2025 RI Math Prelim P2 solutions with comments

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 6 Page 1 of 30 H2 Math Year 6 Preliminary Examination Paper 2: Solutions with comments 1 (a) Without using a calculator, solve the inequality 2 2 5 6 2 3 .4 2 x x x x x     [4] (b) (i) Sketch on the same diagram the graphs of lny x and 5,y x  giving the equations of any asymptotes and the x-coordinates of the points of intersection between the two graphs. [2] (ii) Hence solve the inequality ln 5.x x  [2] Solution Comments (a) [4] 2 2 5 6 2 3 4 2 x x x x x     , 2x ( 2)( 3) 2 3 ( 2)( 2) 2 3 2 3 2 2 x x x x x x x x x x          3 2 3 0 2 x x x     0 2 ( 2) 0 0 or 2 and 2 OR 2 or 0 2 or 2 x x x x x x x x x x            A number of students did not simplify the LHS to 3 2 x x   and so arrive at a more complicated inequality 2( 2) ( 2) 0x x x   . Most of them managed to arrive at the second form of the solution from here. Many students who gave the first form of the solution omitted to exclude 2 from the answer. A few students cross multiplied and did not get any mark for this part.

2025 H2 Math Year 6 Preliminary Examination Paper 2: Solutions with Comments _________________________________________________________________________________ Page 2 of 30 (bi) [2] From GC, x-coordinates of intersection are 6.94 (3sf) or 0.00678 (3sf)x x  The graph of lny x was not well drawn with some students copying directly from the GC to have a hanging graph that does not extend to negative infinity. Some students even put in a horizontal asymptote of ey ! Students are reminded to round off the answers to 3 sf. (bii) [2] For ln 5x x  , 6.94 or 0.00678 0 or 0 0.00678 or 6.94 x x x x        OR 6.94 or 0.00678 0.00678 or 6.94 and 0x x x x      Alternative Method For ln 5,x x  the solution is 0 0.00678 or 6.94x x   . To solve ln 5x x  , we replace x byx . The solution is then 6.94 or 0.00678 0 or 0 0.00678 or 6.94 x x x x        As this part carries 2 marks, one mark is for some explanation of the method used and one mark is for the completely correct answer. Hence students who put down an answer with no explanation will not get any mark unless all the intervals are correct. Students should remember that disjoint intervals should be separated by the conjunction “or” and not a comma. lny x 0 x y 0 lny x x y 1 5

2025 H2 Math Year 6 Preliminary Examination Paper 2: Solutions with Comments _________________________________________________________________________________ Page 3 of 30 2 (a) In a triangle ABC, 2,AB angle CAB x radians and angle π 6CBA radians. (i) Show that 3 2 .ic nos sx xAC  [2] (ii) Given that x is a sufficiently small angle, show that 2,AC a b x c x   where ,a b and c are constants to be determined. [3] (b) It is given that 2 ln ln 3.xy y  Show that 22 2 2 2 d d d(2 ) 4 0.d d d y y yxy y yx x x         Hence find the Maclaurin series for ,y up to and including the term in 2.x [5] Solution Comments (ai) [2] π 5ππ 6 6x xACB      Using Sine Rule: π π 5ππ cos sin6o 2 2 55sin sin c ssin6 6 6 x A xx C       2 1 1cos2 2 3 sin2 A xx C        3 2 co ins sx xAC  (shown) Most students were able to use the sine rule to arrive at the first step. However, there was a number of students who calculated ACB wrongly:  ππ 6 1π 2 1 6 x x ACB       π π 2 6 π 3 x x ACB     Students are reminded to give details in their working for a “show” question and thus they are required to give the following steps: 1 5sin cos6 1cos π 5π cos sin6 1 3 sin2 3o sc in 2 2 s x x x x x x          (aii) [3] When x is sufficiently small, 211 ... 3 ... 2 2 AC x x         This part proves to be the most challenging as a number of students did not get the hint to use “small angle 2

2025 H2 Math Year 6 Preliminary Examination Paper 2: Solutions with Comments _________________________________________________________________________________ Page 4 of 30 2 112 1 3 ... 2x x           2 2 21 12 1 3 3 ...2 2x x x x                   2 2 212 1 3 3 .. 2 2 .2 3 7 x x x x x            where 2, 2 3 and 7a b c   . approximation” when question said “x is a sufficiently small angle”. For those who knew to use small angle approximation, some got the wrong approximation for cosine. After applying small angle approximation correctly, there was a large number of students who did not know how to proceed after that. Students are strongly reminded to look through the suggested solution to remember that small angle approximation and binomial expansion are usually used in this manner for a typical question in this topic. (b) [5] 2 ln ln 3xy y  Differentiate implicitly with respect to x: d 1 d2 2 0d d y yx yx y x   Differentiate implicitly with respect to x again: 22 2 2 2 2 2 22 2 2 2 2 2 2 22 2 2 2 d d d 1 d 1 d2 2 2 0d d d d d Multiply by throughout: d d d d d2 2 2 0d d d d d d d d(2 ) 4 0 (shown)d d d y y y y yx x x x y x y x y y y y y yxy y y yx x x x x y y yxy y yx x x                           When 0x , 3y ,   d 1 d d2 0 2 3 0 18d 3 d d y y y x x x     ,       2 2 2 2 2 2 d d3 4 3 18 18 0 324d d y y x x       , 2 2324 3 18 ... 3 18 162 ...2!y x x x x        Students are reminded to use implicit differentiation for this type of questions as the working will usually be shorter. Students are also reminded to be careful in working out the values to be used in the Maclaurin’s series as there are high accuracy marks for this type of question.

2025 H2 Math Year 6 Preliminary Examination Paper 2: Solutions with Comments _________________________________________________________________________________ Page 5 of 30 3 The terms of the sequence U are given by 1u k and 1 8 14 , 1.1 n n n uu n u   (a) For the following values of ,k describe the behaviour of the sequence U. (i) 3k [1] (ii) 10k [1] (b) Find the possible value(s) of k if the sequence U is a constant sequence. [2] The thn term of the sequence V is given by ,(1 ) 1 n n n a b av b a a n     where a and b are non-zero real constants and 1.a (c) For some values of ,a nv L as .n Find, with justification, the range of values of a for L to exist, and state the value of L in terms of a and .b [3] The nth term of the sequence W is given by when is even, when is odd. n n n u nw v n  It is given that the sequence W converges when the sequences U and V converge to the same limit. The sequence W diverges otherwise. (d) For 10,k by using part (a)(ii) and part (c), find the range of values of b for the sequence W to converge. Hence explain whether 1 r r w    is a convergent series. [3]

2025 H2 Math Year 6 Preliminary Examination Paper 2: Solutions with Comments _________________________________________________________________________________ Page 6 of 30 Solution Comments (a) [2] From GC: (i) For 3k , the terms are increasing and converging to 7. (ii) For 10k , the terms are decreasing and converging to 7. Q3 in general was difficult for most students. In (a), many students were unclear about what the “behaviour of a sequence” referred to, and what to describe. Here, the key features were the monotonicity (increasing/decreasing) and convergence to the limit 7. (b) [2] For sequence to be a constant seq