ASRJC 2026 Prelim P1 Ans

[Turn Over ANDERSON SERANGOON JUNIOR COLLEGE H2 MATHEMATICS JC2 Prelim Paper 1 (100 marks) 9758/01 25 Aug 2026 3 hours Additional Materials: Printed Answer Booklet List of Formulae (MF 27) CANDIDATE NAME CLASS / READ THESE INSTRUCTIONS FIRST Write your name and class in the boxes on the Printed Answer Booklet. Please use a dark blue or black pen and write clearly . An HB pencil may be used for graphs and diagrams only. Do not use staples, paper clips, glue, or correction fluid. Answer all the questions. Write your answers on the Printed Answer Booklet. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use an approved graphing calculator. Unsupported answers from a graphing calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphing calculator are not allowed in a question, you must p resent the mathematical steps using mathematical notations and not calculator commands. You must show all necessary workings clearly. All work must be handed in at the end of the examination. If you have used any additional paper, please insert them inside the Printed Answer Booklet. The number of marks is given in brackets [ ] at the end of each question or part question. Question number Marks 1 2 3 4 5 6 7 8 9 10 11 12 Total This document consists of 13 printed pages and 3 blank pages.

2 1 Solution Method 1 (Reverse) Let g(x) = ln(2 4 ) 2x−− Undo Step IV: A reflection in the y - axis Replace x by – x y = g(– x) Undo Step III: A scale parallel to the x – axis by a scale factor of 2 Replace x by 12 x y = g(– 12 x) Undo Step II: A translation of 1 unit in the positive x – direction Replace x by x – 1 y = g(– 1 ( 1)2 x− ) Undo Step I: A translation of 3 units in the positive y – direction Replace y by y – 3 y – 3 = g(– 1 ( 1)2 x− ) y = 3 + g(– 1 ( 1)2 x− ) y = 3 + ln (2 – 4[ 1 ( 1)2 x−− ]) – 2 y = 1 + ln (2x) Method 2 (Direct) Step I: A translation of 3 units in the negative y-direction Replace y by y + 3 y + 3 = f(x) y = f(x) – 3 Step II: A translation of 1 unit in the negative x-direction Replace x by x + 1 y = f(x + 1) – 3 Step III: A scale parallel to the x-axis by a scale factor of 12 Replace x by 2x y = f(2x + 1) – 3 Step IV: A reflection in the y-axis Replace x by – x y = f(– 2x + 1) – 3 f(– 2x + 1) – 3 = ln(2 4 ) 2x−− f(– 2x + 1) = ln(2 4 ) 1x−+ = ln(2(–2x + 1)) +1 Replace – 2x + 1 by x f(x) = ln(2x) + 1 2 Solution

3 [Turn Over (i) 2 2 12 17 373 10xx xx++ ≤+− 22 2 17 37 ( 3 10) 02 3 10x x xxx x+ +− +− ≤+− 2 2 14 47 03 10xxx x++ ≤+− 2( 7) 2 0( 5)( 2)xxx+−+− ≤ ( ) ( ) ( )( ) 72 72 052 xx xx+ +− − ++ ≤ 72 72 x≤≤−− −+ or 2 5x− << (ii) Replacing x with x : 72 72 x≤≤−− −+ or 2 5x−< < No real solution since 0x ≥ for all x or 02 x≤ < 22 x⇒− < < 3 Solution (a) sin , cos 2 dd cos , 2sin 2 4sin cosdd d d 4sin cosd 4sindd c os d x ty t xy t t tttt y y tt t txxt t = = = = −= − −= = = − Gradient at P = – 4 sin p Equation of normal at P: 1cos 2 ( sin )4siny p xp p−= − (b) Sub (sin q, cos 2q) into the equation of normal, 1cos 2 cos 2 (sin sin )4sinq p qp p−= − ( )22 11 2sin 1 2sin (sin sin )4sinq p qp p− −− = − ( )22 12 sin sin ( sin sin )4sinq p qp p− −= − ( ) 12 sin sin (sin sin ) (sin sin ) 4sin 8sin (sin sin ) 1 qp qp qp p pq p − − += − += − x –5 2 + + + ̶ ̶

4 4 Solution (i) 2 3 3 20x x xy y+ − + −= 2 32 3 xxy x+−= >= − Consider the line yk= . To find the range of y where curve C cannot lie, 2 32 3 xxk x+−= − has no real roots ( ) ( ) 2 3 320x kx k+− + −= has no real roots. 2(3 ) 4(1)(3 2) 0kk⇒− − −< 2 2 9 6 12 8 0 18 17 0 kk k kk ⇒− + − +< ⇒− +< (1 7 ) (1 ) 0 1 17 kk k − −< << Thus y cannot lie between 1 and 17, with a = 1 and b = 17. (ii) 5 Solution (a) 2 02 1 10 e1 d e1 d e1 dx xx x x xx xx− −− −− −− = −− − −−∫ ∫∫ 0222 10 ee 22xx xx xx−− − = −−− − −−−   ( ) ( ) 12 11 e 1 e 22 12 − =−−−−+ −− −−− −  23e + e2 −= + (b) (i) ( )d ln sec tand xxx + 2sec tan sec sec tan xx x xx += + y x O

5 [Turn Over ( )sec tan sec sec tan xx x xx += + sec x= (shown) (ii) 3sec dxx∫ 2sec sec dx xx= ⋅∫ secsec tan dtan tanxxxx xx= −⋅∫ ( ) 2 se csec ctan dse 1xx xxx= −−∫ ( )3 scsec deea ct sn xxx xx= −−∫ ( )3sec ln ssec tan cnd e tax xxxx x += −+∫ ( ) 3 d sec tan ln sec tan '2 sec x xx xx xc= + ++∫ ⇒ ( )( )3 as nec 1 d sec t ln sec tan2x xx x xCx = ++ +∫ (shown) 6 Solution (i) Shortest Distance = AC AB AB ×       ( ) ( ) ( ) 32− ×−= −b a baba 32 7 − ×− ×= ba ab ( ) and ×= ×=aa 0 bb 0 7 ×= ab ( )×= −×ab ba sin 3 7 π = ab 33 2 3 21 147   = = (ii) Let θ be the angle between b and AC . cos AC AC θ = b b      ( )32 32 −= −bb abb a 232 32 −= −b abbb a  secux= 2d secdv xx = d sec tandu xxx = tanvx=

6 ( )( )3 2 3 1 cos 3 32 π− = −bb a cosθ∴ = 0 Hence the angle between b and AC is 90 . 7 Solution 1e xy −= OR ln 1yx= − 1 21d 1 (1 ) ( 1)d2y xyx −= −− 1d 1 1 d2 1 y yx x=− − d2 ln dyyy x =− 2 2 d d 1d d2ln 2 d ddd yy y yy xy x xx  += −   22 2 d 2d d2ln ddd yy yy yx xx += − (i) 232 2 2 32 2 2 2 d d 1 d 4d d d 1 d d2 ln 2 2dd d ddd d dy y y yy y y yy y x yx x xxx x y x    + + +− = −           When 0=x , e=y de d2 y x =− 2 2 d 0 d y x = 3 3 de 8d y x =− Maclaurin’s series of y is 23eee ( ) (0)2 2! 8 3! xxyx  =+− + +− +    3eee 2 48y xx= −− +  (ii) 1 1d d 21 21 e e x x y yy x xx − − = = −= − −− 2 2 2 2 d21 d d 1d d21 dd1d d 2d d d2 ln dd dd yxy x y yyx xxxx y yy yy yx x xx −= − −− = − − −= −

7 [Turn Over ( ) ( ) ( ) ( ) ( ) 1 2 1 1 1 31 2223 32 22 2 e e1 ee 1e1 2 48 2 2 ! e e 13e1 2 48 2 8 e3 eeee 28 24 1ee 8 x x x x xx x x x x xx x x xx x − −− − = ⋅−  −−  = − − + +− − + − +        = −− + ++ +       = ++ −− + = +    Alternative method 1e xy −= 1 2 21 d e 11 eed 2 16xxy xx −−= − = −− +  1 2 1 e1 e + e8xx x−−∴= +  8 Solution (a) As Madilyn runs on alternative days, number of days she runs = 30 Distance for the 30th run = 2 + (30 – 1)(0.1) = 4.9 km Total distance = ( )30 2 4.9 103.52 += km Or Total distance = ( )30 2(2) 29(0.1) 103.52 += km (b) So distance on day n, ( ) ( ) ( ) 222 0.2 0.2 0.9 0.2 0.9 ... 0.2 0.9 nd −=+ + + ++ 10.2(1 0.9 )2 1 0.9 nd −−= + − 1 1 2 2(1 0.9 ) 4 2(0.9 ), where 2. n n k − − = +− = −= Using G.C, Total distance run by Nancy = 220.04 km

8 (c) Since the common ratio is 0.9, which is a number between 0 and 1, the daily running distance converges. From (b), as n→∞ , 10.9 0n− → . Hence the daily running distance tends to 4 km. OR Limit of the daily running distance 0.224 1 0.9= += − (d) Consider ( )( ) 12 1 0.05 4 2(0.9 ) nn −+ − >− , where n is odd. From G.C, On day 41, Madilyn will first run more daily distance than Nancy 9 Solution (a) (b) Since 1f ( 1) f 0 2−= =  , f is not a one – one function and so f –1 does not exist. OR Since the line y = 0 cuts the curve more than once, f is