JPJC 8865 2022 Prelim Solutions vetted

2022 JPJC J2 H1 Maths Preliminary Examinations Solutions: 1 22 4 1 0kx x x k+ − + −  ( ) ( ) 21 4 1 0k x x k+ − + −  Discriminant < 0 and 10k+ ( ) ( )( ) 2 4 4 1 1 0kk− − + −  1k − ( ) ( )( ) 2 2 2 16 4 1 0 20 4 0 50 5 5 0 5 or 5 k k k kk kk − −  − − + −  −  Since 1k − , therefore 5k  { : 5}xx 2(i) 2 d 3 1 4 x x− ( ) ( ) ( ) ( ) ( ) 1 2 1 2 1 2 1 2 2 1 4 d3 142 ,13 42 142 32 1 143 1 14 3 where C is an arbitrary constant xx x C x C xC xC −=−   −=+  −  −=+ −  =− − + =− − +  k

2 2(ii) 2 24 24 5 253e 9e 30ee xx xx  − = − + 449e 30 25exx −= − + ( ) ( ) 2 2 2 44 44 d5 3ede 9 4e 25 4e 36e 100e x x xx xx x − −  − = + − =− Hence, 36, 100pq= =− 3(i) Let the original prices of Set Meals A, B and C be a, b and c respectively 11 51.80 13 2 73.4022 a b c abc a b c + − = + + = + + = Using GC, 14.50, 16.90, 20.40abc= = = Therefore, the original prices of Set Meals A, B and C are $14.50, $16.90 and $20.40 respectively. 3(ii) Total amount they will spend if they use the membership discount to pay ( )0.80 14.50 2 16.90 2 20.40 71.28 = +  +  = As $71.28 < $73.40, the total amount that they need to pay using membership discount is lesser than using the sales promotion. Hence, they should pay for the food that they intend to order using the membership discount. 4(i) 4(ii) ( )5 ln 2yx= + − d1 d2 y xx=− − x y 0 When 0x= , 5 ln 2y=+ When 0y= , ( ) ( ) 5 5 5 ln 2 0 ln 2 5 2e 2e x x x x − − + − = − =− −= =− Alternative method: ( ) ( )( ) ( ) 2 2 2 222 2 2 2 2 44 44 d5 3ede d 3e 5ed 2 3e 5e 6e 10e 2 18e 30 30 50e 36e 100e x x xx x x x x xx xx x x − −− − −  − =− = − + = + − − =− Hence, 36, 100pq= =−

3 When 1x=− , ( ) d 1 1 d 2 1 3 y x =− =−−− , ( )5 ln 2 5 ln 3yx= + − = + Equation of tangent: 1 3y x c=− + , where C is an arbitrary constant Substitute ( )1,5 ln 3−+ : ( )15 ln 3 1 3 14 ln 33 c c + =− − + =+ Therefore, equation of tangent is 1 14 ln 333yx=− + + , where 1 3m=− and 14 ln 33c=+ 4(iii) Area = ( ) 0 3 5 ln 2 d xx − +− = 18.7 unit2 5(i) ( )1 e2 btPa=+ When 0t= , 2P= : ( ) 012 e 32 aa= +  = When 1t= , 1.75P= : ( )11.75 3 e2 b=+ 1e 2 1ln ln 22 b b = = =− 5(ii) Using GC, at 5t= , d 0.0108d P t =− The value means that the population of the endangered birds in the 5th year is decreasing at 0.0108 thousand per year. 5(iii) 5(iv) As t→ , ( )ln 2 e0 t− → , 1.5P→ As observed from the equation and graph, the population decreases to (or approach es) 1.5 thousand in the long run. t P 0 2000

4 5(v) 32 1 5 5 1 1 1110 10 2 2Q t t t t t = + + = + + 3 d 1 1 d 10 Q tt =− At minimum point, 3 3 3 d 1 1 0d 10 11 10 10 2.1544 2.15 Q tt t t t = − = = = = ( ) ( ) 2 1 1 12.1544 0.8231710 2 2 2.1544 Q= + + = t 2 2.1544 3 d d Q t 0.025− 0 0.0630 Outline Therefore ( )2.15,0.823 is a minimum point. OR 2 24 d3 d Q tt = When 2.1544t= , 2 24 d3 0.13926 0d Q tt = =  Therefore ( )2.15,0.823 is a minimum point. 5(vi) 5(vii) From GC, 9.95 10t= 6(i) Case 1 (2 vowels): Number of ways = 63 32 60CC= Case 2 (1 vowel) : Number of ways = 63 41 45CC= Case 3 (0 vowel) : Number of ways = 63 50 6CC= Hence, total number of ways = 60 + 45 + 6 = 111 t Q 0

5 6(ii) Method 1 Choose 4 letters from the remaining 7 letters to be in between S and T: 7 4C ways Arrange the 4 letters in between S and T: 4! ways S and T and be in either order: 2 ways Arrange 3 remaining letter and group of S, T and 4 letters: 4! Ways Hence, total number of ways = 7 4 4! 2 4! 40320C    = Method 2 Case 1: S_ _ _ _ T _ _ _ or T_ _ _ _ S _ _ _ : 7! 2 10080= Case 2: _ S_ _ _ _ T _ _ or _ T_ _ _ _ S _ _ : 7! 2 10080= Case 3: _ _ S_ _ _ _ T_ or _ _ T_ _ _ _ S_ : 7! 2 10080= Case 4: _ _ _ S_ _ _ _ T_ or_ _ _ T_ _ _ _ S : 7! 2 10080= Hence, total number of ways = 40320 7 Let C ~ number of office workers who wish to pick up coding skills, out of 12 office workers C ~ B (12, 0.45) (a) (i) P( 6) P( 5) 0.52693 0.527CC =  =  (ii) P( 4) 1 P( 3) 0.86553 0.866CC = −  =  (b) Method 1 Let X ~ number of groups with less than four office workers who wish to pick up coding skills, out of 6 groups X ~ B (6, 1 0.52693− ) P( 6) 0.0112086 0.01121 (4sf)X = =  Method 2 Required probability = (1 0.52693)− 6 = 0.0112086 0.01121 (4sf) 8(i) Required probability = 1325 53 or 0.883(3sf )1500 60= 8(ii) Required probability = 525 75 or 0.457(3sf )1148 164= 8(iii) Let F and G be the events ‘the student is a foreigner’ and ‘the student is in Year 2 respectively P (F) = 114 19 1500 250= P (G) = 700 7 1500 15= 19 7 13P( ) P( ) or 0.0355250 15 3750FG =  = 52 13P( ) = or 0.03471500 375FG=

6 Since P( ) P( ) P( )F G F G   , F and G are not independent. 8(iv) Required probability = 238 237 1262 3 0.06340(5dp)1500 1499 1498   = 9(i) Red Red Yellow Green Red Yellow Yellow Green Green Red Yellow (ii) P(red, red) + P (yellow, yellow) = 3 2 2 1 4 or 0.2676 5 6 5 15     +  =       (iii) P(red, red, not red) + P( red, not red, red) + P(not red, red, red) 3 2 3 93 or 0.456 5 4 20 =    = 10(i) y, profits 127 267 x, number of customers 22.85 515.18yx− 5765 2325

7 (ii) 0.98588 0.986r= . Since r = 0.986 is close to +1, there is strong positive linear correlation between number of customers (x) and profits generated (y). (iii) 22.8493 515.1844 22.85 515.18y x x= −  − (2dp) (iv) m represents the rate of change of profits with respect to number of customers. For every increase in 1 customer, the average revenue increases by $22.85. c represents the amount of profit when there are no customers. Hence, when there are no customers, the restaurant is making an average loss of $515.18 (v) 22.8493(200) 515.1844 4055y= − = (nearest dollar) The estimate is reliable as: (1) r is near to 1 (2) this estimate is an interpolation. 11(i) Unbiased estimate of population mean, 264 1200 1191.230x −= + = Unbiased estimate of population variance, ( ) 2 2 2641 18462 556.5130 1 30s  −= − = −  Let X = mass of a randomly chosen chicken 01H : 1200 H : 1200= where  is the mean mass of chicken Since 30n= is large, by Central Limit Theorem, 556.51~ N 1200, 30X   approximately Level of significance: 5% Critical region: 1.6449z− Test statistics value, 1191.2 1200 2.0432 556.51 30 z −= =− < -1.6449 From GC, p-value = 0.020517  0.0205 < 0.05 Since p-value is less than level of significance (or test statistics is within critical region), we reject 0H . There is sufficient evidence, at the 5% level of significance to indicate that the mean mass of chicken is less than 1200 g. Hence, farmer’s claim is not supported. (ii) Since the sample size 30n= is large, the sample mean mass of chicken can be approximated to a normal distribution by Central Limit Theorem. Hence it is not necessary to assume that the masses of chicken are distributed normally for this test to be valid. (iii) Let Y = mass of a randomly chosen duck 01H : 1495 H : 1495= where  is the mean mass of ducks Since 40n= is large, by Central Limit Theorem, 200~ N 1495, 40Y   approximately Level of significance:  % p-value = 0.025347 In order for 0H to be rejected,

8 level of significance > p-value 0.025347100 2.53     12 Let A, B, C ~ journey times of Bus Service A, B and C respectively. ( ) 2~ N 27,10A , ( ) 2~