SJI 2025 AMATH PRELIM P2 MS

ST JOSEPH’S INSTITUTION PRELIMINARY EXAMINATION 2025 (YEAR 4) ADDITIONAL MATHEMATICS 4049/02 Qn Solutions 1(a) 2 12 5(2 ) 3x x   Let 2xu 22 5 3 0u u   1(b) (2 1)( 3) 0u u   1 2u or 3u  2 0x  therefore 12 2 x   2 3x  lg 2 lg 3x  lg3 lg 2x = 1.6 (2 sf) 2(a)                3 3 2 2 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1x x x x x x x x                  =   2 2 22 4 4 1 4 1 4 4 1x x x x x         =   22 12 1x  2(b)    22 2 1 12 1 0x x   212 1 0x   or 2 1 0x     2Discriminant 0 4 12 1 = 48 0     No real roots 1 2x is the only real root.

ST JOSEPH’S INSTITUTION PRELIMINARY EXAMINATION 2025 (YEAR 4) ADDITIONAL MATHEMATICS 4049/02 3 (a) ( 3) 1y x x       1 2 1(1) 1 ( 3) 1 1 2 dy x x xdx           31 2 1 xx x     2( 1) 3 2 1 x x x     2 2 3 2 1 x x x     3 1 2 1 x x   3(b) ∫ 3𝑥 − 1 2√𝑥 + 1 3 0 𝑑𝑥 = [(𝑥 − 3)√𝑥 + 1]0 3 3 33 0 00 3 1 1 16 d = 6 ( 3) 1 d 2 1 2 1 2 1 x x x x x x x x                       3 33 0 00 9 1 d = 6 ( 3) 1 d 1 2 1 x x x x x x x                       3 1/2 0 1 12 = 6 (3 3) 3 1 (0 3) 0 1 1 2 x                           1/2 1/2 = 6 (3) 3 1 0 1        = 6 (3) 2 1    = 24

ST JOSEPH’S INSTITUTION PRELIMINARY EXAMINATION 2025 (YEAR 4) ADDITIONAL MATHEMATICS 4049/02 4(a) Let E be a point below C (and to the left of D) In , cos 80 CECDE   80cosCE  Let G be a point below D (and to the right of A) In , sin 185 DGADG   185sinDG  h CE EF  185sin +80cos (shown)h   4(b) 185 sin +80 cosh   185sin +80cos sin( )R     sin 80 cos 185 R R     2 280 185 40625 25 65 R    1 80tan 185 23.385          25 65 sin( 23.4 )h     4(c) To clear the kitchen door, max h is 180cm, When h = 180cm 25 65 sin( 23.385 ) 180    180sin( 23.385 ) 25 65     36sin( 23.385 ) 5 65     Basic angle =63.259 23.385 63.259      39.9  

ST JOSEPH’S INSTITUTION PRELIMINARY EXAMINATION 2025 (YEAR 4) ADDITIONAL MATHEMATICS 4049/02 5(a) 22 5y x  2 22 5 3x x  2 5 0x   ( 5)( 5) 0x x   5 5 x   5(b) 22 5y x  ---- (1) 3y kx  3y kx  ---- (2) 22 5 3x kx   22 2 0x kx   2 4 0b ac  2( ) 4(2)(2) 0k   2 16k  4k   5(c) 22 4 2 0x x   2 2 1 0x x   2( 1) 0x  1x  4(1) 3y  = 7 (1,7)P 6(a) 6 2ax x     Term   6 6 2 r r r ax x           6 6 2 r r r x xx   Powers of x 6 2 2(3 )r r   Since the power/exponent can be expressed as a multiple of 2, therefore all the expansion only contains even powers of x.

ST JOSEPH’S INSTITUTION PRELIMINARY EXAMINATION 2025 (YEAR 4) ADDITIONAL MATHEMATICS 4049/02 6(b) 3 1 x a      2 2 3 31 ... x x a a   6(c) 6 2ax x     Term in 4 1 x 6 2 4rx x  6 2 4r   5r  4 1 x term =   56 45 2 1 192ax a x x                 3 1 x a     Term in 2x = 2 2 3 ( )xa from (b) 2 3192 1728a a  3 3 1728 192a  3 27a  3a  6(d) Term independent of x 2 2 3 3 11 ... 1 3 1(1)(1) ... 3 x x a a x x x                     There is no term independent of x.

ST JOSEPH’S INSTITUTION PRELIMINARY EXAMINATION 2025 (YEAR 4) ADDITIONAL MATHEMATICS 4049/02 7(a) d 2 5 10 20 30 I 875 141 35 4.22 3.89 lg d 0.30 0.70 1.00 1.30 1.48 lg I 2.94 2.15 1.54 0.63 0.59 7(b) The incorrect reading of I is 4.22. From the graph, the correct lg I reading is 0.95. 0.95 lg I 0.9510I   8.91 3 sfI  7(c) nI Ad lg lg lgI n d A  From the graph lg 3.6A 3.610A 3981.071

ST JOSEPH’S INSTITUTION PRELIMINARY EXAMINATION 2025 (YEAR 4) ADDITIONAL MATHEMATICS 4049/02  3980 3sf 0.5 3.6Gradient = 1.52 0     2.0394 2.04 3sf   2.04n 7(c) From the table, 1 2 35 141 I I  = 0.248 (to 3 sf) When the distance doubled, the intensity decrease to 0.248 (one-quarter) of its previous value 8(a) 3 25 e tv   3 25 e 0 t  3 2e 5 t  3 2ln e ln 5t  3 2 ln 5t  0.695t 

ST JOSEPH’S INSTITUTION PRELIMINARY EXAMINATION 2025 (YEAR 4) ADDITIONAL MATHEMATICS 4049/02 8(b) Distance in the first second =     1 0.69528 3 2 3 2 0.69528 0 5 e d 5 e dt t t t     1 0.695283 2 3 2 0.69528 0 e e5 5 2 2 t t t t                         3 2(1) 3 2(0.69528)e e5(1) 5(0.69528)2 2                        3 2(0.69528) 3 2(0)e e5(0.69528) 5(0) 2 2                        = 0.3827356 + 4.06636 = 4.449 = 4.45 m 8(c) 3 25 e tv   as approaches t  , 3 2e t approaches 0 v 5 m/s The particle’s velocity cannot exceed 5 m/s, and this is the greatest velocity attained. 9(a) 3 0 3 0 = 1 tm     Gradient of normal = 1 Equation of normal is  3 1 3y x   6y x  9(b) Equation of normal is 2 15y x  --- (1) 6y x  --- (2)

ST JOSEPH’S INSTITUTION PRELIMINARY EXAMINATION 2025 (YEAR 4) ADDITIONAL MATHEMATICS 4049/02 Sub (1) into (2) 2 1 65 7 75 x x x      5x Centre =  5,1     2 22 3 5 3 1r      8 Therefore equation of circle is     2 2 5 1 8x y    9(c)  5 2 2,1  10 sin 2y x x  d 2cos 2 1d y xx   2cos 2 1 0x  1cos 2 2x Basic angle = 3  2 , 3 3x      Min point 42 3x  2 3x  4 2sin 3 3y       3 2 2 3  

ST JOSEPH’S INSTITUTION PRELIMINARY EXAMINATION 2025 (YEAR 4) ADDITIONAL MATHEMATICS 4049/02 Area of region below shaded area 1 3 2 2 2 2 3 3            22 3 9 6    Area under graph =    2 3 0 sin 2 dx x x   2 2 3 0 1 cos 22 2 xx          2 2 1 2 1 3cos 2 [ cos 2 0 ]2 3 2 2                      21 2 1 4 9 2       22 3 9 4   Area of shaded region 2 22 3 2 3 9 4 9 6           3 3 6 4   (shown)