2022 Ch1 Inequalities APQ Soln (RVHS)

H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Ch 1 Inequalities and Equations (Solutions to Additional Practice Questions) 1 Chapter 1: Equations & Inequalities Additional Practice Questions Solutions 1 2 2121 x xx     2 2 1 2 1 021 x x x x      243 021 xx x        22 1 4 3 0x x x       2 1 4 3 1 0x x x    31 42 x   or 1x (ans) 22 22 2 1 2 2 xx xx  2 2 2 12 2 12 1 x x x     2 3 1 1 42 x    or 2 1 1 x   2 110 2x  or 2 1 1 x   2 2x  or 2 1x   22xx  or or 1 1, 0xx    2. Sketch ( 3)( 3) 16 2 10 21( 3)( 2) ( 3)( 2) x x x xyx x x x x            1 1/2 -3/4

H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Ch 1 Inequalities and Equations (Solutions to Additional Practice Questions) 2 Find 0, From GC, 4 3 or 2 or 1 y x x x       Hence, Let 2 , 4 2 3(n.a.) 2 2 1 2 1 0 1 or 0 x x x x y x x xx              3. (a) 4 3 2 3, 4 3 3 2 11From GC, 2 or 453 xx xx xx          (b)             2 2 2 2 2 3 2,1 1 3 2 1 0 1 2 1 0 1 1 0 1 0,since 1 0 1, 1 x x x x x x x x xx xx xx                    

H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Ch 1 Inequalities and Equations (Solutions to Additional Practice Questions) 3 2 3For 2,let 1 1 1, 1 x yxx x xx        4.  2 22211 xa ay xx     Horizontal asymptote: y = 2; vertical asymptote: x = 1 Therefore, solution set:  : , 1 or 3x x x x   . 5. (i) 32y ax bx cx d    When x = 0, y = −8, we get d = −8 (ii) So, 32 8y ax bx cx    When x = 4, y = 0, we get 64 16 4 8 16 4 2a b c a b c       ---- (1) When x = 3, y = −8 − 2(3) = −14, we get

H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Ch 1 Inequalities and Equations (Solutions to Additional Practice Questions) 4 27 9 3 8 14 9 3 2a b c a b c        ---- (2) 2d 32d y ax bx cx    When x = 2, d 2d y x  , we get 12 4 2a b c   ---- (3) Putting equations (1), (2) and (3) together and using the GC to solve, We get a = 1, b = −3 and c = −2. Therefore, equation of curve is 32 3 2 8y x x x    . 6. 7. (a) Let x = 1, 2, 3, 4 (eg. x = 1 for year 1920), y represents the population in millions. Let the cubic polynomial be 32y ax bx cx