQ k sin k x SQ sin x k(sin x.cos x) SQ k cos x sin x SQ PQ + PS PQ.PS.cos S Pˆ Q k + k.k.k. cos (80 x) k + k cos x k + k (cos x ) 4k cos x SQ k cos x tan y k k tan y SQ cos x 6cos x tan y tan y b. sin A a. sin B S Pˆ Q 80 x (S/R) substitution into correct formula sin x SQ subject sin x. cos x () substitution into correct formula cos x cos x simplification tan ratio k subject and substitution () []
Mathematics/P 0 DBE/04 NSC Grade Exemplar Memorandum QUESTION 8 8. the angle subtended by the chord in the alternate segment correct theorem () 8.. Bˆ Ê 68 (tan chord theorem) Ê 68 () 8.. Ê Bˆ 68 (alt s; AE BC) Bˆ 68 (S/R) () 8.. Dˆ Bˆ 68 (ext of cyclic quad) Dˆ 68 () 8..4 Ê 0 + 68 (ext of ) 88 8..5 Ĉ 80 88 (opp s of cyclic quad) 9 Ê 88 (S/R) () Ĉ 9 () [9]
Mathematics/P DBE/04 NSC Grade Exemplar Memorandum QUESTION 9 9. Dˆ  x 4 9. (tan chord theorem)  Dˆ x ( s opp equal sides) Mˆ x (ext of ) or ( at centre at circum) M Dˆ E 90 (radius tan) Mˆ 90 x Ê 80 (90 + 90 x) x (sum of s in MDE) CM is a tangent (converse tan chord theorem) 9. Mˆ 90 (EM AC) A Dˆ B 90 ( in semi-circle) FMBD a cyclic quad (ext of quad int opp ) E Mˆ C 90 (EM AC) A Dˆ B 90 ( in semi-circle) FMBD a cyclic quad (opp s of quad supp) 9.4 DC MC MD (Theorem of Pythagoras) (BC) (BC) (MB MD radii) 9BC 4BC 5BC 9.5 In DBC and DFM: Dˆ Dˆ 4 x (proven in 9.) Bˆ Fˆ (ext of cyclic quad) Ĉ Mˆ DBC DFM ( ; ; ) 9.6 DM DC ( DBC DFM) FM BC 5BC BC 5  x  Dˆ x (S/R) () Mˆ x (S/R) M Dˆ E 90 (S/R) Ê x Mˆ 90 A Dˆ B 90 (S/R) () E Mˆ C 90 A Dˆ B 90 (S/R) () Th of Pythagoras substitution 9BC 4BC () Dˆ 4 Dˆ Bˆ Fˆ Ĉ Mˆ or ( ; ; ) S () [9]
Mathematics/P DBE/04 NSC Grade Exemplar Memorandum QUESTION 0 0. A D h k E B C Construction: Join DC and BE and heights k and h area ΔADE.AD. k AD (equal heights) area ΔDEB.DB. k DB area ΔADE area ΔDEC.AE. h.ec. h AE EC (equal heights) But Area DEB Area DEC (same base, same height) area ΔADE area ΔADE area ΔDEB area ΔDEC AD DB AE EC construction area ΔADE area ΔDEB area ΔADE area ΔDEC AD DB AE EC Area DEB Area DEC (S/R) area ΔADE area ΔADE area ΔDEB area ΔDEC (6)
Mathematics/P DBE/04 NSC Grade Exemplar Memorandum 0.. 0.. AB BE AC CD CD CD 9 units (Prop Th; BC ED) DG FD (Prop Th; FG EA) GA FE 9 x + x 6 54 6x 9 + x 9x 45 x 5 0.. In ABC and AED:  is common A Bˆ C Ê (corres s; BC ED) A ĈB Dˆ (corres s; BC ED) ABC AED (,, ) BC AC ED AD BC 9 0..4 BC 4 units area ΔABC area ΔGFD AC.BC.sin AĈB GD.FD.sin Dˆ ()( )sin Dˆ 4 ()sin Dˆ 9 6 (corres s; BC ED) AB AC (S/R) BE CD substitution () DG FD (S/R) GA FE substitution simplification  is common A Bˆ C Ê (S/R) A ĈB Dˆ (S/R) or ( ; ; ) BC AC ED AD use of area rule correct sides and angles substitution of values sina ĈB (S/R) sindˆ [] TOTAL: 50