AMKSS Prelim 2024 AMath P2 QP

[Turn Over Candidate Name Form Class Index Number ANG MO KIO SECONDARY SCHOOL PRELIMINARY EXAMINATION 2024 SECONDARY FOUR EXPRESS / FIVE NORMAL ACADEMIC ADDITIONAL MATHEMATICS 4049/02 Paper 2 27 August 2024 2 hours 15 minutes Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your name, index number and class in the spaces at the top of this page. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use paper clips, glue or correction fluid. Answer all questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in case of angles in degrees, unless a different level of accuracy is specified in the question. The use of an approved scientific calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. The total of the marks for this paper is 90. For Examiner’s Use 90 This document consists of 19 printed pages and 1 blank page.

2 AMKSS 4E5N AM PRELIM 4049/02/2024 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation 02 =++ cbxax , a acbbx 2 42 −−= Binomial Expansion ( ) nrn -rn-n-nn bbar nbanbanaba ++     ++     +     +=+ ... .......... 2 1 221 where n is a positive integer and ( ) ( ) ( ) ! 1.............. 1 ! ! ! r rnnn rnr n r n +−−=−=      2. TRIGONOMETRY Identities AAec AA AA 22 22 22 co t1co s tan1sec 1co ssin += += =+ BA BABA BABABA BABABA tantan1 tantan)tan ( sinsinco sco s)co s( sinco sco ssin)sin (   = = = A AA AAAAA AAA 2 2222 tan1 tan22tan sin211cos2sincos2cos cossin22sin − = −=−=−= = Formulae for A B C a2 = b2 + c2 – 2bc cos A Cabsin2 1= C c B b A a sinsinsin ==

3 AMKSS 4E5N AM PRELIM 4049/02/2024 [Turn Over 1 Show that the solution of the equation 1100 10 24 0xx ++ − = is in the form lgxa= , where a is an integer. [4]

4 AMKSS 4E5N AM PRELIM 4049/02/2024 2 Explain why 1x− is a factor of 33 5 2xx−+ . Hence solve the equation 33 5 2 0xx− + = , giving your answers in exact form. [5]

5 AMKSS 4E5N AM PRELIM 4049/02/2024 [Turn Over 3 (a) Given that ( )1 4 1y x x= − + , show that d d y x can be written in the form 41 px q x + + where p and q are constants. [4] (b) Hence evaluate 6 2 d 41 x x x+ . [5]

6 AMKSS 4E5N AM PRELIM 4049/02/2024 4 (a) Given that the curve 23y x px q=− + + is always negative for 2x− or 3x  , find the values of p and q. [3] (b) (i) It is now given that 1p = . Find the values of q for which the line 1y qx=+ is a tangent to the curve 23y x px q=− + + at point R. [4]

7 AMKSS 4E5N AM PRELIM 4049/02/2024 [Turn Over (ii) Using the smaller value of q from part (i), find the coordinates of point R. [3]

8 AMKSS 4E5N AM PRELIM 4049/02/2024 5 (a) The table shows the time, t hours, before a scuba diver wearing a wet suit develops hypothermia when submerged in water of various temperature T C. Water Temperature T 2.2 5.0 7.8 10 12.8 Number of hours t 1.4 1.92 2.63 3.36 4.59 The relationship can be modelled by the formula ( )1.064 kT ta= where a and k are constants. (i) Plot the graph of lg t against T and draw a straight line to illustrate the information. [2] (ii) Use your graph to (a) find time taken for the diver to develop hypothermia when the water temperature is 0C, [2]

9 AMKSS 4E5N AM PRELIM 4049/02/2024 [Turn Over (b) find the value of k. [3] (iii) Explain why this model might not be applicable for another scuba diver. [1] (b) The variables x and y are related in such a way that when 2y is plotted against 2xy , a straight line is obtained. This line passes through the points ( )1 , 1−− and ( )1, 3 . Find an equation connecting x and y. [3]

10 AMKSS 4E5N AM PRELIM 4049/02/2024 6 The diagram shows a cross-section of a symmetrical aircraft wheel chock. AB is in contact with the ground. AB, FC and ED are horizontal. The lengths of BC and CD are 4 cm and 15 cm respectively. Angle AFE is 90 and angle DCF is  , where 0 90   . (a) Show that the vertical height of D from the ground is ( )4cos 15sin cm+ . [2] (b) Express 4 cos 15sin+ in the form ( )cosR − where 0R  and 0 90   . [4]  A B C F D E 15 cm 4 cm