Mathematics Grade 12 Paper 1 North West Mock 2019 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS P1 SEPTEMBER 2019 MARKING GUIDELINES MARKS: 150 This marking guidelines consists of 16 pages. NOTE:

Downloaded from hlayiso.com Mathematics/P1 2 NW/September 2019 NSC – Marking guidelines · If a candidate answers a question TWICE, only mark the FIRST attempt. · Consistent Accuracy applies in ALL aspects of the marking memorandum. QUESTION 1 1.1.1 3 x 2 - 18 x = 0 3 x( x - 6) = 0 P factors 3 x = 0 or x - 6 = 0 P x=0 x=0 x=6 P x=6 (3) 1.1.2 7x - 4x - 5 = 0 2 P standard form - b ± b 2 - 4ac x= 2a P substitution into the 4± ( - 4 ) - 4(7)(- 5) 2 = correct formula 2(7) 4 ± 156 = 14 P x = 1,18 x = 1,18 of x = - 0,61 P x = - 0, 61 (4) 1.1.3 ( x + 5)( x - 2 ) > 0 -5 2 \ x < - 5 or x > 2 P P x < - 5 or x > 2 (2) 26 - 52 x = ( 5 x - 6 ) 2 1.1.4 26 - 52 x = 5 2 x - 12.5 x + 36 P simplification 0 = 2.5 2x - 12.5 + 10 x 0 = 5 2 x - 6.5 x + 5 P standard form 0 = ( 5 - 5 )( 5 - 1) x x P factors/formula 5x = 5 or 5 x = 1 P 5x = 5 or 5x = 1 x =1 5 =5x 0 n.a. x= 0 P x = 1 n.a. or x = 0 n.a. n.a. \ No real solution P No real solution (6) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 3 NW/September 2019 NSC – Marking guidelines 1.2 x - 4y = 5 x = 4y + 5 P x = 4y + 5 3 x - 5 xy + 2 y = 25 2 2 3 ( 4 y + 5) - 5 ( 4 y + 5 ) y + 2 y 2 = 25 2 P substitution 3 (16 y + 40 y + 25 ) - 20 y - 25 y + 2 y = 25 2 2 2 48 y 2 + 120 y + 75 - 20 y 2 - 25 y + 2 y 2 - 25 = 0 30 y 2 + 95 y + 50 = 0 P standard form 6 y + 19 y + 10 = 0 2 -19 ± (19)2 - 4(6)(10) P factors/formula (3 y + 2)(2 y + 5) = 0 OR y= 2(6) 3 y = - 2 or 2 y = - 5 2 5 y=- y=- P both y-values 3 2 7 x= x = -5 P both x-values 3 OR (6) x - 4y = 5 4y = x - 5 x-5 x-5 y= P y= 4 4 3 x - 5 xy + 2 y 2 = 25 2 æ x - 5ö æ x - 5ö 2 3x 2 - 5x ç ÷ + 2ç ÷ = 25 P substitution è 4 ø è 4 ø æ 5 x 2 - 25 x ö æ x 2 - 10 x + 25 ö 3x 2 - ç ÷ + 2 ç ÷ - 25 = 0 è 4 ø è 16 ø æ 5 x 2 - 25 x ö æ x 2 - 10 x + 25 ö 3x 2 - ç ÷+ç ÷ - 25 = 0 è 4 ø è 8 ø 24 x - 10 x + 50 x + x - 10 x + 25 - 200 = 0 2 2 2 15 x 2 + 40 x - 175 = 0 3 x 2 + 8 x - 35 = 0 P standard form -8 ± 82 - 4(3)(- 35) (3 x - 7)( x + 5) = 0 OR x = P factors/formula 2(3) 7 x= or x = - 5 P both x-values 3 2 5 y=- y=- P both y-values 3 2 (6) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 4 NW/September 2019 NSC – Marking guidelines 1.3 x = 12 + 12 + 12 + 12 + ... 2 æ ö ( x) = ç 12 + 12 + 12 + 12 + ... ÷ 2 ç ÷ è ø x 2 = 12 + 12 + 12 + 12 + ... x 2 = 12 + x P substitution x 2 - x - 12 = 0 P standard form ( x - 4)( x + 3) = 0 x = 4 or x = - 3 P x=4 n.a. P x = - 3 n.a. (4) [25] QUESTION 2 2.1.1 Tn = a + (n - 1)d = - 11 + (n - 1)(7) P substituting a and d in = - 11 + 7 n - 7 Tn: -11 + (n - 1)(7) = 7 n - 18 P answer answer only, full marks (2) 2.1.2 Tn = 7 n - 18 T60 = 7(60) - 18 P substitute n=60 into Tn = 402 P answer (2) 2.1.3 n Sn = é 2a + ( n - 1) d ùû 2ë 60 S60 = [ 2(-11) + (60 - 1)(7)] P substitution 2 = 11 730 P answer (2) 2.2 -4 + 3 + 10 + ... + 486 Tn = 7 n - 18 486 = 7 n - 18 P 486 = 7 n - 18 504 = 7 n 72 = n P n = 72 72 å (7n - 18) 72 \ å (7 n - 18) PP n=2 n=2 (4) OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 5 NW/September 2019 NSC – Marking guidelines -4 + 3 + 10 + ... + 486 Tn = a + ( n - 1) d = - 4 + ( n - 1)(7) = - 4 + 7n - 7 = 7 n - 11 P Tn = 7 n - 11 486 = 7 n - 11 P 486 = 7 n - 11 497 = 7 n 71 = n P n = 71 71 P å (7n - 11) 71 \ å (7 n - 11) n =1 n =1 (4) 2.3 2a = 7 7 7 a= = 3,5 P a= = 3,5 2 2 T2 - T1 = 3a + b - 11 = 3(3,5) + b 43 - =b 2 - 21, 5 = b P b = - 21,5 b (- 21,5) 43 n=- =- = = 3, 07 P n = 3,07 2a 2(3,5) 14 P T3 < T4 The 3rd term is the smallest P answer (5) OR T1 T2 T3 T 4 T5 \ / \ / \ / \ / -11 - 4 3 10 T1 - 11 = T2 T1 > T2 P proving T1 > T2 T2 - 4 = T3 T2 > T3 P proving T2 > T3 T3 + 3 = T4 T4 > T3 P proving T4 > T3 T4 + 10 = T5 T5 > T4 P proving T5 > T4 T1 > T2 > T3 & T5 > T4 > T3 \ T3 is the smallest term. P answer (5) [15] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 6 NW/September 2019 NSC – Marking guidelines QUESTION 3 3.1 T2 5(3 x + 1) 2 r= = = 3x + 1 P r = 3x + 1 T1 5(3 x + 1) -1< r <1 -1 < 3x + 1 < 1 P -1 < 3x + 1 < 1 - 2 < 3x < 0 2 - <x<0 P answer 3 (3) 3.2 1 x= - 6 æ 1ö æ æ 1ö ö r = 3ç - ÷ + 1 a = 5 ç 3 ç - ÷ + 1÷ è 6ø è è 6ø ø 1 5 1 = = P r= 2 2 2 a S¥ = 1- r 5 5 P a= 2 = 2 P substitution 1 1- 2 =5 P answer (4) [7] QUESTION 4 4.1 C(2 ; - 9) P P C(2; - 9) (2) 4.2 1 y = ( x - 2) - 9 2 2 1 = (0 - 2) - 9 2 P x = 0 2 =-7 \ B (0; - 7) P y =-7 (2) 4.3 f ( x) = x - 7 PP x - 7 (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 7 NW/September 2019 NSC – Marking guidelines 4.4 At the intersection of the asymptotes, x = 2 \y = 2 - 7 = - 5 P y=-5 \ (2; - 5) a h( x ) = -5 P p = -2 x-2 P q = -5 a P substitution of -7 = -5 0-2 B(0 ; - 7 ) a -2 = -2 4=a P a=4 4 \ h( x ) = -5 x-2 (5) 4.5 æ 4 ö 3h ( x) - 2 = 3 ç - 5÷ - 2 èx -2 ø 12 = - 15 - 2 x-2 x=2 P x=2 y = - 17 P y = - 17 (2) 4.6 4 h( x ) = -5 x-2 4 0= -5 P y = 0 x-2 4 5 = x-2 5( x - 2) = 4 P simplification 5 x - 10 = 4 5 x = 14 14 x= = 2,8 P answer 5 (3) 4.7.1 x Î (- ¥ ; 2) or ( 2; 2,8) P ( - ¥ ; 2) PP ( 2; 2,8) OR x Î ( - ¥ ; 2,8) ; x ¹ 2 PP ( - ¥ ; 2,8 ) P x¹2 OR x < 2,8 ; x ¹ 2 PP x < 2,8 P x¹2 (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 8 NW/September 2019 NSC – Marking guidelines 4.7.2 f -1 ( x - 1) : x - 1 = y - 7 P f -1 ( x - 1) \y=x+6 P y=x+6 f -1 ( x - 1) < 2: x + 6 < 2 P x+6<2 \x < - 4 P answer OR (2, –5) lies on f P (–5 ; 2) lies on f -1 (–5 ; 2) lies on f -1 P (–4 ; 2) lies on (–4 ; 2) lies on f -1( x - 1) f -1( x - 1) Since mf = 1, mf -1 = 1 \ f -1 is an increasing function P increasing function \x < - 4 P answer (4) 4.8 1 ( x - 2) - 9 = x - 7 + k 2 P substitution of g(x) and 2 f (x) 1 2 x - 2x - 7 = x - 7 + k 2 1 2 x - 3x - k = 0 P standard form 2 For equal roots : b 2 - 4ac = 0 æ1ö (- 3)2 - 4 ç ÷ (- k ) = 0 P substitution into è2ø b 2 - 4 ac = 0 2k = - 9 9 P value of k for equal k =- 2 roots \ For two unequal positive roots : - 4,5 < k < 0 PP answer OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 9 NW/September 2019 NSC – Marking guidelines 1 ( x - 2) - 9 = x - 7 + k P substitution of g(x) and 2 2 f (x) 1 2 x - 2x - 7 = x - 7 + k 2 1 2 x - 3x - k = 0 P standard form 2 For equal roots : b 2 - 4ac = 0 æ1ö (- 3)2 - 4 ç ÷ (- k ) = 0 P substitution into è2ø b2 - 4ac = 0 2k = - 9 9 P value of k for equal k =- 2 roots y - intercept of y = x - 7 + k : 9 y= 0-7- 2 y = - 11, 5 \ For two unequal positive roots : -11,5 < - 7 + k < - 7 P - 11,5 < - 7 + k < - 7 - 4,5 < k < 0 P answer OR 1 g ( x) = ( x - 2) - 9 2 2 = ( x 2 - 4 x + 4) - 9 1 2 1 1 2 = x2 - 2x - 7 P x - 2x - 7 2 2 g ( x) = x - 2 / 1 =x-2 P g / ( x) = 3= x gradient of f (x) = 1 1 1 g ( 3) = ( 3 - 2 ) - 9 or g (3) = (3)2 - 2(3) - 7 2 2 2 1 g ( 3) = - 8 1 P g ( 3) = - 8 2 2 y=x+ c 1 -8 =3+c 2 c = - 11, 5 P c = - 11,5 For two positive roots: - 11,5 < c < - 7 P - 11,5 < c < - 7 - 11,5 < k - 7 < - 7 - 4,5 < k < 0 P answer (6) [29] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 10 NW/September 2019 NSC – Marking guidelines QUESTION 5 5.1.1 -x æ5ö h ( x) = ç ÷ è6ø æ6ö x P any answer =ç ÷ (1) è5ø x 5.1.2 æ5ö f ( x) = ç ÷ è6ø y y -1 æ5ö æ5ö f : x= ç ÷ P x= ç ÷ è6ø è6ø y = log 5 x x>0 P y = log 5 x (2) 6 6 5.1.3 x Î ( 0;1] P critical values P notation (only if values OR are correct) 0 < x £1 P critical values P notation (only if values are correct) (2) 5.2 y ( - 2,5 ; 6 ) P turning point P P each x-intercept P shape −6 1 (4) [9] QUESTION 6 6.1 75 75 Loan = ´ 2 000 000 P ´ 2 000 000 100 100 = R 1 500 000 P answer OR 25 ´ 2 000 000 100 = R500 000 P R 500 000 Loan = R 2 000 000 - 500 000 = R 1 500 000 P answer (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 11 NW/September 2019 NSC – Marking guidelines A = P (1 + i ) n 6.2 1 æ 0, 095 ö 1 = 1 500 000 ç 1 + æ 0, 095 ö ÷ P 1 500 000 ç1 + ÷ è 4 ø è 4 ø = R1 535 625 x é1 - (1 + i ) ù -n P= ë û i é 0, 095 0, 095 ö ù -n æ P i= 58 000 ê1 - ç 1 + ÷ ú 4 æ 0, 095 ö 1 êë è 4 ø úû P substitution into the 1 500 000 ç 1 + ÷ = è 4 ø 0, 095 correct formula 4 -n æ 0, 095 ö 0, 6288119612 = 1 - ç 1 + ÷ è 4 ø -n æ 0, 095 ö ç1 + ÷ = 0, 3711880386 è 4 ø log æ 0,095 ö ( 0, 3711880386 ) = - n ç1 + ÷ P correct use of logs è 4 ø log ( 0, 3711880386 ) OR = -n æ 0, 095 ö log ç 1 + ÷ è 4 ø n = 42, 22185853 P n = 42,22 \ 43 quarterly payments P answer (6) 6.3 Outstanding balance x é(1 + i ) - 1ù n = P (1 + i ) n - ë û i P n = 23 éæ ù 23 P 1 535 625 æç1 + 0, 095 ö÷ 23 0, 095 ö 58 000 êç 1 + ÷ - 1ú æ 0, 095 ö 23 êëè 4 ø úû è 4 ø = 1 535 625 ç1 + ÷ - 24 è 4 ø 0, 095 OR 1 500 000 æç1 + 0, 095 ö÷ 4 è 4 ø = R 886 790,15 P éæ 0, 095 ö 23 ù 58 000 êç 1 + ÷ - 1ú êëè 4 ø úû 0, 095 4 P answer (4) OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 12 NW/September 2019 NSC – Marking guidelines Outstanding time = 42,22185853 - 23 P 23 = 19,22185853 P n = 19,22185853 Outstanding balance: x é1 - (1 + i ) ù -n P= ë û i é æ 0, 095 ö - 19,22185853 ù 58 000 ê1 - ç 1 + ÷ ú P substitution êë è 4 ø úû = 0, 095 4 = R 886 790,15 P answer (4) 6.4 Interest charged = (58 000 ´ 23) - (1 500 000 - 886 790,1491) P 58 000 ´ 23 PP = R 720 790,15 1 500 000 - 886 790,1491 P answer (4) [16] QUESTION 7 7.1 f ( x) = - x 2 + 3 x - 7 f ( x + h) = - ( x + h)2 + 3( x + h) - 7 P -( x + h) 2 + 3( x + h) - 7 = - ( x 2 + 2 xh + h 2 ) + 3 x + 3h - 7 P = - x 2 - 2 xh - h 2 + 3 x + 3h - 7 - x 2 - 2 xh - h 2 + 3 x + 3h - 7 f ( x + h) - f ( x) = (- x - 2 xh - h + 3 x + 3h - 7) 2 2 - (- x 2 + 3 x - 7) = - 2 xh - h 2 + 3h P simplification f ( x + h) - f ( x ) f / ( x) = lim h ®0 h -2 xh - h 2 + 3h P substitution into formula = lim h ®0 h h (-2 x - h + 3) = lim P factors h ®0 h = lim (-2 x - h + 3) h ®0 = - 2x + 3 P answer (6) OR Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 13 NW/September 2019 NSC – Marking guidelines f ( x) = - x 2 + 3 x - 7 f ( x + h) - f ( x) f / ( x) = lim h ®0 h - ( x + h) + 3( x + h) - 7 - (- x 2 + 3x - 7) 2 P -( x + h) 2 + 3( x + h) - 7 = lim h ®0 h P substitution into formula - ( x + 2 xh + h ) + 3x + 3h - 7 + x 2 - 3x + 7 2 2 = lim h ®0 h P - x - 2 xh - h + 3x + 3h - 7 + x2 - 3 x + 7 2 2 = lim - x 2 - 2 xh - h 2 + 3 x + 3h - 7 h ®0 h - 2 xh - h + 3h 2 P simplification = lim h ®0 h h(-2 x - h + 3) P factors = lim h ®0 h = lim (-2 x - h + 3) h ®0 = - 2x + 3 P answer (6) 7.2 é 3x 7 + x ù Dx ê15 5 x 4 - ú ë 4 x3 û 4 3x 4 é 4 3x 4 x - 2 ù P 15 x 5 P - = Dx ê15 x 5 - - ú 4 ë 4 4 û x-2 1 P - - 1 -3 4 = 12 x 5 - 3 x3 + x 1 2 - P 12 x 5 1 -3 P -3 x 3 P + x 2 (6) [12] QUESTION 8 8.1 x £ - 3 or x ³ 2 P x £ -3 Note: Accept x < - 3 or x > 2 . P x³2 (2) At the stationary points, the graph is both increasing and decreasing. 8.2 x = -3 P x = -3 (1) 8.3 f ( x) = 1( x + 3)( x - 2) / P ( x + 3)( x - 2) = x2 + x - 6 P answer (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 14 NW/September 2019 NSC – Marking guidelines 8.4 f ( x) = px3 + qx 2 + rx + 10 f / ( x) = 3 px 2 + 2qx + r P f / ( x) = 3 px2 + 2qx + r f / ( x) = x 2 + x - 6 \ 3 p = 1 2q = 1 r = - 6 P 3p = 1 1 1 P 2q = 1 p= q= P r = -6 3 2 (4) 8.5 1 1 f ( x) = x3 + x 2 - 6 x + 10 3 2 f ( x) = x + x - 6 / 2 f / / ( x) = 2 x + 1 P f // ( x) = 2 x + 1 Concave down when f / / ( x) < 0 P f // ( x) < 0 2x + 1 < 0 1 1 x<- P x<- 2 2 (3) OR 1 3 1 2 f ( x) = x + x - 6 x + 10 3 2 f ( x) = x + x - 6 / 2 -3 + 2 -3 + 2 x= P x= 2 2 1 x=- 2 1 f / ( x) is decreasing for x < - 2 1 \ f // ( x) < 0 when x < - P f / / ( x) < 0 2 1 P answer \ Concave down: x < - 2 (3) [12] QUESTION 9 9.1 AH = 4t HB = 1 200 - 4t P HB = 1 200 - 4t BD = 5t P BD = 5t HD 2 = (1 200 - 4t ) 2 + (5t ) 2 Pyth P HD 2 = (1 200 - 4t ) 2 + (5t ) 2 = 1 440 000 - 9 600t + 16t 2 + 25t 2 P 1 440 000 - 9 600t + 16t 2 + 25t 2 = 41t 2 - 9 600t + 1 440 000 (4) HD = 41t - 9 600t + 1 440 000 2 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 15 NW/September 2019 NSC – Marking guidelines ( HD ) (t ) = 82t - 9 600 P ( HD2 ) (t ) = 82t - 9 600 2 / / 9.2 0 = 82t - 9 600 P ( HD ) (t ) = 0 2 / 82t = 9 600 t = 117, 07 seconds \ after 117,07 seconds P answer OR The minimum of HD occurs when HD2 is a minimum. P min of HD = min HD2 - ( - 9 600) P substitute into equation min t = of axis of symmetry 2(41) = 117, 07 seconds P answer (3) 9.3 HD = 41(117, 07)2 - 9 600(117, 07) + 1 440 000 P substitution = 878 048, 7805 = 937, 04 m P answer No, the hunter is outside the shooting range. P No (3) [10] QUESTION 10 10.1 0,18 Win Rain 0,82 Loose 0,45 0,32 0,39 Win P first outcomes Cloudy P second outcomes 0,61 Loose 0,23 0,63 Win Sunny (2) 0,37 Loose 10.2 P(Win) = P(RW) + P(CW) + P(SW) P P(RW) = (0,45)(0,18) = (0,45)(0,18) + (0,32)(0,39) + (0,23)(0,63) P P(CW) = (0,32)(0,39) = 0,3507 P P(SW) = (0,23)(0,63) = 35,07% P answer (4) [6] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P1 16 NW/September 2019 NSC – Marking guidelines QUESTION 11 11.1 9! = 362 880 P 9! OR 362 880 (1) 11.2 8! ´ 2! = 80 640 PP 8! ´ 2! OR 80 640 (2) 11.3 5 ´ 7! ´ 4 P 5 ´ 7! ´ 4 9! P 9! in denominator 5 = or 0,28 P answer 18 (3) 11.4 Total number of ways that 9 horses can be arranged in 4 stables = 9 ´ 8 ´ 7 ´ 6 = 3024 P 3 024 Total number of ways without a mare in the arrangement = 4! = 24 P 24 Total number of ways in which at least 1 mare in the arrangement = 3 024 - 24 = 3 000 ways P 3 000 OR The total number of ways in which there is at least 1 mare in the arrangement: é 12 ù P [(5 ´ 4 ´ 3 ´ 2) ´ 4] + = [(5 ´ 4 ´ 3 ´ 2) ´ 4] + ê(5 ´ 4 ´ 4 ´ 3) ´ ú + é 12 ù ë 2û ê (5 ´ 4 ´ 4 ´ 3) ´ 2 ú [(5 ´ 4 ´ 3 ´ 4) ´ 4] + [5 ´ 4 ´ 3 ´ 2] ë û P [(5 ´ 4 ´ 3 ´ 4) ´ 4] + = 3 000 ways [5 ´ 4 ´ 3 ´ 2] P answer (3) [9] TOTAL: 150 Copyright reserved Please turn over

Downloaded from hlayiso.com COGNITIVE LEVELS MATHEMATICS P1 COGNITIVE LEVELS TOPICS LEVEL 1 LEVEL 2 LEVEL 3 LEVEL 4 (20%) (35%) (30%) (15%) DIFFERENTIATION TOTAL MARKS ROUTINE COMPLEX QUESTION KNOWLEDGE PATTERNS FUNCTIONS PROBABILITY PROBLEM ALGEBRA FINANCE SOLVING PROCEDURES PROCEDURES 1.1.1 3 3 1.1.2 4 4 1.1.3 2 2 1.1.4 6 6 1.2 6 6 1.3 4 4 25 2.1.1 2 2 2.1.2 2 2 2.1.3 2 2 2.2 4 4 2.3 5 5 15 3.1 3 3 3.2 4 4 7 4.1 2 2 4.2 2 2 4.3 2 2 4.4 5 5 4.5 2 2 4.6 3 3 4.7.1 3 3 4.7.2 4 4 4.8 6 6 29 5.1.1 1 2 5.1.2 2 1 5.1.3 2 2 5.2 4 4 9 6.1 2 2 6.2 6 6 6.3 4 4 6.4 4 4 16 7.1 6 6 7.2 6 6 12 8.1 2 2 8.2 1 1 8.3 2 2 8.4 4 4 8.5 3 3 12

TOT 11.1 9.3 9.1 % 11.4 11.2 9.2 QUESTION Pol 11.3 10.2 10.1 28 LEVEL 1 19% 1 KNOWLEDGE 20% (25%) 55 ROUTINE LEVEL 2 37% 2 4 2 3 35% PROCEDURES (30 %) 44 COMPLEX LEVEL 3 29% 3 30% PROCEDURES (30%) 23 PROBLEM LEVEL 4 COGNITIVE LEVELS 15% 3 3 4 15% SOLVING (15%) 25 25 ALGEBRA 25 22 PATTERNS 35 38 FUNCTIONS 15 16 FINANCE TOPICS 35 34 3 3 4 DIFFERENTIATION 3 1 2 3 4 2 PROBABILITY 15 15 150 150 9 6 10 TOTAL MARKS Downloaded from hlayiso.com