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Solutions — Full Answer Key

Ecolint Campus des NationsMathematics
Year 10 · Functions

Solutions · Full Answer Key

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

Pack A — Answers

Bronze
1.f(4)=14f(4) = 14
2.f(3)=10f(3) = 10
3.f(−3)=−1f(-3) = -1
4.x=5x = 5
5.13
6.x=5x = 5
7.{3,6,9,12,15}\{3, 6, 9, 12, 15\}
8.g(0)=−4g(0) = -4, g(3)=5g(3) = 5
9.f(x)=3x+5f(x) = 3x + 5
10.f(x)=4x+7f(x) = 4x + 7
Silver
11.Domain: R\mathbb{R}; Range: R\mathbb{R}
12.f(x)≥3f(x) \geq 3 (i.e. [3,∞)[3, \infty))
13.Range: 1≤f(x)≤131 \leq f(x) \leq 13
14.(f∘g)(4)=11(f \circ g)(4) = 11
15.(f∘g)(x)=2x+7(f \circ g)(x) = 2x + 7
16.(a) 16 (b) 10
17.f−1(x)=x−23f^{-1}(x) = \dfrac{x - 2}{3}
18.f−1(9)=4f^{-1}(9) = 4
19.f(5)=4f(5) = 4; f(x)=0f(x) = 0 at x=3x = 3
20.f(x)=3x+2f(x) = 3x + 2
Gold
21.(f∘g)(x)=3x2+1(f \circ g)(x) = 3x^2 + 1
22.(g∘f)(x)=(2x+3)2=4x2+12x+9(g \circ f)(x) = (2x + 3)^2 = 4x^2 + 12x + 9
23.x≥4x \geq 4
24.x∈R,x≠3x \in \mathbb{R}, x \neq 3
25.x=4x = 4 or x=−1x = -1
26.f−1(x)=2x−3f^{-1}(x) = 2x - 3
27.(f∘f−1)(7)=7(f \circ f^{-1})(7) = 7 ✓
28.11
29.4
30.f−1(x)=x+3f^{-1}(x) = \sqrt{x} + 3 (for x≥0x \geq 0)
Platinum
31.(f∘g)−1(x)=x−72(f \circ g)^{-1}(x) = \dfrac{x - 7}{2}
32.x>4x > 4
33.(a) C(n)=12n+25C(n) = 12n + 25 (b) C−1(c)=c−2512C^{-1}(c) = \dfrac{c - 25}{12} (days for cost cc)
34.(a) Min = 2 at x=3x = 3 (b) Range: y≥2y \geq 2
35.x=±2x = \pm 2
36.a=4a = 4, b=3b = 3
37.x+3x + 3
38.x=−1x = -1
39.10 minutes
40.x≥4x \geq 4

Pack B — Answers

Bronze
1.f(4)=17f(4) = 17
2.f(5)=21f(5) = 21
3.f(−2)=−9f(-2) = -9
4.x=5x = 5
5.17
6.x=7x = 7
7.{4,8,12,16,20}\{4, 8, 12, 16, 20\}
8.g(0)=−9g(0) = -9, g(4)=7g(4) = 7
9.f(x)=4x+2f(x) = 4x + 2
10.f(x)=6x+1f(x) = 6x + 1
Silver
11.Domain: R\mathbb{R}; Range: R\mathbb{R}
12.f(x)≥−2f(x) \geq -2 (i.e. [−2,∞)[-2, \infty))
13.Range: 5≤f(x)≤175 \leq f(x) \leq 17
14.(f∘g)(2)=11(f \circ g)(2) = 11
15.(f∘g)(x)=3x+14(f \circ g)(x) = 3x + 14
16.(a) 36 (b) 18
17.f−1(x)=x+45f^{-1}(x) = \dfrac{x + 4}{5}
18.f−1(13)=5f^{-1}(13) = 5
19.f(4)=3f(4) = 3; f(x)=0f(x) = 0 at x=3x = 3
20.f(x)=4x+3f(x) = 4x + 3
Gold
21.(f∘g)(x)=2x2+5(f \circ g)(x) = 2x^2 + 5
22.(g∘f)(x)=(3x+1)2=9x2+6x+1(g \circ f)(x) = (3x + 1)^2 = 9x^2 + 6x + 1
23.x≥7x \geq 7
24.x∈R,x≠5x \in \mathbb{R}, x \neq 5
25.x=6x = 6 or x=−1x = -1
26.f−1(x)=4x−5f^{-1}(x) = 4x - 5
27.(f∘f−1)(9)=9(f \circ f^{-1})(9) = 9 ✓
28.17
29.5
30.f−1(x)=x+5f^{-1}(x) = \sqrt{x} + 5 (for x≥0x \geq 0)
Platinum
31.(f∘g)−1(x)=x−173(f \circ g)^{-1}(x) = \dfrac{x - 17}{3}
32.x>6x > 6
33.(a) C(n)=15n+40C(n) = 15n + 40 (b) C−1(c)=c−4015C^{-1}(c) = \dfrac{c - 40}{15}
34.(a) Min = −3-3 at x=5x = 5 (b) Range: y≥−3y \geq -3
35.x=±3x = \pm 3
36.a=4a = 4, b=5b = 5
37.8x8x
38.x=1x = 1
39.10 minutes
40.x≥7x \geq 7

Problem-solving — Worked Solutions

1Problem 1
Answer
(a) 32, 212 (b) C(f)=59(f−32)C(f) = \frac{5}{9}(f - 32) (c) 100°C (boiling point) (d) c=−40c = -40
Full working
(a) F(0)=32F(0) = 32; F(100)=95(100)+32=180+32=212F(100) = \frac{9}{5}(100) + 32 = 180 + 32 = 212.

(b) Solve f=95c+32f = \frac{9}{5}c + 32 for cc: f−32=95cf - 32 = \frac{9}{5}c, so c=59(f−32)c = \frac{5}{9}(f - 32). Inverse: F−1(f)=59(f−32)F^{-1}(f) = \frac{5}{9}(f - 32).

(c) F−1(212)=59(180)=100F^{-1}(212) = \frac{5}{9}(180) = 100. Interpret: 212°F = 100°C (boiling point of water).

(d) Set 95c+32=c\frac{9}{5}c + 32 = c: 95c−c=−32\frac{9}{5}c - c = -32, so 45c=−32\frac{4}{5}c = -32, c=−40c = -40. So **−40°-40°C = −40°-40°F**.
2Problem 2
Answer
(a) 5, 6, 17, 27 (b) f∘g=3x2+5f \circ g = 3x^2 + 5; g∘f=(3x−1)2+2g \circ f = (3x-1)^2 + 2 (c) x=±3x = \pm\sqrt{3}
Full working
(a) f(2)=5f(2) = 5; g(2)=6g(2) = 6; f(g(2))=f(6)=3(6)−1=17f(g(2)) = f(6) = 3(6) - 1 = 17; g(f(2))=g(5)=25+2=27g(f(2)) = g(5) = 25 + 2 = 27.

(b) (f∘g)(x)=f(x2+2)=3(x2+2)−1=3x2+5(f \circ g)(x) = f(x^2 + 2) = 3(x^2 + 2) - 1 = 3x^2 + 5. (g∘f)(x)=g(3x−1)=(3x−1)2+2=9x2−6x+3(g \circ f)(x) = g(3x - 1) = (3x-1)^2 + 2 = 9x^2 - 6x + 3.

(c) 3x2+5=14⇒x2=3⇒x=±33x^2 + 5 = 14 \Rightarrow x^2 = 3 \Rightarrow x = \pm\sqrt{3}.
3Problem 3
Answer
(a) f−1(x)=x+75f^{-1}(x) = \frac{x+7}{5} (b) g−1(x)=4x−32g^{-1}(x) = \frac{4x - 3}{2} (c) h−1(x)=x+2h^{-1}(x) = \sqrt{x} + 2 for x≥0x \geq 0
Full working
(a) y=5x−7⇒x=y+75y = 5x - 7 \Rightarrow x = \frac{y + 7}{5}. So f−1(x)=x+75f^{-1}(x) = \frac{x + 7}{5}.

(b) y=2x+34⇒4y=2x+3⇒x=4y−32y = \frac{2x + 3}{4} \Rightarrow 4y = 2x + 3 \Rightarrow x = \frac{4y - 3}{2}. So g−1(x)=4x−32g^{-1}(x) = \frac{4x - 3}{2}.

(c) y=(x−2)2y = (x - 2)^2. Since x≥2x \geq 2, x−2≥0x - 2 \geq 0, so y=x−2\sqrt{y} = x - 2, giving x=y+2x = \sqrt{y} + 2. So h−1(x)=x+2h^{-1}(x) = \sqrt{x} + 2, defined for x≥0x \geq 0.
4Problem 4
Answer
(a) 60−2x60 - 2x m (b) A(x)=60x−2x2A(x) = 60x - 2x^2 (c) 0<x<300 < x < 30 (d) x=15x = 15 m (max area = 450 m²)
Full working
(a) Total fencing used: 2x+(side along wall)=602x + (\text{side along wall}) = 60. So side along wall = 60−2x60 - 2x.

(b) Area = x⋅(60−2x)=60x−2x2x \cdot (60 - 2x) = 60x - 2x^2.

(c) For a valid rectangle: x>0x > 0 and 60−2x>0⇒x<3060 - 2x > 0 \Rightarrow x < 30. So domain 0<x<300 < x < 30.

(d) A(x)=−2x2+60xA(x) = -2x^2 + 60x. Vertex of downward parabola at x=−602(−2)=15x = -\frac{60}{2(-2)} = 15 m. Max area: A(15)=60(15)−2(225)=900−450=450A(15) = 60(15) - 2(225) = 900 - 450 = 450 m².
5Problem 5
Answer
g(x)=2x+3g(x) = 2x + 3
Full working
Let g(x)=ax+bg(x) = ax + b. Then (f∘g)(x)=f(ax+b)=2(ax+b)+1=2ax+2b+1(f \circ g)(x) = f(ax + b) = 2(ax + b) + 1 = 2ax + 2b + 1. Set equal to 4x+74x + 7:
- 2a=4⇒a=22a = 4 \Rightarrow a = 2
- 2b+1=7⇒b=32b + 1 = 7 \Rightarrow b = 3

So g(x)=2x+3g(x) = 2x + 3.

Check: (f∘g)(x)=2(2x+3)+1=4x+7(f \circ g)(x) = 2(2x + 3) + 1 = 4x + 7 ✓.
6Problem 6
Answer
(a) Yes (b) No (one-to-many) (c) Yes (d) No (one-to-many)
Full working
A function must assign exactly one output to each input.

(a) **Function.** Each student → exactly one ID.

(b) **Not a function.** A town has many citizens — one input maps to multiple outputs. (It's a relation but not a function.)

(c) **Function.** For each xx, x2x^2 is a single number.

(d) **Not a function.** Each positive yy has *two* preimages: ±y\pm\sqrt{y}. So one input maps to two outputs.
7Problem 7
Answer
(a) x≠3x \neq 3 (b) f(x)≠0f(x) \neq 0 (c) f−1(x)=1x+3f^{-1}(x) = \frac{1}{x} + 3, domain x≠0x \neq 0
Full working
(a) Denominator cannot be zero: x≠3x \neq 3. Domain: R∖{3}\mathbb{R} \setminus \{3\}.

(b) 1x−3\frac{1}{x - 3} takes every non-zero real value (as x→±∞x \to \pm\infty, f→0f \to 0 but never reaches it; as x→3x \to 3, f→±∞f \to \pm\infty). Range: R∖{0}\mathbb{R} \setminus \{0\}.

(c) Solve y=1x−3y = \frac{1}{x - 3}: y(x−3)=1y(x - 3) = 1, so x=1y+3x = \frac{1}{y} + 3. Inverse: f−1(x)=1x+3f^{-1}(x) = \frac{1}{x} + 3. Domain: x≠0x \neq 0 (matches range of ff, as expected).
8Problem 8
Answer
(a) Approx [−1,4][-1, 4] (b) No (c) Inverse not a function on this domain
Full working
(a) From the given values, yy ranges from −1-1 to 44 (taking values 4,0,−1,1,44, 0, -1, 1, 4). Estimated range: [−1,4][-1, 4].

(b) The value y=4y = 4 appears at both x=0x = 0 and x=6x = 6 — so two different inputs give the same output. **Not one-to-one**.

(c) An inverse function must assign a unique input to each output. If two xx values share an output, you cannot decide which xx to map back to. So a non-one-to-one function does **not have an inverse** unless you restrict its domain.
9Problem 9
Answer
(a) E(p)=1.18pE(p) = 1.18p (b) E−1(e)=e/1.18E^{-1}(e) = e/1.18 (euros to pounds) (c) £423.73
Full working
(a) E(p)=1.18pE(p) = 1.18p.

(b) e=1.18p⇒p=e1.18e = 1.18p \Rightarrow p = \frac{e}{1.18}. So E−1(e)=e1.18E^{-1}(e) = \frac{e}{1.18}. This converts euros back to pounds.

(c) E−1(500)=5001.18≈423.7288≈£423.73E^{-1}(500) = \frac{500}{1.18} \approx 423.7288 \approx \mathbf{\pounds 423.73}.
10Problem 10
Answer
(a) Not 1-1 (b) x\sqrt{x} (c) −x-\sqrt{x} (d) Both equal what they should
Full working
(a) f(2)=f(−2)=4f(2) = f(-2) = 4: two inputs give the same output, so ff is not one-to-one and has no inverse on R\mathbb{R}.

(b) Restricted to x≥0x \geq 0: ff is one-to-one and onto [0,∞)[0, \infty). Inverse: f−1(x)=xf^{-1}(x) = \sqrt{x}.

(c) Restricted to x≤0x \leq 0: outputs again cover [0,∞)[0, \infty) but inputs are negative. Inverse takes positive xx to negative root: f−1(x)=−xf^{-1}(x) = -\sqrt{x}.

(d) For (b):
- (f∘f−1)(9)=f(9)=f(3)=9(f \circ f^{-1})(9) = f(\sqrt{9}) = f(3) = 9 ✓.
- (f−1∘f)(3)=f−1(9)=9=3(f^{-1} \circ f)(3) = f^{-1}(9) = \sqrt{9} = 3 ✓.
11Problem 11
Answer
(a) See working (b) See working (c) f(x)=bf(x) = b (constant — not invertible) or f(x)=−x+bf(x) = -x + b for any bb
Full working
(a) f(f(x))=f(−x)=−(−x)=xf(f(x)) = f(-x) = -(-x) = x. So applying ff twice gives back xx → self-inverse ✓.

(b) f(f(x))=f(1x)=11/x=xf(f(x)) = f(\frac{1}{x}) = \frac{1}{1/x} = x ✓.

(c) Require f(f(x))=xf(f(x)) = x. Compute: f(f(x))=a(ax+b)+b=a2x+ab+bf(f(x)) = a(ax + b) + b = a^2 x + ab + b. Set =x= x: a2=1a^2 = 1 and ab+b=0ab + b = 0, i.e. b(a+1)=0b(a + 1) = 0.

Case 1: a=1a = 1, then b(2)=0⇒b=0b(2) = 0 \Rightarrow b = 0. So f(x)=xf(x) = x (the identity, trivially self-inverse).
Case 2: a=−1a = -1, then b(0)=0b(0) = 0 — any bb. So f(x)=−x+bf(x) = -x + b for any bb.

**Family of self-inverse linear functions: f(x)=−x+bf(x) = -x + b (plus the identity).** Geometrically, these are reflections across the line y=b2y = \frac{b}{2}.
12Problem 12
Answer
g(x)=2x+4g(x) = 2x + 4; unique because ff is one-to-one
Full working
Let g(x)=ax+bg(x) = ax + b. Then f(g(x))=2(ax+b)+5=2ax+2b+5f(g(x)) = 2(ax + b) + 5 = 2ax + 2b + 5.

Set equal to h(x)=4x+13h(x) = 4x + 13:
- 2a=4⇒a=22a = 4 \Rightarrow a = 2
- 2b+5=13⇒b=42b + 5 = 13 \Rightarrow b = 4

So g(x)=2x+4g(x) = 2x + 4.

**Uniqueness.** ff is linear with non-zero gradient → one-to-one → invertible. Given hh, the function gg is forced to equal f−1∘hf^{-1} \circ h, which is unique: g(x)=h(x)−52=4x+82=2x+4g(x) = \frac{h(x) - 5}{2} = \frac{4x + 8}{2} = 2x + 4 ✓. So **gg is unique**.