JPJC 2026 J1 H2 Math_WA 2 (Solution)

Jurong Junior Pioneer College H2 Mathematics JC1 2026 WA 2 (Solution) 1 (a) 2 9 18 2 x x x    2 9 1 08 2 x x x     2 2 2 9 8 2 08 2 8 2 x x x x x x x        2 2 1 08 2 x x x x     Alternatively, 2 2 1 02 8 x x x x     2 1 0( 2)( 4) x x x x     2 2 2 1 08 2 1 0(4 )(2 ) x x x x x x x x         Consider 2 2 1 31 2 4x x x         Since 2 2 1 1 3 0, 02 2 4x x              for all real values of x. (OR coefficient of 2 1 0x   and discriminant, 2 2 4 ( 1) 4(1)(1) 3 0b ac      imply that 2 1 0x x   for all real values of x.) Therefore, we consider ( 2)( 4) 0x x   2 or 4x x   (b) 2 4 2 9 12 8 x x x    2 4 2 9 12 8 x x x      2 2 2 2 9 18 2( ) ( ) x x x       By replacing x with 2x , we have: Using the (a) answer, 2 2 2 or 4x x    2 2x  2 2 0x     2 2 0x x   2 or 2x x  - (No solution since 2 0x  for all real x) – 2 4 x + + – Alternatively, (4 )(2 ) 0x x   2 or 4x x   – 2 4 x  + + + Note: Do not cross multiply. Note: Take note of coefficient of x2 and factorise correctly.

2 (a) Note: (0,6) is transformed to (1,12) which is not required in this answer. (b) 3 (a) 1 ,2 a b     (b)  ,a b ,  ,a b (c)  ,a b y x O (2, 8) (4,0)  y x (2,0)  O

4 (a) Least value of 0k  . (b) Let 2 1 4y x  , 2x . 2 4 1x y y  2 1 4x y  1 4x y  Since 2x , 1 4x y  Therefore, 1 1f ( ) 4 , , 0x x x x       1fD 0,   (c) Relationship: Graphs of f and 1f  are reflections of each other about the line y = x. (d) Given     1f f x x  Consider  f x x 2 1 4 xx  3 4 1 0x x   , where 1, 4, 1a b c   y x y = 0 x = 2 x = 0 y = x y = 2 O Note: “symmetrical” is a property of the 2 graphs and not a relationship. Note: 1. 2 graphs must be symmetrical about the line y = x 2. Line y = x must be drawn.