Section A. A 5 12 sin cm. AC cos100 7 M15/5/MATHL/HP2/ENG/TZ2/XX/M. 1. (a) (M1) A1 [2 marks] (b) (M1) therefore AC 13.

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Darrell Caldwell
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1 7 M5/5/MATHL/HP/ENG/TZ/XX/M Section A. (a) A 5 sin cm (M) A [ marks] (b) AC 5 5 cos00 (M) therefore AC 3.8 (cm) A [ marks] Total [4 marks]. (a) (M)A [ marks] (b) M A [ marks] (c) METHOD 5 number of ways all men MA Note: Allow FT from answer obtained in part (a). [ marks] continued

2 8 M5/5/MATHL/HP/ENG/TZ/XX/M Question continued. METHOD M =35 A [ marks] Total [6 marks] 3. (a) general shape including minimums, cusp correct domain and symmetrical about the middle ( x 5) (b) x 9.6 or x AA A AA [ marks] Total [5 marks]

3 9 M5/5/MATHL/HP/ENG/TZ/XX/M 4. (a) (i) X ~ Po(5) P( X 8) 0.33 (M)A (ii) M days A Note: Accept day. 35 (b) P( Y 9) 0.77 Y~Po( ) (A) (M)A [4 marks] Total [7 marks] 5. (a) (0) b b uv a(0) a ba ba (M)(A) b 4 a b b a c (M) a, b, c4 A Note: Award A for two correct. 4 (b) n 4 (A) [5 marks] 4xy4z 0 (x yz 0) A [ marks] Total [7 marks]

4 0 M5/5/MATHL/HP/ENG/TZ/XX/M 6. (a) EITHER y ln ( xa) bln (5x 0) (M) y ln ( xa) ln c ln (5x0) y ln c( xa) ln (5x 0) (M) OR y ln (5x0) ln 5( x ) (M) y ln (5) ln ( x ) (M) THEN a, b ln5 AA Note: Accept graphical approaches. Note: Accept a, b.6 e (b) V π ln (5x 0) dx 99. e (M) A [4marks] [marks] Total [6 marks]

5 M5/5/MATHL/HP/ENG/TZ/XX/M 7. (a) x y6z 0 4x3y4z 4 x y( ) z attempt at row reduction M R R and R3 R eg x y6z 0 yz 4 3 y( 8) z A R 3R eg 3 x y6z 0 yz 4 ( ) z (i) no solutions if, 0 (ii) one solution if (iii) infinite solutions if, 0 A A A A Note: Accept alternative methods e.g. determinant of a matrix Note: Award AAA0 if all three consistent with their reduced form, AA0A0 if two or one answer consistent with their reduced form. [6 marks] (b) yz 4 y 4z A x y6z z46z 4z4xz A therefore Cartesian equation is x y4 z or equivalent A Total [9 marks]

6 M5/5/MATHL/HP/ENG/TZ/XX/M 8. (a) EITHER area of triangle 3 4 ( 6) A 4 area of sector arcsin 5 (.59 ) A 5 OR d x x MA THEN total area 7.59 m (A) 7.59 percentage 00 44% 40 A (b) METHOD [4 marks] 4 a 6 area of triangle A 4 arcsin a (A) 4 area of sector r a arcsin a A therefore total area 4 a 6 a arcsin 0 a A 4 rearrange to give: a arcsin 4 a 6 40 a AG continued

7 3 M5/5/MATHL/HP/ENG/TZ/XX/M Question 8 continued METHOD 4 0 a x dx0 use substitution 4 arcsin a a a 0 a 4 arcsin a 0 cos d 0 dx xasin, acos d cos d 0 M 4 arcsin a sin 40 a a a 0 4 arcsin a 0 sin cos arcsin a 40 a a a 4 arcsin 4 a 6 40 a (c) solving using GDC a 5.53cm A M A A AG [4 marks] [ marks] Total [0 marks]

8 4 M5/5/MATHL/HP/ENG/TZ/XX/M 9. (a) attempt to set up the problem using a tree diagram and/or an equation, with the unknown x M 4 3 x ( x) A x x x x A (b) attempt to set up the problem using conditional probability M EITHER A OR A THEN 3 A 0 Total [6 marks]

9 5 M5/5/MATHL/HP/ENG/TZ/XX/M Section B 0. (a) (i) P(0 X 30) % (M)A Note: Accept 50 Note: Award MA0 for 0.50 (0.500) (ii) P( X 30) ( ) 0.93 M expected number of turnips 9.3 A Note: Accept 9. (iii) no of turnips weighing more than 30 is Y ~ B(00, 0.93) M PY ( 30) A (b) (i) X ~ N44, P( X 30) (M) PZ (A) 9.33 g A [6 marks] (ii) P( X 50 X 30) P( X 50) P( X 30) M A expected number of turnips 55.7 A [6 marks] Total [ marks]

10 6 M5/5/MATHL/HP/ENG/TZ/XX/M. (a) attempt at implicit differentiation M dy dy x 5x 5y y 0 dx dx AA Note: A for differentiation of y and 7. x 5xy, A for differentiation of dy x5 y (y5 x) 0 dx dy 5y x dx y 5x AG (b) dy 56 dx 56 4 gradient of normal 4 A A equation of normal y 4xc M substitution of (6, ) y 4x 5 A Note: Accept y4( x6) [4 marks] (c) setting 5 y x y 5x M y x A substituting into original equation M x 5x x 7 (A) 7x 7 x points (, ) and (,) A (A) distance 8 (M)A [8 marks] Total [5 marks]

11 7 M5/5/MATHL/HP/ENG/TZ/XX/M. (a) METHOD 9 3 s9t3t d t t t ( c) (M) t 0, s3c 3 (A) (b) t 4 s A METHOD 4 s 3 9t3t dt 0 (M)(A) s A 9 s3 t t 3 (A) correct shape over correct domain maximum at (3, 6.5) t intercept at 4.64, s intercept at 3 A minimum at (5, 9.5) A A A [5 marks] continued

12 8 M5/5/MATHL/HP/ENG/TZ/XX/M Question continued (c) 9.5 abcos 6.5 abcos3 (M) Note: Only award M if two simultaneous equations are formed over the correct domain. 7 a A b 3 A (d) at t : t t 9 (M) t t0 9 t A solving 7 3cos t 3 (M) 5 GDC t 6. A Note: Accept graphical approaches. [4 marks] Total [5 marks]

13 9 M5/5/MATHL/HP/ENG/TZ/XX/M 3. (a) L and L are not parallel, since k 6 R if they meet, then and M solving simultaneously 0 A 46 4 contradiction, R so lines are skew AG Note: Do not award the second R if their values of parameters are incorrect. [4 marks] 6 4cos MA 6 cos (A) ( radians) A (b) [4 marks] (c) (i) (M) 4 0 4i 0 j 3k 3 A continued

14 0 M5/5/MATHL/HP/ENG/TZ/XX/M Question 3 continued (ii) METHOD let P be the intersection of L and L 3 let Q be the intersection of L and L 3 OP OQ 4 6 therefore PQ OQOP 6 4 t M MA M 4t 0 0t 0 63t solving simultaneously (M) 3 8 (0.56), ( 0.4) A 5 5 Note: Award A for either correct or. EITHER therefore OP.56 5 A therefore L3: r A.5 3 continued

15 M5/5/MATHL/HP/ENG/TZ/XX/M Question 3 continued OR therefore OQ A therefore L3: r A Note: Allow position vector(s) to be expressed in decimal or fractional form. [0 marks] METHOD a 4 L 3 : r3 b t 0 c 3 forming two equations as intersections with L and L a 4 b t 0 c 3 a 4 b t 0 c MAA Note: Only award MAA if two different parameters t,t used. attempting to solve simultaneously M 3 8 (0.56), ( 0.4) A 5 5 Note: Award A for either correct or. continued

16 M5/5/MATHL/HP/ENG/TZ/XX/M Question 3 continued EITHER a b. 776 c 656. A therefore L3: r t AA OR a b 56. c 5. A therefore L3: r 3.56 t AA Note: Allow position vector(s) to be expressed in decimal or fractional form. Total [8 marks]

(b) g(x) = 4 + 6(x 3) (x 3) 2 (= x x 2 ) M1A1 Note: Accept any alternative form that is correct. Award M1A0 for a substitution of (x + 3).

(b) g(x) = 4 + 6(x 3) (x 3) 2 (= x x 2 ) M1A1 Note: Accept any alternative form that is correct. Award M1A0 for a substitution of (x + 3). Paper. Answers. (a) METHOD f (x) q x f () q 6 q 6 f() p + 8 9 5 p METHOD f(x) (x ) + 5 x + 6x q 6, p (b) g(x) + 6(x ) (x ) ( + x x ) Note: Accept any alternative form that is correct. Award A for a substitution