2014 H1 A Level Mathematics Solution (SAJC)

1 2014 H1 A Level Mathematics Solution Q1.       6 1 16 2 1 61 2 1 1 14 14 14 2 1 552 dx x x dx x          Q2. (i)   2 2 d2ln 4d4 xxxx   (ii) Gradient of C = K 2 2 2 2 4 2 ( 4) 2 4 0 (shown) x kx x k x kx x k          2 2 4 0kx x k   (iii) For equal roots,   2 2 2 40 4 4 4 0 16 4 1 4 11or22 b ac kk k k kk       Q3. (i)

2 At x axis intercept At y axis intercept 12 12 10 1 1 2 0 1 2 x x e e x x       1 2(0)1 1 ye ye   The coordinates of points of intersection with the axes are ( 0 , 1- e ) and ( 1 2 , 0 ). Asymptote : y =1 (ii) When x = 1, y = 11 e 12 1 d 2d At =1 d2 2d xy ex x y exe     Equation of Tangent to C at x = 1  1211 23 1 23 1 yx ee yx ee mc ee             Q4. (i) By Pythagoras’s Theorem,       222 2 2 2 22 2 2 65 2 4 260 5 2 260 (shown) y x x y xy x x x y xy           (ii) Since the perimeter of the rectangle ABCD is 60 cm,

3 2 6 60 3 30 30 3 yx yx yx        22 22 Using 5 2 260 5 30 3 2 30 3 260 4, 18 8, 6 x y xy x x x x xy xy           Q5. (i) 32 2 7 d 3 2 7d y x kx x c y x kxx        For stationary point, 2 d 0d 3 2 7 0 For 1, 3 2 7 0 5 (shown) y x x kx x k k          Given that A has coordinates ( 1 , 2 ), 32 32 57 2 1 5 7 1 5 7 1 y x x x c c c y x x x              (ii) 2d 3 10 7 0d 11 or 2 3 y xxx xx      32 1For 2 , 3 1 1 12 5 2 7 2 13 3 3 22 27 1 22Co-ordinates of 2 , 3 27 x y B                       

4 (iii) Using GC, co-ordinates of point that crosses the x-axis ( 0.161 , 0 ) (iv) Exact Area of Region = 2 32 1 24 3 2 1 2 5 7 1 57 4 3 2 10 1 5 74 14 2 13 4 3 2 71 units12 x x x dx x x x x                           Q6. Let X be the random variable of the heights of the girls in the school.   2142.2,6XN (i)  P 146 0.73674 0.737 (3 s.f)X    (ii)  P 142.2 5 142.2 5X     P 137.2 147.2 0.595(3 s.f ) X    Q7. [This part is not in syllabus] (i) Systematic Sample To draw a sample of 100 households from a population of 5000 households, we first assign a consecutive number for each household starting from 1. k = 50 ( 5000 100 ). Taking the 1st number randomly from 1 to 50, say 8, we select every 50th household i.e 8th, 58th, 108th until all 100 households are selected. (ii)

5 To obtain a sample of 100 households, we draw random samples from the age-groups with sample size in the same proportion as the size of each age-group: Supermarket Online Under 25 years 10 20 25-60 years 18 32 Over 60 years 16 4 (iii) Stratified Sampling will be preferred over systematic sampling due to the following reasons: 1. When there are clear strata present, this method can provide more accurate estimates than simple random sampling and is therefore more likely to give a good representative sample of the population. 2. Each of the strata can be treated separately, and so the sampling is convenient and more accurate. 8(i) [This part is not in syllabus for 2020] (ii) 0.92582r The value of r shows a strong negative linear correlation between the number of hours spent travelling to and from work, and the number of hours spent watching television. (iii) 0.9021 16.15yx  (iv) When 13.2x , 4.2423 4.24y The estimate is unreliable as the value of 13.2 hours is out of the given data range of the number of hours of television watched by a person, i.e. the estimate is an extrapolation. 9(i) Let X be r.v “Number of cakes containing fruit, out of 6 cakes” 2.4~ (6, ) (6,0.4)6X B B  (a) P( 0) 0.46656 0.467X    (b) P( 2) 0.54432 0.544X    2.2 12.8 x Y 14.8 4.5

6 (ii) Let Y be r.v “Number of packs containing at most two cakes containing fruit, out of 8 packs” ~ (8,0.54432)YB P( 4) 1 P( 3)YY    =0.72867 0.729 (iii) Let A be r.v “Number of packs containing at most two cakes containing fruit, out of 150 packs” ~ (150,0.54432)AB Since n is large, and 81.648 5 68.352 5 np nq   ~ (150 0.54432,150 0.54432 0.45568) ~ (68.352,31.147) AN AN    .P( 75) P( 75.5)ccAA    0.10013 0.100   10(i) Let X be r.v “Length of leaves from beech trees” 0 1 :7 :7 H H     Since n is large, 4.4~ (7, ) 50XN approximately by C.L.T. p-value 0.045946 0.05 We reject 0H and conclude that there is sufficient evidence to show that the population mean length of leaves from beech trees in this forest is less than 7cm. (ii) Unbiased estimate of the population mean 310.4 6.208 6.2150   Unbiased estimate of the population variance 21 310.42209.2 5.759949 50    (iii) 0 1 :7 :7 H H     p-value 0.019623 To reject 0H , p-value % 0.019623 % 1.96  11(i) 48 10 20 12 90P( ) 48 10 20 12 130 15 55 290L xx            10 55 20 15 100P( ) 48 10 20 12 130 15 55 290G xx            L and G are independent P( ) P( ) P( )L G L G    30 90 100 290 290 290x x x   2 30 9000 290 (290 )xx

7 3001 290 10( ) x x shown    (ii) 48 10 20 12 15 130 47P( ) 300 60LT        (iii) 12 130 71P( ') 300 150TG    (iv) 90 3P( | ) P( ) 300 10L G L   (v) P(1 student owns exactly two items) 10 15 12 37 300 300  P(2 students owns exactly two items) 37 36 111 300 299 7475   12(i) 2~ (50, )AN  2 P( 75) 0.0189 75 50P( ) 0.0189 25P( ) 0.9811 25 2.0770 12.037 144.877 145( ) A Z Z shown            (ii)  17 17 ~ (75,64) ... ~ (75 7,64 7) 525, 448 ( ... 500) 0.11877 0.119 BN B B N N P B B           (iii) Let  1 7 1 5... 2 ...X B B A A      ( ) 7 ( ) 2[5 ( )] 25E X E B E A   ( ) 7 ( ) 4[5 ( )] 448 2900 3348Var X Var B Var A     ~ (25,3348) ( 0) 0.66715 0.667 XN PX   