ACJC 2026 JC2 H2 Prelim Paper 1 Markers Report

ANGLO-CHINESE JUNIOR COLLEGE 2026 H2 MATHEMATICS 9758/01 2026 ACJC H2 Math Preliminary Examination Paper 1 Solutions Qn Solution Remarks 1 ( ) 2 59 3625 xx xxxx +   −++ ( ) ( ) ( ) ( ) ( ) 2 2 2 3 2 2 32 2 22 32 2 59Consider 25 59 025 5 9 2 5 025 34 025 Since 2 5 1 4 0 for all 3 4 0 3 4 0 ( 1)( 4) 0 xx xxx xx xxx x x x x x xx x x x xx x x x x x x x x x x x x x + ++ + −++ + − + + ++ − + + ++ + + = + +  − + +  − − −  − + −  Using sign test: 1 or 0 4xx−   Consider 3 6 3 Intersection of both results: 1 or 0 3 xx x xx −  −   Please note the following when solving inequalities: Any common term involving x cannot be cancelled on both sides of inequality We should try to factorise and not expand the terms. The denominator should not just be ignored or stating the quadratic denominator has non-real roots is not enough. Instead, it is necessary to show that the denominator is positive for all x. 32 3 4 0 becomes ( 1)( 4) 0 or ( 1)( 4) 0 x x x x x x x x x − + +  − + −  + −  There are 3 critical points when doing sign test including x = 0. Since question requires you to use an algebraic method, either the sign test or sketch of the curve to arrive at answer should be shown. Intersection of the answers to the two inequalities should be + - + -1 4 - 0

ANGLO-CHINESE JUNIOR COLLEGE 2026 H2 MATHEMATICS 9758/01 Alternative method: ( ) 2 59 3625 xx xxxx +   −++ ( ) 2 59Consider , 25 xx xxx + ++ ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) 2 22 2 2 2 2 2 32 59 25 Since 2 5 1 4 0 for all 5 9 2 5 5 9 2 5 0 5 9 2 5 0 3 4 0 3 4 0 3 4 0 ( 1)( 4) 0 xx xxx x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x + ++ + + = + +  +  + + + − + +  + − + +  − + +  − − −  − + +  − + −  Using sign test: 1 or 0 4xx−   Consider 3 6 3 Intersection of both results: 1 or 0 3 xx x xx −  −   considered to arrive at final answer. + - + -1 4 - 0

ANGLO-CHINESE JUNIOR COLLEGE 2026 H2 MATHEMATICS 9758/01 2(a) ln(2 sin ) x+ ln(2 ) x+ since x is small, sin xx 2 2 ln 2 1 2 ln 2 ln 1 2 2ln 2 22 ln 2 28 x x x x xx  =+   = + +   + − = + − Read question carefully, using standard series from MF27 is expected. Do NOT use the wrong method of repeated differentiation. Students need to revise the laws of logarithms. ln( ) ln lnAB A B=+ 2(b) Equation of the tangent is ln 2 2 xy=+ . Equation of tangent is a line, not curve, which can be inferred from answer in part (a). 3(a) Length of equilateral triangle base 2 tan 6 2 3 (shown) xl lx   =−   =− Students should not complicate the proof by splitting into many triangles and using different angles. 3(b) ( ) ( ) ( ) 2 2 2 1 2 3 sin23 13 2322 3 234 V l x x l x x l x x =−  =−  =− ( ) ( ) ( ) ( )( ) ( )( ) 2d 3 3 2 3 2 2 3 2 3d 4 4 3 2 3 2 3 4 34 3 2 3 6 34 V l x l x xx l x l x x l x l x = − + − −  = − − − = − − Students need to revise on area of triangle 1 sin2 ab C= . l is a fixed length, x is the variable, hence to maximise volume of box, you need to find d d V x . Then set d 0d V x = to solve for x.

ANGLO-CHINESE JUNIOR COLLEGE 2026 H2 MATHEMATICS 9758/01 ( )( ) dWhen 0d 3 2 3 6 34 or (rejected as length of base 0) 6 3 2 3 V x l x l x llx =  − −  = = Students must learn to read the diagram carefully and reject one of the solutions. 4(a) 23 1 ( ) dd3 2( ) 1dd dd3 3 2 2 2( )dd d ( 2 ) 2d xy x y yyx y x yxx yyx y x y x yxx y x y x yx + = +    + = + +       + = + + + − = − d2 d2 y x y x x y −= − (shown) 4(b) Tangent is parallel to the x-axis d 02d y yxx =  = Solving simultaneously with equation of C, ( ) 2 2 2 3 1 ( ) 3 2 1 ( 2 ) 31 1 3 xy x y x x x x x x + = + + = + = = Tangent // to x-axis d 0d y x= , useful to draw a diagram to help visualize that the tangent line is a horizontal line. Note that the gradient is 0, not undefined. When solving d 0d 2 02 y x xy xy = −= − For fraction = 0, numerator = 0. Do not set denominator =0 too! 4(c) When d10, d2 yx x== Acute angle between the tangents to C at ( )0,1 and the x-axis 1 1tan or 0.464 rad (to 3 s.f.)2 −= Recall angle of inclination. Let d1 d2 y mx == , Hence angle of inclination 1tan m−=

ANGLO-CHINESE JUNIOR COLLEGE 2026 H2 MATHEMATICS 9758/01 There is no need to find the equation of tangent here. Please leave angle in radian form for calculus questions. 5(a) 2f: , , 2.2 kxx x x x + −  If ( )7,4− lies on the curve -1f ( ),yx= then ( )4, 7− lies on the curve f ( ),yx= f (4) 7 42 724 3 (shown) k k =− + =−− = Finding the inverse of f is not the expected method. Take note of the property of all inverse functions: -1 If f ( ) , then f ( ) ab ba = = 5(b) 3 2 8For f ( ) = 3 22 When 0, 1. Asymptotes at 2 and 3. xyx xx x y x y += = − + −− = = = =− Graph of ( )g( ) fy x x== : Asymptotes at 2 and 3xy=− =− Reading the question well should help. ( )g : f , for , 0 and 2 xx xx x  −   Graph of function should be sketched for the domain of function only. O x y 1

ANGLO-CHINESE JUNIOR COLLEGE 2026 H2 MATHEMATICS 9758/01 (c) ( )g: f , for 0 and 2.x x x x  −  For 0x , replace byxx − . ( ) ( ) ( ) ( ) 1 32 32 32f = = 2 2 2 2 3 2 22 3 22g 3 x x xyx x x x y xy x yx y xx x − + −+ −+== − − − + + =− + −= + −= + ( )  )1 gg , 3 1, 31 DR or x or x − = = − −   −  To find inverse of g, figure out rule of function g first. From above part, g is the reflection of f in the y axis. So g(x)= f(-x) for for , 0 and 2 xx x  −  Range of g can be read from the graph in part (b). Use of notation :  is union  is intersection 6(a) 1h n= 6(b) Total area of the n rectangles = 1 1 1 2 1f ( ) f ( ) .......... f ( ) n n n n n n n+ + + 1 f ( ) n r h rh A = = Rectangles drawn under the curve with width 1 or hn labelled. The rectangles should be of equal width. Explanation that relates 1 f ( ) n r h rh =  to total area of rectangles (not strips) before concluding that it is less than area of A 0 x y 1 n 2 n 1n n − 1

ANGLO-CHINESE JUNIOR COLLEGE 2026 H2 MATHEMATICS 9758/01 6(c) 1 lim f ( ) n n r h rh → =  1 2 2 0 1 2 0 d1 1 ln 12 x x x e xe e − − − = + = − +  21 [ln(2) ln(1 )]2 e−= − + As the number of rectangles used in part (b) increases the total area of rectangles tends to area of A. The limit as n tends to infinity is the exact area of A which is obtained using definite integration 7(a) ( ) 2 f: 4 4 , 0 ,where is a constant Using GC: 7Greatest value of is or 3.5.2 x x x x   − − −    For 1f− to exist, f must be one to one function. Note that domain of f is given as 0 x  . Use GC to obtain graph of f and find greatest value of  . Use GC “2nd calc” “minimum” to find the x coordinate of the first minimum point as the domain given is 0 x  . You may need to use Zoom box to enlarge the graph first.

ANGLO-CHINESE JUNIOR COLLEGE 2026 H2 MATHEMATICS 9758/01 7(b) ( ) 2 for 2 0,g 2 for 0 13. xxx xx + −  = − −   When sketching such graphs, take note which point is included and which is not. Note that g(0) is 2 and not -2. Your graph should show that. There should be an unshaded circle on (0,-2) to indicate that this point is not part of the graph of g(x). 7(c)     f g fg 1Using GC: ,12 or 0.25,124 2,13 gf exists since R D RD = − − =−  f R and gD should be stated and the term subset should be used ( not f R inside gD ) Use correct notation for subset. 7(d) ( ) 17g or 1.7544 g 12 14 Range of function gf ( ) =[ 14, 2) [1.75, 2] 7or [ 14, 2) , 2 4 714 2 or 2 4 x