ASRJC 8865 2022 Prelim Solutions vetted

ASRJC H1 Math Prelims Solutions 1 (i) ( )( ) 22 3 2 0 2 1 2 0 xx xx − −  + −  Sketch the graph: 12 or 2xx − (ii) 1 < e 22 x− ln 2x 2 (a) 2 2 2Let 3yx x =− 3 4 2 49 12y x x x= − + 3 2 d4 36 18d y xxxx= − − (b) 3 2 d 54 x x− = 1 32 (5 4 ) dxx − − ( ) 2 23 354 32 C (5 4 ) C2 4( 4)3 x x−= + =− − + − where C is an arbitrary constant. 1 2− x 2

ASRJC H1 Math Prelims Solutions 3 (a) 14d 8ed xy x −= When x = 1, y = 8 – 2e– 3 1 4(1) 3d 8e 8ed y x −−== Equation of tangent at the point x = 1: y – (8 – 2e– 3) = 38e ( 1)x− − y = 3 3 38e 8e 8 2ex− − −− + − y = 338e 10e 8x−− −+ where m = 38e− and c = 38 10e −− (b) A = 1 14 1 4 8 2e d x x−− 1 14 1 4 1 8 e 2 xx −=+ 311 8 e 222 −   = + − +       3 115 2 2e=+ 115 and 22pq== 4 (i) y = x4 – 10x3 + 36x2 – 54x + 17 32d 4 30 72 54d y x x xx = − + − d 0d y x = 324 30 72 54 0x x x− + − = Using GC, x =1.5 or x = 3 When x = 1.5, y = – 11.6875 When x = 3, y = – 10 Coordinates of stationary points are (1.5, – 11.6875) and (3, – 10) (ii) x 1.4 1.5 1.6 d d y x – 1.0240044 0 0.7839964 Slope (1.5, – 11.6875) is a minimum point. x 2.9 3 3.1 d d y x 0.0560016 0 0.0640024 Slope

ASRJC H1 Math Prelims Solutions (3, – 10) is an inflexion point. (iii) (iv) x4 – 10x3 + 37x2 – 52x + 18 = 0 x4 – 10x3 + 36x2 – 54x + 17 + x2 + 2x +1 = 0 x4 – 10x3 + 36x2 – 54x + 17 + x2 + 2x +1 = 0 x4 – 10x3 + 36x2 – 54x + 17 = – x2 – 2x – 1 (v) Using GC, x = 0.503 or x = 2.246 (to 3 dp) 5 (a) Let x, y and z be the price of a Brand A, Brand B and Brand C laptop respectively. 16x + 23y + 20z = 109 741 ------------ (1) 20z = 16x + 23 596 – 16x + 20z = 23 596 ------------ (2) 2x + 2y + 2z = 10 994 ------------ (3) Using GC, x = 1399, y = 1799 and z = 2299. 2 equations correct – 1 mark 3 equations correct – 2 marks The total price of two Brand A laptops and three Brand C laptops = $[2(1399) + 3(2299)] y = x4 – 10x3 + 36x2 – 54x + 17 y = – x2 – 2x – 1

ASRJC H1 Math Prelims Solutions = $9695 (b) (i) (ii) P = 1.8 + 0.6ln (9 – 2) = 2.9675 = 2.97 million dollars The value 2.97 million dollars represents the annual profits in the year up to time t = 3 years. (iii) P = 1.8 + 0.6ln (3t – 2) d 1.8 9 ord 3 2 5(3 2) P t t t= −− (iv) For t  1, d 1.8 9 0d 3 2 5(3 2) P t t t= = −− As the years t increases, the annual profits are expected to increase at a decreasing rate. (v) When t = 5, d 1.8 0.13846 0.138d 3(5) 2 P t = = =− million dollars/year P/ million dollars t /years

ASRJC H1 Math Prelims Solutions 6 (i) The customers can be numbered from 1 to 10 000. Using a random number generator, 1000 numbers can be generated. The customers whose number matches the number generated, will be asked to do the survey. (ii) The advantage of random sampling is that ever yone has equal chance of being selected. The disadvantage of random sampling is that it is time consuming. 7 (i) Number of committees that can be formed = 7 2 21C = (ii) Required Probability = 77 33 99 44 70 5 126 9 CC CC+ = = (iii)Number of ways = 4! 3! 2 12−  = 8 (i) ( ) 300 23 277P' 300 300FB − = = (ii) ( ) 23 1P| 23 80 35 6BM == ++ (iii) ( ) 80 4P 300 15ME = = (iv) Since ( ) ( ) ( )138 137P P 0.210 P 300 300M E M E=  =   , M and E are not independent events. (v) Probability = 63 62 237 3 0.10389 0.104300 299 298   =  9 (i) 1P( | ') 2 P( ') 1 P( ') 2 11P( ') 1 0.25 22 BA BA A BA =  =  =  − = P( ) 0.5 0.25 0.75AB = + = (ii) 4P( | ') 9 P( ') 4 1 P( ) 9 AB AB B =  =− Let ( )P A B x= ( ) ( ) ( ) 0.5 4 1 0.25 9 9 0.5 4 4 0.25 0.3 x x xx x − =−+ − = − + =

ASRJC H1 Math Prelims Solutions 10 (i) Let X be the number of trees that will produce pears in the first year after planting, out of 5. ( )~ B 5,0.85X ( ) ( )P 3 P 2 0.026611XX =  = ( )P no replacement needed 1 0.026611 0.97338 0. 973= − =  (ii) 0.026611 3000 79.8 customers (iii) Let Y be the number of customers that require a replacement. ( )~ B 3000,0.026611Y ( ) ( )P 100 =1 P 100 1 0.98847 0.0115YY −  = −  (iv) Since n = 3000 is large, by Central Limit Theorem, 5 0.85 0.15~ N 5 0.85, approximately3000X   ( )P 4.28 0.019795 0.0198X  =  11 (i) 0.2 0.25 0.25 A B 0.3

ASRJC H1 Math Prelims Solutions (ii) Product Moment Correlation Coefficient, r-value is 0.940. There is a strong positive linear correlation between the year and the total cards billing in Singapore in the 1st Quarter. The total cards billing increases year after year. (iii) 0.676071 1349.898929yx=− 0.6761 1349.8989 (4 dec places)yx=− (iv) ( )0.676071 2023 1349.898929 17.79 17.8y= − =  The total cards billing is estimated to be $17.8 thousand million in year 2023. Although the product moment correlation coefficient r-value = 0.940 is close to 1, 2023x= is outside of data range of [2015 to 2022], the linear relationship between the total cards billing and year may not hold (extrapolation). Hence the estimate is unreliable.

ASRJC H1 Math Prelims Solutions 12 (i) Let X be the random variable denoting “the time in minutes taken by Tom to service a randomly chosen car”. Let Y be the random variable denoting “the time in minutes taken by Tom to service a randomly chosen van”. X ~ N(90, 52), Y ~N(130, 102) (i) ( )P 0.95Yt= 146.45 146t= (ii) ( ) ( ) 2 2 22 ~ N 130 2 90,10 2 5 2 ~ N 50, 200 YX YX − −  +  −− ( ) 0.76025 0.760P 60 2 60YX− −   = Let V be the random variable denoting “the time in minutes taken by Tom to take a break”. V~ N(5.5, 1.42) (iii) ( ) ( ) 2 2 2 1 2 1 2 1 2 1 2 ~ N 2 90 130 2 5.5, 2 5 10 2 1.4 ~ N 321,153.92 X X Y V V X X Y V V + + + +  + +   + +  + + + + ( )1 2 1 2P 5 60 0.045259 0.0453X X Y V V+ + + +   = 

ASRJC H1 Math Prelims Solutions (iv) ( ) ( )12 22 22 50 90Let 60 60 5 9 5 9~ N 2 90 130, 2 5 106 6 6 6 ~ N(345, 259.722) C X X Y C C = + +         +    +              ( )P 350 0.37818 0.378C = 

ASRJC H1 Math Prelims Solutions 13 (i) Unbiased estimate of population mean 303.4 4.740625 (exact)64== Unbiased estimate of population variance ( ) ( ) 21 303.41615.9663 64 2.8199 to 5sf 2.82 to 3sf =−  = = (ii) Let X be the random variable denoting “the crop weight per plant” and  be “the population mean crop weight per plant” To test H0 :  = 5 against H1 :  < 5 (horticulturist’s claim) (iii) Under H0, 2.8199~ N(5, ) 64X approximately by Central Limit Theorem since n = 64 is large, Using GC, the test statistics 4.740625x= gives 1.2357calcz =− and p-value = 0.10829 = 0.108(to 3 sig fig) Since p-value = 0.108 > 0.1, we do not reject H0 and there is insufficient evidence to conclude that the mean crop weight per plant is less than 5kg at 10% level of significance. The horticulturist’s claim is not supported by the data. (iv) Reject H0 when p-value < level of significance 0.10829100 10.8%     (v) Level of significance is the probability of concluding the mean crop weight per plant is less than 5 kg when in fact it is not. (vi) Let Y be the random variable denoting “the crop weight per plant in region B” To test H0 :  = 5 H1 :  ≠ 5 Two tail–tailed test at 1% level of significance Under H0, 21.5~ N(5, )20Y H0 is not rejected y lies outside the critical region. 4.14 5.86m (vi) The crop weight per plant in region B follows a normal distribution.