2025 ASRJC JC1 H2 Math Promos Solutions

[Turn Over ANDERSON SERANGOON JUNIOR COLLEGE H2 MATHEMATICS JC1 Promo Paper (100 marks) 9758 7 Oct 2025 3 hours Additional Material(s): List of Formulae (MF 27) CANDIDATE NAME CLASS / READ THESE INSTRUCTIONS FIRST Write your name and class in the boxes above. Please write clearly and use capital letters. Write in dark blue or black pen. HB pencil may be used for graphs and diagrams only. Do not use staples, paper clips, glue or correction fluid. Answer all the questions and write your answers in this booklet. Do not tear out any part of this 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. Where unsupported answers from a graphing calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. All work must be handed in at the end of the examination. If you have used any additional paper, please insert them inside this 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 Total This document consists of 13 printed pages and 3 blank pages.

2 1 (a) 2 1cosec 2 cot 2 d 94 x x x x  − − ( ) 22 1 1cosec 2 cot 2 d 32 1 1 2cosec 2 sin2 2 3 x x x x xxC −  =−  − =− − +   (b) ( ) ( ) 3 1 3 2 2 2 0 0 1 d d dx x x x x x x x x− = − − + −   132 3 3 2 012 3 3 2 1 1 9 1 1 92 3 2 3 2 29 6 x x x x   = − + −            = − + − − −           = 2 (a) Diff wrt x, dd4 3 2 0 dd yyx x y y xx  + + + = ( ) ( )d3 2 4 3 d d 4 3 d 3 2 yx y x y x y x y x x y  + =− + + =− + Gradient of normal at (1,2) 17 10 10 7 =− =− Equation of normal at (1,2): ( )721 10yx− = − 7 13 10 10yx=+ (b) When x = 0, 13 10y= and y = 0, 13 7x=− Area of triangle formed = 1 13 13 169 29 12 7 10 140 140    ==     3 10 13 12xx+ −+

3 [Turn Over ( )( ) ( ) ( ) ( )( ) 3 1 2 10 2 1 012 x x x x xx − + + + − − −+ ( )( ) 23 3 6 10 20 1 012 x x x x xx + − + + − + −+ ( )( ) 2 45 012 xx xx ++ −+ Method 1 ( ) ( )( ) 2 21 012 x xx ++ −+ ( )( ) ( )( ) 21 0 2 1 0 12 xxxx  + +   −+  Method 2 Consider 2 4 5 0xx+ + = Discriminant = ( ) 24 4 5 4− =− AND Coeff of 2x is positive Hence 2 4 5 0 x x x+ +     ( )( ) 1 0 12xx −+ So 21 x−   . Replace x in 10 13 12xx+ −+ by e x− So 2 e 1 e 1 (since e 0 ) x xx x − −− −        0x 4 (ai) Vertical asymptote: 1 2 2 2 2 0x x x=  =  − = d = – 2 (ii) 2 222 2 2 2 ax bx c kyx xx ++= = − +−− 2 (2 2 )(2 2) 2 2 2 2 ax bx c x x k xx + + − − +=−− 22 4 8 4 2 2 2 2 ax bx c x x k xx + + − + − +=−− By comparison, a = – 4 and b = 8. When x = 0, y = 2.5− , −2 1 + − +

4 24(0) 8(0)2.5 2(0) 2 5 c c − + +−= − = (b) 24 4 2 21 xxy x −+= − 121 21yx x= − + − 1Replace by 2 Replace by 1 112 1 1 2 1 1 111 1 xx xx y x y x xx y x y x xx + = − + ⎯⎯⎯⎯⎯ → = − +−− = − + ⎯⎯⎯⎯⎯→ = +− The sequence of transformation is (1) Scaling parallel to x axis by a factor of 2 (2) Translation of 1 unit in the negative x direction 5 (a)(i) 2f( ) xayx xb +== + 2 ( 2) yx by a x x y a by + = + − = − 1 1 f ( )2 f ( ) 2 a byxy y a bxx x − − −== − −= − 22 2,x a a by x b x b +−= = +++  1 ffD R \ 2− = =  ( ) 1f ( ) f ( )xx −=ii 2 2 a bx x a x x b −+ =−+ Comparing, 2b=− 242 a a −  − (b)(i) ( )g 7 3xx= + − , 3x  )gR 7,= ( )fD , 1 ( 1, )= − −  −  Since gfRD  ( )fg x exists. (ii) fg(7) ( ) ( )f 7 4 f 7 2= + = + 2(9) 1 91 1.7 −= + =

5 [Turn Over (iii)  )gR 7,= → New fR = ( ) )f 7 , 2  = fgR = New fD fgR = 1, 2 2 (7) 1 3 7 1 8 , 2 −    = +    6 (a)(i) (ii) y x (1,0) x = −3 x = 0 0 y x x=1 4 0

6 (b) g(2025) g(337 6 3) g(3) 32cos 2 2 24  =  + = = + = − 7 (a) ( ) 22 ln 3 n nS n n= − + ( ) ( ) ( ) 2 1 1 2 1 1 ln 3 n nS n n − − = − − − + ( ) ( ) ( ) ( ) 1 221 , 2 2 ln 3 2 1 1 ln 3 n n n nn u S S n n n n n − − = −  = − + − − − − +  ( ) ( ) ( ) ( ) 1 2 2 12 ln 3 2 2 1 1 ln 3 n n n nn u S S n n n n n − − =−    = − + − − + − − +    1 34 3 ln 3 n nn − = − +  4 3 ln 3n= − + Since ( )11 1 ln 3 4 1 3 ln 3uS= = + = − + , 4 3 ln 3nun = − + for 1n . 1 4( 1) 3 ln 3nun− = − − +    1 4 3 ln 3 4( 1) 3 ln 3 4nnu u n n− − = − + − − − + = (independent of n) Since 1nnuu −− is a constant, the sequence is an arithmetic progression for 1n (bi) Month (n) Amount at the start of month Amount at the end of month 1 1000 1000(1.001) 2 1000(1.001) + 200 1000(1.001)2 + 200(1.001) y x 4 2 4 0 2

7 [Turn Over 3 1000(1.001)2 + 200(1.001) + 200 1000(1.001)3 + 200(1.001)2 + 200(1.001) n 1000(1.001)n + 200(1.001)n -1 + … + 200(1.001)2 + 200(1.001) Amount at the end of n month = 1000(1.001)n + 200(1.001)n-1 + …+ 200(1.001)2 +200(1.001) = 1000(1.001)n + ( ) 1200(1.001) 1.001 1 1.001 1 n− − − = 1000(1.001)n + ( ) 1200200 1.001 1n− − = 1000(1.001)n + ( ) 1200200 1.001 200200n− − =   11.001 1000(1.001) 200200 200200n− +− ( ) 1201201 1.001 200200n−=− (ii) ( ) 1201201 1.001 200200 6500n− − 1 2067001.001 201201 206700ln 2012011 ln(1.001) 27.977538 n n n −  −  On the April 2027 (the 28th Month), Mr A’s account first became greater than $6500. 8 Using ratio theorem, (a) OB  23 5 OA OP+=  ( ) 1 523OP OB OA= −  52 33=− ba (b) AP OP OA= −  52 33 = − − b a a ( )5 5 5 3 3 3= − = −b a b a AC OC OA= −  75 22 = − − b a a 77 22=− ba ( )7 2=− ba Since AP  is parallel to AC  and A is a common point, the points A, C and P are collinear. (c) Exact area of triangle OBC 1 2= cb 1 7 5 2 2 2 = −  b a b A P B 3 2

8 1 7 5 2 2 2=  − b b a b ( )5 4=   =a b b b 0  5 4= 5 ˆsin 6  a b n 5 4= ( )( ) 12 3 1 (1) 2   5 34= (d) Length of projection of c on a • = ca a • 75 22  −= b a a a •• 75 22 − = b a a a a 27 5 5 cos2 6 2  −= b a a a 7 5 5cos2 6 2 =−  ba ( ) ( ) 7 3 5 271 2 3 32 2 2 4 = − − = 9 (a) ( )11 2 4 1 2xx− =− − − 2 1 d41 x xxx − −+ ( ) 22 1 2 4 1 dd2 4 1 23 x xxxx x −=− − −+ −− ( ) ( ) ( ) 2 2311ln 4 1 ln2 2323 xx x c x −−=− − + − + −+ ( ) ( ) 2 2313ln 4 1 ln26 23 xx x c x −−=− − + − + −+ (bi) 11 2 d1 e ed xx xx  =− 

9 [Turn Over (ii) 1 31 2 1  1 de x xx 1 21 2 1 11 de x xxx =−   − 111 211 22 111 deexx xxx =− − − 11 2 1 2 e 2e e x  =− − −   ( ) 2 2 2e 2e (e e ) e=− − − − = (c) Given d3sec 3sec tand xx   =  = 3, sec 1 cos 1 0 16, sec 2 cos 23 x x       = =  =  = = =  =  = ( ) π 23 0 19sec 9 3sec tan d3sec     −  = π 23 0 3 tan d  = π 23 0 3 sec 1 d− =   π 3 03 tan− = π33 3  − = 33 π− ( 3, 3,