Mastery Questions Solution

2020 SAJC JC1 H2 Mathematics Tutorial Tutorial 5: Functions Page 13 of 15 Mastery Questions 11. [N2007/1/2] Functions f and g are defined by f : 𝑥↦ ଵ ௫ିଷ ,f o r 𝑥∈ℝ , 𝑥് 3, g: 𝑥↦𝑥 ଶ,f o r 𝑥∈ℝ (i) Only one of the composite functions fg and gf exists. Give a definition (including the domain) of the composite that exists, and explain why the other composite does not exist. (ii) Find f ିଵሺ𝑥ሻ and state the domain of f ିଵ. [Ans: (i) gf: 𝑥↦ ଵ ሺ௫ିଷሻమ, 𝑥∈ℝ , 𝑥് 3; (ii) f ିଵ: 𝑥↦ 3 ൅ ଵ ௫, , 𝑥∈ℝ , 𝑥് 0] [Solution] (i) 𝑅 ୤ ൌℝ \ ሼ0ሽ ⊆ℝൌ𝐷 ୥ ∴ gf exists. gf: 𝑥↦ ଵ ሺ௫ିଷሻమ, 𝑥∈ℝ , 𝑥് 3 𝑅௚ ൌ ሾ0, ∞ሻ is not a subset of ℝ \ ሼ3ሽ ൌ𝐷 ௙ ∴ fg does not exist. (ii) Let 𝑦ൌ ଵ ௫ିଷ 𝑥െ 3 ൌ ଵ ௬ 𝑥ൌ ଵ ௬ ൅ 3 ∴ f ିଵ: 𝑥↦ 3 ൅ ଵ ௫, , 𝑥∈ℝ , 𝑥് 0 12. [N2008/2/4] The function f is defined by f: 𝑥↦ ሺ𝑥െ 4ሻଶ ൅ 1 for 𝑥∈ℝ , 𝑥൐ 4. (i) Sketch the graph of 𝑦ൌ fሺ𝑥ሻ. Your sketch should indicate the position of the graph in relation to the origin. (ii) Find f ିଵሺ𝑥ሻ, stating the domain of f ିଵ. (iii) On the same diagram as in part (i), sketch the graph of 𝑦ൌ f ିଵሺ𝑥ሻ. (iv) Write down the equation of the line in which the graph of 𝑦ൌ fሺ𝑥ሻ must be reflected in order to obtain the graph of 𝑦ൌ f ିଵሺ𝑥ሻ, and hence find the exact solution of the equation fሺ𝑥ሻ ൌ f ିଵሺ𝑥ሻ. [Ans: (ii) f ିଵሺ𝑥ሻ ൌ 4 ൅ √𝑥െ 1, 𝑥∈ℝ , 𝑥൐ 1; (iv) 𝑦ൌ𝑥 ; 𝑥ൌ ଽା√ଵଷ ଶ ] [Solution]

2020 SAJC JC1 H2 Mathematics Tutorial Tutorial 5: Functions Page 14 of 15 (i) and (iii) (ii) 𝑦ൌ ሺ𝑥െ 4ሻ ଶ ൅ 1 ሺ𝑥െ 4ሻଶ ൌ𝑦െ 1 𝑥ൌ 4 േ ඥ𝑦െ 1 Since 𝑥൐ 4, choose 𝑥ൌ 4 ൅ ඥ𝑦െ 1 f ିଵሺ𝑥ሻ ൌ 4 ൅ √𝑥െ 1, 𝑥∈ℝ , 𝑥൐ 1. (iv) It must be reflected about the line 𝑦ൌ𝑥 In this case, solving fሺ𝑥ሻ ൌ f ିଵሺ𝑥ሻ is equivalent to solving fሺ𝑥ሻ ൌ𝑥 . ሺ𝑥െ 4ሻଶ ൅ 1 ൌ𝑥 𝑥ଶ െ 8𝑥൅ 17 ൌ𝑥 𝑥ଶ െ 9𝑥൅ 17 ൌ 0 𝑥ൌ ଽേඥଽమିସሺଵ଻ሻ ଶ ൌ ଽേ√ଵଷ ଶ Since 𝑥൐ 4, 𝑥ൌ ଽା√ଵଷ ଶ . 13 [N2014/1/1] The function f is defined by 1f: , , 1 , 0 .1xx x x x   (i) Show that     21ff . x x [ 4 ] (ii) Find   3f x in simplified form. [1] [Solution] (i)

2020 SAJC JC1 H2 Mathematics Tutorial Tutorial 5: Functions Page 15 of 15 2 2 2 ff f() f ( f () ) 1f 1 1 11 1 1 1 11 1f: 1 , , 1 , 0 . ( S i n c e D D ) xx x x x x x x x xx x x x                1f fRD ( , 0 ) ( 0 , 1 ) ( 1 , ) Let f ( ) . 1 1 (1 ) 1 1 1 xy y x yx yx y xy y            1 21 1 1 = 1 1 f: 1 , , 1 , 0 . Hence f ( ) f ( ). yx y y xx x x x xx          (ii) 21 31 Since f ( ) f ( ), f() f ( f () ) xx xx x     