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Mathematics 2
MATH00040
Semester 2 2015-2016 Mock Exam Solutions
1.
(a)
1 −3
−2
0
!

−2
5


+  2 −1 
1
0

can’t be performed since the matrices are not the same size.

 
 


2
6
2
0
1
3
2
0

 
 



0  −  −6 −2 
0  − 2  −3 −1  =  −1
 −1
4
0
1 −1
2
0
1 −1


0 −6


= 5
2 .
−3 −1

(b) det
2 1
−2 2
!
2 1
−2 2
= 2 × 2 − 1 × (−2) = 6 6= 0, so
2 1
−2 2
!−1
1
=
6
2 −1
2
2
!
=
1
3
1
3
!
has an inverse.
− 16
1
3
!
.
(c)
1 −3
−2 −1

!
−2
3 0
1 −3 2
!
=
−5 12 −6
3 −3 −2
!
.



0
−3
0





 −1   2  can’t be performed since the number of columns in  −1  is
3
−1
3


−3


not the same as the number of rows in  2 .
−1

1
(d)

2 −1 −1
−2
3


det  −3 −2
3  =2 −3 −2
1 −3 −2

−3
3
− (−1) 1 −2
−3 −2
+ (−1) 1 −3
=2(−2 × (−2) − 3 × (−3)) + 1(−3 × (−2) − 3 × 1)
− 1(−3 × (−3) − (−2) × 1)
=2(13) + 1(3) − 1(11)
=26 + 3 − 11
=18.

2 −2 1 −5


(e) We will row reduce the augmented matrix  1 −1 1 −2 .
−1
2 3
8

R1 ↔ R2
R2 → R2 − 2R1
R3 → R3 + R1
R2 ↔ R3
R3 → −R3
R1 → R1 − R3
R2 → R2 − 4R3
R1 → R1 + R2

1 −1 1 −2


 2 −2 1 −5 
−1
2 3
8


1 −1
1 −2


0 −1 −1 
 0
0
1
4
6


1 −1
1 −2


1
4
6 
 0
0
0 −1 −1


1 −1 1 −2


1 4
6 
 0
0
0 1
1


1 −1 0 −3


1 0
2 
 0
0
0 1
1


1 0 0 −1


2 
 0 1 0
0 0 1
1

Hence the solution is x = −1, y = 2, z = 1.
2
(f) We have
det
"
6 4
2 4
!
−λ
1 0
0 1
!#
= det
6−λ
4
2 4−λ
!
= (6 − λ)(4 − λ) − 4(2)
= λ2 − 10λ + 24 − 8
= λ2 − 10λ + 16.
Hence the characteristic equation is λ2 − 10λ + 16 = 0 or (λ − 2)(λ − 8) = 0. Thus
the eigenvalues are λ = 2 and λ = 8.
We will now find the eigenvectors corresponding to these eigenvalues.
λ = 2:
!
!
!
!
!
6 4
x
x
6x + 4y
2x
We have
=2
, so
=
.
2 4
y
y
2x + 4y
2y
Hence we have the equations 6x + 4y = 2x and 2x + 4y = 2y. Both these!equations
−1
reduce to x = −y, so taking y = 1, say, we obtain the eigenvector
.
1
λ = 8:
!
!
!
!
!
6x + 4y
x
x
6x + 4y
8x
We have
=8
, so
=
.
2x + 4y
y
y
2x + 4y
8y
Hence we have the equations 6x + 4y = 8x and 2x + 4y = 8y. Both these
! equations
2
reduce to x = 2y, so taking y = 1, say, we obtain the eigenvector
.
1
2.
(a)
√
√
√
|z| = |2 + 3i| = 22 + 32 = 4 + 9 = 13, z = 2 + 3i = 2 − 3i,
Re(z) = Re(2 + 3i) = 2, Im(z) = Im(2 + 3i) = 3.
z+w
z−w
zw
z
w
= (2 + 3i) + (5 − i) = (2 + 5) + (3 − 1)i = 7 + 2i
= (2 + 3i) − (5 − i) = (2 − 5) + (3 − (−1))i = −3 + 4i
= (2 + 3i)(5 − i) = ((2)(5) − (3)(−1)) + ((2)(−1) + (3)(5))i = 13 + 13i
2 + 3i
2 + 3i 5 + i
7 + 17i
7
17
=
=
·
=
=
+ i
5−i
5−i 5+i
26
26 26
√
(b) The real part of 1 − 3i is positive and its imaginary part is negative, so we are in
the situation of Figure 5 in √
the Complex Numbers notes.
Hence the argument of 1 − 3i is
√ !
√ π
−1 − 3 3 =− .
θ = −φ = − tan
= − tan−1
1 3
3
q
√
√ 2 √
Also, the magnitude of 1 − 3i is r = 12 + − 3 = 1 + 3 = 4 = 2.
√
Hence, 1 − 3i in polar form is
π π
√
+ i sin −
.
1 − 3i = 2 cos −
3
3
√
To calculate (1− 3i)5 we will use Corollary 2.3.9 from the Complex Numbers notes.
That is we will use
√
(r(cos(nθ) + i sin(nθ)))n = r n (cos(nθ) + i sin(nθ)),
π
with r = 2, θ = − and n = 5.
3
Hence
π 5
√ 5 π + i sin −
1 − 3i = 2 cos −
3 3 5π
5π
= 25 cos −
+ i sin −
3
3
π π = 32 cos
+ i sin
3 !
3
√
1
3
+i
= 32
2
2
√
= 16 + 16 3i.
(c) We will use the fact (see P.12 of the Complex Numbers notes) that the nth roots
are given by
1
θ 2kπ
θ 2kπ
+ i sin
k = 0, 1, . . . , n − 1.
+
+
zk = r n cos
n
n
n
n
√
π
In this case we have r = 2 2 and θ = − and we are looking for the third roots, so
4
we take n = 3.
Thus the roots are
√ 31 −π/4 2kπ −π/4 2kπ
+ i sin
k = 0, 1, 2.
cos
+
+
zk = 2 2
3
3
3
3
That is
π √ π
+ i sin −
2 cos −
12
12
√
7π
7π
z1 = 2 cos
+ i sin
12
12
√
5π
5π
+ i sin
z2 = 2 cos
4
4
z0 =
4
3.
(a) (i) We first have to find the critical points of f . To do this we will differentiate f
and solve the equation f ′ (x) = 0.
However f ′ (x) = 6x2 − 6x − 36, so we solve the equation 6x2 − 6x − 36 = 0. Now
6x2 − 6x − 36 = 0 ⇔ x2 − x − 6 = 0
⇔ (x + 2)(x − 3) = 0
⇔ x = −2 or x = 3.
Thus the critical points are x = −2 and x = 3.
Next, f ′′ (x) = 12x − 6 and we evaluate f ′′ (x) at each of the critical points.
f ′′ (−2) = 12(−2) − 6 = −30 < 0, so the critical point at x = −2 is a local
maximum.
f ′′ (3) = 12(3) − 6 = 30 > 0, so the critical point at x = 3 is a local minimum.
(ii) We first have to find the critical points of f , regarding it as having domain R.
To do this we will differentiate f and solve the equation f ′ (x) = 0.
However f ′ (x) = 6x2 + 6x − 36, so we solve the equation 6x2 + 6x − 36 = 0. Now
6x2 + 6x − 36 = 0 ⇔ x2 + x − 6 = 0
⇔ (x + 3)(x − 2) = 0
⇔ x = −3 or x = 2.
Thus the critical points are x = −3 and x = 2.
We can now find where the global maxima and minima of f occur by evaluating it at the endpoints of the domain and at the critical points that lie in the
domain.
So we evaluate f (x) at x = −1, 1.
f (−1) = 40 and f (1) = −28.
Hence the global maximum of f is 40 attained at x = −1 and the global minimum of f is −28 attained at x = 1.
e−2x cos(−2x)
(b) With f (x) =
we have a product in the numerator, so we have to use
ln(x)
the product rule before we use the quotient rule.
First let us differentiate g(x) = e−2x cos(−2x).
d −2x d
e
cos(−2x) + e−2x (cos(−2x))
dx
dx
−2x
−2x
= −2e
cos(−2x) + e
(2 sin(−2x))
−2x
= 2e (sin(−2x) − cos(−2x)).
g ′ (x) =
We can now use the quotient rule with g(x) = e−2x cos(−2x) and h(x) = ln(x).
Then
f ′ (x) =
=
g ′(x)h(x) − g(x)h′ (x)
h(x)2
2e−2x (sin(−2x) − cos(−2x)) ln(x) − e−2x cos(−2x) ·
(ln(x))2
5
1
x.
3
2
With g(x) = e−x +2x −x−2 sin(x) we will use the chain rule with
u = −x3 + 2x2 − x − 2 sin(x) and y = eu (where we are letting y = g(x)).
Then
dy
dy
=
dx
du
= eu
du
dx
−3x2 + 4x − 1 − 2 cos(x)
3
2
= −3x2 + 4x − 1 − 2 cos(x) e−x +2x −x−2 sin(x) .
4.
·
π
(a) (i) The graph of f (x) = sin(2x) lies above the x-axis between the points x = and
4
π
3π
π
and
. Thus the required
x = and below the x-axis between the points
2
2
4
area is
Z
π
2
π
4
sin(2x) dx −
Z
3π
4
sin(2x) dx
π
2
π2 3π4
1
1
= − cos(2x) − − cos(2x)
2
2
π
π
4
2
π 1
1
1
1
3π
= − cos (π) − − cos
− − cos
− − cos (π)
2
2
2
2
2
2
1
1
1
1
= − (−1) − − (0)
− − (0) − − (−1)
2
2
2
2
1
1
= − −
2
2
= 1.
(ii) Using the formula V = π
Z
b
f (x)2 dx , the volume is
a
V =π
=π
Z
1
−1
Z 1
−e−3x
e−6x dx
2
dx
−1
1
1 −6x
=π − e
6
−1
1 −6(−1)
1 −6(1)
− − e
=π − e
6
6
1
−e−6 + e6
=π
6
π (e6 − e−6 )
=
.
6
6
(b) (i) Here we use integration by parts.
Let f (x) = 2x and g ′(x) = sin(−2x),
1
1
so that f ′ (x) = 2 and g(x) = −
cos(−2x) = cos(−2x).
−2
2
Hence, using the integration by parts formula,
Z
Z
1
1
2x sin(−2x) dx = 2x ·
cos(−2x) − 2 ·
cos(−2x) dx
2
2
Z
= x cos(−2x) − cos(−2x) dx
1
sin(−2x) + c
−2
1
= x cos(−2x) + sin(−2x) + c.
2
= x cos(−2x) −
(ii) Here we use integration by substitution.
du
= 2x.
Let u = x2 − 4, so that
dx
du
du
Then dx =
=
.
du/dx
2x
Also, when x = 2, u = 0 and when x = 3, u = 5.
Hence
Z
2
3
x
√
dx =
x2 − 4
5
x du
√ ·
u 2x
0
Z 5
1 −1
=
u 2 du
0 2
5
1
1
=
2u 2
2
0
h 1 i5
= u2
Z
0
1
2
1
= 5 − 02
√
= 5.
5.
(a) Let A be the event ‘the staff member selected is a nurse’ and let B be the event
‘the staff member selected is male’, so that P (A ∪ B) is the probability that the
staff member selected is either a nurse or is male. Since there are 13 nurses out of
13
a total of 6 + 13 = 19 staff members, P (A) =
. Also, there are 2 male doctors
19
and 13 − 7 = 6 male nurses, so there are a total of 8 male staff members. Thus
8
6
P (B) = . Next, there are 6 male nurses so that P (A ∩ B) = . Hence we have
19
19
that the probability that the staff member selected is either a nurse or is male is
P (A ∪ B) = P (A) + P (B) − P (A ∩ B) =
7
8
6
15
13
+
−
= .
19 19 19
19
(b) Let us call tossing a head a success and let X denote the number of successes
we get in twenty tosses, so that we want to find P (X > 10). In this case the
probability of a success is p = 0.5 and we are tossing a coin twenty times, so n = 20.
Since the table doesn’t directly give us P (X > 10), we have to use the fact that
P (X > 10) = 1 − P (X 6 9). So we look at the c = 9 row and the p = 0.5 column
in the n = 20 block. Hence the required probability is
P (X > 10) = 1 − P (X 6 9) ≃ 1 − 0.412 = 0.588.
(c) We will assume that this is a Poisson process with λ = 9, where λ is the average
number of customers arriving in the 3 minutes between between 5.30pm and 5.33pm
on Fridays. Thus we have to calculate P (X < 9), where X is the number of customers
arriving in the given six minutes. However P (X < 9) = P (X 6 8) and using the
Poisson distribution tables we see that the required probability is
P (X < 9) = P (X 6 8) ≃ 0.4557.
8