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 log5 x3 = 7 for x.
Answer: 57/3
Solution: c.f. §1.3.43
log5 x3 = 7 ⇒ x3 = 57 ⇒ x = 57/3
1 mark
(
x2 − 3 if x 6= 5
(b) Let f (x) =
, find lim f (x).
x→5
20
if x = 5
Answer: 22
Solution: c.f. §2.2
Note that f (5) is of no importance when computing lim f (x).
x→5
1 mark
(c) True or False:
If f (x) and g(x) both have a discontinuity at x = 4, then it so does y = f (x) · g(x).
Answer: False
(
(
1 if x ≤ 4
0 if x ≤ 4
.
, and g(x) =
Solution: Let f (x) =
0 if x > 4
1 if x > 4
1 mark
(d) If C(q) = 50 + ln q, find the marginal cost when production is at 7 units.
Answer: 1/7
Solution: c.f. §3.6.21
dC
= 1q ⇒ C 0 (7) = 17 .
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 = sec x.
Answer: sec x tan 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 = ln x.
Answer: y 0 =
√
Solution: y = ln x = ln x1/2 = 12 ln x =⇒ y 0 =
1 mark
1
2x
1
.
2x
(h) An investment portfolio earns an annual interest rate of 3% compounded continuously.
How long will it take for the investment to triple?
Answer:
ln 3
0.03
years
Solution: c.f. notes: 3P V = P V e0.03t ⇒ ln 3 = 0.03t ⇒ t =
ln 3
0.03
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) = cos2 (x2 ).
Answer: g 0 (x) = −2 cos (x2 )·sin (x2 )·2x
Solution: Use the Chain Rule a couple of times.
2 marks
x2 + 4x − 21
.
x→3 2x2 − 4x − 6
(b) Evaluate lim
Answer: 5/4
Solution: c.f. Quiz 1 #2
Direct substitution yields 0/0, so we simplify first:
x2 + 4x − 21
(x + 7)(x − 3)
x+7
=
=
2
2x − 4x − 6
(2x + 2)(x − 3)
2x + 2
x+7
x2 + 4x − 21
=
lim
= 10/8 = = 5/4 .
x→3 2x + 2
x→3 2x2 − 4x − 6
Hence the limit is lim
2 marks
(x3 − 7)7 − 1
by interpreting as a derivative.
x→2
x−2
Answer: 84
(c) Evaluate lim
Solution: c.f. §3.7.97
(x3 − 7)7 − 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)7 and a = 2. Therefore, f 0 (x) = 7(x3 − 7)6 (3x2 ) and the limit must
be equal to f 0 (2) = 7 · 12 · 3(4) = 84.
2 marks
(d) Suppose the position (in feet) of an object moving horizontally at time t (in seconds)
is s(t) = t sin 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 sin t ⇒ v(t) = ds
= sin t + t cos t ⇒ a(t) =
dt
here see that a(2) = 2 cos 2 − 2 sin 2f t/s2
dv
dt
= cos t + cos t − t sin t. From
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 = 3x .
e
Answer: 1/3
Solution:
x
e3x
e3x − xe3x 3
dy
=
dx
(e3x )2
1 − 3x
=
e3x
y=
Therefore the only critical number occurs when x = 1/3.
2 marks
(
π
(f) Let f (x) =
0
if x = 0
, find lim f (f (x)).
x→0
if x =
6 0
Answer: π
Solution: If x 6= 0, then f (x) = 0. Thus, f (f (x)) = f (0) = π when x 6= 0.
Therefore limx→0 f (f (x)) = limx→0 f (0) = limx→0 π = π.
2 marks
√
7− x
(g) Evaluate lim
.
x→49 x − 49
1
Answer: − 14
Solution:
√ §2.3.47
√
√
7− x
7− x 7+ x
−1
1
√ = lim
√ =− .
lim
= lim
·
x→49 x − 49
x→49 x − 49
14
7 + x x→49 7 + 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 40 years. How fast is the population growing when
there are 1,000,000 people?
Answer: 25, 000 · ln 2
Solution: We know that P = 40, 000ekt will double after 40 years. Thus, 80000 =
40000e40k ⇒ 2 = e40k ⇒ k = ln402 .
We also know that P 0 = 40, 000ekt · k = P · k. Therefore P 0 = 1, 000, 000 · ln402 =
25, 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) = 2x3 + 3x2 + 2x + 3 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) = 3, so we just need to find a negative value.
f (−1) = −2 + 3 − 2 + 3 = 2, which doesn’ help much.
f (−2) = −16 + 12 − 2 + 3 = −3.
There there must by a root in the interval (−2, −1) by the IVT. Note that any
interval contain -3/2 would be sufficient.
Solution: Notice that f (x) = 2x3 + 3x2 + 2x + 3 = x2 (2x + 3) + 1 · (2x + 3) =
(x2 + 1)(2x + 3). From which we can see that only root is x = −3/2, 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 sufficienct.
4 marks
3. Let a, b be real numbers and consider the function

x

if x ≤ 0
b + a
2
g(x) = bx + a(x + 3) if 0 < x ≤ 1

√

7ax − b 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− b + ax = b + 1 = lim+ bx2 + a(x + 3)
x→0
x→0
b + 1 = b + 1 = 3a
From here we see that b = 3a − 1 .
At x = 1 we need:
lim f (x) = f (1)
x→1−
= lim+ f (x)
x→1
√
lim− bx + a(x + 3) = b + 4a = lim+ 7ax − b x
2
x→1
x→1
b + 4a = b + 4a = 7a − b
From here we see that 2b = 3a .
Combining these we get:
2b = 3a
2(3a − 1) = 3a
6a − 2 = 3a
3a = 2
a = 2/3.
Thus, b = 1 and a = 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 + 0.008p = 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 + 0.008p = 4
1 dq 1
·
− + 0.008 = 0
q dp p
p dq
·
− 1 + 0.008p = 0
q dp
E(p) − 1 + 0.008p = 0
E(p) = 1 − 0.008p
Since revenue is maximized at unit elasticity, we must have E = −1, or 1 − 0.008p =
2
−1 ⇒ p = 0.008
= $250.
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 + 0.008p = 4
ln q = 4 + ln p − 0.008p
q = e4+ln p−0.008p
q = e4 · eln p · e−0.008p
q = e4 · ·e−0.008p
Therefore, revenue is given by R = pq = p · (e4 p · e−0.008p ) = e4 · p2 · e−0.008p
Since revenue is maximized when
dR
dp
= 0, we need R0 (p)
R = e4 · p2 · e−0.008p
dR
= e4 · 2p · e−0.008p + e4 · p2 · e−0.008p (−0.008)
dp
dR
= e4 · p · e−0.008p [2 + p(−0.008)]
dp
Since p ≥ 200, it follows that R0 (p) = 0 if and only if −0.008p = −2 ⇒ p = 2000/5 =
$250.
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 6 ft/s, how fast is the boat approaching the dock when it is 25ft from the dock?
Solution: c.f. WW-7.5 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
and a concluding statement.
time; 1pt for solving 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
= −6 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(−6).
dt
dx
−6 √ 2
=
25 + 9
dt
25
Therefore, the boat is approaching the dock at a rate of
6√ 2
25 + 9 f t/s .
25
Mathematics 104:184
4 marks
Page 10 of 10
Student-No.:
x
6. Find the absolute minimum value of f (x) = √
on the interval [4, 12].
x−3
Solution: c.f §4.1.63 Marking scheme: 1pt for f 0 (x), 1pt for finding the critical
number x = 6, 1pt for evaluating f (4), f (6), and f (12), 1pt for explaining why f (6) is
the absolute maximum.
x
Differentiating f (x) = √
give
x−3
√
1
x − 3 − x · 2√x−3
0
√
f (x) =
( x − 3)2
1
2(x − 3) − x
√
=
·
x−3
2 x−3
1
x−6
=
· √
x−3 2 x−3
From which we see that x = 6 is the only critical number. The next step is to evaluate
f (x) a the critical number and the end points:
f (4) = 4
√
6
f (6) = √ = 2 3
3
12
f (12) = √ = 4
9
To finish we note that
√6
3
2
=
36
3
= 12 < 16 so that
Therefore, the absolute minimum is
√6 ,
3
√6
3
< 4.
which occurs when x = 6.