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 differentiating 2 1 c x − + , with some even arriving at a ln function. 3(a) ( ) ( ) 2 2 2 ln ln ln ln 1 ln d 1 1 lnd 1 ln d 1 lnd 1 lnd d 1 ln xyyx y xy x y xxy yy xxy x x yy xyx yxy xy = = =  − =+   − =+ += − Well done. Most realised the need to take ln on both sides before differentiating implicitly. Those who rearranged to ln lny xxy = were most successful, while some who differentiated ln lny xy x= missed out the third term in applying the product rule on the RHS. Some mistakes observed: 1. xy x yx x x 2. ( )ln lnln x xyy − 3. Assuming the base is a constant, e.g. ( ) ( )dd lndd xy xyx x x xyxx    4. Assuming the index is a constant, e.g. ( ) ( ) 1dd dd xy xyx xy xxx −

ANGLO-CHINESE JUNIOR COLLEGE 2025 H2 MATHEMATICS 9758/01 Alternatively, ( ) ( ) ( ) ( ) 2 2 ln ln dd1 ln 1 lndd 1 lnd d1 ln 1 ln 1 ln 1 ln 1 ln xyyx y xy x yy x x y xy x x yxy x xxy yx xy x yx y = = = + + += − += − += − 3(b) Tangent // to y-axis, d d y x is undefined. 1 ln 0 ln 1 e 1ln e eln ln e 1.32 (by GC) y y y x x x x x −= = = =  = = The coordinates of the point is ( )1.32,e . Most equated the denominator of d d y x to zero correctly. Many got stuck solving 1lnxx e= , and they should be reminded to use the GC. A handful did not give the final answer in coordinates, swopped x- and y-coordinates, or gave 11.32, e   as the final answer. 4 Volume ( ) 22ee 222 ee π d π ln dy x x x x== ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) 33 222 232 33 23 3 232 33 23 23 1ln d ln 2ln d33 12 ln ln d33 1 2 1ln ln d3 3 3 3 1 2 1ln ln d3 3 3 3 12 ln ln3 3 3 9 1 9 ln 6ln 227 xxx x x x x x x x x x x x xxx x x x x xx x x x x xxx x x c x x x c =−  =−  = − −   = − −  = − − +  = − + +        ( ) ( )( ) ( ) ( )( ) ( ) ( ) 22 ee 2223 e e 63 3 6 3 3 ππ ln d 9 ln 6 ln 2 27 π e 36 12 2 e 9 6 227 π πe26e 5e 26e 527 27 x x x x x x = − + = − + − − + = − = −  26, 5ab= =− Most students knew the procedure to be carried out, i.e. obtain the correct limits, apply the formula, apply integration by parts, and evaluate the limits. Almost all cited the correct volume formula, with common mistakes (i) using πd yx and 22πd yx ; and (ii) finding the wrong region or limits of integration. Most could apply integration by parts correctly once, but many made sign errors applying it twice. Some made algebraic slips at the outset that made them think that only one integration by parts was necessary. These mistakes include: 1. ( ) ( ) 2 2ln ln 2lnx x x= 2. ( ) ( ) 22 ln d ln dx x x x x x

ANGLO-CHINESE JUNIOR COLLEGE 2025 H2 MATHEMATICS 9758/01 5(a) ( ) 0 0AB  =  −  = = a n b n b a n n AB ⊥ n  AB is parallel to the plane p. Not well done. It is insufficient to say that a and b are coplanar with normal vector n, as the property that rn is a constant for planes is derived from this proof. In writing proofs, students should ensure logical consistency, i.e. 1. It should not insinuate that 0 or 0 =  =a n b n (both are given to be non-zero quantities) 2.  = a n b n does not imply that =ab State geometrical relationships precisely (e.g. “parallel” or “perpendicular”), as opposed to: 1. More colloquial expressions like “lies in” and “contains”, which is also vague here as neither A nor B are on the plane p; 2. “length of projection” when neither vector is unit and is a geometrical interpretation; and 3. “direction vector of the plane” which applies to lines. 5(b) ( ) −r = b a (Since origin lies on p.) Not well done. Common errors are to think that A or B lie on P (and hence on the line), or that the line is parallel to n. Some did not realise that 0=rn meant that the origin is a known point on p. Abuse of notation, e.g. ( )l = + −r a b , was observed. 5(c) ( ) 2 2 2 Substituting into . 0, .0 .0 . .OF     += += += =−   −   r = c n r n c n n c n n cn n cn= c n n Alternatively, ( ) ( ) ( ) 2 2 .ˆˆ. . FC OC OF OC CF ==  = + = − cnn n n n cncn n Not well done. Many could not use the line equation or vector projection formula, probably due to the abstraction of this question. Many arrived at 2 . − cncn n , but went ahead to “cancel” by n to get −=cc n 0n . The final answer mark was withheld for this conceptual error, and students are reminded that this is not a valid vector operation. In a similar vein, 2=n n n was also seen. Some who attempted the line equation approach only considered CFn instead of the equation of line CF. More mistakes were seen by students using the vector projection approach, including:

ANGLO-CHINESE JUNIOR COLLEGE 2025 H2 MATHEMATICS 9758/01 1. Stating ( )ˆˆ.OC nn is CF or OF 2. Using ˆˆ.FC OC= nn 3. Using ˆˆOF OC= nn 6(a) ( ) ( ) 2 22 π 1000π 1000 1000 ( ) ( ) r a h a h a h a r a r a − − = − =  = + −− ( ) ( ) ( )