2025 DHS H2 Math Prelim P1 Soln

1 2025 Year 6 H2 Math Prelim Exam P1 Suggested Solutions Qn Suggested Solutions Comments 1 dd ... (1)dd y y uu y ux x ux x x=  =  = + 22 2 2 2 2 2 2 2 d2 d d1 ... (2)d 2 2 2 yxy x yx y x y x u x u x xy ux u =− − − − = = = Equate (1) with (2): 2 22 d1 d2 d 1 1 3 d 2 2 uuxuxu u u uxux u u −+= −− = − = 2 2 d 1 (shown)1 3 d uu u x x =− 2 2 21 dd13 1 6 1 dd3 1 3 u uxux u uxux =− −−= −   21 ln 1 3 ln ( 0)3 u x c x− − = +  3 2 1ln2 3 3 3 2 3 2 3 2 33 ln 1 3 3ln 3 1 3 e e (where e ) 3 1 1 13 11 1 ( 0)33 cc x u x c AuA x Au x yA xx x A Ay x y xx −− − =− − − =  = = =−    =−          =− − =− −        Use y ux= or yu x= to find d d y x or d d u x (by implicit differentiation) before showing 2 2 d 1 1 3 d uu u x x =− Note: u, x, y are variables Separate variables u & x to find the general solution of y in terms of x by integration Take note that x > 0 & y < 0 Total marks: 6

2 Qn Suggested Solutions Comments 2(a) ( ) 1 1 1 2(4 3 ) 2(4 3 ) 4 3 constant n n n n n n ux ux u xu + + + =− =− =− Since 43 x− is a constant common ratio, the series is a geometric series. For the sum to infinity to exist, 4 3 1 1 4 3 1 5 3 3 51 3 x x x x − −  −  − − −  As it is given that the series is geometric, the general form for common ratio 1n n u u + = constant must be used to justify it. For the sum to infinity to exist, common ratio 1 . (b) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( ) ( )( ) ( ) 1 0 11 00 11 00 1 1 2 11 2 2(4 3 ) 1 2 5 2(4 3 ) 1 2 5 2 1 2 5 (since 1) 2 1 1 2 1 5 2 2 3 2 2 3 2 2 1 2 1 3 162 2 5 23 3 2 6 n r r nn r rr nn rr n r n r nn rr x r r x r r r r x n r r n r r n r r nnn n n n n n n − = −− == −− == = = ==  − + + + = − + + + = + + + = = + − + − + = + + = + +    = + + + + +       = + +           2 5 23,,3 2 6abc= = = Note that 1 0 f ( ) n r r − =  has n terms Replace r as (r – 1) to change the limits for 1 0 f ( ) rn r r =− =  1 cc n r n = = where c is a constant 1 ( 1) 2 n r nnr = += (AP) 2 1 ( 1)(2 1)6 n r nr n n = = + + is applicable for lower limit ‘1’, & the upper limit ‘n’ Total marks: 7

3 Qn Suggested Solutions Comments 3(a) ( )   23 3 2 e sin e sin cos cos sin e sin 1 ... ... 2! 3! 3 xx x x x x x xxxx xxx   + = + =−   = − + + + − +     =− − − + Alternative (using repeated differentiation) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 f ( ) e sin f '( ) e sin e cos e cos df ''( ) e sin e cosd d e cos OR 2e cosd ddf '''( ) e sin e cosdd dd e cosdd OR 2e cos sin x xx x xx xx xx x x xx x x x yx yx x x x y y x xx yyx x x xx yy yxxx xx         =+ = + + + = + + = − + + + = − + + + = − − + + + = − − + + + − + When x = 0, f (0) 0 f '(0) 1 f ''(0) 2 f ''(0) 2 = =− =− =− 23 3 2 22f ( ) ... 2! 3! 3 x x x x xxx −−=− + + + =− − − + Co