8 M13/5/MATME/SP2/ENG/TZ1/XX/M 9 M13/5/MATME/SP2/ENG/TZ1/XX/M. x is σ = var,

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1 8 M/5/MATME/SP/ENG/TZ/XX/M 9 M/5/MATME/SP/ENG/TZ/XX/M SECTION A. (a) d N [ mark] (b) (i) into term formula () eg u (99), 5 + (00 ) u 00 0 N (ii) into sum formula () eg S 00 ( (5) + 99() ), S 00 (5 + 0) S N (c) into term formula () eg ( n ), 50 n+ n 500 N [ marks]. (a) valid approach eg 5 6, 6 + p 5 Total p 9 N [ marks] (b) (i) mean 6.7 A N (ii) recognizing that variance is (sd) eg.0, σ var,.58 σ N. (a) p 5, q 7, r 7 (accept r 5 ) (b) correct working () 5 7 eg ( x) ( 7 ), 79, 4,, (a) (i) coefficient of term in 5 x is N Note: Do not award the final for an answer that contains x. (b) Note: (ii) [ marks] Total A 0 N 0 0 A A N Note: Award for 6, 7 or 8 correct elements. evidence of multiplying by A (in any order) eg X A B, BA, one correct element 9 X 8 (accept x 9, y 8, z.5 ) A.5 Award for two correct elements. Total Total

2 0 M/5/MATME/SP/ENG/TZ/XX/M M/5/MATME/SP/ENG/TZ/XX/M 5. (a) 7. recognizing one quartile probability (may be seen in a sketch) eg P( X < Q ) 0.75, 0.5 finding standardized value for either quartile eg z , z attempt to set up equation (must be with z values) Q 50 x 50 eg 0.67, () Note: Award for approximately correct shape crossing x-axis with < x <.5. Only if this is awarded, award the following: for maximum in circle, for endpoints in circle. one correct quartile eg Q 56.74, Q 4.5 () correct working eg other correct quartile, Q µ valid approach for IQR (seen anywhere) eg Q Q, ( Q µ ) () () (b) (i) t π (exact),.4 N (ii) recognizing distance is area under velocity curve eg s v, shading on diagram, attempt to integrate v IQR.5 N4 valid approach to find the total area eg area A + area B, vt d vt d.4 5, vt d + vt d 0, v.4 correct working with integration and limits (accept dx or missing dt) eg.4.4 vt d + vt d 0, , 5 5 e sin t 0 () distance.95(m) Total [8 marks] 6. (a) (i) k N (ii) p N (iii) q 5 N (b) recognizing one transformation eg horizontal stretch by, reflection in x-axis, A is (, 5) Total

3 M/5/MATME/SP/ENG/TZ/XX/M M/5/MATME/SP/ENG/TZ/XX/M SECTION B 8. (a) evidence of choosing cosine rule eg c a + b abcosc, CD + AD CD ADcos D eg cos04, cos04 AC 5.5(m) N (b) (i) METHOD evidence of choosing sine rule eg sin A sin B sin ACD ˆ sin D, a b AD AC eg sin ACD ˆ sin ACD ˆ 0.0 METHOD evidence of choosing cosine rule eg c a + b abcosc eg (.5)(5.56 )cosc ACD ˆ 0.0 (ii) subtracting their ACD ˆ from 7 eg 7 ACD ˆ, ACB ˆ 4.0 N N N (c) () eg area ADC (8)(.5)sin04 area 44.6 (m ) N [ marks] (d) attempt to subtract eg circle ABCD, πr ADC ACB area ACB (5.56 )(4)sin 4.98 ( ) () correct working eg π(8) (5.56 )(4)sin 4.98, 64π shaded area is 8.4 (m ) Total 9. (a) 00 f (0) (exact),.96 5 N [ mark] (b) setting up equation eg 00 95, sketch of graph with horizontal line at y e 0.x x 4. N [ marks] (c) upper bound of y is 00 () lower bound of y is 0 () range is 0 < y < 00 (d) METHOD setting function ready to apply the chain rule 0.x eg ( ) e evidence of correct differentiation (must be substituted into chain rule) eg ( 0. x ) ( 0. x u v ) e, 50e ( 0.) correct chain rule derivative eg ( 0. x ) ( 0. x f x ) ( ) e 50e ( 0.) correct working clearly leading to the required answer 0.x 0.x eg ( ) f ( x) f ( x) 000e + 50e 000e METHOD 0.x 0.x ( + 50e ) attempt to apply the quotient rule (accept reversed numerator terms) vu uv uv vu eg, v v evidence of correct differentiation inside the quotient rule eg f '( x) 0.x 0.x ( + 50e )( 0) 00( 50e 0.) 0. x ( + 50e ) 0.x 00( 0)e 0, 0.x + 50e ( ) any correct expression for derivative (0 may not be explicitly seen) eg 0.x ( ) 0.x ( + 50e ) 00 50e 0. correct working clearly leading to the required answer 0.x 0.x 0 00( 0)e 00( 0)e eg f ( x), 0.x 0.x + 50e + 50e f ( x) 000e 0.x ( + 50e ) ( ) 0.x ( ) ()() AG ()() () AG N0 N0 continued

4 4 M/5/MATME/SP/ENG/TZ/XX/M 5 M/5/MATME/SP/ENG/TZ/XX/M Question 9 continued (e) METHOD sketch of f ( x) () eg 0. (a) valid approach eg + diameter, + maximum height 5 (m) N [ marks] 60 (b) (i) period.4 period 5 (minutes) AG N0 recognizing maximum on f ( x) eg dot on max of sketch finding maximum on graph of f ( x) eg (9.6, 5), x maximum rate of increase is 5 N METHOD recognizing f ( x) 0 finding any correct expression for f ( x) () eg ( 0. x ) ( 0. x ) ( 0. x )( ( 0. x 50e 00e 000e 50e )( 0e 0. x + + )) 0.x 4 ( + 50e ) finding x maximum rate of increase is 5 N Total [5 marks] (ii) (c) METHOD π b ( 0.08 π ) 5 N [ marks] valid approach 5 eg max 74, a, 74 a 6 (accept a 6) () a 6 N METHOD attempt to substitute valid point into equation for h π.5 eg acos 5 correct equation eg acos( π ), 74 + a a 6 N () continued

5 6 M/5/MATME/SP/ENG/TZ/XX/M M09/5/MATME/SP/ENG/TZ/XX/M+ Question 0 continued SECTION A (d) QUESTION Note: Award for approximately correct domain, for approximately correct range, for approximately correct sinusoidal shape with cycles. Only if this last awarded, award for max/min in approximately correct positions. (e) setting up inequality (accept equation) eg h > 05, acosbt, sketch of graph with line y 05 N4 (a) attempt to substitute into sum formula for AP (accept term formula) 0 0 S0 ( 7) 9 d, or 7 u 0 setting up correct equation using sum formula 7) 9 d 60 d 4 N (b) u (4) () 0 N any two correct values for t (seen anywhere) eg t 8.7, t 6.68, t.7, t 4.68, valid approach t t 8.57 (.5 8.7) eg,,, M p 0.0 N Total

6 - 0 - M09/5/MATME/SP/ENG/TZ/XX/M+ - - M09/5/MATME/SP/ENG/TZ/XX/M+ QUESTION (a) METHOD choosing cosine rule substituting correctly AB.9.9 (.9)(.9)cos.8 AB 6. (cm) N METHOD evidence of approach involving right-angled triangles substituting correctly x sin 0.9, AB.9sin AB 6. (cm) N METHOD choosing the sine rule substituting correctly sin sin.8.9 AB AB 6. (cm) N (b) METHOD reflex AOB ˆ.8 ( 4.48) (A) A (.9) ( ) area 4. (cm ) N METHOD finding area of circle A (.9) ( ) () finding area of (minor) sector A (.9) (.8) (.68...) () subtracting M (.9) 0.5(.9) (.8), area 4. (cm ) N METHOD finding reflex AOB ˆ.8 ( 4.48) (A) finding proportion of total area of circle.8 (.9), r area 4. (cm ) N QUESTION (a) attempt to form any composition (even if order is reversed) x correct composition h( x) g () x x h( x) 4cos 4cos x, 4cos 6 (b) period is 4 (.6) N (c) range is 5 hx ( ) ([ 5, ]) N QUESTION 4 (a) evidence of attempt to find P( X 475) P( Z.5) P( X 475) N (b) evidence of using the complement 0.7, p QUESTION 5 z 0.68 () setting up equation a a 46 evidence of appropriate approach 5 9 s t 5 two correct equations 5s9 t, s 5t, s t attempting to solve the equations one correct parameter s, t P is (,, ) accept

7 - - M09/5/MATME/SP/ENG/TZ/XX/M+ - - M09/5/MATME/SP/ENG/TZ/XX/M+ QUESTION 6 (a) evidence of substituting into formula for n th term of GP u4 r 8 setting up correct equation r 8 r N (b) METHOD setting up an inequality (accept an equation) ( n ) ( n ) 8 40 ; 8 40 ; n 648 evidence of solving graph, taking logs n () n 8 N METHOD if n 7, sum ; if n 8, sum A M M QUESTION 7 (a) appropriate approach tree diagram or a table P(win) P( H W) P( A W) (0.65)(0.8) (0.5)(0.6) (or 0.6) N (b) evidence of using complement p, choosing a formula for conditional probability P( W H) P( HW) P( W ) (0.65)(0.7) P(home) 0.99 [8 marks] n 8 (is the smallest value) A N

8 - 4 - M09/5/MATME/SP/ENG/TZ/XX/M M09/5/MATME/SP/ENG/TZ/XX/M+ QUESTION 8 SECTION B (a) evidence of using mid-interval values (5, 5, 5, 5, 50, 67.5, 87.5) 9.8 (cm) A (b) (i) Q5, Q 40 ()() IQR 5 (accept any notation that suggests the interval 5 to 40 ) (ii) METHOD 60 % have a length less than k () () k 0 (cm) N METHOD () () k 0 (cm) N QUESTION 9 (a) evidence of substituting ( 4,) a( 4) b( 4) c 6a4bc AG N0 [ marks] (b) 6a6b c, 4ab c NN [ marks] (c) (i) (ii) 6 4 A 6 6 ; 4 B NN A A N (c) l 0 cm 70 fish 70 P(small) ( 0.5) 00 N [ marks] (d) Cost $X 4 0 P( X x) (e) (of their p values) into formula for E( X ) () N [ marks] E( X ) 8.07 (accept $8.07) N [ marks] Total [5 marks] (d) (iii) evidence of appropriate method X A B, attempting to solve a system of three equations 0.5 X 0.5 (accept fractions) A f( x) 0.5x 0.5x (accept a0.5, b0.5, c, or fractions) N [8 marks] f( x) 0.5( x).5 (accept h, k.5, a 0.5, or fractions) Total [5 marks]

9 - 6 - M09/5/MATME/SP/ENG/TZ/XX/M M09/5/MATME/SP/ENG/TZ/XX/M+ QUESTION 0 (a) attempt to expand ( xh) x x hxh h N [ marks] (b) evidence of substituting x h ( xh) 4( xh) ( x 4x) f( x) lim h 0 h QUESTION SECTION A (a) 8 N (b) (i) 0 A N (ii) 44 A N simplifying ( x x hxh h 4x4hx 4x) h factoring out h h(x xhh 4) h QUESTION (a) f( x) x 4 AG N0 (c) f () () setting up an appropriate equation M x 4 A N at Q, x, 4 y Q is (, 4) (d) recognizing that f is decreasing when f() x 0 R (b) Description of transformation Horizontal stretch with scale factor.5 Maps f to f( x ) + Diagram letter C D N correct values for p and q (but do not accept p.5, q.5 ) NN p.5, q.5; ; an interval such as.5 x.5 (c) 6 translation (accept move/shift/slide etc.) with vector N A N [ marks] (e) f ( x) 4, y 4, 4, Total [5 marks] QUESTION evidence of appropriate approach x a sketch, writing e 4sinx 0 x 0.7, x.6 M AA NN

10 - 0 - M09/5/MATME/SP/ENG/TZ/XX/M+ - - M09/5/MATME/SP/ENG/TZ/XX/M+ QUESTION 4 (a) METHOD evidence of choosing the cosine formula 7 7 cos ACB ˆ ˆ ACB.8 radians ( 6 ) N METHOD evidence of appropriate approach involving right-angled triangles 6 sin ACB ˆ.5 7 ˆ ACB.8 radians ( 6 ) N (b) METHOD ACD ˆ π.8 ( ) () evidence of choosing the sine rule in triangle ACD sin sin ADC ˆ ADC ˆ ( ) CAD ˆ π ( ) ( 80 ( ) ).54 ( 88.5 ) METHOD ˆ ABC (π.8) (80 6.4) evidence of choosing the sine rule in triangle ABD 6.5 sin sin ADC ˆ () QUESTION 5 7 r (a) (accept ) N r 4 (b) (i) METHOD recognizing a GP 4 u, r, n 7 into formula for sum 4 7 ( ) S7 S METHOD 0 0 recognizing r r r 4 () () recognizing GP with u, r, n 0 () into formula for sum 0 ( ) S0 () r (+ 4+ 8) r (ii) valid reason ( infinite GP, diverging series), and r (accept r > )RR N N4 N4 ˆ ADC ( ) CAD ˆ π ( π.8...) ( (80 6.4) ).54 ( 88.5 ) Note: Two triangles are possible with the given information. If candidate finds ˆ ADC. ( ) leading to ˆ CAD (4.5 ), award marks as per markscheme. [8 marks]

11 - - M09/5/MATME/SP/ENG/TZ/XX/M+ - - M09/5/MATME/SP/ENG/TZ/XX/M+ QUESTION 6 SECTION B (a) gradient is 0.6 A N (b) at R, y 0 (seen anywhere) at x, y ln5 ( ) () gradient of normal () evidence of finding correct equation of normal 5 y ln 5 ( x ), y.67x+c x.97 (accept.96) QUESTION 7 coordinates of R are (.97, 0) QUESTION 8 (a) finding the limits x 0, x 5 () integral expression 5 0 f ( x)dx area 5. N (b) evidence of using formula v πy dx correct expression 4 volumeπ x ( x 5) dx 5 0 volume 4 0 A N (a) attempt to use discriminant, ( k ) 4 k () setting their discriminant equal to zero M ( k ) 4 k 0, k 0k k, k 9 (b) k, k 9 A N (c) a area is x ( a x )d x 0 ax x 0 substituting limits a a a setting expression equal to area of R correct equation a a 5., a 6 5. a 6.79 Total

12 - 4 - M09/5/MATME/SP/ENG/TZ/XX/M M09/5/MATME/SP/ENG/TZ/XX/M+ QUESTION 9 A N(46,0 ) B N( μ, ) (a) P ( A > 60) A N [ marks] (b) correct approach () 60 μ P Z < 0.85, sketch 60 μ () μ 47.6 N (c) (i) route A N (ii) METHOD P( A < 60) valid reason R probability of A getting there on time is greater than probability of B 0.99 > 0.85 METHOD P( B > 60) valid reason R probability of A getting there late is less than probability of B < 0.5 N (d) (i) let X be the number of days when the van arrives before 07:00 5 P( X 5) (0.85) (ii) METHOD () evidence of adding correct probabilities P( X ) P( X ) + P( X 4) + P( X 5) correct values () P( X ) 0.97 METHOD evidence of using the complement P( X ) P( X ), p correct values () P( X ) 0.97 N N Total [ marks] QUESTION 0 (a) Note: Award for approximately sinusoidal shape, for end points approximately correct, ( π, 4),(π, 4) for approximately correct position of graph, (y-intercept ( 0, 4), maximum to right of y-axis). (b) (i) 5 N (c) (ii) π (6.8) N (iii) 0.97 N f( x) 5sin( x+ 0.97) (accept p 5, q, r 0.97 ) (d) evidence of correct approach max/min, sketch of f ( x) indicating roots one s.f. value which rounds to one of 5.6,.5, 0.64,.8 N [ marks] continued

13 - 6 - M09/5/MATME/SP/ENG/TZ/XX/M+ 9 M0/5/MATME/SP/ENG/TZ/XX/M+ Question 0 continued SECTION A (e) k 5, k 5 N [ marks] (f) METHOD graphical approach (but must involve derivative functions) M QUESTION (a) A A N (b) evidence of attempting to solve equation multiply by A (on left or right), setting up system of equations each curve x 0.5 METHOD A g ( x) x + f ( x) cosx 4sin 5cos( x ) x ( ) evidence of attempt to solve g ( x) f ( x) M x 0.5 A N N Total [8 marks] QUESTION X 0 (accept x, y 0, z ) A (a) common difference is 6 N (b) evidence of appropriate approach u n 5 correct working 5 ( n )6, 5 6 n 6 N (c) evidence of 6( 5) S 6, 6 ( 5 6) S (accept 5000 ) N

14 0 M0/5/MATME/SP/ENG/TZ/XX/M+ M0/5/MATME/SP/ENG/TZ/XX/M+ QUESTION (a) evidence of choosing the product rule x ( sin x) cos x f () x cosx xsinx (b) QUESTION 4 (a) fx () (4)... 7(4), fx 46 5x (seen anywhere) evidence of substituting into mean fx f correct equation 46 5 x 4.5, 46 5x 4.5(4 x) 4 x x 4 N (b).54 A N QUESTION 5 (a) (i) evidence of finding the amplitude 7, amplitude 5 p 5 N (ii) period 8 () q N 8 4 (iii) r 7 () r N Note: Award for correct domain, 0 x 6 with endpoints in circles, for approximately correct shape, for local minimum in circle, for local maximum in circle. N4 (b) k (accept y ) N

15 M0/5/MATME/SP/ENG/TZ/XX/M+ M0/5/MATME/SP/ENG/TZ/XX/M+ QUESTION 6 evidence of integrating the acceleration function sin t dt t correct expression lnt cost c evidence of substituting (, 0) 0 ln cos c c 0.64 cos ln or cos () v ln t cost 0.64 ln t cost cos or ln t cost cos ln () v (5).4 (accept the exact answer ln5.5cos0.5cos ) QUESTION 7 (a) evidence of recognizing binomial probability (may be seen in (b) or (c)) 7 4 probability (0.9) (0.), X ~B(7,0.9), complementary probabilities 4 probability 0.00 N (b) correct expression N ( ), 5 ( ) 4 p p p p Note: Award for binomial coefficient accept 7, for 4 p ( p ). (c) evidence of attempting to solve their equation 7 4 ( ) p p, sketch p 0.56, QUESTION 8 SECTION B (a) cos x, cosx AC 4 40cos x AG [ mark] (b) AC 4, AC 4sin x sin x sin0 4sin x AC 8sin x accept N sin0 (c) (i) evidence of appropriate approach using AC M 8sin x 4 40cos x, sketch showing intersection correct solution ,.7... obtuse value.7... () () [ marks] x. to dp (do not accept the radian answer.94 ) N (ii) substituting value of x into either expression for AC AC 8sin. AC 7.45 N (d) (i) evidence of choosing cosine rule a c b cos B ac , 7.45 cos y, cos y y 7 N continued

16 4 M0/5/MATME/SP/ENG/TZ/XX/M+ 5 M0/5/MATME/SP/ENG/TZ/XX/M+ Question 8 continued (ii) into area formula () 4 4 sin7, 8sin7 area 5.4 N Total QUESTION 9 (a) substituting (0, ) into function M 0 A e A A 0 AG N0 [ marks] (b) substituting into f (5).49 k.49 0e, e 5 5k evidence of solving equation sketch, using ln ln k 0.0 accept 5 N (c) (i) f x 0.0x ( ) 0e ( ) 0e x x f x 0.0 (.0e ) Note: Award for 0e 0.0 x, for 0.0, for the derivative of is zero. (ii) valid reason with reference to derivative R N f () x 0, derivative always negative (iii) y N (d) finding limits.895, (seen anywhere) evidence of integrating and subtracting functions correct expression g( x) f( x) dx, x ( 4) (0e ) d x x x area 9.5 A N4 Total

17 6 M0/5/MATME/SP/ENG/TZ/XX/M+ 7 M/5/MATME/SP/ENG/TZ/XX/M QUESTION 0 (a) evidence of appropriate approach 0.85, diagram showing values in a normal curve P( w 8) 0.5 N [ marks] (b) (i) z.64 N (ii) evidence of appropriate approach x , N (c) (i) 68.8 weight 84.4 Note: Award for 68.8, for 84.4, for giving answer as an interval. (ii) evidence of appropriate approach P(.5 z.5), P(68.76 y 84.44) P(qualify) N (d) recognizing conditional probability P( AB ) P( A B) P( B) P(woman and qualify) () P(woman qualify) P(woman qualify) 0.0 Total [5 marks] SECTION A. (a) (i) d 4 N (ii) evidence of valid approach u (4), repeated addition of d from 6 u 8 64 N (b) (i) into sum formula n n Sn { (6) + ( n )(4) }, {7 + 4n 4} evidence of simplifying n {4n + 68} Sn n 4n + AG N0 (ii) 868 N Total. (a) (i) (, 7) or x, y 7 N (ii) evidence of valid approach graph, completing the square, equating coefficients f( x) ( x ) 7 N (b) evidence of valid approach graph, quadratic formula , x -0.95, 4.9 Total

18 8 M/5/MATME/SP/ENG/TZ/XX/M 9 M/5/MATME/SP/ENG/TZ/XX/M. (a) into formula for determinant () ( x)() ( x)( x ) det M x x N [ marks] (b) det N 6 N [ mark] evidence of valid approach x x 6, graph x.4 A Total 4. (a) evidence of valid approach y 0, sin x 0 π k 6.8 N [ marks] (b) attempt to substitute either limits or the function into formula (accept absence of dx) k π ( ) π ( ) π π π 6.8 V f( x) d x, ( x )sin x, y dx correct expression A π π π ( ) π π 6.8 ( x ) sin xd x, ( x )sin x dx (c) V V 69.6 A N [ marks] Total 5. (a) evidence of valid approach finding the inverse of M, MM I p, q (b) evidence of attempt to solve system 7 X M, or correct values, substitution x 0.5, y.5, z.5 A 0 6. (a) Valid attempt to find term in x 7 8 ( )( b ) 7b, ( x ) 07 x correct equation 7 8 ( )( b ) 07 Total b N (b) evidence of choosing correct term 7 th term, r 6 correct expression 8 ( x ) 6 x 6 k 8648 (accept 8600) N Total

19 0 M/5/MATME/SP/ENG/TZ/XX/M M/5/MATME/SP/ENG/TZ/XX/M 7. (a) evidence of recognizing binomial distribution X ~ B(0, 0.57), p 0.57, q 0.4 EITHER 4 P( X ) ( ) () evidence of using complement any probability, P( X 4) P( X ) P( X 4) 0.99 OR summing the probabilities from X 4 to X 0 correct expression or values () (0.57) r (0.4) r, r 4 r P( X 4) 0.99 (b) evidence of valid approach 9 6 three tails in nine tosses, (0.57) (0.4) correct calculation 9 6 (0.57) (0.4) 0.57, th th P(4 tail on 0 toss) N () Total SECTION B 8. (a) (i) p 7, q N (ii) 75 T < 85 N (b) evidence of valid approach adding frequencies P( T < 95) N [ marks] (c) (i) 0 N (ii) 50 N [ marks] (d) (i) evidence of approach using mid-interval values (may be seen in part (ii)) x 79. A (ii) σ 6.4 N (e) evidence of valid approach standardizing, z P( T < 95) 0.8 N [ marks] Total [ marks]

20 M/5/MATME/SP/ENG/TZ/XX/M M/5/MATME/SP/ENG/TZ/XX/M 9. (a) (i) evidence of valid approach choosing cosine rule 6 (5 p) + (4 p) (4 p) (5 p)cos0.7 simplification 6 5 p + 6 p 40 p cos0.7 () p (4 40cos0.7) 6 AG N0 Question 9 continued (d) (i) evidence of valid approach recognize isosceles triangle, base angles equal π (0.95 ) CBD.9 AG N0 (ii) area of sector BCD () 0.5 (.9) (6) (ii) p.86 N Note: Award A0 for p ±.86, i.e. not rejecting the negative value. area of triangle BCD 0.5 (6) sin.9 evidence of subtraction () M (b) BD 6 N [ mark] (c) evidence of valid approach choosing sine rule sin ADB ˆ sin 0.7 4p area 5.94 Total [5 marks] acute ADB ˆ () π ADB. continued

21 4 M/5/MATME/SP/ENG/TZ/XX/M 5 M/5/MATME/SP/ENG/TZ/XX/M 0. (a) (i) evidence of valid approach ship A where B was, B km away distance N Question 0 continued (c) (ii) evidence of valid approach new diagram, Pythagoras, vectors s 5 + () s 6.6 N Note: Award M0A0A0 for using the formula given in part (b). (b) evidence of valid approach a table, diagram, formula d r t distance ship A travels t hours after noon is 5( t ) (A) distance ship B travels in t hours after noon is t () evidence of valid approach M [ ] st ( ) 5( t ) + ( t) correct simplification 5( t t+ ) + t st ( ) 46t 450t+ 5 AG N0 continued Note: Award for shape, for minimum at approximately ( 0.7, 9 ), for domain. (d) evidence of valid approach s () t 0, find minimum of st (), graph, reference to more than 8 km min (accept or more sf) since s min > 8, captain cannot see ship B R N0 Total

22 - 9 - M09/5/MATME/SP/ENG/TZ/XX/M M09/5/MATME/SP/ENG/TZ/XX/M+ QUESTION SECTION A (a) 8 N (b) (i) 0 A N QUESTION (a) (ii) 44 A N QUESTION 4 (a) METHOD evidence of choosing the cosine formula 7 7 cos ACB ˆ ˆ ACB.8 radians ( 6 ) N METHOD evidence of appropriate approach involving right-angled triangles 6 sin ACB ˆ.5 7 ˆ ACB.8 radians ( 6 ) N (b) METHOD ACD ˆ π.8 ( ) () evidence of choosing the sine rule in triangle ACD sin sin ADC ˆ (b) Description of transformation Horizontal stretch with scale factor.5 Maps f to f( x ) + Diagram letter C D A N N ADC ˆ ( ) CAD ˆ π ( ) ( 80 ( ) ).54 ( 88.5 ) METHOD 6 (c) translation (accept move/shift/slide etc.) with vector QUESTION N ˆ ABC (π.8) (80 6.4) evidence of choosing the sine rule in triangle ABD 6.5 sin sin ADC ˆ () evidence of appropriate approach x a sketch, writing e 4sinx 0 x 0.7, x.6 M AA NN ADC ˆ ( ) CAD ˆ π ( π.8...) ( (80 6.4) ).54 ( 88.5 ) Note: Two triangles are possible with the given information. If candidate finds ˆ ADC. ( ) leading to ˆ CAD (4.5 ), award marks as per markscheme. [8 marks]

23 - - M09/5/MATME/SP/ENG/TZ/XX/M+ - - M09/5/MATME/SP/ENG/TZ/XX/M+ QUESTION 5 7 r (a) (accept ) N r 4 (b) (i) METHOD recognizing a GP 4 u, r, n 7 into formula for sum 4 7 ( ) S7 S METHOD 0 0 recognizing r r r 4 () () recognizing GP with u, r, n 0 () into formula for sum 0 ( ) S0 () r (+ 4+ 8) r N4 N4 QUESTION 6 (a) gradient is 0.6 A N (b) at R, y 0 (seen anywhere) at x, y ln5 ( ) () gradient of normal () evidence of finding correct equation of normal 5 y ln 5 ( x ), y.67x+c x.97 (accept.96) QUESTION 7 coordinates of R are (.97, 0) (a) attempt to use discriminant, ( k ) 4 k () setting their discriminant equal to zero M ( k ) 4 k 0, k 0k k, k 9 (b) k, k 9 A N (ii) valid reason ( infinite GP, diverging series), and r (accept r > )RR N

24 - - M09/5/MATME/SP/ENG/TZ/XX/M M09/5/MATME/SP/ENG/TZ/XX/M+ SECTION B QUESTION 8 (a) finding the limits x 0, x 5 () integral expression 5 0 f ( x)dx area 5. N (b) evidence of using formula v πy dx correct expression 4 volumeπ x ( x 5) dx 5 0 volume 4 0 A N (c) a area is x ( a x )d x 0 ax x 0 substituting limits a a a setting expression equal to area of R correct equation a a 5., a 6 5. a 6.79 Total QUESTION 9 A N(46,0 ) B N( μ, ) (a) P ( A > 60) A N [ marks] (b) correct approach () 60 μ P Z < 0.85, sketch 60 μ () μ 47.6 N (c) (i) route A N (ii) METHOD P( A < 60) valid reason R probability of A getting there on time is greater than probability of B 0.99 > 0.85 METHOD P( B > 60) valid reason R probability of A getting there late is less than probability of B < 0.5 N (d) (i) let X be the number of days when the van arrives before 07:00 5 P( X 5) (0.85) (ii) METHOD () evidence of adding correct probabilities P( X ) P( X ) + P( X 4) + P( X 5) correct values () P( X ) 0.97 N N METHOD evidence of using the complement P( X ) P( X ), p correct values () P( X ) 0.97 Total [ marks]

25 - 5 - M09/5/MATME/SP/ENG/TZ/XX/M M09/5/MATME/SP/ENG/TZ/XX/M+ QUESTION 0 (a) Question 0 continued (e) k 5, k 5 N [ marks] (f) METHOD graphical approach (but must involve derivative functions) M Note: Award for approximately sinusoidal shape, for end points approximately correct, ( π, 4),(π, 4) for approximately correct position of graph, (y-intercept ( 0, 4), maximum to right of y-axis). (b) (i) 5 N each curve x 0.5 A N (c) (ii) π (6.8) N (iii) 0.97 N f( x) 5sin( x+ 0.97) (accept p 5, q, r 0.97 ) (d) evidence of correct approach max/min, sketch of f ( x) indicating roots METHOD g ( x) x + f ( x) cosx 4sin 5cos( x ) x ( ) evidence of attempt to solve g ( x) f ( x) M x 0.5 A N Total [8 marks] one s.f. value which rounds to one of 5.6,.5, 0.64,.8 N [ marks] continued

26 9 M0/5/MATME/SP/ENG/TZ/XX/M+ 0 M0/5/MATME/SP/ENG/TZ/XX/M+ QUESTION SECTION A. (a) (i) evidence of appropriate approach ,4+ 5 p 69 N (ii) evidence of valid approach 09 their value of p, 0 ( ) q 40 N (b) evidence of appropriate approach fx substituting into, division by 0 n mean.6 N (c).09 N QUESTION QUESTION 4 evidence of substituting into binomial expansion a + a b+ a b +... identifying correct term for 4 x evidence of calculating the factors, in any order 5 6 4,7 x, ; 0( x ) x x Note: Award for each correct factor. 4 term 080x N Note: Award MMA0 for 080 with working shown. QUESTION 5 (a) (a) evidence of equation for u 7 M 6 u + 6, u7 u+ ( n ), 6 ( 6) u N (b) (i) correct equation 56 + ( n ),59 ( n ) n 50 N (ii) into sum formula 50( + 56) 50( ( ) + 49 ) S50, S50 S 50 5 (accept 00) N QUESTION (a) 6 outcomes (seen anywhere, even in denominator) () valid approach of listing ways to get sum of 5, showing at least two pairs (, 4)(, ), (, 4)(4, ), (, 4)(4, ), (, )(, ), lattice diagram 4 P(prize) 6 9 (b) recognizing binomial probability B8,, binomial pdf, P( prizes) N (b) x., x.68 (accept x.4, x.9 if working in degrees) N (c). < x <.68 (accept.4< x <.9 if working in degrees) A N QUESTION 6 (a). N (b) (i).0 N (ii).59 N (c) q f ( )d 9.96 p N split into two regions, make the area below the x-axis positive RR N

27 M0/5/MATME/SP/ENG/TZ/XX/M+ M0/5/MATME/SP/ENG/TZ/XX/M+ QUESTION 7 (a) 0 n 800e () n 800 N (b) evidence of using the derivative n (5) 7 N (c) METHOD setting up inequality (accept equation or reverse inequality) n () t > 0000 evidence of appropriate approach sketch, finding derivative k () least value of k is 6 N METHOD n (5) 984, and n (6) 08 A least value of k is 6 A N [8 marks] M SECTION B QUESTION 8 (a) appropriate approach 6 8θ AOC 0.75 N [ marks] (b) evidence of substitution into formula for area of triangle area 8 8 sin (0.75) area.8 () evidence of substitution into formula for area of sector area area of sector 4 () evidence of substituting areas r θ absinc, area of sector area of triangle area of shaded region.9 cm N4 (c) attempt to set up an equation for area of sector 45 8 θ COE.4065 (.4 to sf) N [ marks] continued

28 M0/5/MATME/SP/ENG/TZ/XX/M+ 4 M0/5/MATME/SP/ENG/TZ/XX/M+ Question 8 continued (d) METHOD attempting to find angle EOF EOF ˆ (seen anywhere) evidence of choosing cosine rule EF cos0.985 EF 7.57 cm METHOD attempting to find angles that are needed angle EOF and angle OEF EOF ˆ and OEF ˆ (or OFE) ˆ evidence of choosing sine rule EF 8 sin sin.08 () EF 7.57 cm QUESTION 9 (a) (i) (, 4,0) N (ii) choosing velocity vector finding magnitude of velocity vector ( ) + +, () speed.74 ( 4 ) N (b) (i) substituting p 7 B (,7, 7) N (ii) METHOD appropriate method to find AB or BA AO + OB, A B 4 4 AB or BA 7 7 () distance 6. ( 7 4 ) METHOD attempting to find angle EOF π EOF ˆ (seen anywhere) evidence of using half of triangle EOF x 8sin correct calculation x.78 METHOD evidence of applying distance is speed time.74 7 (M) distance 6. ( 7 4 ) METHOD attempt to find AB, AB ( ( ) ) + ( 4 7) + ( 0 7) ), ( ( ) ) + ( 4 7) + ( 0 7) ) EF 7.57 cm Total [5 marks] AB 686, AB 686 () distance AB 6. ( 7 4 ) continued

29 5 M0/5/MATME/SP/ENG/TZ/XX/M+ 6 M0/5/MATME/SP/ENG/TZ/XX/M+ Question 9 continued QUESTION 0 (c) correct direction vectors and ()() a a + 5, a + 8 a a ()() substituting M cos 40 a a + 5 a., a Total (a) (i).5,.5 N (ii) recognizing that it occurs at P and Q x.5, x.5 k., k. (b) evidence of choosing the product rule uv + vu (c) derivative of x is x x derivative of ln (4 x ) is 4 x x x + ln (4 x ) x 4 x 4 x g ( x) + x ln(4 x ) 4 x () () AG N0 N [ marks] (d) w.69, w< 0 A N Total