[WGS] [2023] 4E AM 4049 Prelims P2 MS

Name Index Number Class O-LEVEL PRELIMINARY EXAMINATIONS 2023 LEVEL & STREAM : SECONDARY 4 EXPRESS/ 5 NORMAL ACADEMIC SUBJECT (CODE) : ADDITIONAL MATHEMATICS (4049) PAPER NO : 02 DATE (DAY) : 12 SEPTEMBER 2023 ( TUESDAY) DURATION : 2 HOURS 15 MINUTES 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 staples, paper clips, glue or correction fluid. Answer all the questions. 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. The use of an approved scientific calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks in this paper is 90. DO NOT TURN OVER THE QUESTION PAPER UNTIL YOU ARE TOLD TO DO SO. Student’s Signature Parent’s Signature Date Date This document consists of _18_ printed pages including this cover page Setter : Mr Eric Bay___ 90

2 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ,02 =++ cbxax x = 2 4 2 b b ac a −  − Binomial expansion ,......21)( 221 nrrnnnnn bbar nbanbanaba ++     ++     +     +=+ −−− 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 += cosec 2 A = 1 + cot 2 A BABABA sincoscossin)sin( = BABABA sinsincoscos)cos( = BA BABA tantan1 tantan)tan(  = AAA cossin22sin = AAAAA 2222 sin211cos2sincos2cos −=−=−= A AA 2tan1 tan22tan −= Formulae for ABC C c B b A a sinsinsin == Abccba cos2222 −+= 1 sin2 bc A=

3 1 It is given that ( )f ( ) 2 sin cosxx e x x=− . (a) Show that f '( ) 4 sin xx e x= . [4] ( ) ( ) ( ) ( ) ( ) ( ) f ( ) 2 sin cos f'( ) 2 sin cos 2 cos sin -------------------- M1 2 sin cos cos sin -------------------- M1 =2 2sin =4 sin (shown)------------------- A1 x xx x x x x e x x x e x x e x x e x x x x ex ex =− = − + + = − + + (b) Hence evaluate π 0 sin dxe x x . [4] ( ) ( ) ( ) 4 sin d =2 sin cos -------------------- M1 4 sin d =2 sin cos 1sin d = sin cos ------------------- M12 xx xx xx e x x e x x c e x x e x x c e x x e x x c −+ −+ −+    ( ) ( ) ( ) 0 0 0 1sin d sin cos ------------------- M12 11= sin cos sin 0 cos 022 11 ------------------- A122 xxe x x e x x ee e     =−  − − − =+ 

4 2 (a) Prove that 3sin 3 3sin 4sinx x x=− . [3] ( ) ( ) ( ) 2 23 23 3 LHS sin 3 = sin 2 ------------------- M1 sin 2 cos cos 2 sin 2sin cos cos 1 2sin sinx ------------------- M1 =2sin cos sin 2sin 2sin 1 sin sin 2sin ------------------- M1 2sin 2sin sin x xx x x x x x x x x x x x x x x x x x x x = + =+ = + − +− = − + − = − + − 3 3 2sin 3sin 4sin RHS ------------------- A1 x xx=− = (b) Hence solve the equation 36sin 8sin 1xx−= for 0 120x   . [4] ( ) 3 3 6sin 8sin 1 2 3sin 4sin 1 2sin 3 1 ------------------- M1 1sin 3 = 2 30 ------------------- M1 3 30,180 30 3 30,150 ------------------- M1 10 ,50 ------------------- A1 xx xx x x x x x  −= −= = = =− = =  

5 3 (a) Show that 4 2 23 4 ( 1)( 1)( 4)x x x x x+ − = + − + . [2] ( )( ) ( )( )( ) 42 22 2 LHS = 3 4 1 4 -------------------- M1 1 1 4 -------------------- A1 xx xx x x x +− = − + = + − + (b) Hence express 2 42 37 34 x xx + +− in partial fractions. [6] ( )( )( ) ( ) ( ) ( ) ( )( ) ( )( ) ( )( )( ) 2 22 2 2 2 37 -------------------- M1111 1 4 4 3 7 1 4 1 4 1 1 sub 1 10 (10) 1-------------------- M1 sub 1 10 10 1-------------------- M1 sub 0 7 4 4 7 4 4 x A B Cx D xxx x x x x A x x B x x Cx D x x x B B x A A x A B D D ++ = + ++−+ − + + + = − + + + + + + + − = = = =− =− =− = =− + − = + − ( )( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 2 22 2 1-------------------- M1 compare coeff of , 0 0 1 1 0-------------------- M1 3 7 1 1 1 111 1 4 4 1 1 1 -------------------- A111 4 D x A B C C C x xxx x x x xx x = = + + =− + + = +− = + ++−+ − + + = − +−+ +

6 4 A curve y, is such that 2 2 d 2d y xx = and the point ( )0, 3P − lies on the curve. The gradient of the curve at P is 5. (a) Determine if the curve passes through point ( )3, 21Q . [5] 2 2 2 2 3 3 d 2d d = ------------------ M1d d0, =5 ------------------ M1d 5 d =5d 1 5 ------------------ M13 0, 3 3 1 533 when 3 ------------------ M1 9 15 3 21 the curve passes y xx y xCx yx x C y xx y x x D xy D y x x x y y = + = = + = + + = =− =− = + − = = + − = through the point (3,21) ------------------ A1

7 (b) Explain why the curve has no turning point. [2] 2 2 2 d =5d dfor turning point, 0,d 50 5 5---------------- M1 no Solution, therefore no turning point ---------------- A1 y xx y x x x x + = += =− =− (c) Determine whether the curve is an increasing or decreasing function. [2] 2 2 2 d =5d 0 5 0--------------- A1 Therefore the curve is an increasing function for all value of --------------- A1 y xx x x x +  + 2 dor as 0, 5---------------- M1d dsince can never be zero,d the curve has no turning point. ---------------- A1 yx x y x  

8 5 The points ( )2,1A − , ( )3, 4B − and ( )3,1C lies on a circle. (a) Show that the centre of the circle is 13,.22  − [6] ( ) ( )2,1 , 3, 4 41 3 ( 2) 1-------------------- M1 1 2 3 1 ( 4)int , 22 13, -------------------- M122 31 22 2 2-------------------- M1 AB AB AB m m midpo y mx c c c yx ⊥ −− −−= −− =− = − + + −= =− =+ − = + =− =− ( ) ( )3, 4 , 3,1 41 33 undefine-------------------- M1 0 3midpoint 3, 2 3 02 3 2 3 -------------------- M12 BC BC BC m m y mx c c c y ⊥ − −−= − = = =− =+ − = + =− =− 2 3 2 1 2 13Therefore centre of circle is , ---------- ---------- A122 yx y x =− =− =  −

9 (b) Explain why AB is the diameter of the circle. [1] Midpoint of AB is the center of the circle. Or student show that 1BC ACmm =− -----------B1 (c) Find the equation of the circle. [3] 22 2 22 133 4 -------------------- M122 25 25 44 25 -------------------- M12 1 3 25 -------------------- A12 2 2 r xy    = − + − +       =+ =    − + + =       (d) Show that point ( )2, 2D lies outside the circle. [2] 22 13Distance of from centre 2 2 -------------- ------ M122 9 49 44 29 25 -therefore point lies out side the circ le------------------- A122 D D    = − + +       =+ =

10 56 Solve the following equations. (a) 223 log ( 4) 2log (3 4).xx+ + = − [4] ( )( ) 22 2 22 2 2 2 3 2 2 2 3 log ( 4) 2log (3 4) log (3 4) log ( 4) 3 (3 4)log 3 -------------------- M1( 4) (3 4) 2 -------------------- M1( 4) (3 4) 8( 4) 9 24 16 8 32 9 32 16 0-------------------- M1 4 9 4 0 xx xx x x x x xx x x x xx xx + + = − − − + = − =+ − =+ − = + − + = + − − = − + = 44 or (Rej) ------------------ A19xx= =− (b) 32log log 3 1.yy−= [5] ( )( ) 3 3 3 3 3 2 2 33 2log log 3 1 log 32log 1 -------------------- M1log let log 12 1 -------------------- M1 21 2 1 0 -------------------- M1 2 1 1 0 1 or 12 1log or log 1 2 yy y y xy x x xx xx xx xx yy −= −= = −= −= − − = + − = =− = =− = -------------------- M1 1 or 3 -------------------- A1 3 yy==