ACJC 2025 JC1 H2 Math Promos Solution and Markers Report

ACJC Mathematics Department 2025 H2 Mathematics Promotional Examinations Solutions Qn Solutions Remarks General comments: • If additional pages are used, students should alert the marker by indicating clearly in the original space for the question in the answer booklet. • If the additional pages at the end of this booklet are used, the question number must be clearly indicated. • Please do not change the order of the question number indicated in the answer booklet by cancelling off the original indicated question number. Your answers must be written only on the correct allocated space indicated by the question number in the answer booklet. 1 cos sin 2xxyx = cos sin 2ln( ) ln( )xxyx = (cos )ln (sin 2 )lnx y x x= Differentiate both sides w.r.t x 1 d sin 2( sin )(ln ) (cos )( ) (2cos 2 )(ln )d yxx y x x x y x x− + = + ---- (1) Substitute x = into cos sin 2xxyx = , cos sin 2 1 1 yx y  − = = Hence y = 1 From (1) d0 (cos )(1) 2ln 0d y x+ = + d 2lnd y x =− As the base is not a constant, students should not apply the formula f ( ) f ( ) f '( ) lnxxd a x a adx = Students are expected to leave answers in exact form and not 3s.f.

ACJC Mathematics Department 2025 H2 Mathematics Promotional Examinations Solutions Qn Solutions Remarks 2 ( ) ( ) 1 21 2 12 2 2 22 2 22 2 2 2 111 tan 2 22 1 tan 2 Differentiate w.r.t d22 d 1 4 d1 14d 1 4 d d d 1 4 8d d d d d 8 0 (shown)d d (1 4 ) y x x yx x yy xx yyx xx y y yy x xx x x y y xy x x x − − − − = + −   =+ = + = = ++ + =− + + + = + Students should consider implicit differentiation instead of differentiating it explicity and then trying to substitute in the expressions to prove LHS = RHS. 3 2 2 2 3 2 2 3 2 2 1 2 2 3 3 Differentiate w.r.t x, d d d d d 2 8 2(1 4 ) 8 8(1 4 ) 0d d d d d when 0, 1 tan 0 1 d 1d d 1d d 8 2 1 5d y y y y yy x x x xx x x x x x y y x y x y x −− −  + + + − + + + =  = = + = = =− =− + + =− 2 3 2 3 Maclaurin series for y: 1 5 1 511 2! 3! 2 6y x x x x x x + − − = + − − Most students were able to differentiate to find the third derivative but made careless mistakes.

ACJC Mathematics Department 2025 H2 Mathematics Promotional Examinations Solutions Qn Solutions Remarks 3(a) Given that ( )1 1 11 n r n r r n= =++ , ( ) ( ) 2 1 1 1 [Note: replace by 1]1 1 1 1 11 1 11 (shown) n r n r rrrr rr n n n n n = − = +− = + −= −+ −= =−   Some students who have identified the correct replacement fail to write down the correct start and end values for r. (b) ( ) ( ) 2 22 2 11 21 11 21 1 12 lim 111 2 1since as , 0, 1 12 3 2 r r r rr n rr rr n n n  =  == →  + − =+   −  = + − − → → =+ =   Many students wrongly wrote 1  instead of “ as 1,0n n→ → . ” Students should list 2 1 1 1 1 ...2 4 8 16r r  =  = + + + to observe that it is the sum to infinity of a GP.

ACJC Mathematics Department 2025 H2 Mathematics Promotional Examinations Solutions Qn Solutions Remarks 4(a) ( ) For 90, 4 9 04 4 9 0 ................... 1 y a bx when y x ba ab =− == −= −= ( ) ( ) At 1, 6 6 ................... 2 At 3, 3 3 6 3 3 6 .................. 3 From GC: 9, 4 , 1 x a b m a b m x a b m a b m a b m = − =− + − + = = − + =− + − + + = = = = Most students could substitute (9/4, 0) to form equation 1. Students who squared both sides of the inequality ended up with lengthy calculations that did not lead to a clear conclusion. Some students incorrectly compared the coefficients of 22 6 and ( 1)( 3) 0mx a bx x x− = − − − = y x

ACJC Mathematics Department 2025 H2 Mathematics Promotional Examinations Solutions Alternative Solution: ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ) ( ) ( ) 22 22 6 6 60 6 6 0 660 660 Solution given as { : , 1 or 3} 6 16 6 3 3 3 6 mx a bx mx a bx mx a bx mx a bx mx a bx amx a bx x mb amx a bx x mb x x x x a a b mmb a a b mmb −  − −  − − − −  − + − − − −  −− + − =  = − +− − − =  = +    − =  − + =− + = − + + =+

ACJC Mathematics Department 2025 H2 Mathematics Promotional Examinations Solutions Q n Solutions Remarks 4 (b) ( ) 2 2 2 2 22 2 2 2 2 2 22 2 2 2 1 1( 1) 1 10( 1) 1 ( 1) 0( 1) ( 1) ( 1) 1 0( 1) 1 0 since 1 0( 1) 11 122 0( 1) 13 24 0( 1) kx x k x k kx x k x k kx x k x k x k x k k x k x k x k x k xx kx k x k x x k x k x x k x k + −+ + + + ++ + + + + + + + + + + + + + + + + + + ++  + + + +    + − +       + + +  ++ + + + 2 2 2 2 1OR 0 ( 1) For 1, discriminant 1 4(1) 3 0 and coefficient of >0 xx x k x k xx x ++ + + + ++ = − =−  Therefore, numerator is positive for all values of x. ( )( ) 2 ( 1) 0 10 1 x k x k x x k x k or x  + + +  + +  − − Most students who attempted this question are aware that they should not cross multiply. Full credit is not awarded to students who obtain the correct answer but did not explain why ( ) 2( 1) ( 1) 1k x k x k+ + + + + >0. All necessary working must be shown clearly. Quite a number of students assume that 2 10xx+ +  because discriminant is < 0. They need to include coefficient 2x >0. Quite a number of students used the quadratic formula to factorise 2 ( 1)x k x k+ + + . They did not realise that 2 ( 1) ( 1)( )x k x k x x k+ + + = + + Quite a number who used the quadratic formula did not simplify the expression found and end up with a messy square root expression.

ACJC Mathematics Department 2025 H2 Mathematics Promotional Examinations Solutions Qn Solutions Remarks 5(a) Some students drew a parabola because they sketch for 02  instead of 0 2  . Some wrote the other endpoint as ( 23sec , 2 tan22  ) (b) 23secx = , d 6sec (sec tan )d x    = 2d 2secd y  = d d1 d dd 3tan d y y xx    == Generally well attempted except for some students who do not know how to differentiate or use the wrong notation d d x t or d d y t instead of using d d x  and d d y  . (c)(i) At the point where the tangent 1L is parallel to y axis, 3tan = 0 0= and 23sec 3x == . Hence the equation of tangent 1L parallel to y axis is x = 3. Students need to answer the question by stating clearly that the equation of the tangent 1L is x = 3. Students who merely state x = 3 will not be given full credit as x = 3 may also represent the x- coordinate of a point. (c)(ii) At (3.75, 1), 2 tan 1 = 1tan 2= 1 2 12 3( ) 3 dy dx == Gradient of L2 is 2 3 Students need to state clearly that 1 2 12 3( ) 3= is the gradient or dy dx (d) 2tan( )23  −= Hence 3tan 2 = Some students did not read the question carefully and find angle  instead of tan . (3,0) x y

ACJC Mathematics Department 2025 H2 Mathematics Promotional Examinations Solutions (e) 2 tan( ) 2 tany   = − =− C2 is a reflection of C1 in the x-axis. Note: You can also use a GC to sketch the curve to infer the transformation . Some students thought that transformation involves translation followed by reflection even though the question mention “a transformation.” Students must reali se that even though tan( )− is seen, however, it does not involve a translation because we are plotting y against x not y against  .

ACJC Mathematics Department 2025 H2 Mathematics Promotional Examinations Solutions Qn Solutions Remarks 6(a) (i) ( )2f 1yx=− + Many students failed to notice the existence of two horizontal asymptotes. Curve should be drawn such that it is tending towards the asymptotes. 6(a) (ii) 1 f ( )y x= Some students labelled “y = 0” when the asymptote does not exist in the graph. Students should take note of the shape and position of the graph. y x O O x y

ACJC Mathematics Department 2025 H2 Mathematics Promotional Examinations Solutions