[ET 2018]
Mathematics
15 January 2011
We provide solutions for all four versions, but Version A is solved in much greater detail.
Version A
1. A construction firm has the choice between three different payment schedules:
(a) Pay e60300 in cash.
(b) Pay e10800 for year per 8 years, where the first installment is to be paid at once.
(c) Pay e19800 in cash and thereafter e6300 per year for 12 years, where the first installment is
to be paid after 1 year.
Determine which schedule is least expensive when the interest rate is 11.5% per year.
Sol. Present value for (a) is of course e60300; for (b) it is 10800
−8
10800(1 + 0.115) 1−(1+0.115)
0.115
−12
'
19800 + 6300 1−(1+0.115)
0.115
' 60879. 98; for (c) it is 19800 +
P7
−k
=
k=0 (1 + 0.115)
P12
−k
6300 k=1 (1 + 0.115)
=
59745. 67. Hence, option (c) is the cheapest.
2. Find the quadratic approximation for y about (x,y) = (0,1) when y is defined implicitly as a
function of x by the equation 1 + x3 y + x = y 1/2 .
Sol. By implicit differentiation, we find that the 3x2 y + x3 y 0 + 1 = (1/2)y −1/2 y 0 . Inserting x = 0
and y = 1 gives y 0 = 2. Similarly, a second round of differentiation yields 6xy + 6x2 y 0 + x3 y 00 =
(−1/4)y −3/2 (y 0 )2 + (1/2)y −1/2 y 00 . Inserting x = 0, y = 1 and y 0 = 2 gives y 00 = 2. Hence,
y(x) ≈ 1 + 2x + x2
3. (a) Compute the integral
Z
3
4
2t − 2
dt
t2 − 2t
(b) Solve the equation
Z
3
x
2t − 2
dt = ln
t2 − 2t
2
x−1
3
Sol.
(a) Using the substitution z = t2 − 2t, we get
Z
2t − 2
dt = ln t2 − 2t + C
2
t − 2t
so the definite integral evaluates to
Z 4
t=4
2t − 2
dt = ln t2 − 2t t=3 = ln 8 − ln 3
2
3 t − 2t
(b) Using the indefinite integral, the left-hand side of the equation can be written as
Z x
2t − 3
dt = ln x2 − 2x − ln 3
2
3 t − 3t
so that we need to solve
ln x2 − 2x = ln
This gives two systems of inequalities


 x<0∨x>2



x > 23
∨




 2
x − 4x + 3 = 0
The solution is: {3} ∪
2
x − 1 + ln 3
3











0<x<2
x>
3
2
−x2 + 2x = (2x − 3)
√ 3 .
4. Let
f (x,y) = e2x+y + e2x−y − 4x − y
(a) Find the first and second order partial derivatives of f (x,y).
(b) Show that f (x,y) is convex.
(c) Find its stationary points and classify them.
Sol. (a) The first order partial derivatives are
fx0 (x,y)
=
2e2x+y + 2e2x−y − 4 = 2e2x (ey + e−y ) − 4
fy0 (x,y)
=
e2x+y − e2x−y − 1 = e2x (ey − e−y ) − 1
The second order partial derivatives, arranged as a Hessian matrix, are
2x y
4e (e + e−y ) 2e2x (ey − e−y )
2
D f (x,y) =
2e2x (ey − e−y ) e2x (ey + e−y )
00
(b) We have fxx
= 4e2x (ey + e−y ) > 0 and D2 f (x,y) = 16e4x > 0, so that f is convex.
(c) Stationary points are solutions to:
 2x y
 2x+y
=
 e (e + e−y ) = 2
 e
⇔
 2x y
 2x−y
e (e − e−y ) = 1
e
=
3
2
⇔
1
2

 2x + y = ln 32
2x − y = ln 12

Since f is convex, the only stationary point is a global minimum point.
5. Consider an n × n matrix such that
A − 2A2 − I = 0
(a) Show that A is invertible and find a formula for A−1 .
(b) Calculate A − 2A2 and, if possible, A−1 , when
1
4
A=
1
16
−7
1
4
.
⇔

 x=
1
4
ln 34
y=
1
2
ln 3

Sol. (a) Note that
A − 2A2 − I = 0 ⇔ A(I − 2A) = I.
We compute the determinant of both sides in the last equality∗ and derive
|A| |I − 2A| = |I| = 1
so that
|A| =
6 0, and |I − 2A| =
6 0
In particular. A is invertible and has inverse
A−1 = I − 2A.
(b) Note that
A − 2A2 =
1
4
1
16
−7
1
4
−2
1
4
1
16
−7
2
=
1
4
1
0
0
1
=I
so that A satisfies the equation in (a). Hence A is invertible and has inverse
1
1
1 0
−7
14
4
2
A−1 = I − 2A =
−2
=
1
1
0 1
− 81 12
16
4
6. Consider the two functions:
a) f (x,y,z) =
√
y2 x
z
b) g(x,y,z) = e
√
y2 x
z
(a) Establish by an explicit argument whether f is homogeneous (and, if so, of which degree).
(b) Establish whether g is homothetic. (Is g homogeneous?)
Sol. (a) The function f is homogeneous of degree 32 . Indeed for all t > 0, and for all x, y, z in the
domain of f
√
2√
2
3 y
3
(ty) tx
x
2
=t
= t 2 f (x,y,z)
f (tx,ty,tz) =
tz
z
(b) Note that g(x,y,z) = H(f (x,y,z)), where H(r) = er . Then g is homothetic, as it is the transformation of a homogenous function by means of the strictly increasing (and differentiable) map
H(r) = er .
It is not homogeneous, though. A proof of this fact is the following. If there was a k in R such
that g(tx,ty,tz) = tk g(x,y,z) for all t > 0, and for all x, y, z in the domain of g, then in particular,
for (x,y,z) = (1,1,1) it should be
3/2
g(t,t,t) = tk g(1,1,1), ∀t > 0, that is,
et
But if we choose t = 4, the preceding equality gives k =
contradiction.
7
ln 4 ,
∗ We
recall that if A and B are square matrices, then |A B| = |A| |B|.
= tk e, ∀t > 0.
while for t = 9 we derive k =
26
ln 9 ,
a
Version B
1. A construction firm has the choice between three different payment schedules:
(a) Pay e70350 in cash.
(b) Pay e12600 for year per 8 years, where the first installment is to be paid at once.
(c) Pay e23100 in cash and thereafter e7350 per year for 12 years, where the first installment is
to be paid after 1 year.
Determine which schedule is least expensive when the interest rate is 11.5% per year.
Sol. (a) Present value for (a) is of course e70350; for (b) it is approximately e71026.64; for (c) it
is approximately e69703.28. Hence, option (c) is the cheapest.
2. Find the quadratic approximation for y about (x,y) = (0,4) when y is defined implicitly as a
function of x by the equation 2 + x4 y − x = y 1/2 .
Sol. By implicit differentiation, we find that the 4x3 y + x4 y 0 − 1 = (1/2)y −1/2 y 0 . Inserting x = 0
and y = 4 gives y 0 = −4. Similarly, a second round of differentiation yields 12x2 y + 8x3 y 0 =
(−1/4)y −3/2 (y 0 )2 + (1/2)y −1/2 y 00 . Inserting x = 0, y = 1 and y 0 = −4 gives y 00 = 2. Hence,
y(x) ≈ 4 − 4x + x2
3. (a) Compute the integral
Z
4
5
2t − 3
dt
t2 − 3t
(b) Solve the equation
Z
4
x
2t − 3
dt = ln
t2 − 3t
3
13
x+
4
16
Sol.
(a) Using the substitution z = t2 − 3t, we get
Z
2t − 3
dt = ln t2 − 3t + C
2
t − 3t
so the definite integral evaluates to
Z 5
t=5
2t − 3
dt = ln t2 − 3t t=4 = ln 10 − ln 4
2
4 t − 3t
(b) Using the indefinite integral, the left-hand side of the equation can be written as
Z x
2t − 3
dt = ln x2 − 3x − ln 4
2
4 t − 3t
so that we need to solve
ln x2 − 3x = ln
This gives two systems of inequalities

x<0∨x>3





13
x > − 12
∨




 2
x − 3x = 3x + 13
4
The solution is: − 21 , 13
∪ ∅.
2
13
3
x+
4
16











+ ln 4
0<x<3
x > − 13
12
−x2 + 3x = 3x +
13
4
4. Let
f (x,y) = e3x+y + e3x−y − 5x − y
(a) Find the first and second order partial derivatives of f (x,y).
(b) Show that f (x,y) is convex.
(c) Find its stationary points and classify them.
Sol. The gradient is
∇f (x,y) =
3e3x+y + 3e3x−y − 5
e3x+y − e3x−y − 1
The stationary points are
3x+y
3x+y
3e
+ 3e3x−y − 5 = 0
e
=
⇔
e3x+y − e3x−y − 1 = 0
e3x−y =
The Hessian is
2
D f (x,y) =
4
3
1
3
⇔
9e3x+y + 9e3x−y
3e3x+y − 3e3x−y
3x + y = ln 34
3x − y = ln 13
3e3x+y − 3e3x−y
e3x+y + e3x−y
Determinant: 36e6x > 0, 9e3x+y + 9e3x−y > 0, f convex and
point.
1
6
⇔
x=
y=
1
6
1
2
ln 94
ln 4
ln 49 ; 12 ln 4 is a global minimum
5. Consider an n × n matrix such that
3A − A2 − I = 0
(a) Show that A is invertible and find a formula for A−1 .
2
(b) Calculate 3A − A2 and, if possible, A−1 , when A =
−1
−1
1
.
Sol.
3A − A
2
A−1
2 −1
2 −1
1 0
= 3
−
=
1
−1 1
0 1
1 0
2 −1
1 1
= 3I − A = 3
−
=
0 1
−1 1
1 2
2
−1
6. Consider the two functions:
a) f (x,y,z) =
√
x3 y
z2
b) g(x,y,z) = e
√
x3 y
z2
(a) Establish by an explicit argument whether f is homogeneous (and, if so, of which degree).
(b) Establish whether g is homothetic. (Is g homogeneous?)
Sol. (a) The function f is homogeneous of degree 23 . (b) The function g is homothetic, but it is
not homogeneous.
Version C
1. A construction firm has the choice between three different payment schedules:
(a) Pay e72360 in cash.
(b) Pay e12960 for year per 8 years, where the first installment is to be paid at once.
(c) Pay e23760 in cash and thereafter e7560 per year for 12 years, where the first installment is
to be paid after 1 year.
Determine which schedule is least expensive when the interest rate is 11.5% per year.
Sol. Present value for (a) is of course e72360; for (b) it is approximately e73055.97; for (c) it is
approximately e71694.81. Hence, option (c) is the cheapest.
2. Find the quadratic approximation for y about (x,y) = (0,9) when y is defined implicitly as a
function of x by the equation 3 + x5 y − x = y 1/2 .
Sol. By implicit differentiation, we find that the 5x4 y + x5 y 0 − 1 = (1/2)y −1/2 y 0 . Inserting x = 0
and y = 9 gives y 0 = −6. Similarly, a second round of differentiation yields 20x3 y +10x4 y 0 +x5 y 00 =
(−1/4)y −3/2 (y 0 )2 + (1/2)y −1/2 y 00 . Inserting x = 0, y = 9 and y 0 = −6 gives y 00 = 2. Hence,
y(x) ≈ 9 − 6x + x2
3. (a) Compute the integral
Z
5
6
2t − 4
dt
t2 − 4t
(b) Solve the equation
Z
5
x
2t − 4
dt = ln
t2 − 4t
4
17
x+
5
20
Sol.
(a) Using the substitution z = t2 − 4t, we get
Z
2t − 4
dt = ln t2 − 4t + C
2
t − 4t
so the definite integral evaluates to
Z 6
t=6
2t − 4
dt = ln t2 − 4t t=5 = ln 12 − ln 5
2
5 t − 4t
(b) Using the indefinite integral, the left-hand side of the equation can be written as
Z x
2t − 4
dt = ln x2 − 4x − ln 5
2
5 t − 4t
so that we need to solve
ln x2 − 4x = ln
This gives two systems of inequalities

x<0∨x>4





x > − 17
16




 2
x − 4x = 5 45 x +
1 17 The solution is: − 2 , 2 ∪ ∅.
∨
17
20
17
4
x+
5
20











+ ln 5
0<x<4
x > − 17
16
−x2 + 4x = 5
4
5x
+
17
20
4. Let
f (x,y) = ex+2y + ex−2y − 3x − y
(a) Find the first and second order partial derivatives of f (x,y).
(b) Show that f (x,y) is convex.
(c) Find its stationary points and classify them.
Sol. Gradient is
∇f (x,y) =
ex+2y + ex−2y − 3
2ex+2y − 2ex−2y − 1
Stationary points
ex+2y + ex−2y − 3 = 0
⇔
2ex+2y − 2ex−2y − 1 = 0
Hessian
D2 f (x,y) =
ex−2y + ex+2y
−2ex−2y + 2ex+2y
x = 21 ln 35
16
y = 41 ln 75
−2ex−2y + 2ex+2y
4ex−2y + 4ex+2y
Determinant: 16e2x > 0, ex−2y + ex+2y > 0, f convex and
point.
1
2
1
7
ln 35
16 ; 4 ln 5
3
2
a global minimum
5. Consider an n × n matrix such that
4A − A2 − I = 0
(a) Show that A is invertible and find a formula for A−1 .
2
2
−1
(b) Calculate 4A − A and, if possible, A , when A =
1
.
Sol.
2
1
3
2
4A − A2
=
4
A−1
=
4I − A = 4
−
1
0
2
1
0
1
3
2
2
1
0
=
−
2
1
3
2
0
1
=
2
−1
−3
2
6. Consider the two functions:
1
x4 y 2
a) f (x,y,z) =
z2
1
b) g(x,y,z) = ln
x4 y 2
z2
!
(a) Establish by an explicit argument whether f is homogeneous (and, if so, of which degree).
(b) Establish whether g is homothetic. (Is g homogeneous?)
Sol. (a) The function f is homogeneous of degree 25 . (b) The function g is homothetic, but it is
not homogeneous.
Version D
1. A construction firm has the choice between three different payment schedules:
(a) Pay e63650 in cash.
(b) Pay e11400 for year per 8 years, where the first installment is to be paid at once.
(c) Pay e20900 in cash and thereafter e6650 per year for 12 years, where the first installment is
to be paid after 1 year.
Determine which schedule is least expensive when the interest rate is 11.5% per year.
Sol. Present value for (a) is of course e63650; for (b) it is approximately e64262.20; for (c) it is
approximately e63064.88. Hence, option (c) is the cheapest.
2. Find the quadratic approximation for y about (x,y) = (0,8) when y is defined implicitly as a
function of x by the equation 2 + x5 y − x = y 1/3 .
Sol. By implicit differentiation, we find that the 5x4 y + x5 y 0 − 1 = (1/3)y −2/3 y 0 . Inserting x = 0
and y = 8 gives y 0 = −12. Similarly, a second round of differentiation yields 20x3 y+10x4 y 0 +x5 y 00 =
(−2/9)y −5/3 (y 0 )2 + (1/3)y −2/3 y 00 . Inserting x = 0, y = 8 and y 0 = −12 gives y 00 = 12. Hence,
y(x) ≈ 8 − 12x + 6x2
3. (a) Compute the integral
Z
6
8
2t − 5
dt
t2 − 5t
(b) Solve the equation
Z
6
x
2t − 5
dt = ln
t2 − 5t
5
23
x+
6
8
Sol.
(a) Using the substitution z = t2 − 5t, we get
Z
2t − 5
dt = ln t2 − 5t + C
2
t − 5t
so the definite integral evaluates to
Z 8
t=8
2t − 5
dt = ln t2 − 5t t=6 = ln 4
2
6 t − 5t
(b) Using the indefinite integral, the left-hand side of the equation can be written as
Z x
2t − 5
dt = ln x2 − 5x − ln 6
2
6 t − 5t
so that we need to solve
ln x2 − 5x = ln
This gives two systems of inequalities

x<0∨x>5





x > − 69
20




 2
x − 5x = 6 56 x +
3 23 The solution is: − 2 , 2 ∪ ∅.
∨
23
8
23
5
x+
6
8











+ ln 6
0<x<5
x > − 69
20
−x2 + 5x = 6
5
6x
+
23
8
4. Let
f (x,y) = ex+3y + ex−3y − 6x − 2y
(a) Find the first and second order partial derivatives of f (x,y).
(b) Show that f (x,y) is convex.
(c) Find its stationary points and classify them.
Sol. Gradient is
∇f (x,y) =
ex−3y + ex+3y − 6
−3ex−3y + 3ex+3y − 2
Stationary points
ex−3y + ex+3y − 6 = 0
⇔
−3ex−3y + 3ex+3y − 2 = 0
Hessian
x + 3y = ln 10
3
x − 3y = ln 83
ex−3y + ex+3y
−3ex−3y + 3ex+3y
2
D f (x,y) =
⇔
x = 21 ln 80
9
y = 16 ln 54
−3ex−3y + 3ex+3y
9ex−3y + 9ex+3y
1
2
Determinant: 36e2x > 0, ex−3y + ex+3y > 0, f convex and
point.
1
5
ln 80
9 ; 6 ln 4
a global minimum
5. Consider an n × n matrix such that
5A − A2 − I = 0
(a) Show that A is invertible and find a formula for A−1 .
(b) Calculate 5A − A2 and, if possible, A−1 , when A =
3
1
5
2
.
Sol.
5A − A
A
2
−1
=
=
5
3
1
5
2
−
5I − A = 5
1
0
3
1
0
1
5
2
2
1
0
=
−
3
1
5
2
0
1
=
2
−1
−5
3
6. Consider the two functions:
√
x3 y
a) f (x,y,z) =
z2
b) g(x,y,z) = ln
√ x3 y
z2
(a) Establish by an explicit argument whether f is homogeneous (and, if so, of which degree).
(b) Establish whether g is homothetic. (Is g homogeneous?)
Sol. (a) The function f is homogeneous of degree 23 . (b) The function g is homothetic, but it is
not homogeneous.