Mathematics 100-201
Page 1 of 9
Student-No.:
Mock Midterm 1
Duration: 50 minutes
This test has 5 questions on 9 pages, for a total of 40 points.
• Read all the questions carefully before starting to work.
• Q1 and Q2 are short-answer questions; put your answer in the boxes provided.
• All other questions are long-answer; you should give complete arguments and explanations
for all your calculations; answers without justifications will not be marked.
• Continue on the back of the previous page if you run out of space.
• Attempt to answer all questions for partial credit.
• This is a closed-book examination. None of the following are allowed: documents,
cheat sheets or electronic devices of any kind (including calculators, cell phones, etc.)
First Name:
Last Name:
Student-No:
Section:
Signature:
Question:
1
2
3
4
5
Total
Points:
9
12
7
7
5
40
Score:
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not fall into the traditional, paper-based method, examination candidates shall adhere to any special rules for conduct as established and
articulated by the examiner.
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Mathematics 100-201
Page 2 of 9
Student-No.:
Short-Answer Questions. Questions 1 and 2 are short-answer questions. Put your answer in
the box provided. Full marks will be given for a correct answer placed in the box. Show your
work also, for part marks. Each part is worth 3 marks, but not all parts are of equal difficulty.
Simplify your answers as much as possible in Questions 1 and 2.
9 marks
1. Determine whether each of the following limits exists, and find the value if they do. If a
limit below does not exist, determine whether it “equals” ∞, −∞, or neither.
(a)
t2 − 9
lim 4
t→3 t − 81
Answer: 1/18
Solution: Try direct substitution — get 0/0 so simplify.
t2 − 9
t2 − 9
=
t4 − 81
(t2 − 9)(t2 + 9)
1
= 2
t +9
provided t 6= 3
Hence
t2 − 9
1
1
= lim 2
=
4
t→3 t − 81
t→3 t + 9
18
lim
(b)
√
√
lim ( x2 + 5x − x2 + 4x)
x→+∞
Answer:
1
2
Solution: If we naively take the limit we get +∞ − ∞ which is undetermined.
Mathematics 100-201
Page 3 of 9
Student-No.:
Multiply with the conjugate expression
√
√
√
√
√
√
( x2 + 5x − x2 + 4x)( x2 + 5x + x2 + 4x)
2
2
√
√
lim ( x + 5x − x + 4x) = lim
x→+∞
x→+∞
( x2 + 5x + x2 + 4x)
x2 + 5x − x2 − 4x
√
= lim √
x→+∞ ( x2 + 5x +
x2 + 4x)
x
√
= lim √
x→+∞ ( x2 + 5x +
x2 + 4x)
x
q
= lim q
x→+∞
5
2
( x (1 + x ) + x2 (1 + x4 ))
x
q
q
= lim
x→+∞
|x|( (1 + x5 ) + (1 + x4 ))
x
x≥0
q
q
= lim
x→+∞
5
x( (1 + x ) + (1 + x4 ))
= lim
x→+∞
=
1
q
q
( (1 + x5 ) + (1 + x4 ))
1
2
(c)
2 − cos 3x
x→+∞
3x5
lim
Answer: 0
Solution: cos(3x) doesn’t have a limit as x → ∞ but it is at least bounded and
the denominator goes to infinity. We will use the Squeeze Theorem.
−1 ≤ cos(3x) ≤ 1 ⇒ −1 ≤ − cos(3x) ≤ 1
⇒ −2 + 1 ≤ 2 − cos(3x) ≤ 2 + 1
⇒
1
2 − cos(3x)
3
≤
≤ 5
5
5
3x
3x
3x
Taking limits and employing the Squeeze theorem
2 − cos(3x)
=0
x→+∞
3x5
lim
(d)
√
lim
x→1
√
1+x− 3−x
x−1
Answer:
√1
2
Mathematics 100-201
Page 4 of 9
Student-No.:
Solution: When x → 1 we get 0/0 so something cancels. Multiply by conjugate
to simplify
√
√
√
√
√
√
1+x− 3−x
1+x− 3−x
1+x+ 3−x
√
=
·√
x−1
x−1
1+x+ 3−x
(1 + x) − (3 − x)
√
√
=
(x − 1)( 1 + x + 3 − x)
2(x − 1)
√
√
=
(x − 1)( 1 + x + 3 − x)
2
√
=√
provided x 6= 1
1+x+ 3−x
So the limit is
√
√
2
2
1+x− 3−x
1
√
lim
= lim √
= √ =√
x→1
x→1
x−1
1+x+ 3−x
2 2
2
Mathematics 100-201
12 marks
Page 5 of 9
2. (a) If f (x) = −9x2 + xex +
1
√
,x
2 x
Student-No.:
6= 0, find f 0 (x).
Answer: −18x + ex (x + 1) +
x−3/2 , x 6= 0
−1
4
·
Solution:
1
f (x) = −9x2 + xex + √
2 x
1
= −9x2 + xex + x−1/2
2
By the Product Rule: (xex )0 = 1 · ex + x · ex = ex (x + 1) since (ex )0 = ex , hence
1 −1 −3/2
·
x
2 2
−1 −3/2
= −18x + ex (x + 1) +
·x
, x 6= 0
4
f 0 (x) = −18x + ex (x + 1) +
(b) Let f (t) =
2t
. Determine where f (t) is differentiable.
1 − t2
Answer: t 6= ±1
Solution: Both numerator and denominator are differentiable so by the theorem
in class (about the quotient rule) the function is differentiable but for the points
where the denominator becomes 0, that is, t2 − 1 = 0 ⇒ t = ±1.
(c) Let f (x) be a function differentiable at x = 2 and let g = f (x)/x. The tangent line to
the curve y = f (x) at x = 2 has slope 3 while the tangent line to the curve y = g(x) at
x = 2 has slope 5. What is f (2)?
Answer: -14
Solution: We are told that f 0 (2) = 3 and g 0 (2) = 5.
xf 0 (x) − f (x)
x2
0
2f (2) − f (2)
g 0 (2) =
4
6 − f (2)
5=
4
g 0 (x) =
So f (2) = −14.
(d) Let
h(x) =
x − f (x)
xf (x)
where f (2) = 2, f 0 (2) = −1. Compute h0 (2).
Answer: 1/2
Mathematics 100-201
Page 6 of 9
Solution: Simplify, then use the quotient rule
1
1
−
f (x) x
−f 0 (x) −1
h0 (x) =
− 2
f (x)2
x
1 −1
= 1/2
h0 (2) = −
4
4
h(x) =
Student-No.:
Mathematics 100-201
Page 7 of 9
Student-No.:
Full-Solution Problems. In questions 3–5, justify your answers and show all your work. If
a box is provided, write your final answer there. Unless otherwise indicated, simplification of
answers is not required in these questions.
7 marks
3. Show that the following equation has at least two different solutions:
x2 + 1 = 2 cos(x)
Solution: We use the intermediate value theorem.
• Let f (x) = x2 + 1 − 2 cos(x).
• Since f (x) is the sum of a polynomial and cos(x) it is continuous everywhere and
we can apply the IVT.
• It suffices to show that this function has 2 zeros.
• At x = 0, f (0) = 0 + 1 − 2 cos(0) = −1 < 0.
• At x = π/2:
f (π/2) = π 2 /4 + 1 − 2 cos(π/2)
= 1 + π 2 /4 − 0
> 1 + 9/4 = 13/4 > 0
since π > 3
or at x = π:
f (π) = π 2 + 1 − 2 cos(π)
= 1 + π2 + 2
> 3 + 9 = 12 > 0
since π > 3
• Similarly, at x = −π/2:
f (−π) = π 2 /4 + 1 − 2 cos(−π/2)
= 1 + π 2 /4 − 0
> 1 + 9/4 = 13/4 > 0
since π > 3
or at x = −π:
f (−π) = π 2 + 1 − 2 cos(−π)
= 1 + π2 + 2
> 3 + 9 = 12 > 0
since π > 3
• By the IVT since the function is continuous and is negative at x = 0 and positive
at x = π/2, there exists some 0 < c1 < π/2 so that f (c1 ) = 0.
• By the IVT since the function is continuous and is negative at x = 0 and positive
at x = −π/2, there exists some −π/2 < c2 < 0 so that f (c2 ) = 0.
• Since c2 < 0 < c1 there exist at least two points at which the function is zero and
so two points where the original equation is satisfied.
Mathematics 100-201
√
4 marks
4. (a) Let f (x) =
Page 8 of 9
Student-No.:
x2 + 3 − 2x
. Determine the horizontal asymptotes of the graph y = f (x)
x
Solution: Assume x > 0 then
p
√
x 1 + 3/x2 − 2x
x2 + 3 − 2x
=
x
x
p
1 + 3/x2 − 2
=
1
so as x → +∞
lim f (x) =
x→+∞
√
1 + 0 − 2 = −1
Now assume x < 0 then
p
√
−x 1 + 3/x2 − 2x
x2 + 3 − 2x
=
x
x
p
− 1 + 3/x2 − 2
=
1
so as x → −∞
√
lim f (x) = − 1 + 0 − 2 = −3
x→−∞
So the asymptotes are y = −1, as x → +∞ and y = −3, as x → −∞.
√
3 marks
(b) Now consider the function g(x) =
x2 − 3 − 2x
. Determine where g(x) is continuous.
x
Solution:
• The denominator is a polynomial and so continuous everywhere — however
its zero (at x = 0) may cause a discontinuity.
• The numerator is continuous everywhere on its domain — namely where x2 −
3 ≥ 0.
√
√
(x2 − 3) = (x − 3)(x + 3)
√
√
√
√
Hence we must have either x ≥ 3 or x ≤ 3, since if − 3 < x < 3 then
x2 + 3 < 0 and the function is not defined.
√
√
• Hence g is continuous on (−∞, − 3] ∪ [ 3, +∞).
Mathematics 100-201
Page 9 of 9
Student-No.:
5. Let a, b be a constant real numbers and let
(
a + 2x2
h(x) =
3 + bx − 2x2
2 marks
x≤0
x>0
(a) For what values of a is h continuous at x = 0?
Solution:
We have
h(0) = a
lim− h(x) = lim− a + 2x2 = a
x→0
x→0
lim+ h(x) = lim+ 3 + bx − 2x2 = 3
x→0
x→0
Hence we need a = 3.
3 marks
(b) Using your result above, now use the definition of the derivative to determine which
value of b makes h differentiable at x = 0.
Solution: The definition is
h(x) − h(0)
h(x) − 3
= lim
x→0
x→0
x
x
h0 (0) = lim
The left-limit is
lim−
x→0
h(x) − 3
3 + 2x2 − 3
= lim−
x→0
x
x
2x
= lim−
=0
x→0
1
The right-limit is
lim+
x→0
h(x) − 3
3 + bx − 2x2 − 3
= lim+
x→0
x
x
b − 2x
= lim+
=b
x→0
1
For the limit to exist (and for h to be differentiable at x = 0) we need b = 0.