Mathematics 100 and 180
Page 1 of 17
Student-No.:
Final Examination — December 16th 2015
Duration: 2.5 hours
This test has 12 questions on 17 pages, for a total of 100 points.
• Read all the questions carefully before starting to work.
• Q1-Q3 are multiple choice questions; write your answer in the box provided.
• Q4-Q6 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
6
7
8
9
10
11
12
Total
Points:
4
4
8
8
12
12
6
7
7
14
8
10
100
Score:
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like.
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the examination run forty-five (45) minutes or less, no examination
candidate shall be permitted to enter the examination room once the
examination has begun.
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be articulated by the examiner or invigilator prior to the examination
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similar practices, may be immediately dismissed from the examination
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(i) speaking or communicating with other examination candidates,
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(iii) purposely viewing the written papers of other examination candidates;
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or other memory aid devices other than those authorized by the
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(v) using or operating electronic devices including but not limited to telephones, calculators, computers, or similar devices
other than those authorized by the examiner(s)(electronic devices other than those authorized by the examiner(s) must be
completely powered down if present at the place of writing).
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material, must hand in all examination papers, and must not take any
examination material from the examination room without permission
of the examiner or invigilator.
7. Notwithstanding the above, for any mode of examination that does
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.
8. Examination candidates must follow any additional examination rules
or directions communicated by the examiner(s) or invigilator(s).
Mathematics 100 and 180
Page 2 of 17
Student-No.:
Multiple choice questions:
4 marks
1. Each part is worth 1 mark. Write your choice in the box provided.
ANSWERS HERE
(i)
(ii)
(iii)
(iv)
t−3
t→3 (t + 1)2
(i) Compute lim
1
8
E. 3
F. does not exist
A. −1
3
B. −
16
D.
C. 0
(ii) Compute lim
x→+∞ x2
x+1
.
+ 2x − 8
A. 0
D. 4
E. −∞
F. +∞
B. 1
C. 2
(iii) Compute lim+
x→4
x
(x − 4)2
A. −4
D. 3
B. 0
E. −∞
C. 1
F. +∞
(iv) Describe all points where the function f (x) =
A. (0, ∞)
B. (1, ∞)
C. (−∞, 1) ∪ (1, ∞)
ex (x − 1)
is continuous.
x2
D. (−∞, 0) ∪ (0, ∞)
E. (−∞, 0) ∪ (1, ∞)
F. (−∞, 0) ∪ (0, 1) ∪ (1, ∞)
Mathematics 100 and 180
4 marks
Page 3 of 17
Student-No.:
2. Each part is worth 1 mark. Write your choice in the box provided.
ANSWERS HERE
(i)
(ii)
(iii)
(iv)
(i) Find the derivative of f (x) = tan x.
1
cos(x)
cos(x)
B.
sin(x)
cos2 (x)
sin2 (x)
1
E. − 2
sin (x)
cos(x)
F. − 2
sin (x)
A.
C.
D.
1
cos2 (x)
(ii) Find the derivative of f (x) = x log x. Remember log x = loge x = ln x.
log x
x
log x
B. 1 +
x
C. x log x
D. x − log x
A.
(iii) Find the derivative of f (x) =
E. 1 + log x
F. x + log x
x2
.
x+1
A. 2x
D.
B.
x(x + 2)
x+1
C.
2x
(x + 1)2
x(x + 2)
(x + 1)2
x2 (x + 1)
(x + 2)2
x+1
F.
x(x + 2)
E.
(iv) Find the derivative of f (x) = sin(x2 ).
D. 2x cos(x2 )
A. sin(2x)
B. cos(2x)
E. 2x cos(2x)
2
C. 2x sin(x )
F. 2 sin(x) cos(x)
Mathematics 100 and 180
8 marks
Page 4 of 17
Student-No.:
3. Each part is worth 2 marks. Write your choice in the box provided.
ANSWERS HERE
(i)
(ii)
(iii)
(i) What is the equation of the line tangent to the function f (x) =
1
A. y = x + 2
4
1
B. (y − 2) = (x − 4)
4
1
C. (y − 2) = (x − 2)
2
dy
(ii) Let xy 2 + yx2 = 2. Find
at the point (1, 1).
dx
A. −1
B. −1/3
C. 0
(iv)
√
x at the point (4, 2)?
1
D. (y − 2) = (x − 4)
2
1
E. (y − 4) = (x − 2)
2
F. (y − 4) = 2(x − 2)
D. 1/3
E. 1
F. 2
(iii) Let f (x) = xx . What is f 0 (x)?
A. x · xx−1
B. xx (1 + xx )
C. xx log x
D. 1 + log x
E. xx (1 + log x)
F. xex
Remember log x = loge x = ln x.
(iv) Let T3 (x) = 24 + 6(x − 3) + 12(x − 3)2 + 4(x − 3)3 be the third-degree Taylor polynomial
for some function f (x), expanded about a = 3. What is f 00 (3)?
A. 2
D. 6
B. 3
E. 12
C. 4
F. 24
Mathematics 100 and 180
Page 5 of 17
Student-No.:
Short-answer questions: Put your answer in the box provided. Full marks will be given for a
correct answer placed in the box, while part marks may be given if workings are shown. Please
simplify your answers.
3π
.
2 marks 4. (a) Compute arcsin sin
5
Answer:
2π
5
Solution: Since 3π/5 > π/2 we do not just have 3π/5. Using sin(π − θ) = sin(θ):
sin(3π/5) = sin(2π/5)
arcsin(sin(3π/5)) = arcsin(sin(2π/5))
= 2π/5.
2 marks
x2 − 3x + 2
.
x→2
x−2
(b) Evaluate lim
Answer: 1
Solution:
(x − 1)(x − 2)
x2 − 3x + 2
=
= x − 1.
x−2
x−2
Mathematics 100 and 180
2 marks
(c) Evaluate lim
x→−∞
Page 6 of 17
x+
Student-No.:
√
4x2 − x
.
6x
Answer: − 61
Solution: We can assume x < 0 so
√ q
√
x
+
x2 4 −
2
x + 4x − x
=
6x
q 6x
1−
=
2 marks
4−
6
1
x
1
x
x−x
=
q
4−
1
x
6x
√
−1
4
x→−∞ 1 −
=
−−−−→
6
6
x2 − 1
(d) Find the domain of f (x) = √
.
x2 − x − 6
Answer: (−∞, −2) ∪ (3, ∞)
Solution: We need x2 − x − 6 = (x + 2)(x − 3) > 0. Hence x 6∈ [−2, 3].
Mathematics 100 and 180
3 marks
Page 7 of 17
Student-No.:
5. (a) Consider the curve defined by y 2 + 4xy − 2x2 = 3. Find the x-coordinates of all points
dy
= 0.
on the curve where
dx
Answer: x = −1, +1
Solution: Implicit differentitation gives
2yy 0 + 4y + 4xy 0 − 4x = 0
So when y 0 = 0 we have
4y − 4x = 0
That is x = y. Substituting this back into the original equation gives
x2 + 4x2 − 2x2 = 3
3x2 = 3
Hence x = ±1.
3 marks
(b) Let f (x) be a function differentiable at x = 3 and let g(x) = x · f (x). The line tangent
to the curve y = f (x) at x = 3 has slope 2 while the line tangent to the curve y = g(x)
at x = 3 has slope 5. What is f (3)?
Answer: f (3) = −1
Solution: We know that g 0 (3) = 5, f 0 (3) = 2, so
g 0 (x) = xf 0 (x) + f (x)
5 = g 0 (3) = 3 × 2 + f (3)
Hence f (3) = −1.
Mathematics 100 and 180
3 marks
Page 8 of 17
Student-No.:
(c) Find the coordinates of the inflection point of the function h(x) = x ex .
Answer: (−2, −2e−2 )
Solution:
h(x) = xex
h0 (x) = ex + xex = (x + 1)ex
h00 (x) = (x + 1)ex + ex = (x + 2)ex
We see that h00 (x) = 0 when x = −2 and also that h00 (x) changes sign at x = −2.
Thus x = −2 is the inflection point.
3 marks
log(1 + x) − sin x
. Remember log x = loge x = ln x.
x→0
x2
Answer: -1/2
(d) Evaluate the limit lim
Solution: We see that this is a 0/0 indeterminate form
1
d
numerator =
− cos x
dx
1+x
d
denominator = 2x
dx
Now the limit
lim
1
1+x
x→0
− cos x
2x
so also a 0/0 indeterminate form, so we differentiate again
−1
d
numerator =
+ sin x
dx
(1 + x)2
d
denominator = 2
dx
And the limit
lim
x→0
Hence the original limit is −1/2.
−1
(1+x)2
+ sin x
2
= −1/2
Mathematics 100 and 180
3 marks
Page 9 of 17
Student-No.:
6. (a) A colony of bacteria doubles every 4 hours. If the colony has 2000 cells after 6 hours,
how many cells were present initially? Simplify your answer.
√
√
Answer: P (0) = 1000/ 2 = 500 2
Solution: Assuming exponential growth
P (t) = P (0)ekt
We are told that
P (4) = 2P (0) = P (0)e4k
P (6) = 2000 = P (0)e6k
Hence
e4k = 2
or e2k =
√
2 so
e6k = 23/2
Thus
2000 = P (0)23/2
√
√
P (0) = 1000/ 2 = 500 2.
3 marks
(b) Estimate
√
8 using a linear approximation.
Answer:
Solution: Let f (x) =
√
√
8 ≈ 17/6
x then then the linearisation of f at a = 9 is given by
L(x) = f (9) + f 0 (9)(x − 9).
Since f (9) = 3, f 0 (x) =
1
√
2 x
and f 0 (3) = 1/6, we have
L(x) = 3 + (x − 9)/6.
So a linear approximation of
√
8 is L(8) = 3 − 1/6 = 17/6.
Mathematics 100 and 180
3 marks
Page 10 of 17
(c) What are the critical points of the function f (x) = ex
Student-No.:
3 −9x2 +15x−1
?
Answer: x = 1, 5
3
2
Solution: The derivative of ex −9x +15x−1 is
3 2
3
2
3x2 − 18x + 15 ex −6x +9x−1 = 3(x − 1)(x − 5)ex −9x +15x−1 .
Since eblah is non-zero, the only zeros come from the polynomial factor. Thus the
critical points are x = 1 and x = 5.
3 marks
2
(d) Find a function f (x) such that f 0 (x) = sin(x) + √ and such that f (π) = 0.
x
√
√
Answer: f (x) = 1 − π − cos x + 4 x
√
2
Solution: Every antiderivative of sin(x)+ √ can be written as − cos(x)+4 x+C,
x
√
where C is a constant.
Plugging
x
=
π
gives
C
=
1
−
π. So the function f we
√
√
search is f (x) = 1 − π − cos(x) + 4 x.
Mathematics 100 and 180
Page 11 of 17
Student-No.:
Full-solution problems: 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.
6 marks
7. Is the function
(√
1 + x2 − 1 if x ≤ 0
f (x) =
x2 cos(1/x) if x > 0
differentiable at x = 0? You must explain your answer using the definition of the derivative.
Solution: We must show that the limit
f (x) − f (0)
f (x)
= lim
x→0
x→0 x
x−0
f 0 (0) = lim
exists and is finite. To do this we evaluate both one sided limits.
√
f (x)
1 + x2 − 1
lim−
= lim−
x→0
x→0
x
x
√
√
1 + x2 − 1
1 + x2 + 1
·√
= lim−
x→0
x
1 + x2 + 1
(1 + x2 ) − 1
= lim− √
x→0 x( 1 + x2 + 1)
x
= lim− √
x→0
1 + x2 + 1
= 0.
The other limit is a little harder and we use the squeeze theorem. First note that for
x>0
f (x)
= x cos(1/x)
x
and
−x ≤ x cos(1/x) ≤ x
Since limx→0 x = limx→0 −x = 0, by the squeeze theorem
lim+
x→0
f (x)
= lim+ x cos(1/x) = 0
x→0
x
Since the left and right hand limits agree, the limit exists and f 0 (0) = 0.
Mathematics 100 and 180
Page 12 of 17
Student-No.:
A tank of water in shape of an inverted circular cone of height
8. 16m and diameter 20m at the top is being filled with water at
a rate of 2m3 /minute.
2 marks
(a) Draw a vertical cross-section of the cone and label it with all the lengths that appear
in your solution to part (b)
Solution: Since the cone is inverted, the water in the tank also forms a cone. Let
h be the depth (m) of water and V its volume (m3 ) and x the radius of the circle
formed by the water’s surface (in m). Finally let t be time in minutes.
5 marks
(b) Find the rate of change of the height of the water in the tank when the height of the
water is 10m.
Solution:
By similar triangles x/h = 10/16, so x = 5h/8, so the volume of water is
V =
π 52 3
π 2
x h=
h
3
3 82
Then differentiating with respect to time gives
52 dh
dV
= π · 2 h2
dt
8
dt
We know that
dV
= 2 when h = 10, so
dt
dV
52 dh
= π · 2 h2
dt
8
dt
52
dh
2 = π · 2 · 102
8
dt
2
dh
2×8
=
dt
π · 52 · 102
27
=
π · 4 · 54
25
=
π · 54
metres per minute.
Mathematics 100 and 180
7 marks
Page 13 of 17
Student-No.:
9. A business wants to manufacture a huge closed box with a square base a square top and
rectangular sides. The material used for the four upright sides of the box costs $2 per square
metre, and the material used for the base and top of the box costs $8 per square metre.
Find the dimensions of the box with lowest possible cost that has volume 32m3 .
Solution: Let b be the dimension of the square base (in m), and let h be the height (in
m) and V be the volume. Then V = b2 h = 32.
Let C be the cost of the materials for the box consists of the costs for the four upright
sides, and the base and the top:
C = 4 × bh × 2 + 2 × b2 × 8
Since the box must have volume 32, we know that b2 h = 32 and so we can write h = 32/b2 .
Hence
32
C =8×
+ 16 × b2
b
16
2
= 16
+b
b
To find when C is minimised, we need to know its critical and singular points, so we find
its derivative
dC
16
= 16 − 2 + 2b
db
b
The only singular point is at b = 0, and the only critical point is when
2b = 16/b2
b3 = 8
rearrange
So at b = 2.
We want the global minimum of C(b), so we note that as b approaches 0 or infinity, C(b)
grows without bound. So, b = 2 must be its global minimum. When b = 2, h = 8. So,
the dimensions of the cheapest box are 2 by 2 by 8 .
Mathematics 100 and 180
Page 14 of 17
Student-No.:
10. Let f (x) = (x + 2)(x + 1)e−x .
1 mark
(a) Find the domain of the function.
Answer: All real numbers
Solution: Since the function is the product of a polynomial and an exponential
function, it is continuous everywhere.
2 marks
(b) Find the x-intercepts and y-intercepts of the function.
Answer: y = 2, x = −2, −1.
Solution:
• y-intercept: f (0) = 2 × 1 × e0 = 2.
• x-intercept: f (x) = 0 only when (x + 1)(x + 2) = 0 (since eblah 6= 0). Hence
only when x = −2, −1.
2 marks
(c) Find the horizontal asymptotes of the function (if they exist).
Answer: y = 0 when x → +∞. No others
Solution: As x → −∞, e−x → +∞ while (x + 1), (x + 2) → −∞. Hence no
horizontal asymptote. On the other hand (using l’hopital) as
(x + 2)(x + 1)
2x + 3
2
=
lim
=
lim
=0
x→+∞
x→+∞
x→+∞ ex
ex
ex
lim
So y = 0 is a horizontal asymptote as x → +∞.
1 mark
(d) Find the vertical asymptotes of the function (if they exist).
Answer: None.
Solution: Since the function is continuous everywhere there are no vertical asymptotes.
Mathematics 100 and 180
1 mark
Page 15 of 17
Student-No.:
(e) Find f 0 (x). You must simplify your answer.
Answer: −(x2 + x − 1)e−x
Solution:
f (x) = (x2 + 3x + 2)e−x f 0 (x) = (2x + 3)e−x − (x2 + 3x + 2)e−x = (−x2 − x + 1)e−x
2 marks
(f) Find all critical points and singular points of f (x).
Answer: No singular, critical at x =
√
−1± 5
2
Solution: Since f 0 is continuous there are no singular points. The only critical
points are when x2 + x − 1 = 0. This happens at
√
−1 ± 5
x=
2
3 marks
(g) Find the intervals on which f (x) is increasing. Find the intervals on which f (x) is
decreasing.
√
√
Solution:
We need
√
√ to examine 3 intervals (−∞, (−1− 5)/2), ((−1− 5)/2, (−1+
5)/2), ((−1 + 5)/2, ∞).
√
• On (−∞, (−1 − 5)/2)), f 0 < 0 so function is decreasing.
√
√
• On ((−1 − 5)/2, (−1 + 5)/2), f 0 > 0 so function is increasing.
√
• On ((−1 − 5)/2), ∞), f 0 < 0 so function is decreasing.
2 marks
(h) The second derivative is f 00 (x) = (x − 2)(x + 1)e−x . Find the x-coordinates of any
points of inflection. You must explain your answer.
Solution: The only possible points are x = −1, x = +2. When x < −1 f 00 > 0.
When −1 < x < 2, f 00 < 0 and when x > 2 f 00 > 0. Hence concavity changes at
each point and they really are inflection points.
f (−1) = 0
f (2) = 4 × 3 × e−2
so (−1, 0) is an inflection point.
so (2, 12e−2 ) is an inflection point.
Mathematics 100 and 180
Page 16 of 17
Student-No.:
4 marks 11. (a) If the second degree Maclaurin polynomial for g(x) is T2 (x) = 5 − x/3 + 2x2 find the
second degree Maclaurin polynomial for h(x) = ex · g(x).
Solution: Now
h(0) = e0 · g(0) = 5
h0 (x) = ex · g(x) + ex · g 0 (x) = ex (g(x) + g 0 (x))
h0 (0) = e0 · (5 − 1/3) = 14/3
h00 (x) = ex (g(x) + g 0 (x)) + ex (g 0 (x) + g 00 (x)) = ex (g(x) + 2g 0 (x) + g 00 (x))
h00 (0) = e0 (5 − 2/3 + 4) = 25/3
Hence the 2nd degree Maclaurin polynomial is
h(0) + h0 (0)x +
4 marks
h00 (0) 2
14
25
x = 5 + x + x2
2
3
3
x sin(x) + x2 cos(x)
, show that the absolute value of the error when we
10 − x2
approximate f (1) using the second Maclaurin polynomial is less than 0.04 = 1/25.
(b) If f (3) (x) =
Solution: The remainder is given by
R2 (x) =
1 (3)
f (c)(x − a)3
3!
where x = 1, a = 0 and c ∈ (0, 1). Hence
1 c sin(c) + c2 cos(c) 3
|R2 (1)| = ·1
6
10 − c2
Since 0 < c < 1, and | sin(c)| < 1, cos(c) < 1, we know that
|c sin(c) + c2 cos(c)| ≤ 1 × 1 + 1 × 1 = 2.
Similarly, since 0 < c < 1, 9 ≤ 10 − c2 ≤ 10. Hence
|R2 (1)| ≤
And
1
27
<
1
25
as required.
1 2
2
1
· =
=
6 9
54
27
Mathematics 100 and 180
Page 17 of 17
Student-No.:
4 marks 12. (a) Let g(x) be a continuous function for which g 0 (x) and g 00 (x) exist. If g(x) has at least
three zeros, then how many zeros must g 0 (x) and g 00 (x) have? Explain your answer
carefully.
Solution: Let a < b < c be zeros of g. Then by Rolle’s theorem there is a number
x1 between a and b so that g 0 (x1 ) = 0. Similarly there is a number x2 between b
and c so that g 0 (x2 ) = 0. Thus g 0 has at least two zeros.
Further applying Rolle’s theorem to g 0 we know there is a number x3 between x1
and x2 so that g 00 (x3 ) = 0. Hence g 00 has at least 1 zero.
4 marks
(b) Consider the equation 2x2 − 3 + sin(x) + cos(x) = 0. Show that this has at least two
solutions.
Solution: Let f (x) = 2x2 − 3 + sin(x) + cos(x). Since the function is the sum of a
polynomial and sines and cosines it is continuous. Hence we can apply the IVT.
Evaluating at some points
f (0) = 0 − 3 + sin(0) + cos(0) = −2 < 0.
f (π) = 2π 2 − 3 + sin(π) + cos(π) = 2π 2 − 3 − 1 > 0
f (−π) = 2π 2 − 3 + sin(−π) + cos(−π) = 2π 2 − 3 − 1 > 0
since π > 3
since π > 3
Thus by the IVT there is a zero between 0 and π and another zero between −π and
zero.
2 marks
(c) Show that the same equation cannot have more than two solutions. Part (a) will help
you here.
Solution: Assume that f has three (or more) zeros. Then by part (a), we know
that f 00 (x) must have at least one zero. However when we compute f 00 (x):
f 0 (x) = 4x + cos x − sin x
f 00 (x) = 4 − sin x − cos x
we see that f 00 (x) ≥ 2 (since sin x, cos x ≤ 1). Hence f 00 is nowhere zero, and thus
f cannot have more than two zeros.