Comprehensive Final Evaluation — Chapters 1 through 8
✅ Counts for 30% of Final Course Mark
⏱️ Duration: 3 hours (180 minutes) | Total: /100 marks Materials allowed: Scientific calculator, ruler. No graphing technology during the exam.
Show all work. Answers without supporting work receive partial credit at best.
State the exact value of \( \cos\!\left(\dfrac{7\pi}{6}\right) \).
Solution:
\( \tfrac{7\pi}{6} \) is in Quadrant III; reference angle \( \tfrac{\pi}{6} \). Cosine is negative in Q III, so \( \cos\tfrac{7\pi}{6}=-\tfrac{\sqrt{3}}{2} \).
Question 5 [2 marks]
Identify all vertical asymptotes of \( r(x) = \dfrac{2x+3}{x^2 - 4} \).
Solution:
Set denominator to zero: \( x^2-4=0 \Rightarrow x=\pm 2 \). Numerator non-zero at these points. VAs: \( x=2 \) and \( x=-2 \).
Question 6 [2 marks]
For \( f(x)=2x-1 \) and \( g(x)=x^2 \), evaluate \( (f \circ g)(3) \).
Solution:
\( g(3)=9 \), then \( f(9)=2(9)-1=17 \).
Question 7 [3 marks]
Solve for \( x \): \( 5^{2x-1} = 125 \).
Solution:
\( 125 = 5^3 \), so \( 2x-1=3 \Rightarrow x=2 \).
Question 8 [2 marks]
Identify the amplitude and period of \( y = -4\cos\!\left(\tfrac{\pi}{2}x\right) + 1 \).
A polynomial passes through \( (-2,0), (0,0), (3,0) \) with degree 3 and \( f(1)=-12 \). Determine \( f(x) \) in factored form.
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Part B: Thinking & Inquiry [25 marks]
Question 12 [5 marks]
Solve the inequality \( x^3 - 4x^2 - 11x + 30 > 0 \). Show factoring and a sign analysis on a number line.
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Question 13 [5 marks]
Prove the identity \( \dfrac{1 - \cos 2\theta}{\sin 2\theta} = \tan\theta \). State any restrictions on \( \theta \).
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Question 14 [5 marks]
Solve algebraically on \( 0 \le x < 2\pi \): \( 2\sin^2 x - 3\sin x + 1 = 0 \).
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Question 15 [5 marks]
Solve for \( x \): \( \log_2(x+3) + \log_2(x-3) = 4 \). State and verify any restrictions.
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Question 16 [5 marks]
Given \( f(x)=x^2-1 \) and \( g(x)=\sqrt{x+5} \), determine \( (f\circ g)(x) \), state its domain, and find all \( x \) for which \( (f\circ g)(x)=8 \).
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Part C: Communication [25 marks]
Question 17 [5 marks]
Sketch \( f(x) = -\tfrac{1}{2}(x+3)(x-1)^2(x-4) \) without technology. Label all intercepts, state end behaviour, and indicate behaviour at each zero (cross or bounce).
Sketch one full period of \( y = 3\sin\!\left(2\!\left(x - \tfrac{\pi}{4}\right)\right) - 1 \). Label amplitude, period, phase shift, vertical translation, max value, min value, midline, and one zero of the function.
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Question 19 [4 marks]
Explain, using clear mathematical language, the difference between average rate of change (AROC) and instantaneous rate of change (IROC) of a function. Include a labelled diagram (described in words) showing both as slopes.
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Question 20 [4 marks]
A student writes: "Since \( \log(a+b)=\log a + \log b \), \( \log(2+8)=\log 2 + \log 8 \)." Identify the error, state the correct law of logarithms, and provide a numerical counterexample.
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Question 21 [4 marks]
For the rational function \( r(x) = \dfrac{x^2 - 9}{x - 3} \), explain why \( x=3 \) is a removable discontinuity (hole) rather than a vertical asymptote. State the simplified function and the coordinates of the hole.
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Question 22 [3 marks]
Explain why exponential functions of the form \( y=a\cdot b^x \) (with \( b>1 \)) eventually grow faster than any polynomial function, and contrast this with logarithmic growth.
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Part D: Application [25 marks]
Question 23 [5 marks]
Compound Interest. Maya invests $4000 in an account paying 6.5%/a compounded monthly. (a) Write the function \( A(t) \) giving the value after \( t \) years. (b) Find the value after 8 years. (c) Determine, algebraically using logarithms, the time required for the investment to double.
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Question 24 [5 marks]
Sound Intensity (Decibels). Loudness is given by \( L = 10\log_{10}\!\left(\dfrac{I}{I_0}\right) \) where \( I_0 = 10^{-12} \) W/m². (a) Determine \( L \) for a rock concert with intensity \( I = 10^{-2} \) W/m². (b) How many times more intense is a 110 dB jet engine than a 70 dB vacuum cleaner? Show all logarithmic reasoning.
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Question 25 [5 marks]
Tidal Modelling. The water depth in a harbour varies sinusoidally between a minimum of 1.4 m and a maximum of 7.8 m. One full tidal cycle is 12.4 hours, and high tide first occurs at \( t = 3 \) h after midnight. (a) Write a sinusoidal function \( d(t) \) modelling depth. (b) Predict the depth at 8 a.m. (c) A boat needs at least 4 m of water; for how many hours of a full cycle is it safe to dock?
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Question 26 [5 marks]
Open Box (Optimization-style). A 30 cm × 40 cm rectangular sheet of cardboard is made into an open-top box by cutting equal squares of side \( x \) cm from each corner and folding the sides up. (a) Express volume \( V(x) \) and state the realistic domain. (b) Estimate the value of \( x \) that yields the maximum volume by computing \( V(2), V(4), V(5), V(6), V(8) \), then state the maximum to the nearest cm³.
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Question 27 [5 marks]
Combined Functions — Population Model. The deer population in a park is modelled by \( D(t) = 200 + 50\sin\!\left(\tfrac{\pi}{6}t\right) \) (seasonal variation) and the wolf population by \( W(t) = 30 + 5t - 0.1t^2 \), where \( t \) is months from January 1. (a) Determine \( R(t) = D(t)/W(t) \), the deer-to-wolf ratio. (b) Calculate \( R(0) \) and \( R(6) \). (c) Use these values to estimate the average rate of change of the ratio over the first 6 months and interpret in context.
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Final Exam Evaluation Rubric (per Achievement Chart)
Level
Description
%
4
Thorough understanding; insightful explanations; high degree of effectiveness
80–100%
3
Considerable effectiveness — provincial standard met
70–79%
2
Some effectiveness, approaching standard
60–69%
1
Limited effectiveness
50–59%
R
Insufficient achievement of curriculum expectations
Below 50%
📘 Complete Answer Key & Worked Solutions (click to expand)
Test \( x=2 \): \( 8-16-22+30=0 \). Factor out \( (x-2) \): \( x^3-4x^2-11x+30=(x-2)(x^2-2x-15)=(x-2)(x-5)(x+3) \). Sign chart with critical points \( -3, 2, 5 \): the cubic with positive leading coefficient is positive on \( (-3,2)\cup(5,\infty) \).
Q13 [5] — Identity Proven
Using \( 1-\cos 2\theta=2\sin^2\theta \) and \( \sin 2\theta=2\sin\theta\cos\theta \):
\( \dfrac{2\sin^2\theta}{2\sin\theta\cos\theta}=\dfrac{\sin\theta}{\cos\theta}=\tan\theta \). ✓
Restrictions: \( \sin\theta\neq 0 \) and \( \cos\theta\neq 0 \), i.e. \( \theta\neq \tfrac{n\pi}{2},\ n\in\mathbb{Z} \).
Q14 [5] — \( x = \tfrac{\pi}{6}, \tfrac{5\pi}{6}, \tfrac{\pi}{2} \)
Let \( u=\sin x \): \( 2u^2-3u+1=0 \Rightarrow (2u-1)(u-1)=0 \Rightarrow u=\tfrac{1}{2} \) or \( u=1 \). For \( \sin x=\tfrac{1}{2} \) on \( [0,2\pi) \): \( x=\tfrac{\pi}{6}, \tfrac{5\pi}{6} \). For \( \sin x=1 \): \( x=\tfrac{\pi}{2} \).
\( (f\circ g)(x)=f(\sqrt{x+5})=(\sqrt{x+5})^2-1=x+5-1=x+4 \). Domain inherits from \( g \): \( x+5\ge 0 \Rightarrow x\ge -5 \). Set \( x+4=8\Rightarrow x=4 \) (in domain ✓).
Part C — Communication
Q17 [5] — Polynomial Sketch
Leading term \( -\tfrac{1}{2}x^4 \) → both ends to \( -\infty \). x-intercepts: \( -3 \) (cross, order 1), \( 1 \) (bounce, order 2), \( 4 \) (cross, order 1). y-intercept: \( f(0)=-\tfrac12(3)(1)(-4)=6 \). Sketch: enters from below at left, crosses at \( -3 \), rises through \( (0,6) \), bounces off x-axis at \( 1 \), descends, crosses at \( 4 \), exits below.
Q18 [5] — Sinusoidal Sketch
Amplitude \( =3 \). Period \( =\tfrac{2\pi}{2}=\pi \). Phase shift: \( \tfrac{\pi}{4} \) right. Vertical translation: \( -1 \) (midline \( y=-1 \)). Max \( =-1+3=2 \); Min \( =-1-3=-4 \). One zero: solve \( 3\sin(2(x-\tfrac{\pi}{4}))=1 \) — example zero at \( x=\tfrac{\pi}{4} \) where the curve crosses \( y=-1 \) (midline) ascending.
Q19 [4] — AROC vs IROC
AROC over \( [a,b] \) = \( \tfrac{f(b)-f(a)}{b-a} \), the slope of the secant line through \( (a,f(a)) \) and \( (b,f(b)) \). IROC at \( x=a \) is the limiting slope of secants as the second point approaches \( a \); geometrically the slope of the tangent line at \( (a,f(a)) \). Diagram: curve with secant chord (AROC) and tangent line touching at one point (IROC).
Q20 [4] — Logarithm Error
Error: \( \log(a+b)\neq \log a+\log b \). The correct product law is \( \log(ab)=\log a+\log b \). Counterexample: \( \log_{10}(2+8)=\log_{10}10=1 \), but \( \log_{10}2+\log_{10}8=\log_{10}16\approx 1.204 \neq 1 \).
Q21 [4] — Removable Discontinuity
\( r(x)=\tfrac{x^2-9}{x-3}=\tfrac{(x-3)(x+3)}{x-3}=x+3 \) for \( x\neq 3 \). The factor \( (x-3) \) cancels — both numerator and denominator are zero at \( x=3 \). Because the simplified function is defined and continuous, \( x=3 \) is a hole rather than a VA. Hole at \( (3,6) \).
Q22 [3] — Growth Comparison
Exponential \( b^x \) (with \( b>1 \)) increases by a constant factor over each unit interval, doubling/tripling repeatedly, while a polynomial \( x^n \) increases by amounts that grow only as a polynomial of \( x \). For sufficiently large \( x \), \( b^x \) exceeds any polynomial. Logarithmic functions are the inverse — they grow slower than any positive power of \( x \).
(a) \( L=10\log_{10}(10^{-2}/10^{-12})=10\log_{10}10^{10}=10(10)=100 \) dB.
(b) \( \tfrac{I_{110}}{I_{70}}=10^{(110-70)/10}=10^{4}=10\,000 \). The jet engine is 10 000 times more intense.
Q25 [5] — Tidal Model
Amplitude \( =\tfrac{7.8-1.4}{2}=3.2 \). Vertical shift \( c=\tfrac{7.8+1.4}{2}=4.6 \). Period \( =12.4\Rightarrow k=\tfrac{2\pi}{12.4}=\tfrac{\pi}{6.2} \). High tide at \( t=3 \) so phase shift \( d=3 \).
(a) \( d(t)=3.2\cos\!\left(\tfrac{\pi}{6.2}(t-3)\right)+4.6 \).
(b) \( d(8)=3.2\cos\!\left(\tfrac{\pi}{6.2}(5)\right)+4.6=3.2\cos(2.534)+4.6\approx 3.2(-0.815)+4.6\approx 1.99 \) m.
(c) Solve \( d(t)\ge 4 \Rightarrow \cos\!\left(\tfrac{\pi}{6.2}(t-3)\right)\ge -0.1875 \). The cosine is at least \( -0.1875 \) over a central interval of width \( \tfrac{2}{\pi}\arccos(-0.1875)\cdot 6.2 \approx \tfrac{2(1.7595)(6.2)}{\pi}\approx 6.95 \) hours per cycle.
(a) \( R(t)=\dfrac{200+50\sin(\pi t/6)}{30+5t-0.1t^2} \).
(b) \( R(0)=\tfrac{200+0}{30}=6.667 \); \( R(6)=\tfrac{200+50\sin(\pi)}{30+30-3.6}=\tfrac{200+0}{56.4}\approx 3.546 \).
(c) AROC \( =\tfrac{R(6)-R(0)}{6-0}\approx \tfrac{3.546-6.667}{6}\approx -0.520 \) deer/wolf per month. The number of deer per wolf is decreasing — wolves grow while deer oscillate around a constant mean.