S4 A-Math SJI P1 2024 (with Answer) EXAM555

1 [Turn over Y2024 AM 4049 PRELIM PAPER 1 SOLUTIONS 1 It is given that 32f3 2 5 2 1xx x x   . (a) Solve the equation f0x. [5] 32f 3 25 21xx x x            32 f1 1 3 1 2 5 12 1 0         Therefore 1x is a factor.  32 232 5 2 1 1xx x x a x b x c       By comparing coefficients, 1a 21c 3ba  13 4 b b           32 2 3 25 21 0 14 2 1 0 173 0 xx x xx x xx x             1x 7x 3x

2 [Turn over (b) Hence, solve the equation 3221 25 3 1 0yy y    where 0y. [3] 3221 25 3 1 0yy y    2313 2 5 2 1 0yy y    32 3 11 1132 5 2 1 0y yy y                        Let 1xy, 1 1y 17y 1 3y 1y 1 7y 1 3y

3 [Turn over 2 (a) Given that  2sin cos 2 cos    , show that 1cos 2. [2]  2sin cos 2 cos    221c o s 2 c o s c o s     2cos 1 1cos 2 (b) Hence solve the equation  2sin 2 cos 2 2 cos 2xx x  in the interval 0 x , giving your answers in exact form. [3]  2sin 2 cos 2 2 cos 2xx x  1cos 22x Basic angle 3  02 2x  52,33x  5,66x 

4 [Turn over 3 (a) Find the amplitude and period of 3sinyx  . [2] amplitude = 3 period 2 2   (b) On the grid below, sketch the graph of 3sinyx  for 03x . [4] 3sinyx 

5 [Turn over 4 (a) Given that 2 92 7log logxu , show that 3ux. [3] 2 92 7log logxu 2 33 33 log log log 9 log 27 xu 2 33log log 23 xu 2 33 3log log2 xu 3 33log logxu Therefore, 3ux (b) Hence, solve the equation   22 27 9 2 1log 6 8 log log 27xx x   . [5]   22 27 9 2 1log 6 8 log log 27xx x    2 27 27 27log 6 8 log log 2xx u    23 27 27 27log 6 8 log log 2xx x   2 27 27 3 68log log 2xx x   2 3 68 2xx x  2368 2xx x  32268 0xxx    2 340xx x   41 0xx x   0x (NA), 4x, 1x (NA)

6 [Turn over 5 A circle C has equation 22 24 0xy y   . Find the equations of the two circles centred at point 8, 5 and touching C. [8] 22 24 0xy y        22 01 1 4 0xy          22 01 5xy    Centre = 0,1 Radius = 5 Distance between two centres    22 80 51    80 45 Radius of Circle 1 45 5 35   Equation:    222 85 3 5xy        22 85 4 5xy    Radius of Circle 2 45 5 55   Equation:    222 85 5 5xy        22 85 1 2 5xy   

7 [Turn over 6 (a) Express    2 2 51 1 41 xx xx    in partial fractions. [5]        2 22 51 1 4141 1 xx A B C xxxx x               2251 1 1 4 1 4xx A x B xx C x        When 1x, 2 51 1 11 3 36 2 C C C       When 4x, 22 54 1 14 3 93 6 4 A A A        When 0x, 2 04 1 4 1 2 4 44 1 B B B             2 22 51 1 4 1 2 4141 1 xx xxxx x       

8 [Turn over (b) Hence, find    2 2 10 22 d 41 xx x xx    . [4]    2 2 10 22 d 41 xx x xx       2 2 51 12 d 41 xx x xx        2 41 22 d41 1 xxx x          24122 1 d41 xxxx             1 2124 l n4 l n1 1 xxx c                 48ln 4 2ln 1 1xx c x     

9 [Turn over 7 The following diagram shows the graph of the function 32f1 0 3 0 2 4 0 1 5 0xxx x    , for 57x   (a) State whether the function is increasing or decreasing at 3x. Give a reason for your answer. [2] Increasing. Gradient of the tangent line is positive. y x 3

10 [Turn over (b) Find the coordinates of the turning points. [5] 32f1 0 3 0 2 4 0 1 5 0xxx x    2f3 0 6 0 2 4 0xxx    f0x 230 60 240 0xx   2 28 0xx   24 0xx   2x 4x When 2x 32 f1 0 2 3 0 2 2 4 0 2 1 5 0 430 x         Max point 2, 430 When 4x 32 f1 0 4 3 0 4 2 4 0 4 1 5 0 650 x      Min point 4, 650