AISS 2023 4E Add Math Paper 1 Prelim solution

_____________________________________________________________________________ This document consists of 19 printed pages. AISS PRELIM/4E/4049/P1/2023 [Turn over AHMAD IBRAHIM SECONDARY SCHOOL GCE O-LEVEL PRELIMINARY EXAMINATION 2023 SECONDARY 4 EXPRESS Name: Class: Register No.: SOLUTION ADDITIONAL MATHEMATICS Paper 1 Candidates answer on the Question Paper. 4049/01 7 August 2023 2 hours 15 minutes READ THESE INSTRUCTIONS FIRST Write your name, class and index number on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions. Give non-exact numerical answers 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. 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 number of marks for this paper is 90. For Examiner’s Use /90

2 AISS PRELIM/4E/4049/P1/2023 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation 2 0ax bx c+ + = , 2 4 2 b b acx a −  −= Binomial expansion 1 2 2() 12 n n n n n r r n n n na b a a b a b a b b r − − −     + = + + + + + +           where n is a positive integer and ! ( 1) ( 1) !( )! ! n n n n n r r r n r r  − − +== − 2. TRIGONOMETRY Identities 1cossin 22 =+ AA AA 22 tan1sec += 22cosec 1 cotAA=+ sin( ) sin cos cos sinA B A B A B =  cos( ) cos cos sin sinA B A B A B= tan tantan( ) 1 tan tan ABAB AB = sin 2 2sin cosA A A= 2 2 2 2cos 2 cos sin 2cos 1 1 2sinA A A A A= − = − = − 2 2 tantan 2 1 tan AA A= − Formulae for ABC sin sin sin a b c A B C== 2 2 2 2 cosa b c bc A= + − 1 sin2 bc A=

3 AISS PRELIM/4E/4049/01/2023 [Turn over 1 The variables x and y are related by the equation 2 hy xk= − . The diagram below shows the graph of 1 y against x. Calculate the value of h and of k. [4] 2 hy xk= − 1 2 2 x k k xy h h h −= = − Gradient of line = 1 ( 4) 1 2 10 0 2 h −− ==− 4h= y-intercept at 4 k h− =− 16k = ( )0, 4− x B1: setting up linear form M1: finding gradient A1 A1

4 AISS PRELIM/4E/4049/01/2023 2 The Richter scale measures the intensity of an earthquake using the formula 0 lg IM I =   , where M is the magnitude of the earthquake, I is the intensity of the earthquake, and 0I is the intensity of the smallest earthquake that can be measured. (a) Calculate the magnitude of an earthquake if its intensity is 1000 times the intensity of the smallest earthquake that can be measured. [1] (b) In February 2011, an earthquake with magnitude 6.2 was recorded in Christchurch, New Zealand. Few weeks later, an earthquake with magnitude 9.0 was detected in Fukushima, Japan. How many times stronger in intensity was the Japan’s earthquake as compared to the New Zealand’s earthquake? Give your answer to 2 decimal places. [3] 0 lg IM I =   0 0 1000lg 3 IM I ==   B1 6.2 0 0 6.2 lg 10 NZ NZ I III =  =  M1: substitution and making I the subject 9 0 0 9.0 lg 10 J J I III =  =  9 6.210 630.96J NZ I I −== (2 d.p.) M1, A1 The Japan’s earthquake is 630.96 times stronger than the New Zealand’s earthquake.

5 AISS PRELIM/4E/4049/01/2023 [Turn over 3 The diagram shows a hemispherical bowl of radius 12 cm. Water is poured into the bowl and at any time t seconds, the height of the water level from the lowest point of the hemisphere is h cm. The rate of change of the height of the water level is 0.4 cm/s. (a) Show that the area of the water surface, A, is given by ( )24A h h=− . [2] Let radius of water surface be r cm ( ) 222 12 12rh+ − = ( ) ( )( ) ( )( ) 222 12 12 12 12 12 12 (24 ) rh hh hh = − − = − − + − =− Area, A = 2 (24 )r h h =− (b) Find the rate of change of A when h = 5 cm. Leave your answer in terms of  . [3] d (24 2 )d A hh =− d d d d d d (24 2 ) 0.4 (24 10) 0.4 when 5 cm 5.6 A A h t h t h h    = = −  = −  = = Rate of change of surface area = 5.6 cm2/s M1: use of Pythagoras thm M1: simplification of r2 and getting the result M1: chain rule A1: simplification of r2 and getting the result M1: correct differentiation

6 AISS PRELIM/4E/4049/01/2023 4 (a) Explain why there is only one solution to the equation ( ) ( )5 5log 13 4 log 2xx− = − . [5] M1: correct change of base M1: obtaining quadratic equation M1: correctly solving quadratic eqn A1, A1: explanation of undefined function and concluding only 1 final answer

7 AISS PRELIM/4E/4049/01/2023 [Turn over (b) Solve the simultaneous equations ( ) 34 32 2 , 9 3 10 . x x y xy ++ = += [7] ( ) ( ) 3 2 6 5 5 4 32 2 2 2 2 2 2 6 5 1 ---------- (1) x x y x x y x y x x y yx ++ + + + + = == + = + + =+ ( ) ( ) 2 21 2 9 3 10 3 3 10 3 3 10 3 3 3 10 0 xy xy xx xx + += += += + − = ( )( ) 2 Let 3 : 3 10 0 5 2 0 2 or 5 3 2 or 3 5 (reject since 3 0) x x x x u uu uu uu = + − = + − = = =− = =−  lg 3 lg 2 lg 2 0.631 (3 sf)lg 3 x x = == lg 2 1 1.63lg3y= + = (3 s.f) M1: linear equation relating x and y M1: equation with 1 variable M1: use of substitution M1: solving quadratic equation M1: Taking lg on both sides to find x A1: correct x A1: correct y

8 AISS PRELIM/4E/4049/01/2023 5 (a) Prove the identity 11cot 2 tan2 tan 2xx x=− . [2] 11cot 2 tan2 tan 2xx x=− 2 2 2 cot 2 1 tan 2 1 2 tan 1 tan 1 tan 2 tan 1 tan 2 tan 2 tan 11 tan2 tan 2 LHS x x x x x x x xx xx = = = − −= =− =− (b) Hence solve the equation ( )tan 3 4cot 2 3xx −= for 0 360x   . [5] ( )tan 3 4cot 2 3xx −= 11tan 3 4 tan 32 tan 2xx x  − − =  ( )( ) 2 2 2tan 3 2 tan 3tan 3tan 2 2 tan 3 2 tan 3tan 5 0 tan 1 2 tan 5 0 tan 1 or tan 2.5 basic angle = 45 or 68.199 45 , 225 , 111.8 , 291.8 xx x xx xx xx xx x  − + = − + = + − = − + = = =−  =     M1: double angle formula M1: splitting terms and getting result M1: simplification to trigo quadratic M1: 2 answers M1: correct basic angles A2: 2 pairs correct answers A1: any 1 pair correct

9 AISS PRELIM/4E/4049/01/2023 [Turn over (c) Without further solving, e xplain why there are 6 roots to the equation ( )tan 3 4cot 32 x x−= for 360 720x−    . [2] There are 4 roots to the equation ( )tan 3 4cot 2 3xx −= for 0 360x   from (b). Since the period of ( )tan 3 4cot2 x x− is doubled of ( )tan 3 4cot 2xx − , there will be 4/2 = 2 roots to the equation for 0 360x   . [B1] For 360 720x−    , the graph of ( )tan 3 4cot2 x x− would have repeated 3 cycles, thereby giving 3 x 2 = 6 roots to the equation. [B1] OR ( )tan 3 4cot 32 x x−= , 360 720x−    Let 2 xy= , then ( )tan 3 4cot 2 3yy −= , 180 360y−    y has 4 solutions in the domain 0 360y   , 1 from each quadrant, from (b). [B1] Therefore, for the domain 180 360y−    , the graph would have entered another half a cycle, giving rise to 2 additional roots. [B1] Therefore, there will be 6 solutions for y in the given domain, and thus, 6 roots to the equation.

10 AISS PRELIM/4E/4049/01/2023 6 The curve 2 13xy e x=− intersects the y-axis at the point P. The tangent and the normal to the curve at P meet the x-