Composite functions

Mark scheme and answers · Total marks: 122

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Question 1 [2 marks]

(a) fg(x) = f(g(x)) = 2x + 3.
(b) g(2) = 4, so fg(2) = f(4) = 7.

Question 2 [2 marks]

(a) gf(x) = g(f(x)) = 3x + 3.
(b) f(0) = -1, so gf(0) = g(-1) = 3.

Question 3 [2 marks]

(a) fg(x) = f(x - 2) = 2x + 1.
(b) g(1) = -1, so fg(1) = f(-1) = 3.

Question 4 [2 marks]

(a) gf(x) = g(4 - x) = 9 - 2x.
(b) f(3) = 1, so gf(3) = g(1) = 3.

Question 5 [2 marks]

(a) fg(x) = (x + 1)² = x² + 2x + 1.
(b) g(0) = 1, so fg(0) = f(1) = 1.

Question 6 [2 marks]

(a) fg(x) = 5(x - 3) = 5x - 15.
(b) g(-1) = -4, so fg(-1) = f(-4) = -20.

Question 7 [2 marks]

(a) gf(x) = g(12x + 2) = x + 5.
(b) f(4) = 4, so gf(4) = g(4) = 9.

Question 8 [2 marks]

(a) fg(x) = f(3x) = 3x + 6.
(b) g(2) = 6, so fg(2) = f(6) = 12.

Question 9 [2 marks]

(a) gf(x) = g(2x - 3) = 8 - 2x.
(b) f(1) = -1, so gf(1) = g(-1) = 6.

Question 10 [2 marks]

(a) fg(x) = f(6x - 2) = 2x + 13.
(b) g(1) = 4, so fg(1) = f(4) = 73.

Question 11 [5 marks]

(a) 2x + 4 = 3x + 1 gives x = 3.
(b) fg(x) = f(3x + 1) = 6x + 6.
(c) gf(x) = g(2x + 4) = 6x + 13.

Question 12 [5 marks]

(a) x + 5 = 2x - 1 gives x = 6.
(b) fg(x) = f(2x - 1) = 2x + 4.
(c) gf(x) = g(x + 5) = 2x + 9.

Question 13 [5 marks]

(a) 3x - 2 = x + 6 gives x = 4.
(b) fg(x) = f(x + 6) = 3x + 16.
(c) gf(x) = g(3x - 2) = 3x + 4.

Question 14 [5 marks]

(a) 4x + 1 = 5x - 3 gives x = 4.
(b) fg(x) = f(5x - 3) = 20x - 11.
(c) gf(x) = g(4x + 1) = 20x + 2.

Question 15 [5 marks]

(a) 5 - x = x + 1 gives x = 2.
(b) fg(x) = f(x + 1) = 4 - x.
(c) gf(x) = g(5 - x) = 6 - x.

Question 16 [5 marks]

(a) 7x - 1 = 4x + 5 gives x = 2.
(b) fg(x) = f(4x + 5) = 28x + 34.
(c) gf(x) = g(7x - 1) = 28x + 1.

Question 17 [5 marks]

(a) 2x + 7 = 5x - 2 gives x = 3.
(b) fg(x) = f(5x - 2) = 10x + 3.
(c) gf(x) = g(2x + 7) = 10x + 33.

Question 18 [5 marks]

(a) x + 8 = 3x gives x = 4.
(b) fg(x) = f(3x) = 3x + 8.
(c) gf(x) = g(x + 8) = 3x + 24.

Question 19 [5 marks]

(a) 6x - 4 = 5x + 1 gives x = 5.
(b) fg(x) = f(5x + 1) = 30x + 2.
(c) gf(x) = g(6x - 4) = 30x - 19.

Question 20 [5 marks]

(a) 9 - 2x = x + 3 gives x = 2.
(b) fg(x) = f(x + 3) = 3 - 2x.
(c) gf(x) = g(9 - 2x) = 12 - 2x.

Question 21 [6 marks]

(a) For a one-to-one function on its domain, ff⁻¹(y) = y, so ff⁻¹(4) = 4 (where defined).
(b) g(0) = 3, g(3) = 0, so gg(0) = 0.
(c) fg(x) = f(3 - x) = 6(3 - x) - 2 = 61 - x.

Question 22 [6 marks]

(a) ff⁻¹(-2) = -2.
(b) g(2) = 1, g(1) = 3, so gg(2) = 3.
(c) fg(x) = 8(5 - 2x) + 1 = 86 - 2x = 43 - x.

Question 23 [4 marks]

(a) fg(x) = f(x + 2b) = (x + 2b)² - a.
(b) With b = a = 2: fg(x) = (x + 4)² - 2, so fg(1) = 25 - 2 = 23.

Question 24 [6 marks]

(a) ff⁻¹(5) = 5.
(b) g(-1) = -1, so gg(-1) = g(-1) = -1.
(c) fg(x) = 10(2x + 1) - 3 = 102x - 2 = 5x - 1.

Question 25 [4 marks]

(a) fg(x) = f(x - 1) = (x - 1)² + k.
(b) When k = -5: fg(2) = f(1) = 1 - 5 = -4.

Question 26 [6 marks]

(a) ff⁻¹(1) = 1.
(b) g(1) = 6, g(6) = 1, so gg(1) = 1.
(c) fg(x) = 4(7 - x) + 2 = 49 - x.

Question 27 [4 marks]

(a) gf(x) = g(3x + a) = (3x + a)² - b = 9x² + 6ax + a² - b.
(b) gf(0) = a² - b = 1 - 4 = -3.

Question 28 [6 marks]

(a) ff⁻¹(3) = 3.
(b) g(2) = -2, g(-2) = 10, so gg(2) = 10.
(c) fg(x) = 12(4 - 3x) - 1 = 123 - 3x = 41 - x.

Question 29 [4 marks]

(a) fg(x) = f(x + 3) = (x + 3)² + a(x + 3) = (x + 3)(x + 3 + a).
(b) fg(0) = f(3) = 9 - 15 = -6.

Question 30 [6 marks]

(a) ff⁻¹(-1) = -1.
(b) g(3) = 1, g(1) = -3, so gg(3) = -3.
(c) fg(x) = 9(2x - 5) + 4 = 92x - 1.