ACJC 2025 H2 Prelims Paper 1 Solutions and Markers' Report

ANGLO-CHINESE JUNIOR COLLEGE 2025 H2 MATHEMATICS 9758/01 2025 ACJC H2 Math Preliminary Examination Paper 1 Markers’ Report Qn Solution Remarks 1(a) ( ) ( ) ( )( ) ( )( ) ( )( ) 2 2 2 22 2 022 2 2 2 022 4 022 4 022 x xx x xx x x x xx x xx x xx +− −+− − − + +− −− +− + +− As 2 40x + for all real x, consider: ( )( )2 2 0xx− +  2 or 2xx−  This is a standard inequality question and generally it was quite well done. Method: Move expressions to one side, common denominator, sign test Some common issues: 1. Multiplying -1 to both sides of inequality should result in an inequality sign flip. 2. When considering 2 4x + , it is not sufficient to just look at whether it has roots (it matters whether it is always positive/negative). 3. ( )( )22xx+− is a n-shaped quadratic curve. 4. In the final solution, 2x is often forgotten. 1(b) 2 22 x xx +− Replace x with x , 2 (no solutions) or 2xx−  2 or 2xx−  The modulus function is still quite poorly understood by a significant minority. 2x − has NO real solutions. 2(a) Since vertical asymptote is 1x=− , 1d = . Since the oblique asymptote is 22yx=+ , consider ( )( ) 2 22 1 2 2 1 1 2 4 2 1 kyx x x x k x x x k x = + + + + + += + + + += + Comparing with 2ax bx cy xd ++= + , 2a= and 4b= . Alternatively, Since vertical asymptote is 1x=− , 1d = . ( ) ( )2 11 c b aax bx cy ax b axx −−++= = + − +++ (by long division). Comparing coefficients of ( )y ax b a= + − with 22yx=+ , 2a= and 4b= . Graphing question, generally well- done. For students doing long division, there were still a minority who made careless slips in the procedure, resulting (thankfully for this part) in wrong remainders. Those who got the quotient ( ( )ax b a+− ) would not have been able to show the given result. Also, as this is a ‘show’ question, the presentation of their method should be clear and substantial. 2 + - + -2

ANGLO-CHINESE JUNIOR COLLEGE 2025 H2 MATHEMATICS 9758/01 2(b) 22 4 2 2211 x x c cyx xx + + −= = + +++ , 2c . For the graph to have no stationary points, the graphs will have to be in these 2 segments. As x→ , ( )22yx − →+ Hence 2 0 2cc−    . Alternatively, ( ) 2 d 22d 1 y c x x −=− + Considering d 0d y x = , ( ) ( ) 2 2 2 2 2 1 2 2 1 2 4 4 0 c x cx x x c − = + − = + + + − = For the graph to have no stationary points, the d 0d y x = equation should have no solutions, ( )( )16 4 2 4 0 16 8 0 2 c c c − −  − +   This part proved more challenging than (a). Those who got the remainder ( 2c− ) wrong would not have been able to get the correct answer. Most students considered the discriminant < 0, but a significant percentage of these solutions considered the discriminant of 224x x c++ (which is the numerator of y). This is conceptually wrong. Students should be considering the discriminant of d 0d y x = . There were also many careless mistakes differ