Woodgrove Secondary 4052/02 MS 2025

Name Index Number Class N-LEVEL PRELIMINARY EXAMINATIONS 2025 LEVEL & STREAM : SECONDARY 4 NORMAL ACADEMIC SUBJECT (CODE) : ADDITIONAL MATHEMATICS (4051) PAPER : 02 DATE (DAY) : 01 AUGUST 2025 (FRIDAY) DURATION : 1 HOUR 45 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 three significant figures, or 1 decimal place in the case of angles in degree, 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 answer. 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 70. 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 13 printed pages including this cover page. Setter: Mdm Tan Hwee Lin 70

2 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ,02  cbxax a acbbx 2 42  2. TRIGONOMETRY Identities sin² A + cos² A = 1 sec² A = 1 + tan² A cosec² A = 1 + cot² A sin (A  B) = sin A cos B  cos A sin B BABABA sinsincoscos)cos(  BA BABA tantan1 tantan)tan(   sin 2A = 2sin A cos A cos 2A = cos2 A – sin2 A = 2cos2 A – 1 = 1 – 2sin2 A tan 2A = A A 2tan1 tan2  Formulae for ABC C c B b A a sinsinsin  a² = b² + c²  2bc cos A Abc sin2 1

3 1 Solve the simultaneous equations. [4] Subst. (2) into (1), When , and when 2 (a) State the amplitude and period of the graph of 3sin 2 1y x  . 12sin3 Given that  xy , amplitude  3 period=360°÷2 =180° or π [2] (b) On the axes below, sketch the graph of 3sin 2 1y x  for 0 2 πx  . [3] y x2 4x 5 x y1 0 y x2 4x 5 1( ) x y1 0 Þ y x1  2( ) * x1 x2 4x 5 x 2 3x 4 0 x 4( ) x1( ) 0 x4, x1 x4, y5 x1, y 0 M1 M1 A2 B1 B1 Correct shape – B1 Correct max and min value of y - B1 Correct Period- B1

4 3 Show that the quadratic expression 2 2 1x px p   is always positive for all real values of x. Let 2 2 1y x px p    ( ) ( )( ) ( ) 2 2 2 2 2 discriminant, 4 1 1 4 4 3 4 0 D p p p p p           Since discriminant  0 and coefficient of 2 0,x  2 2 1x px p   is always positive for all real values of x. [3] 4 A curve is defined by equation 23 1y x  . Find the rate at which y is changing when x = 2, given that x is increasing at a rate of 0.5 units per second. 2 6(2) 0.5 6 cm/s 3 1 dy 6dt dy dy dx dt dx dt y x x         [3] M1 M1 A1 M1 M1 A1

5 5 (a) Differentiate ( ) 725 x with respect to x. ( ) ( ) ( ) 62 62 7 5 2 5 d d 14 x x x y x x     [2] (b) Hence, find ( ) 625 x dxx  . ( ) ( ) ( ) ( ) ( ) ( ) ( ) 7 62 2 7 62 2 6 72 2 72 5 5 1 5 514 15 5 14 1 = 5 14 d 14d d d d d x x x x x dx x x C xx xx x x                 [3] 6 The area and length of a rectangle are 2 cm2 and 32 32   cm respectively. Express, in the form 3a b , (a) the breadth, Length AreaBreadth  ( ) ( )( ) ( )( ) ( ) ( ) 2 2 32 2 3 2 3 2 2 3 2 2 3 2 3 2 3 2 3 2 4 4 3 3 2 3 2 7 4 3                   ( )breadth 14 8 3 cm   [3] M1 M1 A1 M1 M1 A1 M1 A1

6 (b) the perimeter of the rectangle. breadth) length (2Perimeter  ( ) ( ) ( ) ( ) ( ) 2 2 32 14 8 32 3 2 3 2 32 14 8 32 3 2 3 2 4 3 32 14 8 34 3 2 7 4 3 14 8 3 2 21 4 3 42 8 3 cm                                                     [3] 7 (a) Factorise 38 125x  . ( )( ) 3 28 125 2 5 4 10 25x x x x     B1 [1] ( b ) Hence, (i) Express 1125 as a product of two integers, 3 3 3 8 125 1125 8 1000 125 5 x x x x      ( )( ) 21125 2 5 5 4 5 10 5 25 = 15 75         [2] M1 M1 A1 M1 A1

7 (ii) Solve ( ) 38 125 39 2 5x x   ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) ( )( ) 3 2 2 2 2 8 125 39 2 5 2 5 4 10 25 39 2 5 2 5 4 10 25 39 2 5 0 2 5 4 10 25 39 0 2 5 4 10 14 0 x x x x x x x x x x x x x x x x                        ( )( ) ( )( )( ) 2 2 5 2 5 7 0 2 5 2 7 1 0 5 7 , 1 2 2 x x x x x x x or           [4] M1 M1 M1 A1

8 8 The equation of a curve is ( )( ) 21 2 5 31y x x x    . (a) Find the coordinates of the stationary points of the curve. ( )( ) 21 2 5 31y x x x    ( ) ( )( ) 2 2 2 2 2 2 d 2 5 31 1 4 5d 2 5 31 4 5 4 5 2 4 5 4 5 31 5 6 6 36 y x x x xx x x x x x x x x x x x x                        For stationary points, ( ) ( )( ) 2 2 2 d 0d 6 6 36 0 6 6 0 6 0 3 2 0 3 or 2 y x x x x x x x x x x x                When x = 3, ( ) ( ) ( ) ( ) 2 3 1 2 3 5 3 31 4 18 15 31 112 y            When x = 2, ( ) ( ) ( ) ( ) 2 2 1 2 2 5 2 31 8 10 31 13 y               ( ) ( )the stationary points are 3, 112 and 2,13 .   [5] (b) Determine the nature of the stationary points. 2 2 d 12 6d y xx   When x = 3, ( ) 2 2 d 12 3 6d 30 0 y x     ( )3, 112 is a minimum point.  When x = 2, ( ) 2 2 d 12 2 6d 30 0 y x      [3] M1 M1 M1 M1 A1 ( )2,13 is a maximum point.  M1 A1 A1

9 9 The diagram shows part of the curve 2 1,y x  the line y = 2x + 4. The curve and the line intersect at the points P and Q. Region S is bounded by the curve 2 1,y x  the line y = 2x + 4, the x-axis and the y-axis. (a) Find the coordinates of P and Q. ( ) 2 4 1y x   ( ) 2 1 2y x   Sub (1) into (2) : 22 4 1x x   ( )( ) 2 2 3 0 3 1 0 3 or 1 x x x x x x         ( ) ( ) ( ) ( ) ( ) ( ) . Substituting 1 into 1 , 2 1 4 2 Substituting 3 into 1 , 2 3 4 10 the coordinates of and are 1, 2 and 3, 10 respectively x y x y P Q            [3] M1 M1 A1

10 (b) Find the area of the shaded region S. Let S  A B. ( )( ) 2 From the diagram, 1Area of 1 2 2 1 unit A  ( ) 0 2 1 03 1 2 Area of 1 d 3 1 13 11 units3 B x x x x                       2 1area of 1 1 3 12 units3 S    [6] For 2 4, when 0, 0 2 4 2, y x y x x       M1 M1 M1 M1 M1 A1