MCV 4U1 Grade 12 - Calculus & Vectors
Introduction to Calculus Practice Test 3
K/U
/37
APP
/16
COM
/16
COM LEVEL
Instructions
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
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Answer all questions in the space provided.
Neat and complete solutions are required for full marks.
Non-graphing calculators are permitted, but may not be shared.
Your communication mark will be determined by how well you
answer all of the questions: including proper mathematical form,
notation and narration (where appropriate).
Good luck!
Knowledge and Understanding [37 marks]
Place the best answer in the space provided to the right of the questions.
Multiple Choice [5 marks]
Questions 1 to 4 refer to the graph of y = f (x) shown below.
1. The function y = f (x):
a) is differentiable at x = 4
b) is differentiable everywhere except at x = 1, x = 4, and x = 7
c) is continuous for all x ∈ ℝ.
d) is discontinuous at x = – 1.
Answer: b)
2. The function y = f (x) has a removable discontinuity at:
a) x = 7
b) x = 4
c) x = 1
d) x = 0
Answer: a)
3. The value of f (7) is:
a) 0
b) – 0.5
Answer: d)
c) – 1
d) – 1.5
4. The value of lim+ 𝑓(𝑥) is:
𝑥→1
a) 2
b) – 0.5
Answer: c)
c) – 1
d) – 1.5
5. Which of the following is not a true statement about limits?
a) A limit can be used to determine the average rate of change between two
points on a graph.
b) A limit can be used to determine the behaviour of a graph on either side of
a vertical asymptote.
c) A limit can be used to determine if a graph is discontinuous.
d) A limit can be used to determine the end behaviour of a graph.
Answer: a)
Short Answer [32 marks]
Show all work in the space provided.
1. Determine the average rate of change between (–5, 9) and (–7, 11). Leave
your answer in exact form Show all your work for full credit. [2]
Solution
11 − 9
2
rav =
= = –1
−7 − (−5)
−2
2. Estimate the instantaneous rate of change at the tangent point indicated on
the graph shown below. Show all your work for full credit. [2]
Solution
4−0
m=
=2
3−1
3. State the limit of each sequence, if it exists. If it does not exist, write “does
not exist” or “D.N.E.” and explain why. No marks will be assigned if there is
no justification for your answer. [3]
1
1
1
1
a) 1, , 1, , 1, , 1, , 1, ...
2
3
4
5
Solution
If a sequence has a limit, then any it’s subsequence has the same limit.
In our example this does not hold:
1
lim 𝑡2𝑛 = lim
=0
𝑛→∞ 𝑛+1
𝑛→∞
lim 𝑡2𝑛 + 1 = lim 1 = 1
𝑛→∞
𝑛→∞
Therefore, the sequence does not have a limit.
b) 3.1, 3.01, 3.001, 3.0001, 3.00001, ...
Solution
This sequence has a limit of 3: lim 𝑡𝑛 = lim (3 +
𝑛→∞
𝑛→∞
1
10𝑛
) = 3.
4. Determine the following limits, if they exist. If they do not exist, write
“does not exist” or “D.N.E.” and explain why. No marks will be assigned if
there is no justification for your answer. [3]
a) lim
𝑥2 + 4
𝑥→−2
𝑥+2
Solution
This limit does not exist:
lim −
𝑥→−2
𝑥2 + 4
𝑥+2
= – ∞, lim +
𝑥→−2
𝑥2 + 4
𝑥+2
=+∞
b) lim √2𝑥 + 1
𝑥→3
Solution
lim √2𝑥 + 1 = √2(3) + 1 = √7
𝑥→3
5. Determine the following limits, if they exist. If they do not exist, write
“does not exist” or “D.N.E.” and explain why. No marks will be assigned if
there is no justification for your answer. [3]
a) Given that lim− f (x) = 5 and lim+ f (x) = 5, then lim f (x) = ?
𝑥 →1
𝑥 →1
𝑥 →1
Solution
lim− f (x) = lim+ f (x) = 5 ∴ lim f (x) = 5
𝑥 →1
𝑥 →1
𝑥 →1
b) Given that lim − f (x) = 1 and lim + f (x) = – 5, then lim f (x) = ?
𝑥 →−3
𝑥 →−3
𝑥 →−3
Solution
lim − f (x) ≠ lim + f (x) ∴ lim f (x) does not exist
𝑥 →−3
𝑥 →−3
𝑥 →−3
6. a) Use first principles to determine the derivative of f (x) = –2x3 + 1. [5]
Solution
Δ f = f (x + h) – f (x) = –2(x + h)3 + 1 + 2x3 – 1
= –2(x + h)3 + 2x3
= –2(x3 + 3x2h + 3xh2 + h3) + 2x3
= – 6x2h – 6xh2 – 2h3
Δ𝑓
f '(x) = lim
= lim (– 6x2 – 6xh – 2h2) = – 6x2
ℎ →0 ℎ
ℎ →0
b) Evaluate each derivative: [2]
(i) f' (–3)
Solution
f' (–3) = – 6(–3)2 = –54
2
(ii) f' ( )
3
Solution
2
2 2
8
3
3
3
f' ( ) = – 6( ) = –
c) Determine the equation of the tangent for x = –3. Show all work for full
credit. [3]
Solution
f (–3) = –2(–3) 3 + 1 = 55, m = f '(–3) = –54
y – y1 = m (x – x1) ⇒ y – 55 = – 54 (x + 3)
y = – 54 x – 107
The equation of the tangent is y = – 54 x – 107.
7. Let
2
1
−
𝑥
if 𝑥 ≥ 0
f (x) = {
2𝑥 − 1 if 𝑥 < 0
[3, 3, 3]
a) Find the following limits, if they exist.
i) lim+ f (x) ii) lim− f (x) iii) lim f (x)
𝑥→0
𝑥→0
𝑥→0
Solution
lim+ f (x) = lim+ (1 – x2) = 1
𝑥→0
𝑥→0
lim f (x) = lim− (2x – 1) = – 1
𝑥→0−
𝑥→0
lim f (x) ≠ lim− f (x) ∴ lim f (x) does not exist
𝑥→0+
𝑥→0
𝑥→0
b) Sketch the graph of 𝑓(𝑥 ). Label carefully.
Solution
c) Determine whether or not f (x) is discontinuous. If f (x) is discontinuous,
explain why, and identify where the discontinuity exists.
Solution
f (x) is discontinuous at x = 0 (jump discontinuity), since lim f (x) does not
𝑥→0
exist.
Application [16 marks]
1. The displacement, s, in meters, of a particle moving at a straight line is
given by s (t) = 5t2 – 6t + 14, where t is measured in seconds. [4, 6]
a) Find the average velocity over the time interval 3 ≤ t ≤ 4.
Solution
s (3) = 5(32) – 6(3) + 14 = 41
s (4) = 5(42) – 6(4) + 14 = 70
rav [3, 4] =
𝑠 (4) − 𝑠 (3)
4−3
=
70 − 41
1
= 29 m/s
b) Find the instantaneous velocity when t = 3.
Solution
s (2.9) = 5(2.92) – 6(2.9) + 14 = 38.65
s (3.1) = 5(3.12) – 6(3.1) + 14 = 43.45
rav [2.9, 3.1] =
𝑠 (3.1) − 𝑠 (2.9)
3.1 − 2.9
=
43.45 − 38.65
0.2
= 24 m/s
s (2.99) = 5(2.992) – 6(2.99) + 14 = 40.7605
s (3.01) = 5(3.012) – 6(3.01) + 14 = 41.2405
rav [2.99, 3.01] =
𝑠 (3.01) − 𝑠 (2.99)
3.01 − 2.99
=
41.2405− 40.7605
0.02
= 24 m/s
s (2.999) = 5(2.9992) – 6(2.999) + 14 = 40.976005
s (3.001) = 5(3.0012) – 6(3.001) + 14 = 41.024005
𝑠 (3.001) − 𝑠 (2.999)
rav [2.999, 3.001] =
3.001 − 2.999
Therefore, r ins (3) = 24 m/s
=
41.024005− 40.976005
0.002
= 24 m/s
2. A spherical balloon is being inflated. Find the rate of change of the surface
area of the balloon with respect to the radius when the radius is 10 cm. [6]
Hint: If you cannot remember the formula for surface area, you may request it
from your teacher for 1 mark.
Solution
SA(r) =4πr2
SA(10) = 400π
SA(10 + h) =4π(10 + h)2 = 4π(100 + 20h + h2) = 400π + 80πh + 4πh2
rins (10) = lim
𝑆𝐴(10 + ℎ) − 𝑆𝐴(10)
ℎ→0
ℎ
= lim
80πℎ + 4πℎ2
ℎ→0
ℎ
2
= lim (80π + 4πh) = 80π cm /s
ℎ→0
The rate of change of the surface area of the balloon with respect to the radius
when the radius is 10 cm is 80π cm2/s.
Communication [16 marks]
1. For the graph shown below, describe and compare the instantaneous rate
of change at the points indicated. Explain your reasoning.
Note: You do not necessarily need to calculate the instantaneous rate of
change at each of these points to answer this question.
i) A, B, and E
Answer:
rins (A) < rins (B) < rins (E) < 0
ii) C and D
Answer:
rins (D) > rins (C) > 0
iii) A, B and C
Answer:
rins (A) < rins (B) < 0 < rins (C) and |rins (A)| > |rins (B)| > |rins (C)|
2. Explain how limits can help to determine if a function is continuous. [3]
Answer: If a limit at a given point does not exist, then the function is
discontinuous at this point. If a limit at a given point exists and equals to the
value of the function at this point, then the function is continuous at this
point.
3. Describe the types of discontinuities that a graph might have. Explain why
their names make sense. [6]
Answer:
Discontinuities can be classified as jump, infinite, removable, endpoint, or
mixed.
Removable discontinuities are characterized by the fact that the limit exists.
In the graphs below, there is a hole at x = a. Removable discontinuities can
be "fixed" by re-defining the function. Below are examples of removable
discontinuities.
The other types of discontinuities are characterized by the fact that the limit
does not exist.
Jump Discontinuities: both one-sided limits exist, but have different values.
Infinite Discontinuities: both one-sided limits are infinite.
Endpoint Discontinuities: only one of the one-sided limits exists.
Mixed: at least one of the one-sided limits does not exist.
Thinking, Inquiry and Problem solving [20 marks]
1. Determine the equations of both lines that are tangent to the graph of y =
x2, and pass through the point (0, –1). Sketch the curve and the required
tangent lines. Label carefully. For full credit, be sure to include a brief
description of each step using mathematical terminology.
Solution
Determine the slope of the tangent line at the point (a, a2):
m = lim
ℎ→0
(𝑎 + ℎ)2 − 𝑎2
ℎ
= lim
ℎ→0
𝑎2 + 2𝑎ℎ+ ℎ2 − 𝑎2
ℎ
= lim
ℎ→0
2𝑎ℎ+ ℎ2
ℎ
= lim (2a + h) = 2a.
ℎ→0
Determine the equation of the tangent line at the point (a, a2):
y – y1 = m (x – x1) ⇒ y – a2 = 2a (x – a)
y = 2ax – a2
Determine the x-coordinates of the tangency point:
x = 0, y = –1 ⇒ –1 = 2a (0) – a2 ⇒ a2 = 1
a = 1 or a = –1
If a = 1, then the equation of the tangent line is y = 2x – 1
If a = –1, then the equation of the tangent line is y = –2x – 1