ASRJC H2 Mathematics ASP Solution

[Turn Over ANDERSON SERANGOON JUNIOR COLLEGE H2 MATHEMATICS JC2 ASP Paper 1 (100 marks) 9758 27 July 2026 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 Solve the simultaneous equations ( )* 1 i * 3z wz + =− − and 2iwz=− + , where z and w are non-zero complex numbers. Leave your answer in the form iab+ where a and b are real numbers. [5] Solution ( ) ( )* 1 i * 3 1z wz + =− −  From ( )2 i 2wz=− +  * 2 * iwz=− − Subst into (1) ( )* i 2 * i 3z z z z+ =− − − − ( )2* 2i| | 3i 3z z z z z+ = − +  Let iz x y=+ From (3), ( ) ( ) ( )222 2i i 3i ix x y x y x y= + − + + + Comparing the real and imaginary parts 23x x y=− − and ( ) 220 2 3 x y y x= + − + xy =− and so ( ) 220 2 3 y y y y= + − − ( )0 4 1yy=− 0 (rejected since z is non-zero complex number or 1yy== 1x =− So 1iz=− + And from (2) ( )2 1 i i 2 iw=− − + + = − 2 A curve is defined by the parametric equations 21, , for 0 1.x y t tt= =   Show that the equation of the normal to the curve at the point P 21 , pp   is 462 2 1p y px p− = − Hence show that the normal at P cuts the curve exactly once. [6] Solution (a) 2 2 1 d 1 d 2dd x y tt xy tt t t == =− =

3 [Turn Over 3 2 d2 21d yt tx t = =− − Equation of normal at point P: 2 3 11 2y p x pp − = −  46 46 2 2 1 2 2 1 p y p px p y px p − = − − = − To check if the normal cuts the curve again, ( ) 4 2 6 12 2 1p t p p t − = − 4 3 62 (1 2 ) 0p t p p t− + − = ( )( ) 422 1 0t p p t At− + + = 6 5 Comparing coefficient of :1 1 2 2 t Ap p Ap − = − = ( )( ) ( ) 4 2 5 4 2 5 2 2 1 0 or 2 2 1 0 1 t p p t p t t p p t p t  − + + = = + + =  For 4 2 52 2 1 0p t p t+ + = , ( ) ( )( ) 254Discriminant 2 4 2 1pp=− 10 448pp=− ( ) 464 2 0 since 0 1p p p= −    (Need to state the reason) Hence (1) has no real solution. Since there is only 1 real solution for t, so the normal at P cuts the curve only once. 3 Andy and his fiancée signed up for a new 4-room flat in Boon Keng. They take up a housing loan of $450,000 provided by BEST bank for the purchase. The couple pay a fixed monthly instalment of $ A on the first day of each month . Interest is charged on the last day of each year at a fixed rate of 1.6% of the remaining loan amount at the beginning of that year. If the first instalment is paid in January 2026, (i) Show that the amount the couple owe the bank at the end of 2027 is  464515.2 24. 2$ 19 A− . [1] (ii) Given that A is 1500, find the date and amount of the final repayment to the nearest cent. [5] Solution (i) Amt owe at the end of 2017 = (1.016)(1.016)(450000) – 1.016(12A) – 12A =  464515.2 24. 2$ 19 A− (ii) year Amt owed at the beginning Amt owed at the end of the year after paying 18000

4 1st 450000 1.016(450000) -18000 2nd 1.016(450000) -12A (1.016)(1.016)(450000) – 1.016(18000) – 18000 3rd (1.0162)(450000) – 1.016(12A) – 12A (1.016)(1.0162)(450000) – (1.016)(1.016)(18000) – (1.016)(18000) – 18000 … … nth (1.016n)(450000) – (1.016n-1)(18000) – (1.016n-2)(18000) – …... – 18000 Amount of money owe at the end of nth year 21450000(1.016) 18000(1 1.016 1.016 ... 1.016 )nn −= − + + + + Consider ( )1 1.016 1 450000(1.016) 18000 0 1.016 1 n n  −−  − ( )450000(1.016) 1125000 1.016 1 0nn− −  Using G.C, 32.2n When n = 32, Amount owe at the end of 32 years ( ) 32 32$450000(1.016) $1125000 1.016 1 $3233.601= − − = Since they will be paying $1500 each month, they will finished the payment on 1 st March 2058. The last payment is $233.60. 4 Without using a calculator, solve the inequality ( ) 2 1 24 2 1 x x x + − + . [4] Hence, solve ( ) 2 2e 4 2 1 e e e x xx x − − + +− . [3] Solution

5 [Turn Over ( ) 2 1 24 2 1 x x x + − + ( ) 2 24 0 2 1 1 x x x + +− − ( )( ) ( ) ( ) ( ) 2 2 2 4 1 2 0 12 xx x x x + + − + −  − ( ) ( ) ( ) 22 2 2 6 4 4 4 0 21 x x x x xx + + − − +  −+ ( ) ( ) ( ) 2 10 0 21 xx xx +  −+ 10 1 or 0 2 or 2x x x−  −    OR 10 1 or 0, 2x x x−  −   ( ) 2 2e 4 2 1 e e e x xx x − − + +− ( ) 2 2 1 1e 2e 4 e x x x − − − + +− Replace x with e x− 10 e 1 (no solution as e 0 for all real val ues of )xx x−−−  −  or 0 e 2 e 0 and e 2 x xx − −−     or e 2 or ln 2 x x −  − and ln 2 ln 2 xx x  −   −   \ ln 2x  −  5 The rate at which the number of people infected with Middle East Respiratory Syndrome (MERS), x, is varying at any time t, is proportional to the difference between the number of infected people and the number of deaths due to MERS. It is also known that the number of deaths due to MERS is proportional to the square of the number of people infected with MERS. Initially there w ere 10 people infected and the number infected remains constant when it reaches 100. Show that ( ) 2d 100d 100 xk xxt =− , where k is a constant. Hence find x in terms of k and t in the form 1e kt px q −= + where p and q are constants to be determined. [7] -1 0 2 + + + + +

6 Solution Let D represent the number of deaths due to MERS. ( )d d x xDt − ( )d d x k x Dt =− ( ) 2d d x k x Axt =− When d100, 0d xx t== ( )( ) 2 0 100 100kA=− 1 100A= ( ) 22d1 100d 100 100 xk k x x x xt = − = − , shown Integrating 2 1 dd100 100 kxtxx= − ( ) 1 dd100 100 kxtxx =− 1 1 1 dd100 100 100 kxtxx+= − 11 dd100 x k txx+= − ln ln 100 Cx x kt− − = + ln C100 x ktx =+− e100 ktx Bx =− where eCB= 100 e ekt ktx B Bx=− e 100 ekt ktx Bx B+= ( )1 e 100 ekt ktx B B+= 100 e 1e kt kt Bx B= + When 0, 10tx== 10010 1 B B= + 10 10 100BB+= 90 10B= 1 9B=

7 [Turn Over 1100 e9 11e 9 kt kt x = + 100e 9e kt kt= + 100 100 1 9e9e e ktkt kt −== + +  , where p = 100 and q = 9 6 The functions f and g are defined as follows: 2f : 2x x x−+  , 0ax , g : 1xx −+  , 1x− . (i) State the l east value of a for the inverse function of f to exist. Hence, find f −1 in similar form. [4] For the following parts, use t