2015 H1 A Level Mathematics Solution for 8865 (SAJC)

1 2015 H1 A Level Mathematics Solution Q1. For   22 4 2x k x k    to be always negative, Discriminant < 0 and coefficient of 2 20x        2 2 2 4 4 2 2 0 ( 8 16) 16 0 8 16 0 kk k k k kk            2( 4) 0k But 2( 4) 0k for all real values of k. Therefore there is no solution for 2( 4) 0k and there are no values of k for which   22 4 2x k x k    is always negative. Graph of 2( 4)yx Q2. (i)         5 45 d 3 24 3 4 2 2 1d 2 1 2 1 xx xx       (ii) 2 1 1 2 2x dxx   1 2 1 2 2 1 22 1 2 13 1 2 44 44 443 76 24 x dx x x x dx x x x             

2 Q3. (i) 2 2 12 8 d 12 16d x x y x e y ex     For stationary point, 2 2 2 d 0d 12 16 0 16 12 12 3 16 4 x x x y x e e e        2 3ln ln 4 3 1 32 ln ln 4 2 4 xe xx                 (ii) Area of Region bounded by C =     2 0 22 0 22 2 2 2 12 8 64 6 4 0 4 6 4e + 4 units a x ax a a x e dx xe ae a             Q4. y 2 y 2 2 2 By Pythagoras' Theorem, 3 24 yyy  2 x x 2 2 2 By Pythagoras' Theorem, 3 24 xxx  F A D P D B A B

3 Perimeter of the remaining shape PQRSTU = 30 cm  3 2 3 30 10 10 y x x yx yx           Method 1: Area of the remaining shape PQRST, A = Area of triangle FDE – 3 ( Area of triangle PDQ)         22 22 22 2 2 22 2 11 sin 60 3 sin 6022 1 3 1 3 32 2 2 2 3 34 3 10 34 3 100 20 34 3 50 102 ooyx yx yx xx x x x xx                Method 2: Area of the remaining shape PQRSTU, A = Area of triangle FDE – 3 ( Area of triangle PDQ)             22 2 22 2 1 3 1 3 32 4 2 4 1 3 3 3102 2 2 2 3 1 3 100 202 2 2 1 3 50 10 (shown)2 yxyx x x x x x x xx                                            

4     21 3 50 102 d1 3 10 2d2 A x x A xx     For stationary value of A, d 0d 10 2 0 5 A x x x    2 2 d 20d A x    A is maximum when x = 5. Maximum value of   221 753 50 10(5) (5) 3 cm22A              Q5. (i) At x axis intercept (y = 0) At y axis intercept (x = 0)         0.5 ln 1 0 0.5 ln 1 Using GC, 0.786 x x x x x          0 0.5 ln 1 1y   The coordinates of points of intersection with the axes are ( 0 , 1 ) and (0.786, 0 ). Asymptote : x = -1 (ii) When x = 0.5, y = 0.30164 0.5 d 1.1568 1.157d x y x    ( 3 dec. places) (iii) Equation of normal is not in syllabus. But we can change the question to the following: y x O x = – 1  0.786,0

5 The tangent to C at P meet the x-axis at A and y-axis at B. Find the length AB. [5] Equation of tangent to C at P    0.30164 1.1568 0.5 1.1568 0.88004 yx yx       When y = 0, x = 0.760754, point A is ( 0.760754 , 0 ) When x = 0, y = 0.8804, point B is ( 0 , 0.8804 ) By Pythagoras’ Theorem, Length of AB 220.88004 0.760754 1.16328 1.16 (3 sig. fig)    Q6 Let X be the r.v the masses of peaches sold by a shop. 2~ ( , )XN  P( 40) 0.2 40P( ) 0.2 40 0.8416212335 40 0.8416212335 (1) X Z             P( 60) 0.25 60 0.674487495 60 0.674487495 (2) X          (2) – (1) 20 1.51611 13.2 51.1       A ( 0.760754 , 0 ) B ( 0 , 0.8804 ) O ( 0 , 0 )

6 Mean of distribution = 51.1, Variance of distribution = 213.2 174 Q7. Not in syllabus Q8 (i) P( ) P( ) P( ) P( ) 0.42 2 0.03 0.15 (shown) A B A B A B pp p          (ii) P( ') P( ) P( )' 0.15 (1 0.42) 0.73 A B A A B        (iii) P( ') P( ) P( ) 0.15 0.03 0.12A B A A B       P( ) P( ') 0.15(1 2(0.15)) 0.105AB    P( ') P( ) P( ')A B A B   They are not independent events. Q9. Let X be the random variable of the number of 6’s out of 8 throws of a fair die. 1~ (8, ) 6XB (i) P( 3) 0.104X  (ii) P( 4) P( 3) 0.969XX    (iii) Not in syllabus 10(i)& (iii) NOT in 2020 ‘A’ Level (ii) 0.922r 1.60 1.95 h w 110 90 58.0325 6.3005wh A B

7 The value of product moment correlation, r , shows a strong positive linear correlation between the height and weight of members of a rowing club : as height increase, the weight of members of a rowing club increases linearly (at a constant rate). (iii) 58.0325 6.3005wh 58.0 6.30wh (iv) When 1.66h , 58.03250082(1.66) 6.300468972w = 90.0 (3s.f) The estimated weight is 90.0 kg. The estimate is reliable as the value of r is close to 1 and 1.66 metres is within given data range of the height of the members of the rowing club, i.e. the estimate is an interpolation. 11 Let M and W be the random variables denoting the mass of a man and the mass of a woman in kg 22~ (77,9.8 ) ~ (62,10. 6)M N W N (i) P( 2 2) P(77 2 77 2) P(75 79) 0.162 P( 2) P( 2 77 2) P(75 79) 0.162 M M M OR M M M                           (ii) Let 1 2 3 1 4( ) ( ... )X M M M W W          22 3 77 4 62 17 3 9.8 4 10.6 737.56 ~ ( 17,737.56) EX Var X XN            1 2 3 1 2 3 4 1 2 3 1 2 3 4 P( ) P(( ) ( ) 0) = P( 0) 0.266 M M M W W W W M M M W W W W X                (iii) Let 1 2 3 1 4( ) ( ... )Y M M M W W          22 3 77 4 62 479 3 9.8 4 10.6 737.56 ~ (479,737.56) EY Var Y YN           P( 460) 0.242Y 

8 12 (i) ` (ii) P ( student will succeed in acct qualification) = 3 2 1 3 2 3 4 5 4 4 5 8     or 0.375 (iii)     P succeed failed at 1st attemptP(succeed failed at 1st attempt) P failed at 1st attempt 1 3 2 34 4 5 or 0.31 10 4    (iv) Let X be the random variable denoting the number of students who qualify out of 5 students. 3~ (5, ) 8XB ( 2) 1 ( 1)P X P X    = 0.619 13(i ) Let X be the random variable denoting the length of fish in a particular lake in cm and  be the population mean length of fish. 0 1 : 15.2 : 15.2 H H     Under oH , 22.1~ (15.2, ) 30XN Using a two-tailed z-test at 5% level of significance, 14.5x  gives 1.83calz  and p-value 0.0679 0.05 We do not reject Ho and conclude that there is insufficient evidence at 5% level of significance that the mean length of fish in the lake is not 15.2cm. (ii) Unbiased estimate of the population mean , x 32 18 17.240    Part I (1st attempt) Part I (2nd attempt) Part II Pass Fail Fail Fail Fail Pass Pass Pass 3/4 3/4 1/4 1/4 2/5 3/5 2/5 3/5

9 Unbiased estimate of the population variance, 2s 21 ( 32)325 7.6769 7.6839 40      (iii) A sample statistic T is an unbiased estimator of a population parameter  if E(T) =  .i.e. E( )X  and 22E( )S  Or: A sample statistic is an unbiased estimate if it contains no systematic bias i.e. it does not tend to overestimate or underestimate the population parameter. (iv) Let Y be the random variable denoting the length of fish from the second lake in cm and  be the population mean length of fish. 0 1 : 18 : 18 H H     Under oH , 7.6769~ (18, ) 40YN approximately by Central Linit Theorem since 40n is large. Using a 1-tailed z test at % level of significance, 17.2x  gives p-value 0.033917 Since Ho is rejected, p-value % 0.033917 100  3.39  The set values of   is : 3.39 100     