1. Additional Mathematics Functions

2. Questions 1.Given function f : x  mx + 4, x=n. x – n If f(2) = 10 and f(8) = 4, find a) the values of m and n b) the image of 3 c) the object of 11 AnswerAnswer 4.The graph shows the absolute values function f(x) = I hx – k I a) Find the possible values of h and k b) If the value of f(m) = 6, find the value of m Answer 2.Given that f(x) = 6 x = -b,where ax + b a a and b are constants, f(1) = 2 and f(5) = 6/7, find the value of a and b AnswerAnswer 3. Given that f:x  1, x= k x – k and f(3) = 1, Find a) the value of k b) the values of a such that f(a²) = f(3a) Answer 5.Sketch the graph of the function g:x  I3x – 5I for the domain 0≤x≤4 and state the range of g(x) corresponding to the domain. Answer

3. Questions 6.Given the functions f:x  ax + b and g : x  5 – x. if fg(x) = 19 – 3x, find a) the values of a and b b) the value of x such that fg(-x) = f(7) AnswerAnswer 9.Given that f(x) = 3x – 8 and fg(x) = 3x² - 2, find the function g. Answer 7.Given that f:x  2x + 1 and g:x  3 x x = 0, find the composite function a) fg b) gf c) f² d) g² AnswerAnswer 10.Given that g(x) = 2 – x, fg(x) = 2x² + x + 8, find a) the function f b) the value of g(x) if fg(x) = 9 Answer 8.Given that h:x  2x + 1 and k:x  2x +3 3 Find a) the value of x if hk(x) = x b) the value of x if kh(x) = -3 c) the value of x when hk(x) = 3 kh(x) Answer 11. f(x) = 3x + 2 and gf(x)= 9x+12x², find a) g(x) b)fg(2) AnswerAnswer

4. Questions 12.Given that h(x) = 3 + x and hk(x) = 1, x ≠3,find 3 – x a) k(x) b) the value of k(x) such that x = 4 Answer Answer 15. Given that f:x  3x – 1 and gf:x  2x² - 6x + 4, find a) the function g b) the values of x such that f(x) = 9g(x) Answer Answer 13) Given that f:x  2 +x, g(x)  ax + b, where a and b are constants, and gf(x) = 6 + 10x, find a) the value of a and b b) the value of x if g(x) = gf(x) Answer Answer 14) Given that f(x) = 2x + x² and g(x) = 4 – 3x, Find a) the values of x such that f(x) = 8 b) the composite function fg c) the values of x if f(x) = 3g(x) Answer

5. Solution for Question 1(a) Subtracting (1) and (2) Where 2m + 10n = 16 - 2m + n = 7 9n = 9 9n = 9 n = 1 Substitute n = 1 to (2) 2m + 1 = 7 2m = 6 m = 3 Given f(x) = mx + 4 x – n f(2) = 10 and f(8) = 4 f(2) = 10  (substitute x = 2, f(x) = 10) 2m + 4 = 10 2 – n 2m + 4 = 20 – 10n 2m + 10 = 16 ……. (1) f(8) = 4 8m + 4 = 4  (Substitute x = 8, f(x)=4) 8 – n  (bring 8 – n to the right) 8m + 4 = 32 – 4n 8m + 4n = 28  (Simplify by dividing with 4) 2m + n = 7 ……(2) Therefore, f(x) = 3x + 4 x – 1 Menu

6. Solutions for Question 1(b) To find the object of 11 f(x) = 11 3x + 4 = 11 x – 1 3x + 4 = 11x – 11 8x = 15 x = 15 8 (answer) Solution for Question 1(c) to find the image of 3, x = 3 f(3) = 3(3) + 4 3 – 1 = 13 2 (answer) Menu

7. Solution for Question 2 Given f(x) = 6. ax + b f(1) = 2 and f(5) = 6/7 f(1) = 2  ( substitute x = 1, f(x) = 2) 6 = 2 ax + b  (bring a + b to the right) 6 = 2a + 2b a + b = 3 …… (1) Then f(5) = 6/7  ( substitute x = 5, f(x) = 6/7) 6 = 6 ax + b 7  (eliminate 6 on both sides) 1 =1 5a + b 7 after we eliminate 6, we cross multiply therefore, 7 = 5a + b 5a + b = 7 …… (2) Subtracting (1) from (2) 5a + b = 7 - a + b = 3 4a = 4 a = 1 Substitute a=1 into (1) 1 + b = 3 b = 2 Therefore, a = 1, b = 2 (answer) Menu

8. Solution for Question 3(a) b) find the values of a such that f(a²) = f(3a) 1 = 1 a² - 2 3a – 2  (cross multiply for both sides) 3a – 2 = a² - 2 a² = 3a  (both sides cancel out one a) a = 3 (answer) Solution for question 3(b) a) find the value of k f(x) = 1 x – k f(3) = 1  (substitute x = 3, f(x) = 1) 1 = 1 3 – k  ( bring 3 – k over the right) 3 – k = 1 k = 2 Therefore f(x) = 1 x – 2 (answer) Menu

9. Solution for Question 4(a) If k = 8, 0 = I 4h – 8 I 4h = 8 h = 2 Therefore, h = -2 or 2 Then f(x) = I -2x + 8 I or f(x) = I2x – 8I (answer) a) Find the possible values of h and k f(x) = I hx – k I Given f(0) = 8, f(4) = 0 8 = I h(0) – k I I-kI = 8 -k = 8 or -k = -8 when modulus sign, I I is removed, the resulting value can be negative or positive. therefore, k = -8 or 8 When f(4) = 0 x = 4 and f(x) = 0 If k = -8, 0 = I 4h + 8 I -4h = 8 h = -2 Menu

10. Solution for Question 4(b) or 6 = 2x – 8 2x = 14 x = 7 From the graph, m>4 so, x = 7 (answer) b) If the value of f(m) 6, find the value of m Take either one of the function, Let f(x) = I 2x – 8 I Given f(m) = 6 6 = I 2x – 8 I -6 = 2x – 8 or 6 = 2x – 8 when modulus sign, I I is removed, the resulting value can be negative or positive Obtain answers separately -6 = 2x – 8 2x = 2 x = 1 Menu

11. Solution for Question 5 When g(x) = 0 0 = I 3x – 5 I 3x – 5 = 0 3x = 5 x = 5/3 Steps to sketch the graph Given g(x) = I 3x – 5 I -First, find the value of g(x), when x = 0 -Second, find the value of g(x), when x = 4. This is to check the maximum value of g(x) -Finally, find the value of x, when g(x) = 0. This is to check the point where the turning point exists. When x = 0, g(0) = I 3(0) – 5 I = I -5 I = 5 When x = 4, g(4) = I 3(4) – 5 I = 7 Menu

12. Solution for Question 6(a) Substitute a = 3 5(3) + b = 19 15 + b = 19 b = 4 Therefore, a = 3, b = 4 f(x) = 3x + 4 (answer) b) Find the value of x such that fg(-x) = f(4) 19 + 3x = 3(4) + 4 19 + 3x = 16 3x = 3 x = -1( answer) a)the values of a and b Given f(x) = ax + b g(x) = 5 – x fg(x) = 19 – 3x ……(1) fg(x) = f[g(x)] = a(5 – x) + b = 5a – ax + b  (factorize a) = (5a + b) – ax……(2) Compare the x-coefficient of (1) and (2) -a = -3 a = 3 Compare the constant of (1) and (2) 5a + b = 19 Solution for Question 6(b) Menu

13. Solution for Question 7(a) a)fg fg(x) = f[g(x)] substitute g(x) = 3 x = f(3/x) = 2(3/x) + 1 =6 + 1 (answer) x b) gf gf(x) = g[f(x)]  (substitute f(x) = 2x + 1) = g(2x + 1) = 3 (answer) 2x + 1 c) f² f²(x) = ff(x) = f[f(x)]  (substitute f(x) = 2x + 1) = f(2x + 1) = 2(2x + 1) + 1 = 4x + 3 (answer) d) g² g²(x) = g[g(x)]  (substitute g(x) = 3/x) = g(3/x) = 3 3/x = x (answer) Solution for Question 7(b) Solution for Question 7(c) Solution for Question 7(d) Menu

14. Solution for Question 8(a) a)Find the value of x, if hk(x) = x First, find hk(x) hk(x) = h[k(x)] = h(2x + 3) = 2(2x + 3) + 1 3 = 4x + 6 + 1 3 = 4x + 7 3 From hk(x) = x, so 4x + 7 = x 3 4x + 7 = 3x x = -7 (answer) b) Find the value of x if kh(x) = -3 First, find kh(x), kh(x) = k[h(x)] = k ( 2x + 1) 3 = 2 ( 2x + 1 ) + 3 3 = 4x + 2 + 3 3 kh(x) = -3 4x + 2 + 3 = -3 3 4x + 2 + 9 = -9 4x = -20 x = -5 (answer) Solution for Question 8(b) Menu

15. Solution for Question 8(c) c) find the value of x when hk(x) = 3kh(x) From (a) From (b) hk(x) = 3kh(x) 4x + 7 = (4x + 2 + 3) 3 3 3 4x + 7 = 3(4x + 2 + 9) 4x + 7 = 12x + 6 + 27 -26 = 8x x = - 13/4 (answer) fg(x) = 3x² - 2 f[g(x)] = 3x² - 2 Substitute f(x) = 3x – 8 3g(x) – 8 = 3x² - 2 3g(x) = 3x² + 6 g(x) = x² + 2 (answer) Solution for Question 9 Menu

16. Solution for Question 10(a) a)fg(x) = 2x² + x + 8 f(2-x) = 2x² + x + 8 Let y = 2 – x Hence x = 2 – y f(y) = 2(2 – y)² + (2 – y) + 8 = 2(4–4y+y²)+ 2– y + 8 = 8 – 8y+2y² + 2 – y + 8 = 2y² - 9y + 18 f(x) = 2x² - 9x + 18(answer) b) fg(x) = 9 2x² + x + 8 = 9 2x² + x – 1 = 0 (2x – 1)(x + 1) = 0 2x – 1 = 0 or x + 1 = 0 x = ½ x = -1 When x = ½ g( ½ ) = 2 – ½ = 3/2 When x = -1 g(-1) = 2 – (-1) = 3 So, g(x) = 3/2 or 3 (answer) Solution for Question 10(b) Menu

17. Solution for Question 11(a) a)Find g(x) gf(x) = 9x + 12x² g[f(x)] = 9x + 12x² (substitute f(x) = 3x + 2) g(3x + 2) = 9x + 12x² Let y = 3x + 2 3x = y – 2 Hence x = y – 2 3 Let u = y – 2 3 g(u) = 9 (y – 2) + 12 (y – 2)² 3 3 =3y – 6 + 12 (y²- 4y +4) 9 = 4y² - 7y -2 3 so, g(x) = 4x² - 7x – 2 3 (answer) b) fg(x) = f[g(x)] = 3(4x² - 7x – 2 ) + 2 3 = 4x² - 7x – 2 + 2 = 4x² - 7x (answer) Solution for Question 11(b) Menu

18. Solution for Question 12 (a) a) Find k(x), hk(x) = 1 (substitute h(x) = 3 + x) 3 – x h[k(x)] = 1 3 – x 3 + k(x) = 1 3 – x [3 + k(x)][3 – x] = 1 9 – 3x + 3k(x) – k(x)x = 1 So, 8 – 3x = k(x)x – 3k(x) 8 – 3x = k(x) (x – 3) k(x) = 8 – 3x x – 3 (answer) b) Find the value of k(4) k(4) = 8 – 3(4) 4 – 3 = 8 – 12 = -4 Solution for Question 12 (b) Menu

19. Solution for Question 13(a) a)Find the values of a and b gf(x) = 6 + 10x g[f(x)] = 6 + 10x (substitute f(x) = 2 + 5x) 9(2 + 5x) + b = 6 + 10x 2a + 5ax + b = 6 + 10x 5ax + 2a + b = 6 + 10x Compare the x-coefficient 5a = 10 a = 2 Compare the constant 2a + b = 6 (replace a = 2) 2(2) + b = 6 4 + b = 6 b = 2 ( answer) b) the value of x if g(x) = gf(x) g(x) = gf(x) 2x + 2 = 6 + 10x 8x = -4 (Bring the 8 to the right) x = - 4 = - 1 8 2 Solution for Question 13(b) Menu

20. Solution for Question 14(a) a)Find the values of x such that f(x) = 8 f(x) = 8 2x + x² = 8 x² + 2x – 8 = 0 (x + 4)(x – 2) = 0 So, x + 4 = 0 x = -4 OR x – 2 = 0 x = 2 b) Find the composite fg fg(x) = f[g(x)] ( substitute g(x) = 4–x) = f[ 4 – 3x] = 2(4 – 3x) + (4 – 3x)² = 8 – 6x + 16 + 9x² - 24x = 9x² - 30x + 24 (answer) c)Find the values of x if f(x) = 3g(x) f(x) = 3g(x) 2x + x² = 3(4 – 3x) x² + 2x = 12 – 9x x² + 11x – 12 = 0 So, (x + 12)(x – 1) = 0 (x + 12) = 0 OR (x – 1) = 0 x = -12 x = 1 Solution for Question 14(b) Solution for Question 14(c) Menu

21. Solution for Question 15(a) g[3x – 1] = 2x² - 6x + 4 Let y = 3x – 1 3x = y + 1 Hence x = y + 1 3 Let u = y + 1 3 g(u) = 2(y + 1)²-6(y + 1)+4 3 3 =2(y²+2y+1)-2y–2+ 4 9 = 2y²+4y+2–2y–2+ 4 9 = 2y²+4y+2–18y + 18 9 = 2y² - 14y + 20 9 g(x) = 2x² - 14x + 20 9 b) f(x) = 9g(x) 3x – 1 = 9(2x²-14x+20) 9 3x – 1 = 2x²-14x+20 2x²-17x+21=0 (2x-3)(x-7) =0 2x – 3 = 0 or x – 7 = 0 x=3 x = 7 2 Solution for Question 15(b) Menu

22. End of Questions