YI 2026 H2 Math JC2 Prelim Paper 2 (Solutions with comments)

1 ©YIJC 9758/02/JC2PE/2026 Solutions for 2026 JC2 H2 MA Preliminary Examination Paper 2 1 (a) ( ) 22 1 3i 13 2 z =− =+ = Basic angle, 1 3tan 13  − −== Since z is in 4th quadrant, ( )arg π or 603 z =− =− −  Read the question, calculators are not allowed and hence clear workings need to be shown. Common mistake for z : ( ) 221 3i+ *Know the formula for z . While correct, it is a long-winded way to do the following: ( )( )1 3i 1 3iz = + − Common mistakes for argz : 1. ( ) 1 1arg tan 3z −=− 2. Taking basic angle as a negative angle ( 1 3tan 1 − =−  *Basic angle is the acute angle 1tan yx −= . Then based on the quadrant, we can then find ( )arg z While still getting the correct answer, workings such as ( ) 1 3arg tan 1z − =−  may not have clear understanding of finding argument.

2 ©YIJC 9758/02/JC2PE/2026 (b) The vertical axis should be the Im axis and the horizontal axis should be the Re axis. To show the geometrical relationship: 1. AOB should be labelled with 90 2. OA and OB should be labelled as equal length 3. B and C should be horizontally aligned. (BC parallel to Real axis and C must be on the right of B) 4. BC should be indicated as length 2 There is no need to find the cartesian form of each of the complex numbers. You can utilise the z and arg z together with the geometrical relationships to label the points. To label A, ensure that the modulus 2 and the arg z of 3 − are labelled. arg z should be labelled with a directed arrow. Common mistake: - A and B are drawn as reflections of each other in the Real axis. This is for conjugates which is not the case for A and B. - B and C are not horizontally aligned - AOB does not look like a right angle. Im Re O 𝐴 ≡ 𝑧 𝐶 ≡ i𝑧 + 2 𝐵 ≡ i𝑧 2 2 2 3 −

3 ©YIJC 9758/02/JC2PE/2026 (c) arg(i ) 2 3 6z   =−= Since OBCD is a rhombus, ( ) ( )1arg i 2 arg i 2 1 26 or 1512 zz   += =  = Majority did not use the diagram from (b) to find ( )arg i 2z+ which is not answering question requirement of “Hence”. Workings must be shown to indicate understanding of the use of diagram to find the respective angles. Knowledge of property of Rhombus where the diagonal bisects the angle. (OBCD is a Rhombus because OB = BC and we can then proceed to indicate a point D which will form the rhombus) At no point in the working should there be the use of arg __ = ( ) 1tan− . 2(a) ( )n = a c a b for some (non-zero) constant n ( ) n n n  =   −  =  − = a c a b a c a b 0 a c b 0 Given non-zero a, n−=c b 0 or //n−c b a n−=c b 0 or nm−=c b a for some constant m 0 Hence mn=+c a b , m and n are constants. It is important to add in the scalar multiple n since ac and ab are parallel. Many students did the following ( ) ( )   =a c a b 0 And then wrongly manipulated in the following manner : ( )+=a c b 0 . Alternative (For marker’s reference only) ab is perpendicular to a and b, and ac is perpendicular to both a and c If ab is parallel to ac , then the vector c is parallel to the plane with ab as the normal vector Then the vector c would be a linear combination of a and b, hence mn=+c a b where m and n are constants Im Re O 𝐴 ≡ 𝑧 𝐶 ≡ i𝑧 + 2 𝐵 ≡ i𝑧 2 2 D 2 where did this wrongly come from?

4 ©YIJC 9758/02/JC2PE/2026 2(b) Let θ be the acute angle between vectors a and b 1 2tan 2.5 4tan (shown)5   − = =  222 2.5 41 4 41 2 =+ = = a Some students missed out on showing 𝜃 = tan−1 (45). Note that the question asks for the ‘exact’ value of a . a b θ ȁ𝐚 ∙ 𝐛ȁ ȁ𝐚 × 𝐛ȁ

5 ©YIJC 9758/02/JC2PE/2026 Alternative Let θ be the acute angle between vectors a and b sin=a b a b or sina , cos=a b a b  or cosa 1 2 2.5 sin 4 cos 5 4tan 5 4tan (shown)5     −  = = = =  ab ab a a  ( ) ( ) ( ) 2222 22 2 2 2 2 sin cos 2 2.5 sin cos 10.25 41 2    + = + + = + = = a b a b a a a a a  Alternatively, 45tan cos 5 41=  = 2.5 41cos 2.5 25 / 41 =  = =aa Alternatively, 44tan sin 5 41=  = 2 41sin 2 24 / 41 =  = =aa Some students missed out on showing 𝜃 = tan−1 (45). Note that the question asks for the ‘exact’ value of a .

6 ©YIJC 9758/02/JC2PE/2026 3(a) Students are advised to label the asymptote 0y= to leave no doubt that the x-axis is one of the asymptotes. They should also realise that the graph is symmetrical, so the turning point is exactly at 1.5x= . (b) Rf = (−∞, −4] ∪ (0, ∞) There were several errors in writing the interval notation. If students are not confident, they can use alternative forms e.g. { : 4 or 0}y y y −   . (c) (  ( )ffR , 4 0, D {1, 2} \= − −   =  2 ffSince R D , f does not exist.  Many students thought the explanation has to do with the horizontal line test. If they knew that 2f is the composite function ff  , then they should use the usual ‘range of first function must be a subset of the domain of the second function’ idea. (d) Largest value of a = 3 2 This is the largest a that still leaves a one -one function. However, many students were ‘overthinking’ and chose the value 1 due to the vertical asymptote there. They forgot that the original domain already omitted 1x= . x = 2 y x y = 0 x = 1 (1.5, –4)

7 ©YIJC 9758/02/JC2PE/2026 (e) Let ( )fyx= 2 2 2 2 1 32 132 3 9 1 224 3 1 1 24 3 1 1 24 3 1 1 24 y xx xx y x y x y x y x y = −+ − + =  − − + =  − = + − = + =  + Reject f 3 1 1 3 as D ,2 4 2x y = + + = −   ( ) 1 3 1 1f 24x x− = − + (  ( )1fD , 4 0,− = − −   Alternatively Let ( )fyx= 2 2 2 1 32 132 13 2 0 y xx xx y xx y = −+ − + = − + − = ( ) 13 9 4 2 2 3 1 4 122 yx y − −  − − = =  + Reject f 3 1 4 31 as D ,2 2 2x y = + + = −   The majority chose to complete the square while the rest used the quadratic formula. There were a handful of students who committed algebraic errors such as ( )( ) 111 2 1x x x yy− − =  − = or 12x y−= ; these students need to learn ab c a c=  = or bc= only when 0c= . Students must also learn that 2a c a c=  = ; a significant number of them omitted the  sign. The rejection of the ‘ + ’ case must then be accompanied by a reason, especially since this is a ‘hence’ question. Finally, 1 ffDR− = , which is generally recognised by the students.

8 ©YIJC 9758/02/JC2PE/2026 ( ) 1 3 1 4f1 22x x− = − + (  ( )1fD , 4 0,− = − −   4 (a) 22 2 22 2 1e d 4 e d4 1 e4 xx x x x x x C = =+  Many students did not recognise the standard form f ( )f '( )e d xxx and wrongly used “by parts” instead. (b) d 2dw xx =  Do not write but this is useful for the substitution: d 2 dw x x= ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 3 2 2 2 2 2 2 1cos d cos 2 1 cos 2 11 sin (1)sin d22 11 sin cos22 s 2 d d 11 in cos22 x x x x x ww w w w w w w w C x x x C xx w = = =− = − − + = + +    ( ) ( ) ( ) ( ) ( ) ( )( ) 2 3 2 2 3 3 2 3 2 2 2 4 2 2 2 4 cos 11cos d d22 1 cos d2 1 1 1 1 sin cos2 2 2 4 1 2 sin 2cos8 (shown) x x x x x x x x x x x x x x x x x x C + = =+ = + + = + + +     Common errors/misconcepts: Substitution: 1) did not change dx or change wrongly like dx → 2xdw 2) did not change back to x By parts: 3) chose the wrong “u” function 4) integrated cos w wrongly like cos sindv w v wdw = → =− last part 5) Some did not use the “hence…” method, but used substitution w = x2 first .

9 ©YIJC 9758/02/JC2PE/2026 (c) ( ) ( )( ) ( ) 2 2 2 2213 222 00 2 3 2 2 0 0 2 2 2 2 4 00 22 0 22 Volume 1e d cos d 2 1e d cos d 2 11e 2 sin 2cos48 e 1 0 2 0 2 048 e 1 ( 4 )48 e4 x x x x x x x x x x x x x x x x