PAPER A numerical answers 1 Proof by forming quadratic >0 then sh0w quadratic has no solutions using discriminant b 4ac < 0 or similar method 9a 51 + 04px + 4608 p x + 576 p x + a 5y + 9x 1 = 0 9b p = 1 4 b Area = 1 9c 88 90 10, 190, 70 10a QR = 14.7 km 4 x = 10b figure bearing 68 5a p = 5 11a Show that 5b ( 16 ) or 16i 0j 11b 0 max V = 000 ~6160 9 6a 1644 or 1640 11c Show that nd derivative is < 0 6b 100 is the initial population of bacteria 1a A(-, 0) and B(4, 0) 6c 4 hours 1b Area = 148 7 1a = 5 ± 5 m = 9 ± 9 10 5 Or m = 9± 9 10 8a ( 6 q 7 ) or 6i + (q 7)j 1b Sketch graph points (0, -0) (5 + 5, 0) (5 5, 0) on curve q Points (0, ) (6, 0) on p 8b q = or q = 11 1c Intersection (, 11 ) and (-, 4) 4 1d x < or x > In set notation x < x >
PAPER B numerical solutions 1a e.g. should have subtracted 150 from 60 first 9a Graphs y=x+5 y+x=6 y= Shaded triangle 1b x = 75 and x = 105 9b.45 1 65 10a -.7 must be negative p = 44 61, q = 14 61 10b p = 10 q = 4a Prove that expand the expression and 10c Magnitude R R = 6 1 cancel terms to get result 4b Proof by counter example 11a Sketch graph check with geogebra e.g. x= -1 5 h(x) = 6x 5/ 80 x 1 11b (6, 0) or ( 4, 0) 6 x = 48., 60, 11.8, 00 11c ( 7, 11 ) 7a 1 + 4x + 5x + 151x + 1a A is (, 9) B is (7, 1) 7b 1.667 (5sf) 1b Area = 5.5 8a 8b Asymptote at x=-a (1 mark) points (0, log 9 (x + a)) and ( a + 1, 0) ( marks) correct shape with asymptote one x-intercept (mark) Graph of y = log 9 (x + a) is a stretch parallel to y-axis scale factor, of the graph of y = log 9 (x + a) 1c Area = 95 = 1.7 1d Area = 0.8 1a 1b centre (4, 5) radius 40 Y coordinate y = with explanation why rejecting y = 7 1c y = x + 7 1d y = 1 x + 1 1e Area = 0
PAPER C numerical solutions 1 1 mark use correct formula 1 mark three correct steps 1 mark simplify correctly and using limit 9a Accept answer that states 115 (m) is the height of the cliff plus the height of the person who is ready to throw the stone or similar sensible comment. a Correct Graph point (0, 1) 9b 1.656 4.9(t 1.5) b 1 9c T= 6.5 secs i) Translation 1 unit right by 0 i) ii) 0 ii) T = 1.5 Translation 5 units up by 5 a Show that PQ = AB 10a Show that 5 b 6 cm 10b 0.6 4 Correct graph points (0, 1 ) (6, 0) 11a a = 1 a = 1 5 9 Asymptotes x=6 and y=5 5 (x+)(x-)(x+1) 11b Graph points (0, 10) ( 5, 0) 6a p 5 + 5p 4 q + 10p q + 10p q + 5pq 4 + q 5 6b 5 51 11c Greater than 1 1a 7a A = 17.9 cao 1b r = 150 7b Area = 7 1a 8a Show that 1b 8b y = 11 x + 8c x y = 0 1d 15 157 or 15000 1e T=10 1c mark eqn for perimeter rearranged to x= marks substitute to area eqn and rearrange in at least steps π log 4 V = 1 10 t + log 440000 a = 40000 b = 4 1 10 A is the initial value of the car B is the annual proportional decrease the car loses 1% of its value each year
1a 1b PAPER D numerical answers Fully complete graph with all points labelled. ( 1/, 0), (1, 0), (0, ), (1/, 4) Fully complete graph with all points labelled. (, 0), (1, 0), (0, ), (1, 4) Fully correct integration. 9 + + o.e. 5x 0x x C 4 x = = 1. 7 x = =. 4a Sine curve has a period of (can be implied by 5 complete cycles) and passes through (1,0), (,0),..., (10,0). 8a Centre (, q) Radius = 18 + q 8b possible values of q are and 10 9a 9b a = 1600 Show that 1 b = 10a =.46 4b 10 times. 10b Graphical explanation 4c 10 c) solutions to the equation For example: A buoy would not move up and down at exactly the same rate during sin( x 60 ) tan x 0 each oscillation; The period of oscillation the two curves intersect. is likely to change each oscillation; The maximum (or minimum) height is likely to d) solutions to the equation change with time.; Waves in the sea are not sin( x 60 ) tan x 0 uniform. + = will occur where + = will occur where the two curves intersect. 5 7 x < or x > 5 11a dy x x 1 dx 6 s = 816 11b y= 7x 10 7a k 16k+ 60 11c A is (0, 0) 7b K= 10 k=6 1a P= - q=-40 7c complete the square / draw graph to explain a squared term is always greater than or equal to zero, so one more than a square term must be greater than zero 1b (x-5)(x+4)(x+) 1a MN = a+ b 5 1b ST = a b 5 1c Show that
PAPER E numerical answers 1a Observation or measurement of every member of a population 6a Displacement of P: s =.8t + 0.06t Displacement of Q: ( ) ( ) s =.4 t + 0.1 t 1b Two from: takes a long time/costly; difficult 6b Show that to ensure whole population surveyed; cannot be used if the measurement ; process destroys the item ; can be hard to manage and analyse all the data. 1c The list of unique serial numbers. 6c s = 19.6... (m). 1d A circuit board. 7a t = 1 a 7b s = 108 (m). 7c) States that the cyclist is not a particle, or states that the resistive force is unlikely to be constant. b i) = 16 180 = 4 45 = 0.0889 ii) = 17 180 =0.0944 c 0.1 a 6.709 b Mean = 9.6041 σ = 16.5515 c Any sensible reason linked to the shape of the distribution: distribution is (positively) skewed; A few large distances (values) distort the mean d 1 st two marks= Comparison of the two means nd two marks = Comparison of the two standard deviations 4a X ~ B(0, 0.05) binomial 4b P(X = 0) = 0.58 4c = 0.006 5a i) h = 1.7 (m) ii) h =.47 5b x = 4.84 States that the ball will be called in 5c. (m s 1 ) or. (m s 1 )
PAPER F numerical answers 1a a) A collection of people or items b) Opportunity / convenience 1c Systematic 6a 1d 1e Two from: Not random; the electoral register may have errors; there may not be enough (500) households on the register Random sampling from people buying kitchen cleaners in a large stor to avoid bias OR quota sampling from people based on chosen set of ages and genders, continuing until all quotas are filled, 5c 6b 4.0 7a 1 st mark =Not in critical region therefore insufficient evidence to reject H 0. nd mark =There is insufficient evidence at the 5% level to suggest that the value of p is not 0.. 84.75 + 10.86 15.6 (m) distance the boat travelled a Tree diagram with two branches labelled G (0.85 and A (0.15) then nd set branches 0.0, 0097, 0.06, 0.94 b 0.9655 Three comparisons in context: For example: Very much warmer in Beijing than Perth. Both consistent in the temperatures. Less rainfall in Beijing. Less likely to have high rainfall in Beijing. Rainfall in Beijing is consistently less than in Perth AND for 4 th mark Evidence of use of a statistic from the boxplots; For example: Medians; Measure of a difference in medians; Mention of a particular outlier AND for 5 th mark For accurately reading data from boxplots 4a X ~ B(15, 0.5) Binomial 4b i) from calculator P(X = 8) = 0.1968 ii) 0.98 or 50 51 5a Critical region is X 1 11 X ( 0) 5b Significance level = = 0.047 or.47% 7b Top speed 88 m s 1 T = 10.5 (s). 8a a) a = 10 b) R = 0 c) a = 10 d) t = 8 6 9 t = 5.. t = 5 5 v = or 0.9 (m s 1 ). 7
PAPER G numerical answers 1a to obtain a representative sample; large number of students compared to staff so would be unfair to take same numbers of both. 5a 5. ms-1 1b A list of the names of staff and students. 5b - 4.8 ms-1 1c member of staff or a student 5c -0- m 6a 1d Find proportions for different strata out of 60 Select at random using a random number generator 1e absence on the day of the survey; sampling frame may contain errors; a Q = 54 b Q +1.5(Q Q 1 )=80 Patients F (40) and B (90) are outliers 6b T = 0 secs a 1/6 = 0.167 7a R = 640 N b 14/0 = 0.467 7b T = 800 N c 0 7c S = 195.1 m (195 m) No student reads both A and C d 9/0 7d acceleration of the car will be equal to the acceleration of the trailer e Yes 8 Integrate twice and then find two constants of integration by substitution (C=-4 and D = 6) S = 6 m 4a Show that 4b Show that 4c Table of probabilities 4d 5/16 = 0.15
PAPER H numerical answers 1a Quota sampling 5a 680 =6.1 ms-1 1b (+) easy to get sample size; inexpensive; fast 5b b).5( ) ; can be stratefies if required; c) ignore the value of friction between the (-)Not random ; could be biased; hockey puck and the ice 1c 1. Allocate each of the males a number from 1 to 00;. calculator or number generator to generate 50 different random numbers from 1 to 00 inclusive;. Select males corresponding to those numbers 1d 1. 00 50 = 6. Use a random number generator to select the first name as a starting point and then select every 6th name thereafter to get 50 names 5d 1400 kg m 6 Integrate twice, finding constants of integration (C=0, D=0) S = 70/ =. m a All points correctly plotted 7a AB = 500 (m). b The points lie reasonably close to a straight 7b Show that line c d e f a f is the response variable 6 to 1 inclusive It is reliable because it is interpolation No, it is not sensible since this would be extrapolation Correct tree structure. All labels correct. All probabilities correct b i) 1/0 ii) 4/5 4a Two from: Each bolt is either faulty or not faulty; The probability of a bolt being faulty (or not) may be assumed constant. ; Whether one bolt is faulty (or not) may be assumed to be independent (or does not affect the probability of) whether another bolt is faulty (or not) ; There is a fixed number (50) of bolts; A random sample. 4b Critical Region is X 6 X 0 7c Correctly states they will pass each other 440 (m) from A or 60 (m) from B
PAPER I numerical answers 1a All readers of the online newspaper 5a v = 4 (m s 1 ). Accept 4.7 (m s 1 ). 1b 5b Find t = 0.5 (s). = 0.5 (s). A list of readers who subscribe to the extra find t =.91 (s). =.9 (s) content States that the ball is above 85 m for.56 secs 1c The subscribers 5c Finds s =.7...(m). =.7 (m) 1d Advantage: accuracy of the data, 6a unbiased Disadvantage: difficult to get a 100% response to a survey a = 4 b = 5 1e 1. Natural variation in a small sample. Bias a i) 7 minutes ii) Upper quartile or Q b 1. Outliers.. Sensible interpretation: e.g Observation that are very different from the other observations (and need to be treated with caution). Possible errors. These two children probably walked/took a lot longer c Maximum value =55 < 80 6b R = 85 or R = 9.1 N 6c θ = 77.5 ( ). 7a a = 14 5 = 0.4 minimum value = 5 > 0 No outliers (with evidence) X = 14 ms-1 d Correct graph 7c S = 455 m e Three comparisons in context from Comment on comparing averages Comment comparing consistency of times Comment on comparing symmetry Comment on comparing outliers a P(Takes longer than 18 mins)=1/ b P(Takes less than 0 mins) = 9/10 4 P(X 10) = 1 P(X 9) = 0.071 >0.05 insufficient evidence to reject H 0. insufficient evidence at the 5% level to suggest that the book has underestimated the probability 7b Use t1+t=5 to find t1=0 and t=15
PAPER J numerical answers 1a 1b 1c very large number of bags Bags are tested to destruction there would be no bags left One value is less than 1 kg therefore claim is not reliable Different samples can lead to different conclusions due to natural variations Only a small sample taken so unreliable 6a 6b 6c X=15/1 = 10.4 m H = 5.4 m The pole vaulter can leave the ground between.6 m and 6.8 m from the bar 1d Larger sample 6d i) Allows the person to be treated as a single mass and allows the effects of rotational forces to be ignored ii) effects of air resistance can be ignored Mean y =.5 Mean x = 749.165 Sd y = 6.594 Sd x = 15.9 a P(Less than 17 cm) =67/10 = 0.475 7b b P(Between 1 cm and 18 cm) =67/10 = 0.558 Assumption: foot lengths between 17 and 19 are uniformly distributed. 7a 7c a = 45 16 (m s )=.815 (m s ). T = 1677 00 F = 17 00 (N) = 8.85 (N (N) = 6.15 (N). 4a = 0.5 7d the total distance travelled will be 1. m (0.9 + 0.) 4b P( 1 X < ) = P( 1) + P(0) + P(1) = 0.6 4c P(X >.) = 0.85 5a P(X = ) = 0.176 5b P(X 4) = 0.980