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ENG1090MOCKEXAM1
1. Functions and graphs2
Question 1.13
Identify the domain of fpxq “
?
x2´4
x2´4x`34
A) Dompfq “ Rzt1, 3u5 >B) Dompfq “ p´8,´2s Y r2, 3q Y p3,8q6
C) Dompfq “ p´8, 1q Y p3,8q7
D) Dompfq “ p´8,´2s Y p1, 3q Y p3,8q8
9
SOLUTION: ANSWER: B10
Since we cannot take the square root of a negative number, x ě 2 or x ď ´2. Also we cannot divide by11
zero, so the roots of x2´ 4x` 3, 1 and 3, must be excluded. This means Dompfq “ p´8,´2s Y r2, 3q Y12
p3,8q.13
Question 1.214
Identify the domain and range of fpxq “
b
x2`2x´3
x´1 .15
A) Dompfq “ r´3,8qzt1u and Ranpfq “ r0,8qzt2u16
B) Dompfq “ p´8,´3s Y p1,8q and Ranpfq “ r0,8q17
C) Dompfq “ p´8,´3s Y r1,8q and Ranpfq “ r0,8q18
D) Dompfq “ r´3, 1q and Ranpfq “ r0,8q19
SOLUTION: ANSWER: A20
Since division by zero is not defined, x ‰ 1. And if x ‰ 1,
b
x2`2x´3
x´1 “
?
x` 3. Therefore the domain21
of f is x ě ´3 and x ‰ 1, Dompfq “ r´3,8qzt1u. The range of ?x` 3 is r0,8q, but we cannot have22
x “ 1, so we have to exclude ?1` 3 “ 2, Ranpfq “ r0,8qzt2u.23
Question 1.324
Let f : RÑ R, fpxq “ 1x2`1 . Identify the inverse function f´1pxq.25
A) f´1pxq “
b
1´x
x with domain Rzt0u26
B) f´1pxq “
b
1´x
x with domain p0, 1s27
C) f´1pxq “
b
1´x
x with domain R28
D) f´1pxq does not exist29
SOLUTION: ANSWER: D30
Since f is not one-to-one, the inverse of f does not exist.31
Question 1.432
Let f : p´8, 0s Ñ R, fpxq “ 3x´2 . Then the equation f´1pxq “ fpxq has solution x “(placeholder).33
SOLUTION: Since the graph of f´1 is the graph of f reflected across the y “ x diagonal, the intersection34
of the graphs of f and f´1 must be on the line y “ x. Therefore we can solve for fpxq “ x. Solving35
3
x´2 “ x we get x “ 3,´1. Since the domain of f is negative reals, x “ ´1 is the only solution.36
Question 1.537
Let fpxq “ 6 sinp2x` pi3 q ´ p3
?
3` 4q cosp2xq. Suppose f written in single sine function in phase-angle38
form, fpxq “ A sinp2x` αq.39
1
2 ENG1090 MOCK EXAM
a) Determine the period of fpxq.40
b) Find the amplitude A.41
c) Which quadrant is the phase angle α in?42
d) Find the phase angle α.43
SOLUTION:44
a) The period is 2pik “ 2pi2 “ pi.45
b) By addition of angle formula,
6 sinp2x` pi
3
q ´ p3?3` 4q cosp2xq “ 6 sinp2xq cosppi
3
q ` 6 cosp2xq sinppi
3
q ´ p3?3` 4q cosp2xq
“ 3 sinp2xq ´ 4 cosp2xq.
Therefore A “a32 ` p´4q2 “ 5.46
c) We have 5 sinp2xq cospαq`5 cosp2xq sinpαq “ 3 sinp2xq´4 cosp2xq. Therefore sinpαq “ ´4{547
is negative and cospαq “ 3{5 is positive. This means α is in the 4th quadrant.48
d) tanpαq “ sinpαqcospαq “ ´43 . Since α is in the fourth quadrant α “ Arctanp´43 q “ ´Arctanp 43 q, as49
Arctan is an odd function.50
Question 1.651
Which of the following identities are true for all real numbers x?52
1 cosppi2 ´ xq “ sinpxq53
2 Arcsinpsinpxqq “ x54
3 cosp2xq “ 2 cos2pxq ´ 155
4 sinppi ´ xq “ ´ sinppi ` xq56
A) 3 and 4 only57
B) All 1, 2, 3, and 458
C) 1 and 2 only59
D) 1,3, and 4 only60
61
SOLUTION: ANSWER: D62
cosppi2 ´ xq “ sinpxq is the complementary angle formula; cosp2xq “ 2 cos2pxq ´ 1 is the double angle63
formula; sinppi´xq “ ´ sinppi`xq follows from sinp´xq “ ´ sinpxq and sinpx`2piq “ sinpxq. However64
Arcsinpsinpxqq “ x is NOT true for all x, since Arcsin only returns values between ´pi{2 and pi{2.65
Question 1.766
The value of a house has appreciated exponentially from 2000 to 2020. If the house is worth 30000067
dollars in 2000 and 500000 dollars in 2010, then it is worth (placeholder) dollars in 2020 (round your68
answer to the nearest dollar).69
SOLUTION: V ptq “ Aect where t is number of years since 2000. Using initial value we see A “70
300000. At year 2010 we have 500000 “ 300000e10c, or e5c “ 53 . At year 2020 the value is V p20q “71
300000e20c “ 300000pe10cq2 “ 300000ˆ p 53 q2 “ 833333 dollars.72
ENG1090 MOCK EXAM 3
2. Complex numbers73
Question 2.174
Simplify the complex number z “ p3` iqp2` 3iq
2´ i .75
z “(placeholder) ` (placeholder)i.76
SOLUTION:
p3` iqp2` 3iq
2´ i “
p3` iqp2` 3iqp2` iq
p2´ iqp2` iq “
´5` 25i
5
“ ´1` 5i.77
Question 2.278
Suppose z “ 5` 12i, find |z¯|.79
|z¯|=(placeholder).80
SOLUTION: z¯ “ 5´ 12i, |z¯| “a52 ` p´12q2 “ 13.81
Question 2.382
Let z “ ?2cisppi6 q and w “ 12e
3pii
4 . Then
w
z3
“(placeholder) ` (placeholder)i.83
SOLUTION: In polar form
w
z3
“ 12cisp
3pi
4 q
2
?
2cisp 3pi6 q
“ 6?
2
cisppi4 q “ 3` 3i.84
Question 2.485
Use De Moivre’s Theorem or otherwise, calculate z “ p1´?3iq5 in polar form with principal argument.86
z = (placeholder) cisp(placeholder)pi{(placeholder)q.87
SOLUTION: 1´?3i “ 2cisp´pi3 q, z “ p1´
?
3iq5 “ 25cisp´ 5pi3 q “ 32cisppi3 q.88
Question 2.589
Which of the following is a 4th-roots of z “ 1´?3i?90
A) 2cisp 5pi12 q91
B) 2
1
4 cisp´7pi12 q92
C) -2
1
4 cisp 7pi12 q93
D) -2
1
4 cisp´5pi12 q94
SOLUTION: ANSWER: B95
The fourth roots of 1´?3i are 21{4cisp´ pi12 q, 21{4cisp´ 7pi12 q, 21{4cisp 5pi12 q, and 21{4cisp 11pi12 q.96
Question 2.697
Let z be the square root of 3`4i with positive real part. Find z by solving the equation pa` biq2 “ 3`4i,98
where a and b are real numbers. z “ (placeholder) ` (placeholder)i.99
SOLUTION: pa` biq2 “ pa2´ b2q` 2abi “ 3` 4i. Setting a2´ b2 “ 3 and 2ab “ 4, we get the solution100
a “ 2,b “ 1.101
Question 2.7102
The solutions of the equation z2 ´ 3iz “ 3` i are z1 “ 1` (placeholder)i and z2 “ (placeholder)`i.103
4 ENG1090 MOCK EXAM
SOLUTION: Substitute z “ 1`ai into the equation z2´3iz “ 3`iwe get p1´a2`3aq`p2a´3qi “ 3`i.104
Hence a “ 2. Substitute z “ b`i into the equation z2´3iz “ 3`iwe get pb2´1`3q`p2b´3bqi “ 3`i.105
Hence b “ ´1. Note that both real and imaginary part equations have to be satisfied, so only one of the106
solutions of each quadratic is acceptable.107
3. Vectors108
Question 3.1109
Which of the following identities is NOT true?110
A) p~a`~bq ¨ p~a´~bq “ |~a|2 ´ |~b|2111
B) ~a ¨~b “ ~b ¨ ~a112
C) p~a ¨~bq ¨ ~c “ ~a ¨ p~b ¨ ~cq113
D) p~a`~bq ¨ ~c “ ~a ¨ ~c`~b ¨ ~c114
115
SOLUTION: ANSWER: C116
All the identities are properties of dot product except p~a ¨~bq ¨~c is not defined, since ~a ¨~b is a scalar, and we117
cannot take the dot product of a scalar with a vector.118
Question 3.2119
Let p “ 2ˆi´3jˆ`6kˆ and q “ 3ˆi´ jˆ`2kˆ. Then 2p´q “ (placeholder) iˆ` (placeholder) jˆ` (placeholder)120
kˆ.121
SOLUTION: 2p2ˆi´ 3jˆ ` 6kˆq ´ p3ˆi´ jˆ ` 2kˆq “ iˆ´ 5jˆ ` 10kˆ.122
Question 3.3123
Let p “ 2ˆi´ 3jˆ ` 6kˆ and q “ 3ˆi´ jˆ ` 2kˆ. Then p ¨ q “(placeholder).124
SOLUTION: p2ˆi´ 3jˆ ` 6kˆq ¨ p3ˆi´ jˆ ` 2kˆq “ 6` 3` 12 “ 21.125
Question 3.4126
Let p “ iˆ´ jˆ ` 2kˆ and q “ 2ˆi` 2jˆ ` 3kˆ.Which of the following statements is correct?127
A) p and q are perpendicular to each other128
B) p and q are parallel to each other129
C) The angle between p and q is less thanpi2130
D) The angle between p and q is greater thanpi2131
E) The angle between p and q is pi2132
SOLUTION: ANSWER: C133
Since p ¨ q “ 6 is positive, the cosine of the angle between them is positive. Therefore the angle between134
p and q is less than pi{2.135
Question 3.5136
Let q “ 3ˆi´ jˆ ` 2kˆ. Then the direction angle of q in the jˆ direction is137
A) pi ´ Arccosp 1?
14
q138
B) Arccosp 37 q139
ENG1090 MOCK EXAM 5
C) Arccosp 2?
14
q140
D) Arccosp 27 q141
SOLUTION: ANSWER: A142
The direction angle of q in ~j direction is Arccosp´1|q| q “ Arccosp´1{
?
9` 1` 4q “ pi´Arccosp1{?14q,143
using the supplementary angle formula for cosine.144
Question 3.6145
Let p “ 2ˆi´ 3jˆ` 6kˆ and q “ 3ˆi´ jˆ` 2kˆ. The scalar projection of q in the direction of p is (placeholder).146
SOLUTION: The scalar projection is q ¨ pˆ “ q¨p|p| “ 21?4`9`36 “ 3.147
Question 3.7148
Two forces F1 of 9 Newtons in the direction of 2ˆi´2jˆ` kˆ, and F2 of 5 Newtons in the direction of 3ˆi`4kˆ,149
are acting on a particle and displaced it from p2, 4,´1q to p1, 1, 0q. The work done on the particle by the150
forces is (placeholder) Joules.151
SOLUTION: Displacement is p1, 1, 0q ´ p2, 4,´1q “ p´1,´3, 1q. Work done is
p 9p2ˆi´ 2jˆ ` kˆqa
22 ` p´2q2 ` 12 `
5p3ˆi` 4kˆq?
32 ` 42 q ¨ p
ˆ´i´ 3jˆ ` kˆq “ 16J.
4. Limits and differentiation152
Question 4.1153
Let f : RÑ R be a piecewise function defined by fpxq “ ?x´ 1 for x ě 1 and fpxq “ |x|´1 forx ă 1.154
Determine all values of x where f is continuous.155
A) R156
B) Rzt0u157
C) Rzt1u158
D) Rzt0, 1u159
SOLUTION: ANSWER: A160
f is continuous at every point except possibly 1, by addition and subtraction of continuous functions. At 1161
the left limit is 0 and the right limit is also 0, and fp1q “ 0. Therefore f is continuous for all R.162
Question 4.2163
Let f : R Ñ R be the function fpxq “ ?x´ 1 for x ě 1 and fpxq “ |x| ´ 1 forx ă 1. Determine all164
values of x where f is differentiable.165
A) R166
B) Rzt0u167
C) Rzt1u168
D) Rzt0, 1u169
SOLUTION: ANSWER: D170
limxÑ0´
fpxq ´ fp0q
x´ 0 is ´1, but limxÑ0`
fpxq ´ fp0q
x´ 0 is 1, so f is not differentiable at 0.171
6 ENG1090 MOCK EXAM
limxÑ1´
fpxq ´ fp1q
x´ 1 is 1, but limxÑ1`
fpxq ´ fp1q
x´ 1 diverges to infinity. Therefore f is also not differ-172
entiable at 1.173
Question 4.3174
Evaluate the limit175
lim
xÑ0
ex
2 ´ 1
1´ cospxq
Write“N/A” if the limit does not exist.176
SOLUTION: The limit is of the type 00 , therefore we can apply l’Hopital’s rule.
lim
xÑ0
ex
2 ´ 1
1´ cospxq “ limxÑ0
2xex
2
sinpxq .
This is still of the type 00 , therefore we can apply l’Hopital’s rule again.
lim
xÑ0
2xex
2
sinpxq “ limxÑ0
2ex
2 ` 4x2ex2
cospxq “ 2.
Question 4.4177
The derivative of fpxq “ lnpxq cosplnpxqq is178
A)
cosplnpxqq ´ lnpxq sinplnpxqq
x
179
B) ´ sinplnpxqq
x
180
C)
cosp 1x q
x
181
D) ´ cosplnpxqq
x sinplnpxqq182
E) None of others183
SOLUTION: ANSWER: A
Using the product and chain rule, we get
d
dx
lnpxq cosplnpxqq “ 1
x
cosplnpxqq ´ lnpxq sinplnpxqq 1
x
“ cosplnpxqq ´ lnpxq sinplnpxqq
x
.
Question 4.5184
The derivative of xx
2
is185
A) xx
2`1186
B) 2xx
2`2187
C) plnpxq ` 1qxx2188
D) p2 lnpxq ` 1qxx2`1189
SOLUTION: ANSWER: D Using logarithmic differentiation we see lnpxx2q “ x2 lnpxq, thus
d
dx
x2 lnpxq “ 2x lnpxq ` x.
Therefore
d
dx
xx
2 “ xx2p2x lnpxq ` xq “ p2 lnpxq ` 1qxx2`1.
ENG1090 MOCK EXAM 7
Question 4.6190
The gradient of the tangent line to the curve191
sinp4xyq ` y2 “ 1
at the point p0, 1q is (placeholder).192
SOLUTION: Using implicit differentiation we see
cosp4xyqp4y ` 4xdy
dx
q ` 2y dy
dx
“ 0.
At p0, 1q this becomes
4` 2dy
dx
“ 0.
So
dy
dx
“ ´2.193
Question 4.7194
Find the area of the largest rectangle that can be inscribed in a circle of radius 1.195
SOLUTION: Let the coordinate of the top right corner be px, yq. Then x2 ` y2 “ 1 and the area of the
rectangle is A “ 4xy. Therefore A “ 4x?1´ x2. To find the critical point, set
0 “ dA
dx
“ 4
a
1´ x2 ´ 4x
2
?
1´ x2 “
4´ 8x2?
1´ x2 , 0 ď x ď 1.
The critical point is x “
?
2
2 , y “
?
2
2 . Since f
1pxq ą 0 for x ă
?
2
2 , and f
1pxq ă 0 for x ą
?
2
2 , this is a196
local maximum. Therefore the maximum area is 4
?
2
2
?
2
2 “ 2197
Question 4.8198
Suppose a function f : R Ñ R has only two critical numbers, ´1 and 2, and f2p´1q “ 2, f2p2q “ ´2.199
Which of the following statements are true?200
1 f has a global maximum at 2.201
2 f has a local minimum at ´1.202
3 fpxq “ 0 has a solution between ´1 and 2.203
4 f is increasing between ´1 and 2.204
5 f has a point of inflection between ´1 and 2.205
A) 2, 4 and 5 only206
B) 2 only207
C) 1 and 2 only208
D) 3 and 4 only209
E) all of them are true210
SOLUTION: ANSWER: A211
By the second derivative test f has a local minimum at ´1 and local maximum at 2. However f may212
not have a global maximum (for example a cubic curve). f 1 is positive between ´1 and 2, therefore f is213
increasing between ´1 and 2, and since the second derivative changed from negative to positive, there is a214
point of inflection between ´1 and 2 as well.215
8 ENG1090 MOCK EXAM
5. Integration and applications216
Question 5.1
An antiderivative of
x2 ´ 1
5x
` 1
is217
A) 13x
3 ´ 1p5xq2 ` x218
B) 13x
3 ´ 5` x219
C) 13x
3 ´ 1x ` x` 3220
D) 13x
3 ´ ln |5x| ` x221
E) None of the others222
SOLUTION: ANSWER: E223
An antiderivative of 15x is
1
5 ln |x|, hence none of the expressions are correct.224
Question 5.2225
Let fpxq “ esin2p2pixq. Then ş1
0
f 1pxq dx “(placeholder).226
SOLUTION: By the Fundamental Theorem of Calculus,ż 1
0
f 1pxq dx “ fp1q ´ fp0q “ esin2p2piq ´ esin2p0q “ 1´ 1 “ 0.
Question 5.3
Calculate the integral ż 2
0
x` 5
x2 ´ 2x´ 3 dx
A) ´3 lnp3q227
B) lnp7q ´ lnp5q228
C) ´ 23229
D) 115230
E) None of the others231
SOLUTION: ANSWER: A
Using partial fractions,
x` 5
x2 ´ 2x´ 3 “
2
x´ 3 ´
1
x` 1 .
Therefore ż 2
0
x` 5
x2 ´ 2x´ 3 dx “ r2 ln |x´ 3| ´ ln |x` 1|s
2
0 “ ´3 lnp3q.
Question 5.4
Calculate the indefinite integral ż
e2x
1` e2x dx
A) lnp?1` e2xq ` c232
B)
1
1` e2x ` c233
ENG1090 MOCK EXAM 9
C) lnp1` e2xq ` c234
D)
e2x
p1` e2xq2 ` c235
E) None of the others236
SOLUTION: ANSWER: A
Let u “ 1` e2x, then du “ 2e2x dx. Thereforeż
e2x
1` e2x dx “
1
2
ż
1
u
du “ 1
2
ln |u| ` c “ 1
2
lnp1` e2xq ` c “ lnp
a
1` e2xq ` c.
Question 5.5237
The volume of the solid of revolution formed by rotating the curve y “ Arcsinpx2 q, 0 ď x ď 2 about the238
y-axis is (placeholder)pi2 cubic units.239
SOLUTION: We have x “ 2 sinpyq, therefore
V “ pi
ż pi{2
0
x2 dy “ pi
ż pi{2
0
4 sin2pyq dy “ pi
ż pi{2
0
2p1´ cosp2yqq dy “ pi2.
Question 5.6240
Let fpxq “ x34 , x ě 0. Find the area enclosed by fpxq and f´1pxq, the inverse function of f .241
A “(placeholer) square units.242
SOLUTION: By symmetry, the area enclosed by f and f´1 is twice the area enclosed by f and the line
y “ x. x34 and x intersects at 0 and 2. Therefore the area is
2
ż 2
0
x´ x
3
4
dx “ 2
„
x2
2
´ x
4
16
2
0
“ 2.
Question 5.7
Calculate the integral ż 1{2
0
1
p1´ x2q 32 dx
A) 1?
3
243
B) 12244
C) pi6245
D) pi3246
E) None of the others247
SOLUTION: ANSWER: A
Use the substitution x “ sinptq, then dx “ cosptq dt.ż 1{2
0
1
p1´ x2q 32 dx “
ż pi
6
0
1
cos3ptq cosptq dt “
ż pi
6
0
sec2ptq dt “ rtanptqspi60 “
1?
3
.
6. Differential equations and related rates248
Question 6.1249
An ice cube is melting at a rate of 300cm3{s, retaining its shape in the process. What is the rate of change,250
10 ENG1090 MOCK EXAM
in cm{s, of the side length of the cube when its volume is 125cm3?251
Rate of change = (placeholder).252
SOLUTION: Let V be the volume of the cube and x its side length. Then V “ x3 and dV
dx
“ 3x2. Thus
dV
dt
“ dV
dx
dx
dt
.
At V “ 125, x “ 5, so
´300 “ p3ˆ 52q dx
dt
dx
dt
“ ´4
Question 6.2
The equation
sinpxyq
ˆ
dy
dx
˙3
` 2x “ 0
is a253
A) First order differential equation254
B) Second order differential equation255
C) Separable differential equation256
D) Third order differential equation257
E) None of the others258
SOLUTION: ANSWER: A259
This is a first order differential equation, since the higher order derivative that appears in the equation is260
the first derivative.261
Question 6.3
Let y be the solution to the initial value problem
dy
dx
“ 3x2 ` 1, yp0q “ 2.
Then yp1q=(placeholder).262
SOLUTION: Since
dy
dx
is a function of x, we can integrate to find y “ x3 ` x ` c. Using the initial263
condition yp0q “ 2, we find c “ 2. Therefore yp1q “ 13 ` 1` 2 “ 4.264
Question 6.4
Find the general solution of the differential equation
sinpxqdy
dx
` cospxqy “ 2 cospxq
A)
A
cospxq ` 2265
B)
A
sinpxq ` 2266
C)
eA
cospxq267
ENG1090 MOCK EXAM 11
D)
eA
cotpxq ´ 2268
E) None of the others269
SOLUTION: ANSWER: B
Write the differential equation in standard form,
dy
dx
“ p2´ yqcospxq
sinpxq .
This is a separable equation, therefore ż
1
2´ y dy “
ż
cospxq
sinpxq dx
´ ln |2´ y| “ ln | sinpxq| ` C
We can absorb the ˘’s from absolute value signs and C into a single constant A,
1
2´ y “ A sinpxq
y “ A
sinpxq ` 2
Question 6.5270
A tank initially contains 200 litres of salt water with salt concentration 8 grams per litre. Fresh water is271
pumped into the tank at a rate of 3 litre per second. At the same time, the mixture in the tank is leaking at272
1 litres per second through a hole in the bottom of the tank. What is the concentration of the salt water in273
the tank, in grams per litre, after 5 minutes? Concentration=(placeholder).274
SOLUTION: First we set up the differential equation. Let S be the amount of salt in the tank. Then
dS
dt
“ ´1 ¨ concentration of salt in the tank.
Since the total amount of liquid in the tank is increasing by 2 litres per second (3 incoming and 1 outgoing),
the concentration at time t is
S
200` 2t . Therefore
dS
dt
“ ´ S
200` 2t .
This is a separable equation, thus
´
ż
1
S
dS “ 1
2
ż
1
100` t dt
´ ln |S| “ 1
2
ln |100` t| ` C
1
S
“ A?100` t
S “ A?
100` t .
Using the initial condition Sp0q “ 1600 we get A “ 16000. Therefore at t “ 300 second, the concentra-
tion is
S
200` 2t “
16000?
100` 300p200` 600q “ 1
12 ENG1090 MOCK EXAM
Question 6.6
A particle moves with velocity
vptq “ ´6 sinp3tqi` 6 cosp3tqj
Find the magnitude of the acceleration at t “ pi3 .275
Magnitude of acceleration = (placeholder).276
SOLUTION: The acceleration is aptq “ v1ptq “ ´18 cosp3tqi´ 18 sinp3tqj. At t “ pi3 , a´ 18 cosppiqi´277
18 sinppiqj, |a| “ 18.278
Question 6.7
A particle moves with velocity
vptq “ ´6 sinp3tqi` 6 cosp3tqj
Suppose the initial position of the particle is p2, 3q, then the position of the particle at t “ pi is (placeholder,279
placeholder).280
SOLUTION: We integrate vptq to get the position vector function.
xptq “ p2 cosp3tq ` c1qi` p2 sinp3tq ` c2qj
Using the initial condition we see c1 “ 0, c2 “ 3. At time t “ pi, the position is 2 cosp3piqqi`p2 sinp3piq`281
3qj “ ´2i` 3j.282

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