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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 11 · 11.3 Functions

Solutions · Full Answer Key

Pack A answers · Pack B answers · Problem-solving worked solutions

Pack A — Answers

Bronze
1.13
2.x=4x = 4
3.x∈R,  x≠3x \in \mathbb{R},\; x \neq 3
4.x≥1x \geq 1
5.6
6.Yes — passes the vertical line test.
7.R\mathbb{R}
8.f(−2)=−8f(-2) = -8
9.x>4x > 4
10.f(n)=3n+1f(n) = 3n + 1
Silver
11.17
12.2x2−12x^2 - 1
13.f−1(x)=x−52f^{-1}(x) = \dfrac{x - 5}{2}
14.Domain x≠2x \neq 2; range y≠0y \neq 0
15.x=1x = 1
16.Domain R\mathbb{R}; range y≥0y \geq 0
17.x≥3x \geq 3
18.x=−2x = -2 or x=3x = 3
19.3
20.f(−3)=−5f(-3) = -5; f(2)=4f(2) = 4
Gold
21.f−1(x)=x−52f^{-1}(x) = \dfrac{x - 5}{2}; domain R\mathbb{R}
22.(f∘g)(x)=x−4(f \circ g)(x) = \sqrt{x - 4}; domain x≥4x \geq 4
23.y≥2y \geq 2
24.f(2.7)=2f(2.7) = 2; f(−1.4)=−2f(-1.4) = -2
25.Restrict to x≥0x \geq 0; f−1(x)=xf^{-1}(x) = \sqrt{x}
26.Yes — exponential decay with horizontal asymptote y=0y = 0 and f(0)=4f(0) = 4.
27.(g∘f)(x)=12x+1(g \circ f)(x) = \dfrac{1}{2x + 1}; domain x≠−12x \neq -\dfrac{1}{2}
28.x=−1x = -1 or x=5x = 5
29.(a) CHF 0.50 cost per extra bottle; (b) CHF 200 fixed cost.
30.y=xy = x
Platinum
31.(g∘f)(x)=2x+12x(g \circ f)(x) = \dfrac{2x + 1}{2x}; domain x≠0x \neq 0
32.(f∘f)(x)=x−12−x(f \circ f)(x) = \dfrac{x - 1}{2 - x}; defined for x≠1,2x \neq 1, 2
33.f−1(x)=x+3x−2f^{-1}(x) = \dfrac{x + 3}{x - 2}; domain x≠2x \neq 2
34.x<−2x < -2 or x>2x > 2
35.a=−1a = -1
36.y≥3y \geq 3
37.f−1(x)=x−13f^{-1}(x) = \dfrac{x - 1}{3}
38.x=3.5x = 3.5
39.f−1(x)=2+xf^{-1}(x) = 2 + \sqrt{x}; domain x≥0x \geq 0
40.Not a function: a vertical line e.g. x=0x = 0 meets the circle at (0,5)(0, 5) and (0,−5)(0, -5). The two function pieces are f1(x)=25−x2f_1(x) = \sqrt{25 - x^2} (top) and f2(x)=−25−x2f_2(x) = -\sqrt{25 - x^2} (bottom).

Pack B — Answers

Bronze
1.11
2.x=4x = 4
3.x∈R,  x≠−2x \in \mathbb{R},\; x \neq -2
4.x≥5x \geq 5
5.10
6.No — a vertical line, e.g. x=4x = 4, meets x=y2x = y^2 at (4,2)(4, 2) and (4,−2)(4, -2).
7.R\mathbb{R}
8.f(3)=−1f(3) = -1
9.x>−1x > -1
10.f(n)=3n−1f(n) = 3n - 1
Silver
11.31
12.3x2+53x^2 + 5
13.f−1(x)=x+13f^{-1}(x) = \dfrac{x + 1}{3}
14.Domain x≠−3x \neq -3; range y≠0y \neq 0
15.x=4x = 4
16.Domain R\mathbb{R}; range y≥0y \geq 0
17.x≤5x \leq 5
18.x=−3x = -3 or x=2x = 2
19.−2-2
20.f(−1)=−1f(-1) = -1; f(3)=9f(3) = 9
Gold
21.f−1(x)=x+34f^{-1}(x) = \dfrac{x + 3}{4}; domain R\mathbb{R}
22.(f∘g)(x)=2x+1(f \circ g)(x) = \sqrt{2x + 1}; domain x≥−12x \geq -\frac{1}{2}
23.y≤5y \leq 5
24.f(3.9)=3f(3.9) = 3; f(−2.1)=−3f(-2.1) = -3
25.Restrict to x≤0x \leq 0; f−1(x)=−xf^{-1}(x) = -\sqrt{x}
26.Yes — f(0)=1+3=4f(0) = 1 + 3 = 4, decreasing to y=1y = 1.
27.(f∘g)(x)=2x+1(f \circ g)(x) = \dfrac{2}{x} + 1; domain x≠0x \neq 0
28.x=−5x = -5 or x=1x = 1
29.(a) CHF 0.80 per bottle; (b) CHF 150 fixed cost.
30.Domain of f−1f^{-1} = range of ff (and range of f−1f^{-1} = domain of ff).
Platinum
31.(f∘g)(x)=3x−1x−1(f \circ g)(x) = \dfrac{3x - 1}{x - 1}; domain x≠1x \neq 1
32.(f∘f)(x)=2−x3−2x(f \circ f)(x) = \dfrac{2 - x}{3 - 2x}
33.f−1(x)=2x+13−xf^{-1}(x) = \dfrac{2x + 1}{3 - x}; domain x≠3x \neq 3
34.x≥1x \geq 1 with x≠3x \neq 3
35.a=−1a = -1
36.y≤4y \leq 4
37.f−1(x)=x+13f^{-1}(x) = \dfrac{x + 1}{3}
38.x=69x = \dfrac{6}{9} (i.e. 23\frac{2}{3})
39.f−1(x)=−1−xf^{-1}(x) = -1 - \sqrt{x}; domain x≥0x \geq 0
40.Not a function. Top: f1(x)=21−x2/9f_1(x) = 2\sqrt{1 - x^2/9}; bottom: f2(x)=−21−x2/9f_2(x) = -2\sqrt{1 - x^2/9}.

Problem-solving — Worked Solutions

1Problem 1
Answer
(a) 13. (b) 1. (c) x=4x = 4.
Full working
(a) f(5)=2(5)+3=13f(5) = 2(5) + 3 = 13. (b) f(−1)=2(−1)+3=1f(-1) = 2(-1) + 3 = 1. (c) 2x+3=11⇒x=42x + 3 = 11 \Rightarrow x = 4.
2Problem 2
Answer
(a) x∈R,x≠3x \in \mathbb{R}, x \neq 3. (b) x≥1x \geq 1. (c) x<4x < 4.
Full working
(a) Denominator ≠0\neq 0. (b) Radicand ≥0\geq 0. (c) Radicand must be **strictly** positive (also in denominator).
3Problem 3
Answer
(a) Yes. (b) No. (c) No. (d) Yes.
Full working
(a) Each xx gives exactly one yy. (b) x=4x = 4 gives y=2y = 2 or −2-2 — fails VLT. (c) Circle — vertical lines −3<x<3-3 < x < 3 give two yy. (d) Each xx gives exactly one non-negative yy.
4Problem 4
Answer
(a) 17. (b) 2x2−12x^2 - 1. (c) f−1(x)=x−52f^{-1}(x) = \frac{x - 5}{2}; domain R\mathbb{R}.
Full working
(a) g(3)=6g(3) = 6; f(6)=17f(6) = 17. (b) f(g(x))=2(x2−3)+5=2x2−1f(g(x)) = 2(x^2 - 3) + 5 = 2x^2 - 1. (c) y=2x+5⇒x=y−52y = 2x + 5 \Rightarrow x = \frac{y - 5}{2}, so f−1(x)=x−52f^{-1}(x) = \frac{x - 5}{2}. Domain of f−1f^{-1} = range of ff = R\mathbb{R}.
5Problem 5
Answer
(a) 5−x2\sqrt{5 - x^2}. (b) −5≤x≤5-\sqrt{5} \leq x \leq \sqrt{5}. (c) 0≤y≤50 \leq y \leq \sqrt{5}.
Full working
(a) f(g(x))=5−x2f(g(x)) = \sqrt{5 - x^2}. (b) Need 5−x2≥0⇒x2≤55 - x^2 \geq 0 \Rightarrow x^2 \leq 5, so −5≤x≤5-\sqrt{5} \leq x \leq \sqrt{5}. (c) Max when x=0x = 0: 5\sqrt{5}. Min at endpoints: 0. Range [0,5][0, \sqrt{5}].
6Problem 6
Answer
(a) −3,0,4,9-3, 0, 4, 9. (b) Linear piece, then upward parabola from (0,0)(0,0) to (3,9)(3,9), then horizontal at y=9y = 9. (c) Range: f≥−5f \geq -5, but with the linear piece dipping arbitrarily low as x→−∞x \to -\infty; on the stated domain the range is [−5,9][-5, 9].
Full working
(a) Pick the correct piece for each xx. (b) Three sections joined; check continuity at x=0x = 0 (yes, both give 0) and x=3x = 3 (yes, both give 9). (c) On −3≤x≤5-3 \leq x \leq 5: minimum at x=−3x = -3: f(−3)=−5f(-3) = -5. Maximum 9 reached at x=3x = 3 and beyond. Range [−5,9][-5, 9].
7Problem 7
Answer
(a) x=−2x = -2 or x=5x = 5. (b) −3<x<5-3 < x < 5, i.e. (−3,5)(-3, 5). (c) V-shape with vertex at (2,−1)(2, -1); xx-intercepts (1,0)(1, 0) and (3,0)(3, 0); yy-intercept (0,1)(0, 1).
Full working
(a) 2x−3=±7⇒x=52x - 3 = \pm 7 \Rightarrow x = 5 or x=−2x = -2. (b) ∣x−1∣<4⇔−4<x−1<4⇔−3<x<5|x - 1| < 4 \Leftrightarrow -4 < x - 1 < 4 \Leftrightarrow -3 < x < 5. (c) V-shape shifted right 2 and down 1. Vertex (2,−1)(2, -1). Set y=0y = 0: ∣x−2∣=1⇒x=1|x - 2| = 1 \Rightarrow x = 1 or 33. yy-intercept ∣0−2∣−1=1|0 - 2| - 1 = 1.
8Problem 8
Answer
(a) C(d)=cd+FC(d) = cd + F. (b) c=0.2c = 0.2, F=55F = 55. (c) CHF 155.
Full working
(b) System: 200c+F=95200c + F = 95, 350c+F=125350c + F = 125. Subtract: 150c=30⇒c=0.20150c = 30 \Rightarrow c = 0.20. Then F=95−40=55F = 95 - 40 = 55. (c) C(500)=0.20(500)+55=155C(500) = 0.20(500) + 55 = 155.
9Problem 9
Answer
(a) c=3c = 3; a+b+c=6a + b + c = 6; 4a+2b+c=134a + 2b + c = 13. (b) a=2a = 2, b=1b = 1, c=3c = 3. (c) f(−1)=4f(-1) = 4.
Full working
(a) Substitute each point. From f(0)=3f(0) = 3: c=3c = 3. Then a+b=3a + b = 3 and 4a+2b=104a + 2b = 10. (b) From 4a+2b=104a + 2b = 10: 2a+b=52a + b = 5. Subtract a+b=3a + b = 3: a=2a = 2. Then b=1b = 1. (c) f(−1)=2(1)+(−1)+3=4f(-1) = 2(1) + (-1) + 3 = 4.
10Problem 10
Answer
(a) (x−2)2+3(x - 2)^2 + 3. (b) Min value 3 at x=2x = 2. (c) Range y≥3y \geq 3.
Full working
(a) Half of 4 is 2: (x−2)2=x2−4x+4(x - 2)^2 = x^2 - 4x + 4, so f(x)=(x−2)2+3f(x) = (x - 2)^2 + 3. (b) Minimum 3 at x=2x = 2. (c) Range [3,∞)[3, \infty).
11Problem 11
Answer
(a) f(x)=12x+2f(x) = \frac{1}{2}x + 2. (b) f−1(x)=2x−4f^{-1}(x) = 2x - 4. (c) ff and f−1f^{-1} are reflections of each other in y=xy = x.
Full working
(a) Gradient =4−14−(−2)=12= \frac{4 - 1}{4 - (-2)} = \frac{1}{2}. Through (−2,1)(-2, 1): 1=12(−2)+c⇒c=21 = \frac{1}{2}(-2) + c \Rightarrow c = 2. So f(x)=x2+2f(x) = \frac{x}{2} + 2. (b) y=x2+2⇒x=2y−4⇒f−1(x)=2x−4y = \frac{x}{2} + 2 \Rightarrow x = 2y - 4 \Rightarrow f^{-1}(x) = 2x - 4. (c) The two lines are reflections of each other across y=xy = x.
12Problem 12
Answer
(a) Anya: 3, Bao: 3, Cara: 3. (b) x≠1x \neq 1. (c) Not strictly equal — they agree everywhere except at x=1x = 1. (d) Cara — she states the domain restriction explicitly.
Full working
(a) Anya: 4−12−1=3\frac{4 - 1}{2 - 1} = 3. Bao: 2+1=32 + 1 = 3. Cara: 2+1=32 + 1 = 3 (with x≠1x \neq 1, OK). (b) Anya's denominator must not be zero, so x≠1x \neq 1. (c) Bao's formula is defined at x=1x = 1 (giving 2), but Anya's gives 00\frac{0}{0} — undefined. So they have different natural domains. (d) Cara — she matches Anya's natural domain and clears up the ambiguity.