ACJC 2025 JC2 H2 Prelim Paper 2 Solutions and Markers' Report

ANGLO-CHINESE JUNIOR COLLEGE 2025 H2 MATHEMATICS 9758/02 2025 ACJC H2 Math Preliminary Examination Paper 2 Markers’ Report Qn Solution Remarks 1(a) 5 :4 3 OBl OB   = = −  r Since R lies on the line OB, 5 4 3 OR   =−  . 53 41 33 5 5 3 5 4 0 4 1 4 0 3 3 3 3 25 15 16 4 9 9 0 2 5 AR OR OA AR           − = − = − −  − −            − =  − −  − =      −      − + + + − = = 52 45 3 OR  =−  21 5 3 5 52 425 3 5 3 1 5412 4 1 21 55 3 3 3 OA OAOR OA OR OA − − + = = −             = − = − − = = −       −      Alternatively (for finding foot of perpendicular from point to line), ( ). 3 5 5111 . 4 4 50 503 3 3 3 5 5 5 51 15 4 9 21 . 4 4 4 450 50 53 3 3 3 3 OR=            = − −                    −+         = − − = − = −                   a b b Generally well done. Students understood that the procedure involves 0AR OB= or applying the projection vector formula to find OR and using the midpoint theorem to find OA . However, some students failed to recognize that points O, R, and B are collinear, hence, OR OB= . This oversight led to unnecessarily complicated expressions and careless mistakes in the calculations. eg 3 1 3 x AR y z − =− − or 55 44 33 OR         = − + −       .

ANGLO-CHINESE JUNIOR COLLEGE 2025 H2 MATHEMATICS 9758/02 2(a) ( ) 1 2 3 3 511 2 2 2 2 2 2 2 ( ) ( )( )2 2 2 2 3 2! 3! 2 4 6 1f( ) 1 4 14 11 ( 4 ) ( 4 ) ( 4 ) ...2 1 2 6 20 ... xx x x x x x x x − − − − − − = = − − = − − + − + − + = + + + + Most students knew the procedure to be carried out, i.e. express f(x) in power form and use the maclaurin expansion formula ( )1 n x+ to obtain the first 4 non-zero terms. A handful of students used the repeated differentiation method; however, no marks were awarded as this did not follow the instructions of the question. Common mistakes: 1. Expressing f(x) in an incorrect power form due to carelessness, eg. ( ) 1214 x − − or ( ) 1 2 214 x− . 2. Omitting the negative sign in the expansion, eg 31 22()2 2 2 2! 11 (4 ) (4 ) ...2 xx −− − + + 3. Taking out 4 and then applying the Maclaurin expansion formula directly, which is a misapplication of the formula. 2(b) 2 4 6 2 1 3 5 7 1 3 5 7 1 d 1 2 6 20 d 14 6 2012sin 22 3 5 7 404 12sin 2 2 3 5 7 x x x x x x x x x x x c x x x x x d − − = + + + − = + + + + = + + + +  1sin 0 0 0 d− =  = 1 3 5 7 404 12sin 2 2 3 5 7x x x x x− = + + + Most students recognized that 1 2 d2sin 2d 14 xx x − = − ; however, many either forgot to include the factor of 2 or were unsure how to use the results from 2(a) to proceed. Several students attempted repeated differentiation, which is a lengthy and tedious process. While some successfully obtained the result by integrating the previous answer, they forgot to include the constant of integration C, and consequently were unable to explain why C disappears in the final solution.

ANGLO-CHINESE JUNIOR COLLEGE 2025 H2 MATHEMATICS 9758/02 3(a) (i) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 32 1 1 1 22 22 22 2 2 2 2 1 2 1 4 43 1 2 1 1 2 12 1 2 ... 6 1 1 2 1 12 3 1 4 1 2 1 6 112 ( 1) 3 1 4 2 1 612 ( 1) 3 3 8 4 612 ( 1) 3 11 102 1 1 2 12 n n n r r r r r r r r n n n n n n n n n n n n n n n n n n n n nn n r r r n nn nn n n n n nn nn n = = = ++ + + + = + + + + +   + + += + + + + + += + + + += + + + + += + + + + += + + + = + + = +=    ( )( )2 3 5 12 n++ Mostly well done, except that some students were unfamiliar with the AP formula for 1 n r r =  . Students who combined the three fractions by expansion had difficulty arriving at the final result; a better approach is to factorize instead. 3(a) (ii) ( )( ) 1 2 5 2 3 n r r r − = + + Replace with 2rr − ( )( ) ( )( ) ( ) ( ) ( )( )( ) ( )( ) ( )( ) ( )( ) ( )( )( )( ) 21 25 1 7 1 2 2 6 22 11 13 1 11 1 2 3 3 1 5 12 6 6 1 6 2 3 6 5 12 1 2 3 3 8 6442 22 1 rn r n r n rr r r rr r r rr n n n n n n n n − = − −= + = + == −+ =+ = + − + + + + + += + + +− + + + +=− −+   Many students tried replacing with 2rr + in the expression from part (i) but failed to recognize the connection to part (ii). A more straightforward approach is to replace with 2rr − in the expression given in part (ii). Additionally, some students made careless mistakes when applying the change of limits formula or substituting values into the result from part (i). 3(b) 2 1 112u = =−− ( ) 3 11 1 1 2u ==−− 4 1 211 2 u == − 2025 3 1 2uu == Very well done—almost all students were able to find the three terms. However, some students failed to recognize the pattern and were therefore unable to find 2025u .

ANGLO-CHINESE JUNIOR COLLEGE 2025 H2 MATHEMATICS 9758/02 4(a) Let 2 1 65y xx= ++ . ( ) ( ) 2 2 2 165 134 134 13 4 since 3 xx y x y x y xx y + + = + − = + = + =− + + − ( ) 1 1f 3 4x x − =− + + Most students were able to recognise the need to make x the subject in order to find the rule of the inverse function. Students need to be reminded that “±” must be considered on taking the square root, together with the reason for rejecting the negative root. The expression inside the square root must be simplified. The following mistakes should NOT be made: ( )( ) 11 1 5 1 1 or 5 x x x yy x y + + =  + = += ( ) 2 113 4 3 2xx yy + = +  + = + 4(b) Let 2 1 65y xx= ++ . 2 2 6 5 1 6 5 1 0 yx yx y yx yx y + + = + + − = This quadratic equation must have real roots for fRy . Discriminant 0 ( ) ( ) ( ) 2 2 36 4 5 1 0 16 4 0 16 4 0 1 or 0 as 04 y y y yy yy y y y − −  + + −   ( )f 1R , 0, 4 = − −    Many students had no idea how to approach this question algebraically. Some students did partial fractions just to be able to do differentiation to obtain the stationary point. There were no tests to determine that this is a maximum point. There were also missing working for the interval ( )0, , though some students tried to explain graphically. Some students were not familiar with the set notation, especially the “  ” sign and the interval notation. It is WRONG to write ( ),0 . Some students did not handle inequalities with care, as “4y” should be included as a factor in 2 16 4 0yy+ . Thus, the result from 4 1 0y+ was wrong and incomplete. 4(c) ( ) ( ( ) g 0.25 f 1R , 0, 0,e 1, 4 − = − −   →    ( ( ) 0.25 gfR 0,e 1, − =    Some students gave the wrong rule for the composite function gf, which should be ( ) 2 1 65gf e xxx ++= . As exact range of gf is required, ( ) 1 4gf 3 e −−= or 1 4ge 1 4 − − = or ( )g0 1= in exact form is expected.

ANGLO-CHINESE JUNIOR COLLEGE 2025 H2 MATHEMATICS 9758/02 5(a) ( ) ( )( ) ( ) 22 22 2 2 2 2 22 22 22 1 d tan 1d sec d sec dtan tan 1 sec d tantan sec sec dtan 1 cosec d 1 ln cosec cot sin 1 ln 1 ln , 0 x x a x x a xaa a a a xaa a a a a xca xa x a a cax x c x aa x a a               = + == + == = = =− + + = + ++=− + = +  ++         It’s very strange to see solutions expanding 2 sec  instead of simplifying 2 2 2 tanaa  + to 22 seca  . Some students were not able to simplify sec tan   to cosec , while some others were not able to use MF27 for cosec d ln cosec cot c   = − + + and the modulus sign was often missing. Many steps were wasted to convert cosec cot+ in terms of tan for x a instead of using the right- angled triangle to do so. As this is a “show” question, it would be good to include the explanation that ,0xa  for converting the modulus sign to round brackets at the end. 5(b) (i) d d x xyt = and 2d d y xt = 2d d y x x x xy y = = 22 22 d ddd 22 yy x y y x xx yx c y x d  =  =  = +  =