2023 CGS Sec 4 AM Prelim P1 Solutions

Name _____Solutions_____________ ( ) Class 4 ______ ADDITIONAL MATHEMATICS 4049/01 Paper 1 24 August 2023 2 hours 15 minutes Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your name, class and index number 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 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. This question paper consists of 20 printed pages. Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Marks 3 4 4 3 4 7 8 6 11 11 9 8 8 4 Table of Penalties Qn. No. Presentation –1 Accuracy/ Units –1 Parent’s/ Guardian’s Signature For Examiner’s Use CRESCENT GIRLS’ SCHOOL SECONDARY FOUR PRELIMINARY EXAMINATION 90

2 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation 02 =++ cbxax , a acbbx 2 42 −−= 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) !( )! ! n n n n r n r r r n r r  − − +== − 2. TRIGONOMETRY Identities 1cossin 22 =+ AA AA 22 tan1sec += 22cosec 1 cotAA=+ BABABA sincoscossin)sin( = cos( ) cos cos sin sinA B A B A B= tan tantan( ) 1 tan tan ABAB AB = sin 2 2sin cosA A A= 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 Given that 2 2 3k =− , without using the calculator, express 23k k− in the form 23ab c − , where a, b and c are integers. [3] ( ) ( ) 2 2 3 223 3 2 2 3 2 2 3 2 2 2 3 6 2 3 3 83 42 6 2 3 3 2 3 55 26 2 17 3 5 26 17 = 2 355 k k k =− − = − − − + = − − − = − − − −= − 2 The straight line 20y kx=+ intersects the curve 23 2 21y kx=− at the points A and B whose x-coordinates are ‒3 and 4.5 respectively. Find the value of k. [4] ( )( ) 2 2 22 2 3( 20) 2 21 2 3 81 0 812 3 81 1.5 0 2 3 and 4.5 are solutions 3 4.5 0 1.5 13.5 0 by comparison 81 13.52 3 kx kx kx kx kx kx x x k xx xx k k + = − − − = − − = − − = − + − = − − = − =− =

4 3 Express 23 12 4xx− + − in the form 2()a x h k−+ , where a, h and k are integers. Hence state the coordinates of the turning point of the curve 23 12 4y x x=− + − . [4] 22 2 2 2 2 3 12 4 3( 4 ) 4 3( 4 2 2 ) 4 3( 2) 8 x x x x x x x − + − =− − − − − + − − =− − + Turning point (2, 8) 4 Integrate 2tan 2 x with respect to x. [3] 22tan 2 d = sec 2 1 d tan 2 ( is a constant)2 x x x x x x c c − = − + 

5 5 Express ( ) ( ) 2 2 6 5 5 1 2 xx xx −+ −+ in partial fractions. [4] ( )( ) ( ) ( )( ) 2 22 22 2 22 6 5 5 = 1212 6 5 5 = ( 2) ( ) 1 Sub 1, 63 2 Sub 0, 52 1 Sub 2, 28 4 6 5 5 2 4 1 = 1212 x x A Bx C xxxx x x A x Bx C x x A A x AC C x B B x x x xxxx − + + +−+−+ − + + + + − = = = = =− =− = = = − + −+ −+−+

6 6 ( ) ( ) pxpxx n ++−= 22 1f , where n and p are positive integers. (a) Show that ( )1+x is a factor of ( )xf for all values of p. [2] ( ) ( ) ( )( ) 22 f 1 1 1 1 1 ( 1)(1) 0 n pp pp − = − − + − + = − + + = Therefore (x + 1) is a factor (b) Given 4=p , (i) find the value of n for which ( )2−x is a factor, [2] ( ) ( )( ) 22 2 f 2 0 0 2 4 1 2 4 0 = 2 16 2 n n n = = − + + − = (ii) hence, solve ( ) 0f =x . [3] ( )( )( ) ( )( )( ) ( )( )( )( ) 4 2 2 2 5 4 1 2 by observation 2 substitute 1, 1 5 4 2( 1)(1 2) 1 1 2 2 0 1 2 1 2 0 1,1, 2, 2 x x x x x ax b b x a a x x x x x x x x x − + = + − + + =− = − + = − + − = + − + − = + − − + = =− −

7 7 For π,0  x f ( ) 3sinx nx= , where n is a positive integer, and g( ) 4cos 2 1xx=+ . (i) Given that 6 π satisfies the equation ),(g)(f xx = show that smallest value for .3=n [2] 3sin 4cos 2 166 sin 1 6 smallest 3 n n n      =+        = = (ii) State the amplitude of g(x). [1] Amplitude = 4 (iii) Sketch, on the axes below, the graphs of )(f xy= and ).(g xy= [4] (iv) State, in terms of π, the other roots of the equation )(g)(f xx = for π.0  x [1] f(x) g(x) 5,26  y x 5- 3- 0- ‒3- y = g(x) y = f(x)

8 8 A piece of wire, 100 cm in length, is bent to form the figure as shown. Given that angle angle 60ABC EFG= =  , angle angle 90CDE GHA= =  , AB = BC =EF = FG = x cm and CD = DE = GH =HA = y cm. (a) Show that the area of the figure, P cm2, is given by 23 1 2 (25 2 50) 6252P x x= + − + − + . [4] 4( ) 100 25 25 xy xy yx + =  += =− yyyCE 222 =+= P = 2 Area of ABC + 2 Area of CDE + Area of rectangle ACEG = 22sin 60 ( 2 )x y x y + + = 223 (25 ) 2 (25 )2 x x x x+ − + − = 2 2 23 625 50 25 2 22 x x x x x+ − + + − = 23 1 2 (25 2 50) 6252 xx + − + − + 60° y 60° H G F E D C B A y x x y y x x

9 (b) Find the value of x for which P has a stationary value. [2] d3 2 1 2 25 2 50d2 P xx = + − + − 0=dx dP 32 1 2 50 25 22 x + − = − 50 25 2 16.2 32 1 22 x −==  +−  (3 s.f.)

10 9 The diagram shows a kite ABCD with ADAB= and .CDCB= The diagonals intersect at M. The point A lies on the y-axis, the point B is (‒9, 2) and the equation of AC is 2x + y = 9. (i) State the coordinates of A. [1] (ii) Find the equation of BD. [2] A (0, 9) Gradient of AC = ‒2 Gradient of BD = 1 2 Equation BD : 12 ( ( 9))2 11 622 yx yx − = − − =+ M D C A B O (–9, 2) x y