2022 SGSS AMath Prelims P1 Ans

1 2022 AM 4E5N Prelim Paper 1 Marking Scheme Solutions: 1 3 33(1 ) 2 3(1 ) 2d d1 1 1 x x x xx x x             2 23(1 ) d 1x x x        33(1 ) 2ln(1 )(3)( 1) x x c    3(1 ) 2ln(1 )x x c     2 3 2 2 1 22 1 2 1 2 p q       3 3 2 2 2 2(2) 1 2     7 5 2 1   7 5 2  7, 5p q   3 34 π3V r 2d 4πd V rr  d d d d d d V V r t r t  2 d20 4 4 π d rr t   2 d 4 d π r t r When r = 11, 2 d 4 d π(11) r t  ≈ 0.0105 cm/s (correct to 3 s.f.) 4 2 22 5 48 0x xy y    --- (1)

2 Solutions: 8 2y x  --- (2) From (2): 2 8y x  --- (3) Put (3) into (1): 2 22 5 (2 8) (2 8) 48 0x x x x      2 2 22 10 40 (4 32 64) 48 0x x x x x       2 2 22 10 40 4 32 64 48 0x x x x x       28 8 16 0x x   2 2 0x x   ( 1)( 2) 0x x   1 or 2x x  Put 1x  into (3): 2( 1) 8y   10 Put 2x  into (3): 2(2) 8y  4  Coordinates of the points of intersection are (–1, –10) and (2, –4). 5 Let 2 2 2 8 3 1 (2 1) 2 1 (2 1) x x A B C x x x x x        Multiplying throughout by the denominator, 2 28 3 1 (2 1) (2 1)x x A x Bx x Cx       When 0x  , 1A When 1 2x  , 2 1 1 18 3 12 2 2 C              3 1 2 2 C 3C  When 1x  , 2 28(1) 3(1) 1 (1)(3) 3 3(1) B     12 9 3 3 B   2B

3 Solutions: 2 2 2 8 3 1 1 2 3 (2 1) 2 1 (2 1) x x x x x x x        6(a) (b) 1 24 3 2x x   2 2(2 ) 3 2 (2 )4 x x  2(2 ) 16(2 ) 12 0x x    Let 2xu . 2 16 12 0u u   2( 16) ( 16) 4(1)( 12) 2(1)u       16 304 2u  0.7177978871 or 16.71779789u u  2 0.7177978871 or 2 16.71779789x x  (no solutions) * ln16.71779789 ln 2x  ≈ 4.06 ** (correct to 2 d.p.) 7(a) (b) 2C( ) 40 480 1442x x x   240( 12 ) 1442x x   2 2 2 12 1240 12 1442 2 2x x                     240 ( 6) 36 1442x      240( 6) 2x   Minimum value of C(x) = 2 Corresponding value of x = 6

4 Solutions: (c) Minimum average cost per chair = $2 Number of chairs produced = 6000 8(a) (b) (c) (Alternate Segment Theorem) ( s in the same segment) ABC CAF DEF     Since ABC DEF  , by the Alternate Angles Property, AB is parallel to ED. (proven) In ACF and BAF , (common ) (Alternate Segment Theorem) AFC BFA CAF ABF       Triangle ACF is similar to triangle BAF. (proven) AF CF BF AF 2AF BF CF  (shown) 9(a) (b) L.H.S. cos2 1 3 3sin     21 2sin 1 3(1 sin )      22 2sin 3(1 sin )     22(1 sin ) 3(1 sin )     2(1 sin )(1 sin ) 3(1 sin )       2(1 sin ) 3  = R.H.S. (proven) cos2 1 1 3 3sin 2    

5 Solutions: 2(1 sin ) 1 3 2   31 sin 4  1sin 4  Basic angle,  = 1 1sin 4  o14.47751219 o o o 14.47751219 , 180 14.47751219    o o14.5 , 165.5 (correct to 1 d.p.) 10(a) (b) d cosln(sin )d sin xxx x  cot x coty x --- (1) 2y  --- (2) (1) = (2): cot 2x  1tan 2 x 1 1tan 2 x  Area of shaded region 1 π 2 1tan 2 cot dx x     1 π 2 1tan 2 ln(sin )x   1π 1ln sin ln sin tan2 2               0 ( 0.5493061443)  

6 Solutions: ≈ 0.549 units2 (correct to 3 s.f.) 11(a) (b) 3 2f ( 2) ( 2) 8( 2) ( 2) 18a b        0 8 32 2 18a b    8 50 2a b  25 1 4 4a b  --- (1) 3 2f ( 1) ( 1) 8( 1) ( 1) 18a b        4 8 18a b    22a b  --- (2) (1) = (2): 25 1 224 4 b b   21b Put 21b into (2): 22 21a   = 1 3 2f ( ) 8 21 18x x x x    By long division, 2 3 2 3 2 2 2 6 9 2 8 21 18 2 6 21 18 6 12 9 18 9 18 0 x x x x x x x x x x x x x x             3 2f ( ) 8 21 18x x x x    2( 2)( 6 9)x x x    2( 2)( 3)x x   12(a) 10 3x y  --- (1) 2y x  --- (2) Put (1) into (2): 10 3 2y y  

7 Solutions: (b) (c) 2y Put 2y into (1): 10 3(2)x  = 4 Coordinates of A are (4, 2). Let the coordinates of C be (x, y). 4 2(7.5, 3.5) , 2 2 x y     47.5 2 x and 23.5 2 y 11x  5y  Coordinates of C are (11, 5). Gradient of AC 2 5 4 11   3 7 Equation of AC is 32 ( 4)7y x   3 2 7 7y x  (or 7 3 2y x  ) Area of triangle ABC 4 7 11 41 2 1 5 22  1 (4 35 22) (14 11 20)2      1 (61 45)2  = 8 units2 13(a) Since 2πperiod 1 b  ,

8 Solutions: (b) (c) (d) 2π4π 1 b  2b (shown) At ( π , 2), π2 sin 2a c       2 a c  2a c  --- (1) At (4 π , –1), 4π1 sin 2a c        1c Put 1c into (1): 2 ( 1)a    = 3 1sin 2 12 π 3 x x      3sin 12 4 π x x      3sin 1 22 4 π x x      

9 Solutions: Draw the line of 24π xy   . The x-coordinates of the points of intersection of the graphs of f ( )y x and 24π xy   give the solutions to the equation 1sin 2 12 π 3 x x      . 14(a) (b) 2 150r r   150 2r r   150 2r   Area of sector OAB, A 21 2 r  21 150 22 r r       275r r  cm2 (shown) d 75 2d A rr   For stationary value, d 0d A r  75 2 0r  37.5r  When 37.5r  , 275(37.5) (37.5)A  = 1406.25 2 2 d 2d A r  Since 2 2 d 0d A r  , 1406.25A is maximum when 37.5r  .