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 −−=− + + + =− − − + Common mistake: When expanding ( )sin x + using Maclaurin series, some students wrongly substitute ( )x + directly into the sine expansion: ( )sin x + = ( ) ( ) ( ) 35 ...3! 5! xxx  +++ − + − Note that expansion of ( ) 3 x + and ( ) 5 x + both contain constant term, x term and x2 terms. This means every term in the infinite expansion contributes to the first few powers of x. You cannot cleanly stop at the x2 term, so this defeats the purpose of using Maclaurin series. ➔ Do not substitute constants (like π) directly into Maclaurin standard series expansions. Always simplify first. Students should use addition formula for ( )sin ab+ or trigonometric identity ( )sin sinxx + =− before applying standard Maclaurin series (b) ( ) ( )( ) ( ) 2 2 23 11 1 ... 2! 1 ...2 c ccax bx ax cbx bx ab c cax abcx x −+ = + + +  −= + + + Comparing with answer in (a), ( ) 2 1 11, 1, 23 ab c ca abc −=− =− =− Solving, Students should apply Maclaurin standard series expansion of ( )1 n x+ which is more efficient than using repeated differentiation Students should apply result from comparing coefficient of x and x2 term 1 and 1a bc=− = to simplify the equation for the x3 coefficient ( ) ( ) 2 11 22 ab c c b c− − −= before

4 ( )1111 and 23 bcbc b c −=  = = 11 23 3 3 2 3 c c cc c − = −= = Therefore, 1 3b= ( ) 3 111 3 c ax bx x x + =− +  The x4 term ( ) 3 33 43 1113 3 27x x x − =− =−  The coefficient of x4 is 1 27− . solving. This will avoid many careless mistakes made. Total marks: 8 Qn Suggested Solutions Comments 4(a) 1 52 23 n n n xx x + += + As n → , ,nxl → 1 .nxl+ → 1 2 52 23 52 3 5 22 23 n n n xx ll l lll x + += + + = += + + 2 2 2 2 0 10 2ll ll − −= −= − 2( 1) ( 1) 4(1)( 1) 15 2(1) 2 1 5 1 5 or 22 l ll − −  − − − == +− = = Given that a sequence (generated by a recurrence relation) converges, letting nxl → and 1nxl+ → as n →  is the standard method to find the limit l of the sequence – do familiarise yourself with this technique. Some students assumed that n →  implies that nx → and got confused with the method of finding the limit of a sequence defined by a formula for the nth term (in terms of n, i.e. f ( )nun= (e.g. 35 2 n nu n −= − ) – do note the difference. A few students erroneously assumed that the sequence is an AP/GP from the start – not every sequence given is an AP/GP.

5 (b) The sequence is decreasing and converges to 15 2 + (or 1.62 (to 3sf)). or This part could have been done without (a) – students can use the GC to observe the behaviour of the sequence. In general, there are two parts to describing the behaviour of a sequence: • Trends (increasing / decreasing, constant, alternating) • Long-run behaviour of an infinite sequence (convergent or divergent) For example: The sequence is a _____________ sequence and diverges/converges to _____. (c) 1 52 23 n n n xx x + += + By making nx the subject, ( ) ( ) 1 11 1 1 2 3 5 2 2 5 2 3 23 25 n n n n n n n n n x x x x x x xx x + ++ + + + = + − = − −= − Starting with 5 3503 2158x = , we can use the GC to compute 5 4 5 35032323 6193 5632158 ... 1.6366335032 5 3784 34425 2158 xx x − − = = = = = =−  − 1 2.9x= Students can try to use the recurrence relation (subbing in 4n= ) to first find 4x using 4 5 4 52 23 xx x += + ; from this, it can be observed that we need to make 4x the subject (and hence in general, nx the subject). This will allow us to use the calculator to help compute 32,xx and finally 1x as seen from the solution. Quite a number of students did this on their own, using the recurrence relation again and again (repeating the process and making 3x the subject then 2x the subject, etc) and managed to obtain 1x successfully – this is ok too for this question. But do note that this method may not be feasible if say 50x was given instead of 5x and we need to obtain 1x - we will need the help of the calculator to do this then. x5 x4 x3 x2 x1

6 Alternative Make use of the SEQ function in the GC, using the recurrence relation, 1 1 23 25 n n n xx x + + −= − and 5 3503 2158x = we can key in nx as nu , 1nx + as 1nu − and 5 3503 2158x = as 1u (we are just making use of the GC to help us compute “backwards”, the values using the recurrence relationship – in the GC, we cannot input nx in terms of 1nx + . 1 2.9x= Total marks: 7 =x5 =x4 =x3 =x2 =x1

7 Qn Suggested Solution Comments 5(a) Max pt at 2x=− , Min pt at 1.5x= . (some explanation below; no need to show) By looking at the graph of f '( )yx= , we can use the usual first or second derivative test to determine the nature of the stationary points on the graph of f( )yx= : First Derivative Test x 2−− 2