2025 JPJC Prelims Math Paper 2 Solutions

Jurong Pioneer Junior College H2 Mathematics JC2-2025 Preliminary Exam Paper 2 Solution Q1 (i) ( )d ,0d x k P x kt = −  1 ddx k tPx =− ( )ln P x kt C− − = + since Px ( )ln P x kt C− =− − e kt CPx −−−= eeC ktPx −−−= e , ekt CP x A A −−− = = e ktx P A −=− When 0, 0t x A P= =  = e ktx P P −=− ( )1e ktxP −=− (Shown) (ii) When 112, 2t x P== ( ) 121 1e2 kPP −=− 12 1e 2 k− = 112 ln ln 22k− = =− 1 ln 212k = ln 2121e t xP −=−   When 0.8xP= , ln 2120.8 1 e t PP −=−   ln 212e 0.2 t− = ln 2 ln 0.212 t−= 12ln 0.2 1n2t −= 27.9 28hours (nearest hour)t = (iii)

2 Q2 (i) Method 1: Equation of l: ,= + r a e Since Q lies on l, , for some value of = + q a e ( ) ( ) (Shown)   −  = + −  = = q a e a e a e ee 0 Method 2: AQ=−qa Since, //AQ e , ( )−  =q a e 0 . (Shown) (ii) ( ) AP−  = p a e e ( )−p a e represent the perpendicular distance from P to l. OR area of parallelogram with adjacent sides parallel and equal in magnitude to AP and e (iii) t x O • A • Q e • O q a l 1 • A • Q e • P 2 a l • A e •P l

3 Method 1: Area of triangle APQ = ( )1 2 AQ−p a e ( ) ( ) ( ) ( ) ( ) ( ) 1 322 2 2 2 (Using (i)) 2 = −  = + −  =  + −  =  + = q a e q q a e q e q a e q e 0 qe Method 2: Area of triangle APQ = ( )1 2 AQ−p a e ( ) ( ) ( ) ( ) ( ) 1 322 3 (Since ) 2 2 2     = −  = − +  = +  = − =  +  =  + = q a e q q e e q a e a q e q e e e q e 0 qe Method 3: Area of triangle APQ = 111 or or222   AQ QP AQ AP AP PQ ( ) ( ) ( ) ( ) ( ) 1 2 1 3 (Since )2 1 22 2 (Since AQ =2, =2 =2) q a p q e q q q a e q a e eq q e e      = −  − =  − = +  − = = =   (iv) Note: AQ q a e = − = Since AQ =2 , AQ 2 or 2ee=− Hence, =2

4 Method 1: Area = 23 =qe ( )( ) 2 sin 3 2 2 1 sin 3 3sin 4 48.590 48.6 (1dp)     = = = =   qe Acute angle between l and PQ = Acute angle between q and e = 48.6 Method 2: Area = 1 sin 32 AQ QP  = ( ) ( ) 1 2 sin 32 2 sin 3 Since 3 2 sin 3 3sin 22 48.590 48.6      −= == = = =   pq q p q q Acute angle between l and PQ = Acute angle between q and e = 48.6 Q3 (i) sin 1x =+ and 3 cos 1y =− d cosd x  = and d 3 sind y  =− d 3 sin 3 tand cos y x   −= =− (Shown) (ii) At ( )sin 1, 3 cos 1T t t +− , t = Tangent at T makes an angle of 3π 4 with the positive x-axis: d3 π3 tan tan 1d4 y tx =− = =−

5 1tan 3 π 6 t t = = Hence, π π 3 1sin 1, 3 cos 1 ,6 6 2 2T   + − =       Gradient of normal at T = 1 Equation of normal at T: ( )13 122yx − = −  1yx=− (iii) For tangent to C to be parallel to the x-axis, d 3 tan 0 0d y x =− =  = Hence, 3 cos 0 1 3 1y= − = − (iv) At Q, 1 3 1 3 x x − = − = Area of quadrilateral ( )( ) 1 3 1 3 12 1 = + − = Q4 (a) Method 1 Since polynomial has only real coefficients, if 3i− is a root, 3i+ is another root. ( ) ( ) ( ) ( ) ( ) 2 2 2 [ 3 i ][ 3 i ] [ 3 i][ 3 i] 3i 6 10 ww ww w ww − − − + = − + − − = − − = − + ( )( ) 3 2 2 3 2 2 5 68 6 10 5 5 30 6 50 10 w pw w q w w w a w aw w aw w a + + + = − + + = + − − + + C x y O (Normal at T) T (Tangent parallel to x-axis) Q

6 Comparing coefficients of 2 : 30 (1) : 6 50 68 (2) Constant : 10 (3) w a p wa aq − = −−− − + = −−− = −−− From (2) : 6 50 68 3 Subt into (1): 3 30 33 Subt into (3) : 30 aa pp q − + =  =− − − =  =− =− ( )( ) 3 2 25 33 68 30 6 10 5 3 0w w w w w w− + − = − + − = The other roots are 3i+ and 3 5 . Method 2 ( ) ( )( ) 222 32 3i 3 i 9 6i i 8 6i 3 i 8 6i 24 18i 8i 6i 18 26i w w w =− = − = − + = − = − − = − − + = − ( ) ( ) ( ) 325 68 0 5 18 26i 8 6i 68 3 i 0 90 130i 8 6 i 204 68i 0 w pw w q pq p p q + + + = − + − + − + = − + − + − + = Comparing real and imaginary parts: Real : 90 8 204 0 (1) Im : 130 6 68 0 33 pq pp + + + = −−−−−− − − − =  =− Sub into (1) 90 8( 33) 204 0 30 qq+ − + + =  =− Since polynomial has only real coefficients, if 3i− is a root, 3i+ is another root. ( ) ( ) ( ) ( ) ( ) 2 2 2 [ 3 i ][ 3 i ] [ 3 i][ 3 i] 3i 6 10 ww ww w ww − − − + = − + − − = − − = − + ( )( ) 3 2 25 33 68 30 6 10 5 3 0w w w w w w− + − = − + − = The other roots are 3i+ and 3 5 . Use GC polyroot finder to check the answer.

7 O Re (z) Im (z) (b) (i) 1 2 1 2i 2 i 2 i 4i 2 i2 i 2 i 5 z z − + − − + + +=  = =+− (ii) 1 12 2 iiz zzz =  = Rotate 2Z by 2  radians anti-clockwise about O to get 1Z (1) 12Z OZ is a right angle. (2) 12OZ OZ= (since 12 izz= or since 1 1 2i 5z = − + = and 2 2 i 5z = + = ) (3) Also, since 3 1 2OZ OZ OZ=+ , 1 3 2OZ Z Z is a parallelogram. From (1) + (2) + (3), we can deduce that 2 3 1OZ Z Z is a square.

8 Q5 (i) Number of ways to draw 4 orbs = 12 4 495C = (ii) Number of ways to draw at least 2 colours = 54 44495 CC−− = 489 (iii) Case 1: 2R, 1B, 1G Number of ways = 5 4 3 2 1 1 120C C C  = Case 2: 1R, 2B, 1G Number of ways = 5 4 3 1 2 1 90C C C  = Case 3: 1R, 1B, 2G Number of ways = 5 4 3 1 1 2 60C C C  = Total number of ways = 120 + 90 + 60 = 270 Q6 (i) The team should conduct a 1-tail test as they are verifying the claim that users are spending more than 15 minutes. (ii) Central Limit Theorem states that sample means will follow a normal distribution approximately when the sample size is more than 30. Since the sample size is 80, the team is able to carry out a hypothesis test without knowing anything about the distribution of the times spend by the users. (iii) X: time spent per visit  : population mean time spent per visit 0H : 15 = 1H : 15  Under H0, since n = 80 is large, by Central Limit Theorem, 2 ~ N 15, 80X   approximately Test statistic, 2 2 2 16 15 1 80 80 xz n     −−= = =

9 For 5% level of significance, reject 0H if 1.6449z In order for platform’s claim to be valid, 0H must be rejected, hence, 2 2 1 1.6449 80 0.607939680 0 5.44       (iv) There is a probability of 0.05 of concluding that the population mean time spent per visit is more than 15 minutes when the population mean time spent is 15 minutes. Q7 (i) 23 1 1 1 1 1 1 11P( 0) 2 2 4 2 4 2 32aX         = = =  +  +  =                    23 3 2 1 1 1 1 5P( 2) 4 2 4 2 32b X C       = = =  +   =                Or 11 15 1 5P( 2) 1 32 32 32 32bX= = = − − − = 11 32a= , 5 32b= (ii) Probability distribution of X: x 0 1 2 3 ( )P Xx= 11 32 15 32 5 32 1 32 ( ) ( ) ( ) ( ) ( ) ( )11 15 5 1 7E = P 0 1 2 3 32 32 32 32 8X x X x        = = + + + =               ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 222 11 15 5 1 11E = P 0 1 2 3 32 32 32 32 8X x X x        = = + + + =               ( ) ( ) ( ) 2 22 11 7 39Var E E = 8 8 64X X X = − − =    (Shown) 1 2 0 head 0 head 1 head 1 head 2 heads 0 head 1 head 2 heads 3 heads 3

10 (iii) Let X = number of heads obtained in one game X = mean number of heads obtained per game Since n =50 is large, by the Central Limit Theorem, 39647 7 39~ N , N ,8 50 8 3200X     =       approximately ( )P 1 0.12876 0.129X  =  Q8 (i) (1)The condition of a portable speaker is independent of the condition of any other portable speakers. Or Whether a portable speaker is faulty is independent of the condition of any other portable speakers. (2) The probability that any portable speaker is faulty is constant. (ii) Let X = number of faulty portable speakers out of 24 ( )~ B 24,0.02X ( )P 1 0.917387 0.917X  =  (iii) Method 1: