MATH 100, Section 921
Final Examination — June 25, 2015
Page 1 of 17
Final Examination
Duration: 2.5 hours
This test has 8 questions on 17 pages, for a total of 106 points.
• Read all the questions carefully before starting to work.
• Q1 and Q2 are short-answer questions; put your answer in the boxes provided.
• For all other questions you should give complete arguments, calculations, and explanations;
answers without justification will not receive marks.
• 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.)
Last Name:
First Name:
Student-ID:
Signature:
Question:
1
2
3
4
5
6
7
8
Total
Points:
15
33
10
9
7
8
7
17
106
Score:
Student Conduct during Examinations
1. Each examination candidate must be prepared to produce, upon the
request of the invigilator or examiner, his or her UBCcard for identification.
2. Examination candidates are not permitted to ask questions of the
examiners or invigilators, except in cases of supposed errors or ambiguities in examination questions, illegible or missing material, or the
like.
3. No examination candidate shall be permitted to enter the examination
room after the expiration of one-half hour from the scheduled starting
time, or to leave during the first half hour of the examination. Should
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.
4. Examination candidates must conduct themselves honestly and in accordance with established rules for a given examination, which will
be articulated by the examiner or invigilator prior to the examination
commencing. Should dishonest behaviour be observed by the examiner(s) or invigilator(s), pleas of accident or forgetfulness shall not be
received.
5. Examination candidates suspected of any of the following, or any other
similar practices, may be immediately dismissed from the examination
by the examiner/invigilator, and may be subject to disciplinary action:
(i) speaking or communicating with other examination candidates,
unless otherwise authorized;
(ii) purposely exposing written papers to the view of other examination candidates or imaging devices;
(iii) purposely viewing the written papers of other examination candidates;
(iv) using or having visible at the place of writing any books, papers
or other memory aid devices other than those authorized by the
examiner(s); and,
(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).
6. Examination candidates must not destroy or damage any examination
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).
MATH 100, Section 921
Final Examination — June 25, 2015
Page 2 of 17
Short-Answer Questions. Put your answer in the box provided but show your work also.
Each question is worth 3 marks, but not all questions are of equal difficulty. Full marks will
be given for correct answers placed in the box, but at most 1 mark will be given for incorrect
answers. Simplify your answers as much as possible.
15 marks
1. Find the following limits if they exist. In case of a vertical asymptote, determine whether
the limit tends to ±∞. Write “DNE” if the limit does not exist for any other reason.
x18 − x10 + 2x9 − x + 1
(a) lim
x→−1
x9 + 1
Answer: 1
Solution: Plugging in −1 we get
division to get:
0
0
which means more work. Use polynomial long
x18 − x10 + 2x9 − x + 1
= lim x9 − x + 1 = 1
x→−1
x→−1
x9 + 1
lim
√
(b) lim
t→0
1 + t3 −
t3
√
1 − t3
Answer: 1
Solution: Plugging in 0 we get
√
lim
x→0
1 + t3 −
t3
√
1 − t3
0
0
which means more work. Use root conjugate:
√
√
1 + t3 + 1 − t3
2t3
√
√
·√
= lim √
1 + t3 + 1 − t3 x→0 t3 ( 1 + t3 + 1 − t3 )
2
√
=1
= lim √
3
x→0
1 + t + 1 − t3
MATH 100, Section 921
Final Examination — June 25, 2015
(c) If 2 tan(x) ≤ g(x) ≤
8
π
Page 3 of 17
arcsin(sin(x)) for all x, evaluate lim g(x).
x→9π/4
Answer: 2
Solution: Since tan(9π/4) = 2 and
follows by the squeeze theorem.
(d) Recall that
0
0
8
π
arcsin(sin(9π/4)) =
8
π
·
π
4
= 2, the result
is known as an indeterminate form. List three more indeterminate forms.
Answer:
∞
,
∞
∞ · 0, ∞0 , 1∞ , ∞ − ∞, 00
1
(e) lim (ex + x) x
x→∞
Answer: e
Solution: Plugging in ∞ we get ∞0 which is an indeterminate form requiring more
work.


ln(ex + x)
 lim

1
1
x +x)
x→∞
x
ln(e
x
x
=e
lim (e + x) = lim e x
x→∞
x→∞
x
by continuity of e . But
ln(ex + x)
ex
ex
ex + 1
= lim x
= lim x = 1
= lim x
x→∞
x→∞ e + 1
x→∞ e
x→∞ e + x
x
lim
by L’Hôpital’s rule. So
1
lim (ex + x) x = e1 = e.
x→∞
MATH 100, Section 921
33 marks
Final Examination — June 25, 2015
2. (a) Find the domain of f (x) =
Page 4 of 17
ln(ln(x))
.
x−3
Answer: x ∈ (0, 3) ∪ (3, ∞)
Solution: The numerator requires ln(x) > 0 or x > 1. The denominator requires
x 6= 3.
(b) If f (x) is a function whose derivative is f 0 (x) = π2 x + arcsin(x). Find
f (1 + x) − f (1)
.
x→0
x
lim
Answer: π
Solution: The stated limit is the definition of the derivative of f (x) at x = 1, that
is, g(1) = π2 + π2 = π.
(c) Find the derivative of y = xln x
Answer: y 0 =
2 ln x
y
x
=
Solution: By logarithmic differentiation
ln y = ln x · ln x
1 0 2 ln x
y =
y
x
2 ln x ln x
x
x
= 2 ln x · xln(x)−1
MATH 100, Section 921
Final Examination — June 25, 2015
(d) Simplify sec(arctan(2)).
Answer:
Page 5 of 17
√
5
Solution: Draw a triangle from y = arctan(2) ⇔ tan(y) = 2 and use Pythagoras.
(e) Find the derivative of y = logπ (x).
Answer: y 0 =
1
ln(π)x
Solution: Since logπ (x) is the inverse function of π x , we get
πy = x
By implicit differentiation:
ln(π) · π y · y 0 = 1
so
y0 =
1
ln(π)x
(f) Find the derivative of f (x) = x sin(4)
Answer: sin(4)
Solution: The function f (x) is just a line with slope sin(4)—a constant!
MATH 100, Section 921
Final Examination — June 25, 2015
Page 6 of 17
(g) As shown in the figure below, the tangent line to the graph of f (x) at x = a intersects
the x-axis at x = b. Which of the following expressions gives the value of b?
Answer: (A)
Solution: Either note that this is the definition of Newton’s method or work out
the x-intercept by using the point-slope formula of a line.
(h) What is the solution of the initial value problem
dy
= 5y,
dt
y(0) = 3
Answer: y(t) = 3e5t
Solution: This is the only differential equation in the whole course you are supposed to know the solution of. In general, y 0 = kt has solution y(t) = Aekt . Work
out A by using the initial condition.
(i) Which of the following functions are solutions of the differential equation x2 y 0 +xy = 1?
(A) y = 0
(B) y = ln x
(C) y = x
ln x
(D) y =
x
1
(E) y =
x
(F) None of these.
Answer: (D)
Solution: Check each option separately.
MATH 100, Section 921
Final Examination — June 25, 2015
Page 7 of 17
(j) A new planet is uniformly gaining volume from orbiting material at a rate of 1012 m3 /s.
How fast is its radius increasing when its diameter is 6 · 106 m?
Answer:
dr
dt
=
1
36π
Solution: Assume that the new planet is spherical with volume V = 43 πr3 . We
know dV
and want dr
at a certain radius. Differentiating the volume formula we
dt
dt
obtain
dV
dr
= dt 2
dt
4πr
Now evaluate with the given values.
(k) Let f be continuous and differentiable everywhere. If f (0) = 0 and f 0 (x) ≥ 2 for
0 ≤ x ≤ 3 what is the smallest value f (3) can be?
Answer: f (3) ≥ 6
Solution: Using the MVT on the interval [0, 3], we obtain f (3) ≥ 6.
MATH 100, Section 921
Final Examination — June 25, 2015
Page 8 of 17
Full-Solution Problems. In the following questions, 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.
3. A person was found dead in a park. The police arrived on the scene at 3am and measured
the body temperature to be 33◦ C. One hour later it was 23◦ C. The outside temperature
remained at a constant 13◦ C. Normal body temperature is 37◦ C. Assume that the body
cools after death according to Newton’s Law of Cooling.
5 marks
(a) How much time before the police arrived did the homeless person die? Leave your
answer in terms of logarithms.
Solution: Fit the exponential function T (t) = Aekt + C through the given data
points (0, 33), (1, 23), and C = 13 (the ambient temperature).
T (0) = A + 13 = 33
T (1) = Aek + 13 = 23
So A = 20 and k = ln(1/2). So T (t) = 20eln(1/2)t + 13 = 20
equal to 37 and solve for t to get
t=
1 t
2
+ 13. Now set T (t)
ln(24/20)
ln(6/5)
=
ln(1/2)
ln(1/2)
Note that this is a negative number so technically we should take the absolute value.
Marking scheme:
• Formula for general exponential or solution to Newton’s law — 1 mark
• Determining A, k, and C — 1 mark each
• Determining time since death (no need for absolute value) — 1 marks
2 marks
(b) Did the death occur before or after 2am?
Solution: Since ln(2) > ln(6/5), we have 1 >
t=
ln(6/5)
ln(1/2)
=
ln(6/5)
ln(2)
ln(6/5)
.
ln(2)
Thus, the time since death,
is less than 1 and the death occured after 2am.
Marking scheme:
• 1 mark for anything remotely sensible
MATH 100, Section 921
3 marks
Final Examination — June 25, 2015
Page 9 of 17
(c) Using a step size h of 6 minutes, write down the first step in Euler’s method applied
to Newton’s Law of Cooling as it appears in this question starting at the time the
police arrived. Your final answer must be a number but you may leave it in terms of
logarithms. Hint: Make sure the step size units agree with the units used in the earlier
parts of this problem.
Solution: The differential equation describing the temperature of the cooling body
is Newton’s law of cooling
dT
= k(T − E)
dt
with initial condition T (0) = 33. Euler’s method is thus
T (t + h) = hk(T (t) − 13) + T (t)
with E the ambient temperature and k = ln(1/2) determined in part (a) in terms
of hours. Converted into hours, h = 1/10 and the first step in Euler’s method is
t=0:
T (1/10) =
1
ln(1/2)20 + 33 = 2 ln(1/2) + 33
10
where we used T (0) = 33.
Marking scheme:
• Newton’s law of cooling — 1 mark
• Solving for T (t + h) a.k.a. Tk+1 — 1 mark
• Substituting in correct values of k, h, and T (0) — 1 mark
• If student writes T 0 = k(E − T ), then he/she must also use k = ln 2 instead.
-1 mark if this is not the case.
4. In this question we investigate the solution of the equation
ln x = −x2 + 3.
3 marks
(a) Show that the equation has at least one solution.
Solution: Let f (x) = ln x + x2 − 3. Then f (x) is defined and continuous for
x > 0 and we can attempt to use the IVT. Since f (1) = 0 + 1 − 3 < 0 and
f (e) = 1 + e2 − 3 > 0, the IVT indeed implies that there is a solution on the interval
[1, e].
Marking scheme:
• Creating a function — 1 mark
• Mentioning continuity — 1 mark
• Finding x-values producing opposite sign y-values and mentioning the IVT —
1 mark
MATH 100, Section 921
3 marks
Final Examination — June 25, 2015
Page 10 of 17
(b) Show that the equation cannot have more than one solution.
Solution: Continuing with f (x) = ln x+x2 −3, note that f (x) is also differentiable
for x > 0 and we can attempt to use the MVT. If there were two solutions to f (x),
then by the MVT we would have f 0 (c) = 0 for some c. However, f 0 (x) = x1 + 2x is
never 0 for x > 0. This contradiction implies that there cannot be a second solution
to f (x).
Marking scheme:
• Mentioning differentiability — 1 mark
• Invoking MVT — 1 mark
• Showing derivative is never 0 — 1 mark
• 2 of 3 marks for inconsistent logic in argument but OK steps
3 marks
(c) Use Newton’s method to approximate the solution of the equation by making an initial
guess and writing down the first approximation.
Solution: Newton’s method is
xk+1 = xk −
f (xk )
f 0 (xk )
As initial guess we could take x0 = 1 but other (less nice) ones work too. We have
f (x) and f 0 (x) from the previous part. Then
x1 = x0 −
f (x0 )
−2
5
=1−
=
0
f (x0 )
3
3
Marking scheme:
• Newton’s method formula or finding x-intercept of tangent — 1 mark
• Reasonable initial guess — 1 mark
• Correct first approximation — 1 mark
MATH 100, Section 921
Final Examination — June 25, 2015
Page 11 of 17
5. Let g(x) be a function that is continuous at x = 0 and g(0) = 8 and let

 g(x) sin2 (x)
if x 6= 0,
f (x) =
x
0
if x = 0.
4 marks
(a) Show that f (x) is continuous at x = 0.
Solution: Need to check that lim f (x) = f (0). Have f (0) = 0, and
x→0
g(x) sin2 (x)
xg(x) sin2 (x)
= lim
x→0
x→0
x
x2
lim f (x) = lim
x→0
sin2 (x)
=0·8·1=0
x→0
x2
= lim x · lim g(x) · lim
x→0
x→0
sin x
= 1 and that g(x) is continuous. Since this limit
x→0 x
is equal to 0, the function value at x = 0, f (x) is continuous at 0.
Marking scheme:
where we have used that lim
• Definition of continuity — 1 mark
• Correct f (0) — 1 mark
• Correct limit — 1 mark
• Invoking continuity of g — 1 mark
3 marks
(b) Find f 0 (0) or show that it does not exist.
Solution: Need to check differentiability at x = 0:
f (h) − f (0)
f (h)
f (0) = lim
= lim
= lim
h→0
h→0 h
h→0
h
0
g(h) sin2 (h)
h
h
sin2 (h)
=8·1=8
h→0
h2
= lim g(h) · lim
h→0
sin x
= 1 and that g(x) is continuous.
x→0 x
where we have again used that lim
Marking scheme:
• Definition of derivative — 1 mark
• Correct limit — 1 mark
• Invoking continuity of g — 1 mark
MATH 100, Section 921
Final Examination — June 25, 2015
Page 12 of 17
6. You are given the hyperbola x2 − y 2 = 1 and the point P = (0, 2).
4 marks
(a) Find all points on the hyperbola at which the tangent line to the hyperbola also passes
through P .
Solution: Note that P does not lie on the hyperbola and let (x, y) be a sought
after point on the hyperbola. The slope of a line passing through (x, y) and P can
be calculated in two ways:
• It is the derivative of the curve at (x, y) which is found by implicit differentiation to be
x
2x − 2yy 0 = 0 ⇔ y 0 =
y
• It is also the slope of the secant which is
m=
rise
y−2
=
run
x−0
Equating these two
x2 = y(y − 2) = y 2 − 2y ⇔ x2 − y 2 = −2y
and using the equation of the hyperbola
1
1 = −2y ⇔ y = − .
2
Substituting this back into the equation for the hyperbola and solving for x:
r
√
5
5
x=±
=±
.
4
2
√
√
The sought after points are thus 25 , − 12 , − 25 , − 12 .
Marking scheme:
• Correct implicit derivative — 1 mark
• Correct slope — 1 mark
• Solving for y — 1 mark
• Solving for x — 1 mark
• Full marks even if ± is overlooked.
4 marks
(b) Find the point(s) on the hyperbola closest to P . Be sure to justify that your result is
indeed closest.
Solution: The distance s between P and a closest point (x, y) on the hyperbola is
given by Pythagoras:
s2 = (x − 0)2 + (y − 2)2 = x2 + y 2 − 4y + 4 = 1 + y 2 + y 2 − 4y + 4 = 2y 2 − 4y + 5
MATH 100, Section 921
Final Examination — June 25, 2015
Page 13 of 17
where the equation of the hyperbola was used. Note that it suffices to minimize the
distance squared. To do so, differentiate s2 with respect to y and equate the result
to 0:
ds2
= 4y − 4 = 0 ⇔ y = 1.
dy
Substituting into the equation for the hyperbola and solving for x:
√
x=± 2
√ √ Thus, the critical points are
2, 1 , − 2, 1 . Note that a hyperbola has no
bounds in the y-direction so these are the only possible extreme points. Since the
second derivative of s2
d ds2
= 4 > 0,
dy dy
the second derivative test implies that the two critical points are indeed local minima
for the distance function.
Marking scheme:
• Correct use of Pythagoras distance formula — 1 mark
• Finding y = 1 — 1 mark
√
• Finding x = ± 2 — 1 mark
• Justifying minima — 1 mark
• Full marks even if ± is overlooked.
MATH 100, Section 921
3 marks
Final Examination — June 25, 2015
Page 14 of 17
√
7. (a) Provide
a
second
degree
Maclaurin
approximation
of
2 using the function f (x) =
√
1 + x. Express your answer as a single simplified fraction.
√
Solution: We have f (x) = 1 + x, n = 2, a = 0, and 1 + x = 2 ⇔ x = 1. The
(k)
1
required polynomial is T2 (x) = c0 +c1 x+c2 x2 with ck = f k!(0) . Since f 0 (x) = 2√1+x
,
1
00
f (x) = − 4(1+x)3/2 , we get c0 = 1, c1 = 1/2, and c2 = −1/8. Thus,
T2 (1) = 1 + 1/2 − 1/8 = 11/8.
Marking scheme:
• Taylor’s formula — 1 mark
• Correct coefficients (needs correct f , n, and a) — 1 mark
• Correct answer — 1 mark
4 marks
(b) Find an upper and lower bound for the error incurred in the above approximation and
state whether the above is an over- or underestimate.
Solution: The Lagrange remainder formula with the above parameters is
E2 (1) =
f 000 (c)
f 000 (c)
(1 − 0)3 =
,
3!
6
The third derivative f 000 (x) =
minimized when x = 1. Thus,
3
8(1+x)5/2
0 ≤ c ≤ 1.
is maximized on [0, 1] when x = 0 and
3
f 000 (1)
f 000 (0)
3
1
1
=
=
≤
E
(1)
≤
=
=
2
5/2
5/2
16 · 2
6·8·2
6
6
8·6
16
and since E2 (1) is positive, it is an underestimate.
Marking scheme:
• Lagrange remainder formula — 1 mark
• Correct upper bound — 1 mark
• Correct lower bound — 1 mark
• Underestimate — 1 mark
• 3 of 4 marks for getting the bounds the wrong way around
MATH 100, Section 921
Final Examination — June 25, 2015
Page 15 of 17
8. The aim of this question is to sketch
f (x) =
1 mark
x3 + 1
x
(a) Find the domain of f (x).
Solution: The only issue is division by 0. Thus, the domain of f (x) is (−∞, 0) ∪
(0, ∞).
2 marks
(b) Find the x- and y-intercepts of f (x), or explain why they don’t exist.
Solution: Since f (0) is not defined, there is no y-intercept. Solving x3 + 1 = 0
gives x = −1 as x-intercept.
2 marks
(c) Find any vertical asymptotes of f (x), if they exist.
Solution: As x → 0, the numerator x3 + 1 always goes to 1 while the denominator
goes to 0 but is positive as x → 0+ and negative as x → 0− . Thus,
lim f (x) diverges to ∞,
x→0+
lim f (x) diverges to −∞.
x→0−
MATH 100, Section 921
4 marks
Final Examination — June 25, 2015
Page 16 of 17
(d) Find the intervals where f (x) is increasing respectively decreasing and determine the
x- and y-coordinates of any local maxima, local minima, or saddle points.
Solution: Since
3x3 − (x3 + 1)
2x3 − 1
=
=0
x2
x2
1
when 2x3 − 1 = 0, there is only one critical number: x = √
(x = 0 is not a critical
3
2
number because f (0)
is not
defined). By the first derivative
test, f (x) is decreasing
1
1
from (−∞, 0) and 0, √
and is increasing from √
3
3 , ∞ . It follows that there is
2
2
√
3
1
1
3
√
a local minimum at x = √
·
2
with
y-coordinate
f
=
3
3
2
2
2
f 0 (x) =
Marking scheme:
• Correct derivative — 1 mark
• Correct critical number (no loss of marks if 0 is mistaken as a critical number)
— 1 mark
• Correct intervals and local minimum (x and y) — 2 marks
4 marks
(e) Determine the concavity of f (x) and find the x- and y-coordinates of any inflection
points.
Solution: Since
f 00 (x) =
2x3 + 2
6x4 − 2(2x3 − 1)x
=
=0
x4
x3
when 2x3 + 2 = 0, there is only one candidate for an inflection point: x = −1 (x = 0
is not a candidate because f (0) is not defined). By checking the sign of f 00 (x) on
(−∞, −1), (−1, 0), and (0, ∞), f (x) is found to be concave down on (−1, 0) and
concave up on (−∞, −1) and (0, ∞). It follows that there is an inflection point at
x = −1 with y-coordinate f (−1) = 0.
Marking scheme:
• Correct second derivative — 1 mark
• Correct x-candidate for inflection point (no loss of marks if 0 is mistaken as
a candidate) — 1 mark
• Correct intervals and inflection point (x and y) — 2 marks
MATH 100, Section 921
2 marks
Final Examination — June 25, 2015
Page 17 of 17
(f) Determine the asymptotic behavior of f (x) as x → ±∞.
Solution: By breaking up the fraction,
1
x3 + 1
= lim x2 + lim
= lim x2 .
x→±∞
x→±∞ x
x→±∞
x→±∞
x
lim
Equivalently,
lim
x→±∞
x3 + 1
− x2
x
= 0.
Thus, f (x) has a parabolic asymptote y = x2 in both directions. The same result
can be obtained by polynomial long division.
2 marks
(g) Sketch the graph of f (x) highlighting all features determined in the previous parts of
this question.
Solution:
Marking scheme:
• 1 of 2 marks for generally correct shape with minor errors