2023 JSS AMATH P1 PRELIM MS

4049/1/GE/23 [Turn Over] O JURONG SECONDARY SCHOOL 2023 GRADUATION EXAMINATION SECONDARY 4 EXPRESS/ SECONDARY 5 NORMAL (ACADEMIC) CANDIDATE NAME CLASS INDEX NUMBER ADDITIONAL MATHEMATICS 4049/01 PAPER 1 22 August 2023 Candidates answer on the Question Paper. 2 hours 15 minutes Additional Materials: Writing Paper (1 sheet) 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 on both sides of the paper. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, 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. 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 This document consists of 19 printed pages including this page. 90

2 4049/1/GE/23 1. ALGEBRA Quadratic Equation For the equation Binomial expansion , where n is a positive integer, and 2. TRIGONOMETRY Identities Formulae for 02 =++ cbxax a acbbx 2 42−−= nrrnnnnn bbar nbanbanaba ++    ++     +     +=+ −−− ......21)( 221 ! )1)...(1( )!(! ! r rnnn rnr n r n +−−=−=      1cossin 22 =+ AA 22sec 1 tanAA=+ 22cosec 1 cotAA=+ ( ) BABABA sincoscossinsin = ( ) BABABA sinsincoscoscos = ( ) BA BABA tantan1 tantantan  = AAA cossin22sin = AAAAA 2222 sincossin211cos22cos −=−=−= A AA 2tan1 tan22tan − = ABC C c B b A a sinsinsin == Abccba cos2222 −+= Abcsin2 1=

3 4049/1/GE/23 [Turn Over] 1 A curve has an equation 225y x x= − + . (a) Express 225y x x= − + in the form of ( ) 2 a x b c−+ . Hence state the coordinates of the turning point. [3] 2 2 22 2 22 2 2 25 25 2 1125 2 4 4 112 2 5 2 4 4 1 392 48 y x x xx xx xx x = − + = − +     = − + − +            = − + − +        = − + 1 39Turning point: , 48   M1: Factor out the 2 A1 B1: FT from wrong completed square form (b) The line 27yx=+ intersects the curve at points A and B. Find the distance AB. [3] ( ) ( ) ( )( ) ( ) ( ) 2 2 2 2 2 5 1 2 7 2 2 5 2 7 2 3 2 0 2 1 2 0 1 OR 22 11When , 2 7 622 When 2, 2 2 7 11 1Coordinates of intersection are ,6 or 2, 112 1Required distance 2 11 2 y x x yx x x x xx xx xx xy xy = − + −−− = + −−−  − + = + − − = + − =  =− = =− = − + = = = + = −  = + + − ( ) 2 556 units2= M1: Solve quadratic equation (FT) A1 A1: Accept 5.59

4 4049/1/GE/23 2 Express 32 32 3 10 1 3 x x x xx + + + + in partial fractions. [6] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 2 2 3 2 3 2 22 3 2 2 2 22 22 2 2 3 10 1 1By long division, 3 33 11 3 3 3 1 3 3 When 0 : 13 1 3 11 3 3 3 When 3: 9 3 1 3 79 7 9 171 3 3 39 x x x x x x x x x x x x x A B C x x x x x x x x x Ax x B x Cx x B B x x Ax x x Cx x C C C x x Ax x x x + + + + + =+++ + + + += = + ++ + +  + + = + + + + = = = + + = + + + + =− − + = − = = + + = + + + + ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 22 32 3 2 2 When 1: 171 1 1 1 3 1 3 1 39 4734 39 2 9 1 2 1 7 3 9 3 9 3 3 10 1 2 1 7 33 9 3 9 3 x A A A xx x x x x x x x x x x x x x = + + = + + + + = + + = ++ = + +++ + + + = + + +++ B1: With long division working M1: Correct form (FT) M3: Substitution or comparing coefficients correctly per unknown A1

5 4049/1/GE/23 [Turn Over] 3 A curve has equation 321 3y x x kx= + + , where k is a constant and 1k . Explain why the curve does not have a stationary point. [4] ( ) ( )( ) 32 2 2 2 2 1 3 d 2d Method 1: dTo find stationary point, 0 :d 20 Assume on the contrary that curve has at least one stationary point. There are real roots to 2 0. 2 4 1 0 4 4 0 1 How y x x kx y x x kx y x x x k x x k k k k = + + = + + = + + =  + + =  −  −  ( ) ( ) ( ) ( ) ( ) 22 2 ever, 1. Curve does not have a stationary point Method 2: 2 1 1 1 1 1 0 1 d 0 for all values of d Graph is strictly increasing for all values of Curve does not have a stationa k x x k x k x k k k y xx x   + + = + + − + + −  −      ry point M1: Find d d y x M1: Find discriminant of d 0d y x = A1: 1k A1: Conclusion M1: Completing the square M1: Establishing the inequality A1: Conclusion

6 4049/1/GE/23 4 The diagram shows a circle. The line PC is the tangent to the circle at P. A and B are points on the circle such that PAB is a straight line. Prove that (a) triangle BPC is similar to triangle CPA, [3] ( ) ( ) Common angle Angles in alternate segment By AA similarity test, is similar to . BPC CPA CBP ACP BPC CPA  =  =    B1 B1 B1 (b) 2PA PB PC= . [2] 2 :BPC BP PC BC CPA CP PA CA BP PC CP PA BP PA CP PC PA PB PC == =   =    = M1: Ratio of corresponding sides of similar triangles A1

7 4049/1/GE/23 [Turn Over] 5 The equation of a curve is ( ) 3 2 24 13y x k x k x=− − + − , where k is a constant. (a) Find the range of values of k for which y is always decreasing. [4] ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) 3 2 2 22 22 22 22 2 2 2 2 2 22 4 13 d 4 2 1d For to be strictly decreasing, d 0d 4 2 1 0 4 2 1 0 For 4 2 1 to be always positive, 40 2 1 4 4 0 1 4 0 1 2 0 1 2 1 2 0 y x k x k x y x k x kx y y x x k x k x k x k x k x k b ac kk kk kk k k k k =− − + − =− − + −  − − + −  + + +  + + + − + −  + −  + −  + + + −  ( )( ) ( )( ) 3 1 1 0 3 1 1 0 1 OR 13 kk kk kk + − +  + −   −  M1: Differentiation and set d 0d y x  B1: 2 40b ac− M1: Solve quadratic inequality A1 (b) Given that y has three distinct roots, find the range of values of k. [2] ( ) ( ) ( ) ( )( ) ( ) ( )( ) 22 22 2 2 2 2 2 to have three distinct roots, has two turning points. d 0 has two real roots.d 4 2 1 0 has two real roots. 4 2 1 0 has two real roots. 40 2 1 4 4 0 1 4 0 3 1 1 0 y y y x x k x k x k x k b ac kk kk kk = − − + − =  + + + =  −  + −  + −  + −  − 1 13 k M1: 2 40b ac− A1

8 4049/1/GE/23 6 A curve is such that 2 2 d 3sin 4cos 2d y xxx =− . The curve passes through ( )0,1A and ( )π,3B . Find the equation of the curve. [7] ( ) ( ) ( ) 2 2 d 3sin 4cos 2d d 3sin 4cos 2 dd 4sin 23cos 2 3cos 2sin 2 3cos 2sin 2 d 2cos 23sin 2 3sin cos 2 y xxx y x x xx xxc x x c y x x c x xx cx d x x cx d =− =− =− − + =− − + = − − + −=− − + + =− + + +   ( ) When 0, 1: 1 3sin 0 cos 0 0 11 0 3sin cos 2 When π, 3: 3 3sin π cos 2π π 31 π 2 π 23sin cos 2 π xy cd d d y x x cx xy c c c y x x x == =− + + + =+ = =− + + == =− + + =+ =  =− + + M1: Integration M1: Integration A2: Minus one mark per mistake M2: Substitute the two conditions A1

9 4049/1/GE/23 [Turn Over] 7 For the curve 22yx= , the tangent at point P where xa= , intersect the y-axis at A. The normal to the curve at point P intersects the y-axis at B. Given that 0a , show that the area of triangle ABP is ( ) 216 1 8 aa + . [7] ( ) ( ) 2 2 2 2 2 ,2 2 d 4d dWhen , 4 d Gradient of tangent at 4 1Gradient of normal at 4 Finding equation of tangent at : 4 When , 2 : 24 2 Equation of tangent at : 4 2 P a a yx y xx yx a a x Pa P a P y ax c x a y a a a a c ca P y ax = = = == =  =− =+ == =+  =−  = − ( ) ( ) ( ) ( ) ( )