2017 H1 A Level Mathematics Solution (SAJC)

1 2017 H1 A Level Mathematics Solution Section A Q1. For     2 4 7 0x k x k     for all real values of x, Discriminant :         2 2 2 2 40 4 4 7 0 ( 8 16) 4 28 0 4 12 0 2 6 0 b ac kk k k k kk kk                  The set of values of k is  : , 2 6k k k    . Q2. (i)         13 22 3 2 d 1 d 1 5 5 2 5 2 5d d 2 52 2 5 2 xxxx x x             (ii)       22 4 2 22 22 2 1 4 4 1 dd 4 4 d x x x xxxx x x x           341 43 x x c x    where c is an arbitrary constant y y = 2 4 12kk (-2, 0) (6, 0) O x

2 Q3. (i) Area of the sign = Area of 2 half circles with diameters AB and DC + Area of rectangle ABCD =   2 2 ADxx  To find the length of AD, Perimeter of the sign =  2 2 ADx  = 10 (Given) AD =  1 10 2 52 xx   Area of the sign =     2 2 2 2 5 10 10 cmx x x x x x x         (Shown) (ii) For a non-calculator method, 210 d 10 2d y x x y xx     For stationary points, d 0d 10 2 0 10 5 2 y x x x      For 5x  2 2 d 2d y x  ( < 0) Hence the stationary value at 5x  gives a maximum area.

3 Required Area 2 2 2 10 5510 50 25 25 units xx                   The maximum value of this area is 225 units . Q4. (i) At x axis intercept (y = 0)   0 ln 4 5 0 4 5 e 4 5 1 3 2 x x x x     The coordinates of point of intersection with the axes are 3 ,02   . Asymptote : x = 5 4 .[Found by setting 4 5 0x ] x y O x = 5 4 3 ,02    ln 4 5yx

4 (ii) When x = 2.5, y = ln 5.  ln 4 5 d1 4d 4 5 yx y xx    [where  d 4 5 4d xx  ] 2.5 d 4 4 d 4(2.5) 5 5x y x    Equation of Tangent to the curve at the point x = 2.5,   45ln 5 52 4 2 ln 55 5 4 10 5ln 5 4 5 5ln 5 10 yx yx yx xy                 where 4, 5, 5ln 5 10a b c    (iii) When y = 0, 10 5ln 5 4x  , point A is 10 5ln 5 ,04   or  0.48820,0 When x = 0, 5ln 5 10 ln 5 25y    , point B is  0,ln 5 2 or  0, 0.39056 P ( 0.48820 , 0 ) Q ( 0 , - 0.39056 ) O ( 0 , 0 )

5 By Pythagoras’ Theorem, Length of PQ   2 2( 0.3908 0 0.3905 8 0.62520 0.625(3 sig. fig) 6 0.48820 0)     Q5. (i) Let x, y, z be the manufacturing cost of a pair of Supers, Runners and Walkers respectively. 2 ...............(1) 10 7 50 ...............(2) 2 6 4 481 ...............(3) xz yx x y z      Rearranging 2 0 ...............(1) 7 10 50 ...............(2) 2 6 4 481 ...............(3) xz xy x y z        Using GC, 55,x 43.5,y 27.5z The manufacturing cost of a pair of Runners is $43.50. (ii) Profit for 100 pairs of Supers = 100 [(80 – 55)] = $2500 Profit for 100 pairs of Runners =100[(80 – 43.50)] = $3650 Profit for 100 pairs of Walkers =100[(80 – 27.5)] = $5250 (iii) The coordinates of the points where the curve crosses the x-axis are ( 0, 0 ) and ( 0, 60.494 ). y P = (0, 60.494) O x

6 (iv) Using GC, The maximum point is ( 15.123, 13.611 ). Hence, the maximum value of P is $13.61 when the value of x is $15.12. (v) Using GC, when x = 55, P = 2.4134. Selling price is $55 + $2.41 (Profit) = $57.41 The selling price of a pair of Extremes is $57.41. (vi) When x = 65, P = -2.064 which indicates that the company will lose $2.06 for every pair of Extremes produced. Hence, we will not advise the company to produce Extremes when the cost of a pair of Extremes increased to $65.

7 Section B Q6. Let X be the random variable of the height in meters of an adult male in a particular country.  ,~ NX Given that 30P( 1.75) 100X  and 20P( 1.6) 100X      30P 1.75 100 30P 1.75 1 100 1.75 70P 100 1.75 70P 100 1.75 0.524405101 0.524405101 =1.75 (1) X X X Z                     20P 1.6 100 1.6 20P 100 1.6 20P 100 1.6 0.8416212335 0.8416212335 =1.6 (2) X X Z                Solving equations (1) and (2), 1.6924 1.69  and 0.10981 0.110  and 2 0.012058 0.0121  . The mean of the distribution is 1.69 and the variance of the distribution is 0.0121. Q7 Let X be the r.v. number of the ink cartridges that will last for one week or more, out of 8 cartridges.  ~ 8,0.7XB (i)  5 0.25412184 0.254PX    Let Y be the r.v. number of the ink cartridges that will less than one week or more, out of 8 cartridges.  ~ 8,0.3YB (ii) Required probability =      4 1 4 1 3 0.19410435 0.194P Y P Y P Y         Let W be the r.v. number of boxes of ink cartridges in stock out of 6 boxes of ink cartridges.  ~ 6,0.19410435WB (iii) Required probability =  2 0.90823 0.908PW    µ

8 Q8 (i) Number of codes that can be formed = 68 33P P 40320 (iia) P (contains the digit 5 exactly once and the letter H exactly twice) = 55 12 33 3! 3! 3! 7 15752! 2! 0.01426 8 110592 CC               Strategy for solving this part: Case 1: Number of ways where the code chosen at random where it contains the digit 5 exactly once with another two identical digits (eg. 511, 522, etc) = 5 1 3! 2!C  Case 2: Number of ways where the code chosen at random where it contains the digit 5 exactly once with another two different digits (eg. 512, 513, etc) = 5 2 3!C  Case 3: Number of ways where the code chosen at random where it contains the letter H exactly twice with another one other letter (eg. HHA, HHB, etc) = 3!7 2! Alternative solution: P (contains the digit 5 exactly once and the letter H exactly twice) = 1 5 4 1 5 1 1 1 73 3 36 6 6 6 6 6 8 8 8             = 0.0142 Explanation notes: Case 1: Probability of the code chosen at random where it contains the digit 5 exactly once with another two identical digits (eg. 511, 522, etc) = 1 5 1 3666 Case 2: Probability of the code chosen at random where it contains the digit 5 exactly once with another two different digits (eg. 512, 513, etc) = 1 5 4 36 6 6   Case 3: Probability of the code chosen at random where it contains the letter H exactly twice with another one other letter (eg. HHA, HHB, etc) = 1 1 7 38 8 8  

9 (iib) P (has 2 as its first character) = 1 6 P (has H as its sixth character) = 1 8 P (has 2 as its first character and H as its sixth character) = 11 6 8 48 P (has 2 as its first character or H as its sixth character, but not both) = 1 1 1 1 26 8 48 4    . Alternative solution: Case 1: 2 as its first character but not H as its sixth character P (has 2 as its first character but not H as its sixth character) = 1 7 7 6 8 48 Case 2: Not 2 as its first character but H as its sixth character P (not 2 as its first character but H as its sixth character) = 5 1 5 6 8 48 P (has 2 as its first character or H as its sixth character, but not both) = 7 5 12 1 48 48 48 4   . Q9 (ii) The product moment correlation coefficient, r is 0.978. y x 10 15 20 25 30 35 40 45 2 4 6 8 10 12 (14,4.9) (40,9.8) (11,5.2)

10 The weekly earnings, y hundred dollars and the employees from the production line who have been with the company for x years has a strong positive linear relationship as r = 0.978 is close to 1. Moreover this is supported by the the scatter diagram which shows