MCR3U Final Exam — Functions, Grade 11 (OF Learning)

📑 Answer Key — Click to Expand (After You're Done!)

Q1

$ f(-2) = -23 $

Q2

(b) $ \{x \in \mathbb{R} \mid x \ne \pm 4\} $ — denominator $ x^2 - 16 $ is zero at $ \pm 4 $.

Q3

Swap and solve: $ x = \tfrac{y + 5}{2} \Rightarrow y = 2x - 5 $, so $ f^{-1}(x) = 2x - 5 $.

Q4

(a) $ (3x + 1)(2x - 5) $.

Q5

$ 2x^2 - 8x + 3 = 2(x^2 - 4x) + 3 = 2((x - 2)^2 - 4) + 3 = 2(x - 2)^2 - 5 $. Vertex $ (2, -5) $.

Q6

$ 32^{3/5} = 2^3 = 8 $.

Q7

$ x = 3 $.

Q8

$ t_{20} = 6 + 19(5) = 101 $.

Q9

$ t_6 = 4 \cdot 3^5 = 972 $.

Q10

Q11

Amplitude 3, period $ 180° $, range $ [-4, 2] $.

Q12

From roots $ -1 $ and $ 4 $: $ y = a(x + 1)(x - 4) $. Sub $ (1, -12) $: $ -12 = a(2)(-3) = -6a \Rightarrow a = 2 $. So $ y = 2(x + 1)(x - 4) = 2x^2 - 6x - 8 $.

Q13

Discriminant $ = 0 $: $ k^2 - 4(1)(25) = 0 \Rightarrow k^2 = 100 \Rightarrow k = \pm 10 $.

Q14

$ 3^{2x} \cdot 3^{3(1-x)} = 3^{-x} \Rightarrow 2x + 3 - 3x = -x \Rightarrow 3 = 0 $, contradiction. Recheck: $ 2x + 3 - 3x = -x \Rightarrow -x + 3 = -x \Rightarrow 3 = 0 $. No solution. (Or $ 2x + 3 - 3x = -x \Rightarrow 3 = 0 $: inconsistent — equation has no solution.)

Q15

$ t_{12} - t_5 = 7d = 65 - 23 = 42 \Rightarrow d = 6 $. Then $ t_5 = a + 4(6) = 23 \Rightarrow a = -1 $. So $ t_n = -1 + (n - 1)(6) = 6n - 7 $.

Q16

$ \sin B = \tfrac{14 \sin 38°}{9} \approx \tfrac{8.62}{9} \approx 0.957 $. Two values: $ B \approx 73.1° $ or $ B \approx 106.9° $. Both yield $ A + B < 180° $, so 2 triangles exist.

Q17–Q22

Communication / explanation questions — assess against rubric below.

Q23

(a) $ t_{\max} = -19.6/(2(-4.9)) = 2 $ s; $ h(2) = -4.9(4) + 39.2 + 1.5 = 21.1 $ m. (b) Solve $ -4.9t^2 + 19.6t + 1.5 = 0 \Rightarrow t \approx 4.075 $ s. (c) $ -4.9t^2 + 19.6t - 13.5 = 0 \Rightarrow t \approx 0.83 $ s or $ t \approx 3.17 $ s.

Q24

(a) $ N(t) = 200 \cdot 2^{t/4} $. (b) $ N(20) = 200 \cdot 2^5 = 6400 $. (c) Need $ 200 \cdot 2^{t/4} \geq 100000 \Rightarrow 2^{t/4} \geq 500 \Rightarrow t/4 \geq \log_2 500 \approx 8.97 \Rightarrow t \approx 35.9 $ h. So at least 36 hours.

Q25

$ i = 0.005, n = 60 $. (a) $ FV = 300 \cdot \tfrac{1.005^{60} - 1}{0.005} \approx 300 \cdot 69.77 \approx \$20{,}931 $. (b) Total deposits = $ 300 \cdot 60 = \$18{,}000 $. (c) Interest $ \approx \$2{,}931 $.

Q26

Distances: $ a = 72, b = 88 $ km. Angle between bearings: 42° + 30° = 72°. $ c^2 = 72^2 + 88^2 - 2(72)(88)\cos 72° = 5184 + 7744 - 12672(0.309) \approx 12928 - 3916 = 9012 $. $ c \approx 95 $ km.

Q27

Amplitude = 25, midline = 27, period 90 s. (a) $ h(t) = -25\cos(\tfrac{360°}{90}t) + 27 = -25\cos(4°t) + 27 $ (using degree mode). (b) $ h(30) = -25\cos(120°) + 27 = -25(-0.5) + 27 = 39.5 $ m. (c) Solve $ -25\cos(4t) + 27 = 50 \Rightarrow \cos(4t) = -0.92 \Rightarrow 4t \approx 156.93° \Rightarrow t \approx 39.2 $ s.