MGS 2023 Sec 4 Prelim AM P1 Solutions

Class Index Number Name : ___Mark Scheme_________________ This question paper consists of 17 printed pages and 3 blank pages. METHODIST GIRLS’ SCHOOL Founded in 1887 PRELIMINARY EXAMINATION 2023 Secondary 4 Friday ADDITIONAL MATHEMATICS 4049/01 18 August 2023 Paper 1 2 h 15 min Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your class, index number and name in the spaces at the top of this page. Write in dark blue or black pen You may use a HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions. Give non-exact numerical answers correct to 3 significant figure, 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 for this paper is 90. 90

Page 2 of 20 Methodist Girls’ School Additional Mathematics Paper 1 Sec 4 Preliminary Examination 2023 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the quadratic equation , Binomial Expansion , where n is a positive integer and . 2. TRIGONOMETRY Identities 22sin cos 1AA+= 22sec 1 tanAA=+ 22cosec 1 cotAA=+ cos2A = cos2 A- sin2 A = 2cos2 A-1= 1- 2sin2 A Formulae for ABC sin sin sin a b c A B C== a2 = b2 + c2 − 2bc cos A  = 2 1 bc sin A 02 =++ cbxax a acbbx 2 42 −−= ( ) nrrnnnnn bbar nbanbanaba ++     ++     +     +=+ −−− 221 21 ( ) ! ( 1)...( 1) ! ! ! n n n n n r r r n r r  − − +== − BABABA sincoscossin)sin( = BABABA sinsincoscos)cos( = BA BABA tantan1 tantan)tan(  = AAA cossin22sin = A AA 2tan1 tan22tan −=

Page 3 of 20 Methodist Girls’ School Additional Mathematics Paper 1 Sec 4 Preliminary Examination 2023 1 A calculator must not be used for this question. In the diagram, AB is parallel to DC, and AC meets BD at X. Given that AB = ( )1 3 3+ cm, CX = ( )53− cm and CD = ( )6 2 3+ cm, find the exact length of AX. [5] A B D C X 1 3 3+ 6 2 3+ 53− ( ) 1 3 3 5 3 6 2 3 1 3 3 53 6 2 3 5 3 15 3 9 6 2 3 14 3 4 6 2 3 7 3 2 3 3 3 3 3 3 21 3 21 6 2 3 93 23 3 27 23 9 36 6 2 AX AX or += −+ +=  − + − + −= + −= + −−= +− − − += − −=−

Page 4 of 20 Methodist Girls’ School Additional Mathematics Paper 1 Sec 4 Preliminary Examination 2023 2 A curve has the equation 12 3 .x xy e − −= (i) Show that the curve is an increasing function of x for 5 .2x [5] (ii) Given that y is decreasing at a constant rate of 2 units per second, find the rate of change of x when x = 0. [2] ( ) 1 2 1 2 212 12 12 ( 1) (3 ) ( 2) 1 6 2 = 52 = xx x x x dy e x e dx e x e x e −− − − − − − − −= − + − − For 5 2x , 12 0xe −  5 2 2 5 0 5 2 0 50 for 2 5Hence, curve is an increasing function for 2 x x x dy xdx x  − −     12 522 52 2 1.09(3sf)5 x dy dy dx dt dx dt x dx e dt dx e dt dx e dt − = −− =  − =  −= =−

Page 5 of 20 Methodist Girls’ School Additional Mathematics Paper 1 Sec 4 Preliminary Examination 2023 3 The function 32f ( ) 2 6x x ax bx= + + + , where a and b are constants, is exactly divisible by x + 2. When f (x) is divided by x – 2, the remainder is –12. (i) Find the value of a and of b. [4] (ii) Factorise f (x) completely and hence solve f (x) = 0. [3] 32 32 ( 2) 0 2( 2) ( 2) ( 2) 6 0 4 2 10 2 5 --------- (1) (2) 12 2(2) (2) (2) 6 12 4 2 34 2 17 ------ (2) (2) into (1), 2 2 5 17 4 12 3 11 f ab ab ba f ab ab ab aa a a b −= − − + − + − + = −= =− =− + + + =− + =− + =− + − =− =− =− =− 3 2 2 2 2 2 3 11 6 ( 2)(2 3) compare coeff of , 43 7 ( 2)(2 7 3) 0 ( 2)(2 1)( 3) 0 12, or 32 x x x x x px x p p x x x x x x x − − + = + + + + =− =− + − + = + − − = =−

Page 6 of 20 Methodist Girls’ School Additional Mathematics Paper 1 Sec 4 Preliminary Examination 2023 4 It is given that 1 3sin 4yx=− , for 0 180x . (i) State the amplitude and period of y. [2] (ii) Sketch the graph of 1 3sin 4yx=− . [3] (iii) Hence, state the number of solutions for 3sin 4 2x= where 0 180x . [1] Amplitude = 3 Period = 90o 4

Page 7 of 20 Methodist Girls’ School Additional Mathematics Paper 1 Sec 4 Preliminary Examination 2023 5 (a) The equation of a curve is 1 ln( 3)3y px=+ , where p is a constant to be determined. The gradient of the tangent to the curve at 1 2x=− is parallel to 3y = x. Find the equation of the normal to the curve at 1 2x=− . [4] (b) (i) Express 22 4 5xx− + − in the form 2()a x p q++ . [2] (ii) Hence, determine, with explanation, if the graph of 22 4 5y x x=− + − will intersect the x–axis. [2] 11 33 1 1 333 2 1 32 3 32 2 11, ln 223 grad of normal = 3 11ln 2 332 313 ln 223 dy pdx px p p pp p p xy yx yx =  + =−+ =− + = = =− = − − =− +  =− − + ( ) ( ) 2 2 2 2 2 4 5 522 2 52 1 1 2 2 1 3 xx xx x x − + − =− − + =− − − + =− − − 2Since coefficient of <0, graph has maxim um turning point at (1, 3). Maximum value of is 3 which is less tha n 0. OR The maximum point is below the -ax is. Hence, graph does not intersect -axis. x y x x − − 11 33 1 1 333 2 1 32 3 32 2 11, ln 223 grad of normal = 3 11ln 2 332 313 ln 223 dy pdx px p p pp p p xy yx yx =  + =−+ =− + = = =− = − − =− +  =− − +

Page 8 of 20 Methodist Girls’ School Additional Mathematics Paper 1 Sec 4 Preliminary Examination 2023 6 (a) In the expansion of ( )2 n x+ , where n > 0, the coefficient of x2 is twice the coefficient of x. Find the value of n. [4] ( ) ( ) ( ) 21 21 1 2 2 2 2 2 2 221 ( 1) 2 2 22 42 2 8 90 ( 9) 0 0 or 9 () nn nn n n nn nn n n nn n n n nn nn nn NA −− −− − −     =       − = −= −= −= −= == (b) Find the value of the term that is independent of x in the expansion of 15 4 12 4x x  − . [3] ( ) 1 2 22 2 2 2 ... 12 n n n n nnx x x −−   + = + + +       ( ) 15 4 15 15 4 15 15 5 3 12 general term 15 12 4 15 12 4 15 12 4 15 5 0 3 15 1value = 2 291203 4 r r r r r r r rr xr x xxr xr r r − − − − −−  =−   =−   =−  −= =  − =− 

Page 9 of 20 Methodist Girls’ School Additional Mathematics Paper 1 Sec 4 Preliminary Examination 2023 7 It is given that log ( ) 2log (4 3) 1ppy px x= + − − , where p is a positive integer. (i) Write down the values of x for which y is defined. [1] (ii) Show that y can be written as 32log (16 24 9 )p x x x−+ . [3] (iii) Find the value of x for which 9 1 log x y p= . [3] 3 4x ( ) 2 2 2 32 log ( ) 2log (4 3) 1 log ( ) log (4 3) log (4 3)log log 16 24 9 log (16 24 9 ) pp p p p p p p y px x px x p px x p x x x x x x = + − − = + − − −= = − + = − + ( ) 32 9 32 32 32 32 2 1log (16 24 9 ) log loglog (16 24 9 ) 1 log 9 log (16 24 9 ) log 9 16 24 9 9 16 24 0 8 (2 3) 0 30 NA or 2 p x p p p pp x x x p px x x x x x x x x x x x xx xx xx − + = − + =  − + = − + = −= −= ==

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