2020 HCI H2 Prelim P1 Soln_edited

2020 HCI H2 Mathematics Preliminary Examination P1 Solutions Page 1 of 18 Qn Solution 1 22 d 1d 12, 1dd d d d dd d xy t ttt y y t xx t = += − ∴= d d y x ( ) ( ) ( ) 2 2 2 2 2 2 2 2 11 12 1 21 132122 21 13 2 22 1 13 2 22 1 t t t t t t t t − = + −= + +− = + = − + = − + 22 2 1 2 21 1 1 2 3 2 tt t +− + −−− − (ii) since 0t ≠ ( ) ( ) ( ) 2 2 2 2 2 11 110 222 1 33 02 22 1 1311 22 22 1 t t t t +> << + − <− < + −< − < + ∴ d11 d2 y x−< <

2020 HCI H2 Mathematics Preliminary Examination P1 Solutions Page 2 of 18 Qn Solution 2 ( ) ( ) ( )( ) ( )( )( ) 5 5 5 5 23 1 2 2 21 2 56 56715 1 ...32 2 2! 2 3! 2 ax ax ax ax ax ax − − − + = + = +   −− −−− = −+ + +    Coefficient of 3x = ( )( )( ) 3 35671 35 32 3! 2 256 a a−−− ×= − ( )( ) 1 2 211 22 4 21 4 12 1 ...2 4 2! 4 ax ax ax ax − = −  − = +− + − +    Coefficient of 2x = 21 64 a− 32 32 35 1 5256 64 35 20 7 4,since 0 4 7 aa aa aa a −= −  = = ≠ =

2020 HCI H2 Mathematics Preliminary Examination P1 Solutions Page 3 of 18 Qn Solution 3(i) 2 2 22 32 4 2 26 22 26 8 x x xx xx x x + − −+ − + − ( ) ( ) 2 21 8f 2233 xxx xx −= = ++−− The asymptotes are y = 2x + 2 and x = 3. 3(ii) y x y x

2020 HCI H2 Mathematics Preliminary Examination P1 Solutions Page 4 of 18 Qn Solution To have an odd number of roots, there are 7 points of intersection between the two curves, hence 2 3ak= + 4(a) Algebraic Method: ( )( ) ( )( ) 2 2 2 2 2 2 2 2 1 ( 1) 01 22 01 22 01 ( 1) 3 011 ( 1) 3 ( 1) 3 011 xx x xx x xx x x xx xx xx +− − ≤− −++ ≤− −− ≥− −− ≥−+   −− −+   ≥−+ ( 1) 3 ( 1) 3 ( 1)( 1) 0x x xx  −− −+ − +≥   + – + – + 1 or1 3 1 or 1 3x xx− −≤ ≥< +< 4(b) 23 2ay bx cx+ += Differentiate wrt x, 2 0d23 d yay bx cx+ += … (*) Sub ( )1, 3 into eq C and obtain 3 2 ..... (1)abc++= Sub ( )1,1− into eq C and obtain 2 ..... (2)abc−−= Sub ( )1,1− into eq * and obtain 3 3 0 ..... (3)a bc− + += –1 13− 1 13+

2020 HCI H2 Mathematics Preliminary Examination P1 Solutions Page 5 of 18 Qn Solution Using GC, 1, 2, 3ab c= = =− Eq of C is 23 2 32y xx+ −= 5 Note that 2 22l hr= + . Total surface area of cone, 22 2rh r r kπ ππ ++ = 22 2rh r k r⇒ += − ( ) 2 22 2 24 2r h r k kr r⇒ += − + 2 24 2 24 2r h r k kr r⇒ +=− + ( ) 2 2 2 2 kr hk ⇒= + (Shown) From above equation, 2 2 2 2 kr hk= + ( ) 22 2 2 2 11 332 32 k khV rh hhk hk πππ = = = + + ( ) ( ) ( ) 22 22 22d d3 2 h k hhVk h hk π +− = + ( ) ( ) 22 22 2 32 k kh hk π − = + At stationary point, d 0d V h = ( ) 22 0 2 0kh h k h−=⇒= >  h ( )2k − ( )2k ( )2k + d d V h +ve 0 -ve Alternatively, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 22 2222 42 2 222 32 22 32 2 (2) 2 2 2 2d d3 2 2 222 3 2 26 0 when 23 2 hk h k hhk hVk h hk h k khkh hk k