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 quadrila