JPJC 2026 Prelim P2 Solutions

Jurong Pioneer Junior College H2 Mathematics JC2-2026 Preliminary Exam Paper 2 Solution Q1 2 2 3 4 10 22 3 x x x x     2 2 2 3 4 10 2( 2 3) 02 3 x x x x x x        2 2 4 02 3 x x x    ( 2)( 2) 0( 1)( 3) x x x x     Using a sign test, 3 or 2 1 or 2x x x     Replace x by | |x : 3 or 2 1 or 2x x x     1 1 or 2 or 2x x x     Q2 (a) A vector equation of L is 2 0 1 1 a b                    r , where  . (b) 0 4 0 4 4 1 a                   An equation of 3p is 0 4 1 a        r . (c) Since L lies in 4p , 4L p and (2, , 1)b  lies in 4p . 1 0 2 0 1 3 a                   and 2 1 2 3 1 3 b                    3a and 2b Q3(a) 2ln( 2 ) ln 1 xk x k k           2 2ln ln 1 2 1 2ln 2 xk k x xk k k                  2 2 2 2lln( 2 ) n xx xk k k k   

2 (b) Expansion in part (a) is valid for 21 1 x k   . 2 2 2 2 k k x k k x       (c) 2( 1)(1 ) 1 ( ) ( ) 2 n n nmx n mx mx       Comparing the first three terms in both series, we have ln 1k  , 2nm k , ----------(1) 2 2 2 ( 1) ( ) ( ) 2 2 2 n n m nm nm m k     . ----------(2) Sub (1) into (2): 2 2 2 4 2 4 8 4 2 m k k k km k k          ek  and 4 em . Q4 (i) From graph, greatest value of k is 2 (ii) Method 1 2 4 7y x x   2 2 ( 2) 3 ( 2) 3 y x x y       2 3 x y   2 4 7y x x  

3 Method 2   2 2 4 7 4 7 0 y x x x x y        2( 4) ( 4) 4(1)(7 ) 2(1) 4 4 12 2 2 3 yx y y            (Reject positive since 2)x 1f : 2 3, , 3x x x x      (iii) (iv) Since f [3, )R   and g ( , )D    , f gR D . Hence gf exists. (v) Method 1: To find range of gf using graph of gf: 2 2 10gf( ) 2 , 2 ( 4 7) 1x x x x      From graph, gf (2,3]R  Method 2: Mapping: fD ( , 2] f  fR [3, ) g  gfR (2, 3] (refer to graph below)

4 Q5(a)(i) Amount owed at the end of n months       1 2 5000 1.025 400 400 1.025 ...1. 400(1.025) 4025 00 n nn              2 1400 1 1.025 1.025 ... 1.025 1.025 1 400 1.025 5000 1.025 5000 1.025 1 n n n n                      165 000 000 1.0 1.0225 5 1nn   16000 (110 1. 50) 20 0 n  (a)(ii) To repay his loan 16000 (1100 1.00) 25 0n  Method 1 1.025(11000) 16000n  16000ln 11000 ln1.025n     15.174n

5 Method 2 GC table n 16000 (110 1.00) 025n 15 68.72 > 0 16  329.6 < 0 17  737.8 < 0 He would take 16 complete months to repay his loan. (b) (i) For the first n weeks: First term = 100, common difference = 10 [2(100) ( 1)(10)]2 (190 10 )2 n nS n n n      (b)(ii) For the remaining (20 n) weeks: First term 100 ( 1)(10) 10 80 10n n      Common difference =  10 20 (20 ) [2(80 10 ) (20 1)( 10)]2 (20 ) (30 30)2 n nS n n n n          Since total number of steel rods delivered is 3180, (20 )(190 10 ) (30 30) 31802 2 n n n n     2 2 2 2 2 95 5 300 300 15 15 3180 10 410 300 3180 10 410 3480 0 41 348 0 n n n n n n n n n n n                 ( 12)( 29) 0n n   n = 12 or n = 29 (reject since n < 20) Q6 (i) P(X = –0.5) = P(BG or RG) = 4 2 4 2 2 210 9 10 9     = 16 45 4 2 4 2 1 1 1 1 10 2 C C C Cor C         (ii) P(X = –1) = P(BR or RB) = 4 4 210 9  4 4 1 1 10 2 C Cor C       = 16 45 P(X = m) = P(BB or RR) = 4 3 4 3 10 9 10 9   4 4 2 2 10 2 C Cor C       = 4 15

6 P(X = m + 0.5) = P(GG) = 2 1 10 9 2 2 10 2 Cor C       = 1 45 The probability distribution of X is (iii) E(X) = 16 1 16 4 1 1 45 2 45 15 2 45m m                             = 13 47 45 90m For game to be favourable 13 47 045 90m  1.8077m i.e. least m = 1.81 Q7(a) 3y ax b  is most appropriate because as x increases, y increases at an increasing rate. (b) 3 30.28668 20.702 0.287 20.7y x x    r = 0.999905 (6dp) (c) x =11 lies within the data range, hence it is an interpolation r is close to +1, which indicates strong positive linear correlation between y and 3x (d) When x = 11, 30.28668(11) 20.702 402y   The original turbine would have generated 402 kW Since 450 > 402, the 2nd turbine appears to perform better x –1 –0.5 m m + 0.5 P(X = x) 16 45 16 45 4 15 1 45

7 Q8 (a)(i) Number of teams = 5 4 3 2 2 1C C C  180 (a)(ii) Case 1: 1 goalkeeper, 2 attackers, 2 defenders 3 5 4 1 2 2 180C C C   Case 2: 1 goalkeeper, 1 attacker, 3 defenders 3 5 4 1 1 3 60C C C   Case 3: 1 goalkeeper, 0 attacker, 4 defenders 3 5 4 1 0 4 3C C C   Total number of ways = 12 5 792C  Hence probability that team has at most 2 attackers = 180 60 3 792   = 243 or 0.307792 (b) Req probability = P(AAAD) + P(AADA) + P (ADAA) + P (DAAA) = 3 5 4 49 9            2000 or 0.3056561 (c) Method 1: Req probability = P(AAA_) + P(_AAA) - P (AAAA) 3 3 4 5 5 5 9 9 9                   1625 or 0.2486561 Method 2: Req prob = P(AAAD) + P(DAAA) + P (AAAA) 3 3 4 5 4 4 5 5 9 9 9 9 9                            1625 or 0.2486561 Q9(a)(i) Let X and Y be the masses of oranges from Farm A and Farm B respectively. X ~ N(245, 82) Y ~ N(212, 72). Then X – Y ~ N(33, 113)      P 30 P 30 P 30 0.61111 0.611 X Y X Y X Y         

8 (ii) W = X1 + X2 + X3 + Y1 + Y2 W ~ N(3(245) + 2(212), 3(64) + 2(49)) W ~ N(1159, 290) P(W > 1180) = 0.10876  0.109 (iii) The mass of every orange is independent of one another. (b) (i) X ~ N(,  2)  P 0.25 P 0.25 mX m Z              2P 2 P P 0.25 mX m Z mZ                     (ii) 2 ~ N , X n       P( 0.01 ) 0.4X     0.01P 0.4Z n                  P 0.01 0.4Z n   P 0.01 0.6Z n  0.01 n < 0.25335 n < 641.85 greatest n = 641 Q10(a) ( 150) 150 153.49 (exact)30 xx      2 2 2 2 1 1 1 104.73367.5 103.52 10429 30 x s x n n                 (b) Let X = mass of meat in a serving prepared by Josh.  is the population mean mass of meat in a serving prepared by Josh. 0H : 150  1H : 150

9 Under H0, since n = 30 is large, by Central Limit Theorem, 103.52~ N(150 , ) 30X approximately. 1.88 103.52 30 xz   z = 1.88 p-value = 0.0301 Level of significance = 5% Since p-value < α, we reject 0H . There is sufficient evidence, at the 5% level, to indicate that the population mean mass of meat is more than 150 g. Hence manager ’s suspicion is justified . (c) Let Y = mass of meat in a serving dispensed by machine  is the population mean mass of meat in a serving dispensed by machine 0H : 150 1H : 150  Under 0H , ~ N(150,100)Y 100~ N(150, )30Y Test statistics 150 100 30 y Level of Significance: 0.05 Critical Region: 1.95996 or 1.95996z z  150 1501.95996 or 1.95996 100 100 30 30 y y    146.42 or 153.58y y  (d) Since the mass of meat dispensed by the machine is normally distributed, a sample of fewer than 30 servings could still be used. (e) If the sample is not chosen randomly, it may be biased /no