2024 ZHSS AMATH P1 PRELIM MS

ZHONGHUA SECONDARY SCHOOL PRELIMINARY EXAMINATION 2024 SECONDARY 4 EXPRESS / 5 NORMAL ACADEMIC Candidate Name Class Register Number Solution ADDITIONAL MATHEMATICS 4049/01 Paper 1 28 August 2024 2 hours 15 minutes Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your index number and name on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a pencil for any diagrams or graphs. Do not use staples, paper clips, glue, or correction fluid. Answer all the questions. Omission of essential working will result in loss of marks. The use of an approved scientific calculator is expected, where appropriate. If the degree of accuracy is not specified in the question, and if the answer is not exact, give the answer to three significant figures. Give answers in degrees to one decimal place. For π, use either your calculator value or 3.142, unless the question requires the answer in terms of π. At the end of the presentation, 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. Setter: Ms Lee Sock Kee Vetter: Mr Francis Tan and Mr Lionel Ang This question paper consists of 22 printed pages (including this cover page). [Turn over 90

2 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation 02 =++ cbxax a acbbx 2 42 −−= Binomial expansion nrrnnnnn bbar nbanbanaba ++     ++     +     +=+ −−− 221 21)( , where n is a positive integer and ! )1()1( )!(! ! r rnnn rnr n r n +−−=−=       2. TRIGONOMETRY Identities 1cossin 22 =+ AA AA 22 tan1sec += 22cosec 1 cotAA=+ 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 −+= Abcsin2 1=

3 [Turn over 1. (a) Solve the simultaneous equations 2 4 0xy− + = [3] 22 24x y x+ = + 24xy=− (1) 22 24x y x+ = + (2) Sub (1) into (2), 22 22 2 2 2 (2 4) 2 4 4 16 16 4 8 4 5 20 20 0 4 4 0 ( 2) 0 2 y y x y y y y yy yy y y − + = + − + + = − + − + = − + = −= = Sub y = 2 into (1), 2(2) 4 0 x x =− = 0, 2xy = = (b) Explain the geometrical meaning of your answer in (a). [1] The line 2 4 0xy− + = is a tangent to the circle at (0, 2).

4 2. (a) Express 23 8 2y x x= − − in the form 2()y a x b c= + + and hence state the maximum value of y. [3] 2 2 22 2 2 2 8 3 2( 4 ) 3 2[( 2) 2 ] 3 2( 2) 8 3 2( 2) 11 y x x y x x yx yx yx =− − + =− + + =− + − + =− + + + =− + + Maximum value of y = 11 (b) Show that there are no values of p for which the curve ( ) ( ) 23 2 1y p x px p= − + + + is always positive. [3] Always positive, 30p− and 2 40b ac− 3p and 2(2 ) 4( 3)( 1) 0p p p− − +  22 22 4 4( 2 3) 0 4 4 8 12 0 8 12 0 3 2 p p p p p p p p − − −  − + +  + − For the curve y to be always positive, 3p and 3 2p− . There are no values of p for which y is always positive.

5 [Turn over (c) A quadratic equation is given by hx2 – 2kx + 6k – 9h = 0, where h and k are constants and h ≠ 0. (i) Show that the equation has real roots for all values of h and k. [3] 2 2 22 22 2 4 ( 2 ) 4( )(6 9 ) 4 24 36 4( 6 9 ) 4( 3 ) 0 b ac k h k h k hk h k hk h kh − = − − − = − + = − + = −  for all values of h and k. Therefore, the roots are real (shown). (ii) In the case where the equation has two real and equal roots, express h in terms of k. [2] 2 2 2 40 4( 3 ) 0 ( 3 ) 0 3 3 b ac kh kh kh kh −= −= −= = =

6 3. Given that cos A 2 11 = where 180 360A , find, without the use of a calculator, the value of (a) tan A , [2] A lies in the 4th quadrant. 2 2 2( ) ( 11) ( 2) 3 3 ( ) opp opp or rej =− =− 3 3 2tan 22 A or=− − (b) ( )sin 90A− , [2] sin( 90 ) sin cos90 cos sin 90 sin (0) cos (1) cos 2 11 A AA AA A − = −  =− =− =− OR sin( 90 ) sin[ (90 )] sin(90 ) cos 2 11 A A A A − = − − = − − =− =− A -3

7 [Turn over (c) 1 sec 2A . [2] 1 sec 2A 2 2 cos2 2cos 1 221 11 4 111 7 11 A A = =− =−  =− =− 4. (a) Factorise 3 327 8 yx − completely. [2] 3 327 8 yx − 3 3 2 2 (3 ) 2 339 2 2 4 yx y xy yxx =−  = − + +   OR 3 327 8 yx − ( ) ( )( ) 33 33 22 1 2168 1 (6 )8 1 6 36 68 xy xy x y x xy y =− =−  = − + +

8 (b) Express 32 2 8 7 4 3 (2 )(2 1) x x x x x x − + − −− in partial fractions. [6] 22(2 )(2 1) (2 1)x x x x x− − = − 3 2 3 2 2 4 4 8 7 4 3x x x x x x− + − + − 32(8 8 2 )x x x− − + 2 23xx+− 3 2 2 22 8 7 4 3 2 3 2(2 )(2 1) (2 1) x x x x x x x x x x − + − + − =+− − − 2 22 23 (2 1) 2 1 (2 1) x x A B C x x x x x +− = + +− − − 22 2 3 (2 1) (2 1)x x A x Bx x Cx+ − = − + − + When 1 2x= , 11 1342 c+ − = 71 42 7 2 C C −= =− When 0x= , 3A=− When 1x= , 71 2 3 3 2 130 2 13 2 B B B + − =− + − =− + = 2 3 13 72 2(2 1) 2(2 1)x x x− + − −−

9 [Turn over 5. A curve is such that 2d2 d3 y x ax x −= , where a is a constant. (a) Given that the curve has a turning point at ( )3,7 , show that the value of a is 2 3 . [1] At turning point, d 0d y x = 22 03 x ax− = 220x ax−= When x = 3, 22(3) (3) 0 69 a a −= = 2 3a= (shown) (b) Find the range of values of x for which y decreases as x increases. [3] y decreases as x increases, d 0d y x  222 3 03 xx−  2220 3xx− 26 2 0xx− 2 (3 ) 0xx − 03x or x 0 3

10 (c) Find the equation of the curve. [4] 222 3 3 xx y dx − = 2 23 23 12 233 1 2 2 3 2 3(3) 2 3 27 y x x dx xxyc xxyc =−   = − +  = − +  When x = 3, y = 7, 233 2(3)7 3 27 7 3 2 6 c c c = − + =−+ = 23 2 63 27 xxy= − +