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 incl