Mathematics 104-184
Page 2 of 10
Student-No.:
1. Very Short-Answer Questions. Put your answer in the box provided. Full marks will be
given for a correct answer placed in the box. Each part is worth 1 mark, but not all parts are
of equal difficulty.
1 mark
(a) Solve log7 x3 = 5 for x.
Answer: 75/3
Solution: c.f. §1.3.43
log7 x3 = 5 ⇒ x3 = 75 ⇒ x = 75/3
1 mark
(
x2 − 3 if x 6= 4
(b) Let f (x) =
, find lim f (x).
x→4
10
if x = 4
Answer: 13
Solution: c.f. §2.2
Note that f (4) is of no importance when computing lim f (x).
x→4
1 mark
(c) True or False:
If f (x) is not differentiable at x = 4, then it is not continuous at x = 4.
Answer: False
Solution: c.f. §3.2
Consider f (x) = |x − 4|.
1 mark
(d) If C(q) = 50 + ln q, find the marginal cost when production is at 8 units.
Answer: 1/8 = $0.125
Solution: c.f. §3.6.21
dC
= 1q ⇒ C 0 (8) = 18 .
dq
Mathematics 104-184
Page 3 of 10
Student-No.:
1. Cont. Very Short-Answer Questions. Put your answer in the box provided. Full marks
will be given for a correct answer placed in the box. Each part is worth 1 mark, but not all
parts are of equal difficulty.
1 mark
(e) Differentiate y = tan x.
Answer: sec2 x
Solution: c.f. §3.5 Thm 3.13.
1 mark
(f) Evaluate lim−
x→2
|x − 2|
.
x−2
Answer: -1
Solution: c.f. §2.2
x < 2 ⇒ |x − 2| = −(x − 2) so that lim−
x→2
1 mark
|x − 2|
−(x − 2)
= lim−
= −1
x→2
x−2
x−2
(g) Differentiate y = x ln x.
Answer: y 0 = ln x + 1
Solution: c.f. §3.9 Example 1
y 0 = ln x + x · x1 = ln x + 1.
1 mark
(h) An investment portfolio earns an annual interest rate of 13% compounded continuously.
How long will it take for the investment to triple?
Answer:
ln 3
0.13
years
Solution: c.f. notes: 3P V = P V e0.13t ⇒ ln 3 = 0.13t ⇒ t =
ln 3
0.13
Mathematics 104-184
Page 4 of 10
Student-No.:
2. Short-Answer Questions. Each question is worth 2 points. Put your answer in the box
provided and show your work. No credit will be given for the answer without the correct
accompanying work.
2 marks
(a) Differentiate g(x) = sin2 (x2 ).
Answer: g 0 (x) = 2 sin (x2 ) · cos (x2 ) · 2x
Solution: g 0 (x) = 2 sin (x2 ) · cos (x2 ) · 2x
2 marks
2x2 − 4x − 6
.
x→3 x2 + 4x − 21
(b) Evaluate lim
Answer: 4/5
Solution: c.f. Quiz 1 #2
Direct substitution yields 0/0, so we simplify first:
(2x + 2)(x − 3)
2x + 2
2x2 − 4x − 6
=
=
2
x + 4x − 21
(x + 7)(x − 3)
x+7
2x2 − 4x − 6
2x + 2
= lim
= 8/10 = = 4/5 .
2
x→3 x + 4x − 21
x→3 x + 7
Hence the limit is lim
2 marks
(x3 − 7)8 − 1
by interpreting as a derivative.
x→2
x−2
Answer: 96
(c) Evaluate lim
Solution: c.f. §3.7.97
(x3 − 7)8 − 1
f (x) − f (a)
lim
= lim
= f 0 (a) from which we can clearly see that
x→2
x→a
x−2
x−a
f (x) = (x3 − 7)8 and a = 2. Therefore, f 0 (x) = 8(x3 − 7)7 (3x2 ) and the limit must
be equal to f 0 (2) = 8 · 12 · 3(4) = 96.
2 marks
(d) Suppose the position (in feet) of an object moving horizontally at time t (in seconds)
is s(t) = t cos t. Determine the acceleration of the object at t = 2s.
Answer: a(2) = −2 cos 2 − 2 sin 2 f t/s2
Solution: c.f. §3.6
s(t) = t cos t ⇒ v(t) = ds
= cos t − t sin t ⇒ a(t) =
dt
From here we see that a(2) = −2 cos 2 − 2 sin 2f t/s2
dv
dt
= − sin t − sin t − t cos t.
Mathematics 104-184
Page 5 of 10
Student-No.:
2. Cont. Short-Answer Questions. Each question is worth 2 points. Put your answer in
the box provided and show your work. No credit will be given for the answer without the correct
accompanying work.
x
2 marks
(e) Find the critical point(s) of y = 2x .
e
Answer: 1/2
Solution: c.f. §4.1
x
e2x
dy
e2x − xe2x 2
=
dx
(e2x )2
1 − 2x
=
e2x
y=
Therefore the only critical number occurs when x = 1/2.
2 marks
(
e if x = 0
(f) Let f (x) =
, find lim f (f (x)).
x→0
0 if x 6= 0
Answer: e
Solution: If x 6= 0, then f (x) = 0. Thus, f (f (x)) = f (0) = e when x 6= 0.
Therefore limx→0 f (f (x)) = limx→0 f (0) = limx→0 e = e.
2 marks
√
8− x
(g) Evaluate lim
.
x→64 x − 64
1
Answer: − 16
Solution:
√ §2.3.47
√
√
8− x
8− x 8+ x
−1
1
√ = lim
√ =− .
lim
= lim
·
x→64 x − 64
x→64 x − 64
16
8 + x x→64 8 + x
2 marks
(h) Some time in the future a human colony is started on Mars. The colony begins with
40,000 people and doubles every 50 years. How fast is the population growing when
there are 1,000,000 people?
Answer: 20, 000 · ln 2
Solution: We know that P = 40, 000ekt will double after 50 years. Thus, 80000 =
40000e50k ⇒ 2 = e50k ⇒ k = ln502 .
We also know that P 0 = 40, 000ekt · k = P · k. Therefore P 0 = 1, 000, 000 · ln502 =
20, 000 ln 2 when P = 1, 000, 000.
Mathematics 104-184
Page 6 of 10
Student-No.:
2. Cont. Short-Answer Questions. Each question is worth 2 points. For parts (i) and (j),
your graphs should be reasonably accurate.
2 marks
(i) Sketch the graph of the inverse function.
Solution: It’s a mirror image across the line y = x.
2 marks
(j) The function f (x) = 3x3 + 4x2 + 3x + 4 has one root. Find an interval containing it.
Solution: Since f (x) is a polynomial, it’s continuous everywhere and we may apply the
Intermediate Value Theorem freely.
f (0) = 4, so we just need to find a negative value.
f (−1) = −3 + 4 − 3 + 4 = 2, which doesn’ help much.
f (−2) = −24 + 16 − 6 + 4 = −10.
Therefore, there must be a root in the interval (−2, −1) by the IVT. Note that any
interval contain -4/3 would be sufficient.
Solution: Notice that f (x) = 3x3 +4x2 +3x+4 = x2 (3x+4)+1·(3x+4) = (x2 +1)(3x+4).
From which we can see that only root is x = −4/3, so any interval containing it would
be sufficient.
Mathematics 104-184
Page 7 of 10
Student-No.:
Full-Solution Problems. In questions 3–6, justify your answers and show all your work.
Each problem is worth 4 points. Please place a box around your final answers. Calculator-ready
answers are sufficient.
4 marks
3. Let a, b be real numbers and consider the function

x

if x ≤ 0
a + b
2
g(x) = ax + b(x + 3) if 0 < x ≤ 1

√

7bx − a x
if x > 1.
Find the values of a and b such that g(x) is continuous for all real x. Make sure you show
all your work and justify your claims.
Solution: c.f. §2.6.85, WW2.9 Marking scheme: 1pt for some mention that f (a)
must equal to lim f (x), 1pt for each box.
x→a
At x = 0 we need:
lim f (x) = f (0) = lim+ f (x)
x→0−
x→0
lim− a + bx = a + 1 = lim+ ax2 + b(x + 3)
x→0
x→0
a + 1 = a + 1 = 3b
From here we see that a = 3b − 1 .
At x = 1 we need:
lim f (x) = f (1)
x→1−
= lim+ f (x)
x→1
√
lim− ax + b(x + 3) = a + 4b = lim+ 7bx − a x
2
x→1
x→1
a + 4b = a + 4b = 7b − a
From here we see that 2a = 3b .
Combining these we get:
2a = 3b
2(3b − 1) = 3b
6b − 2 = 3b
3b = 2
b = 2/3.
Thus, a = 1 and b = 2/3 .
Mathematics 104-184
4 marks
Page 8 of 10
Student-No.:
4. Shark Inc. has determined that demand for its newest netbook model is given by
ln q − ln p + 0.005p = 4,
where q is the number of netbooks Shark can sell at a price of p dollars per unit. Shark has
determined that this model is valid for prices p ≥ 200. What price will maximize revenue?
Solution: c.f. WW6-9 Marking scheme: 2pt for correct differentiation; 1pt for
setting E=-1; 1pt for solving for p.
ln q − ln p + 0.005p = 4
1 dq 1
·
− + 0.005 = 0
q dp p
p dq
·
− 1 + 0.005p = 0
q dp
E(p) − 1 + 0.005p = 0
E(p) = 1 − 0.005p
Since revenue is maximized at unit elasticity, we must have E = −1, or 1 − 0.005p =
2
−1 ⇒ p = 0.005
= $400.
Solution: c.f. WW6-9 Marking scheme: 2pt for correct differentiation; 1pt for
setting R0 = 0; 1pt for solving for p.
ln q − ln p + 0.005p = 4
ln q = 4 + ln p − 0.005p
q = e4+ln p−0.005p
q = e4 eln p e−0.005p
q = e4 pe−0.005p
Therefore, revenue is given by R = pq = p(e4 pe−0.005p ) = e4 p2 e−0.005p
Since revenue is maximized when
dR
dp
= 0, we need R0 (p)
R = e4 p2 e−0.005p
dR
= e4 2pe−0.005p + e4 p2 e−0.005p (−0.005)
dp
dR
= e4 pe−0.005p [2 + p(−0.005)]
dp
Since p ≥ 200, it follows that R0 (p) = 0 if and only if −0.005p = −2 ⇒ p = 2000/5 =
$400.
Mathematics 104-184
4 marks
Page 9 of 10
Student-No.:
5. A boat is pulled into a dock by a rope attached to the bow of the boat and passing through
a pulley on the dock that is 3 ft higher than the bow of the boat. If the rope is pulled in at
a rate of 5 ft/s, how fast is the boat approaching the dock when it is 25ft from the dock?
Solution: c.f. WW-7.5 Marking scheme: 1pt for a decent diagram, defined notation,
and extracting information. Note that a decreasing value must have a negative derivative;
1pt for linking the variables; 1pt for differentiating with respect to time; 1pt for solving
and a concluding statement.
for dx
dt
Let x represent the distance between the boat and the dock. So we are looking for
when x = 25 ft.
dx
dt
Let h represent the distance between the pulley and bow of the boat. Thus, we are given
dh
= −5 ft/s. Note this is negative since the distance is decreasing.
dt
Notice that x and h are related by the Pythagorean Theorem:
x2 + 32 = h2 ,
so that
2x
dx
dh
= 2h .
dt
dt
When x = 25, we have
25 ·
∴
dx √ 2
= 25 + 9(−5).
dt
dx
−1 √ 2
=
25 + 9
dt
5
Therefore, the boat is approaching the dock at a rate of
1√ 2
25 + 9 f t/s .
5
Mathematics 104-184
4 marks
Page 10 of 10
Student-No.:
x
6. Find the absolute minimum value of f (x) = √
on the interval [3, 6].
x−2
Solution: c.f §4.1.63 Marking scheme: 1pt for f 0 (x), 1pt for finding the critical
number x = 4, 1pt for evaluating f (3), f (4), and f (6), 1pt for explaining why f (4) is the
absolute maximum.
x
Differentiating f (x) = √
give
x−2
√
1
x − 2 − x · 2√x−2
0
√
f (x) =
( x − 2)2
1
2(x − 2) − x
√
=
·
x−2
2 x−2
1
x−4
=
· √
x−2 2 x−2
From which we see that x = 4 is the only critical number. The next step is to evaluate
f (x) a the critical number and the end points:
f (3) = 3
√
4
f (4) = √ = 2 2
2
6
f (6) = √ = 3
4
To finish we note that
√
1.5 ⇒ 2 2 < 2(1.5) = 3.
√4
2
2
=
16
2
= 8 < 9 so that
Therefore, the absolute minimum is
√4 ,
2
√4
2
< 3. Alternatively,
which occurs when x = 4.
√
2 ≈ 1.4 <