SJI 2025 AMATH PRELIM P1 MS

ST JOSEPH’S INSTITUTION PRELIMINARY EXAMINATION 2025 (YEAR 4) ADDITIONAL MATHEMATICS 4049/01 Qn Answer 1 18 72 3x x   18 3 72x x    18 3 72x     72 18 3 x        18 372 18 3 18 3 x      = 1296 216 18 3   = 36 6 6 15  12, a 6 b  2 2 5 d2x x x      = 2 2 255 d 4x x x       = 3 2553 4 x x c x   3 2 3 6x xy x   ----- (1) 2 2y x  1 12y x  ----- (2) Sub (2) into (1); 2 13 1 62x x x x       2 2 3 3 6 02x x x x     2 22 3 4 12 0x x x    (5 6)( 2) 0x x   6 5x or 2x

4 2 2 92 6 9 2 3 2x x x x          2 2 2 3 3 92 3 2 2 2x x                   2 3 92 2 4x           2 3 92 2 2x            2 3 92 2 2x            Since 2 3 02x     2 32 0 2x      and 2 3 9 92 2 2 2x                   for all real values of x. Therefore 29 6 2x x   is negative for all real values of x. 5(a) Let x = Angle QP A angle APB = x (AP bisects angle QPB) Angle ABP = x (alt. seg thm) Since angle APB = angle ABP = x Triangle ABP is isosceles 5(b)(i) Angle ACP = = x ( in the same seg) Angle ACP = angle APD = x Angle CAP = angle P AD (common  ) Triangle ACP and Triangle APD (AA Similarity Test)

5(b)(ii)   2 AD AP AP AC AP AC AD AP AB    ∵ 2AB AC AD  6 4 3 2 2 2 9 7 3 ( 1)( 5) x x x x x x       Long division to get the quotient and remainder 4 3 2 2 2 2 9 7 3 ( 1)( 5) 2 82 1 ( 1)( 5) x x x x x x xx x x            2 2 2 8 ( 1)( 5) 1 5 x A Bx C x x x x       22 8 ( 5) ( )( 1 )x A x Bx C x      Let 1x 22( 1) 8 (( 1) 5)A     6 6 A 1A Let 0x  8 1(5) (1)C  3C Let 2x   22(2) 8 1(2 5) 2 3 (2 1)B      2 3 1B  1B 2 2 2 82 1 ( 1)( 5) 1 32 1 ( 1) ( 5) xx x x xx x x         

7(a)     23 22 10 1 2x ax bx x x c      Comparing constant, 10c Comparing x2, 10 4a   – 14 Comparing x, 20 2b   22 7(b) 3 2 1 1 12 14 22 10 02 2 2 k                    9 4k 8(a)   234750 3 x x y     2 34750 3x y x     3 2 4750 3 x y x     2 750 4 3 xy x  8(b) 2 2 750 44 2 3 xS x x x         2 2 2 1500 84 3 x xx x      24 1500 3 x x 

8(c) 24 1500 3S x x  2 d 8 1500 d 3 S xx x   2 8 1500 03 x x   2 8 1500 3 x x  38 4500x  3 4500 8x  (=5.6362) Therefore, 2 3 3 4 4500 1500 3 8 4500 8 S          399.201 = 399 cm2 9(a) 3a Sub 2 ,03      into 6cosy a bx  20 3 6cos 3b       2 1cos 3 2b     2 2 4 (rej),3 3 3b       2b

9(b)(i) 21 xy   3 siny x   9(b)(ii)  2 3 sinx x   2 1 3sinx x     23sin 1 xx    No. of solution(s) = 1 10(a) 8 4 8 4 12 6k k     4 4 12 6k k    ( 2)( 6) 16k k   2 4 12 16 0k k    2 4 4 0k k   2( 2) 0k  2k 10(b) Let the coordinates of E be ( , )x yE E   82 , 2, 42 2 yx EE       ( , ) ( 6,0)x yE E   2 6 2 212 96 8 0 82 a a   y x 3 – 3 0

 6 16 48 2 96a a a       24 48a 2a  D (4, 2) Or 4 8 4 2 2 2 6 a a     4 12 2 a a    4 2 2a a   2a (4, 2)D  11(a) cos 2cot 1 sin (cos sin )       cos 2RHS= sin (cos sin )     2 2cos sin= sin (cos sin )        (cos sin )(cos sin )= sin (cos sin )           (cos sin )= sin     cot 1  RHS (proven) 11(b) cos4 1 sin2 (cos2 sin2 ) 2      1cot 2 1 2  

1cot 2 2  tan2 2 180 180    360 2 360    Basic angle = 63.4349 2 360 63.4349 or 180 63.4349     or 63.4349 or 180 63.4349 148.3 , 58.3 ,31.7 ,121.7       12(a) (i)   (40)90 100 4 k   ( 40)0.9 4 k (40)lg0.9 lg4 k lg 0.9 40 lg 4k  0.00190003866k 0.00190k (3sf) 12(a) (ii)   0.00190003866( 263)100 4M  50.02 Therefore, % remaining is 50.0%. 12(b) 5 25log log ( 2) 1x x   5 5 5 log ( 2)log 1 log 25 xx   5 5 log ( 2)log 1 2 xx   2 5 5log log ( 2) 2x x   2( 2) 25x x   3 2 2 25 0x x  

13(a)   1 2Area 3 sin2 3x x  = 23 3 4 x d 3 3 d 2 A xx  d d d d d d A A x t x t  =  3 3 4 0.52  = 23 3 cm /s 13(b) kF r  Sub 0.5F , 9r , 0.5 9 k 1.5k 2 1.5r F       2 3 2 1.5d d r F F , sub 0.5F     2 3 2 1.5d d 0.5 r F  = –36 units per unit increase in F

14(a) Sub 2 2y x  into 2y x 2 2 2 x x    24 2 4 x x  2 2 0x x     2 1 0x x   2x or 1x (rej)     1 2 d 1 2 2 1d 2 y xx       =   1 22 x   Sub 2x ,   1 2 d 2 2d y x    = 1 2 2m Let R = (a, 0) 4 02 2 a    42 2a   4a  4,0R  14(b) y-coord of P = 4 Distance OP = 2 24 2 20  Distance PR =   2 22 4 2 20   Since OP = PR = 20 , therefore OPR is an isosceles triangle.