quiz for class 1 answers

8/12/2019 Quiz for Class 1 Answers

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Chapter 2

Quiz for Class 1

2.1 Topics for Class 1 Quiz

2.1.1 Topics from the Notes on Vectors

Inner Products (questions 1-4)

Length (questions 5-7)

The Least Squares Problem (questions 8-10)

Inequalities for Vectors (questions 11-12)

The Angle Between Two Vectors (questions 13-15)

Vectors, Lines, Planes and Hyperplanes (questions 16-17)

2.1.2 Topics from the Notes on Open Sets, Closed Sets

and Continuous Functions

Boundaries, Open and Closed Sets (questions 18-20)

Closed Sets, Bounded Sets, Compact Sets and Continuous Functions (ques-tions 21-24)

Continuity for Consumer Theory (question 25)

2.1.3 Topics from the Notes on Introduction to Multivari-ate Calculus

Derivatives and Approximations (questions 26-27)

Level Sets and Vectors of Partial Derivatives (question 28)

5


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6 CHAPTER 2. QUIZ FOR CLASS 1

2.2 Class 1 Quiz

2.2.1 Inner Products (essential topic, questions 1-4)

Class 1 quiz question 1

The inner product (sometimes called the dot product) of the n vectors x and yis defined as:

x0y=nX

i=1

xiyi.

Ifx and y are n vectors is the statement x0y6=y0x

1. true for alln vectors x and y

2. true for some but not all n vectors x and y

3. true for no n vectors x and y

4. I dont know.

Answer: x0y =Pn

i=1xiyi =Pn

i=1yixi = y0x, therefore the statement is

true for non vectors xand y.


Ifxand y are n vectors and k is a real number is the statement

k(x0y) = (ky)0x= x0(ky)

1. true for all real numbers k and all n vectors x and y

2. true for some but not all real numbers k and all n vectors x and y

3. true for no real numbersk and all n vectors x and y

4. I dont know.

Answer: ifk is a scalar

k(x0y) = k

nXi=1

xiyi

!=

nXi=1

(kxi)yi = (kx)0y=

nXi=1

xi(kyi) = x0 (ky)

Therefore the statement is true for all real numbers k and all n vectors xand y.


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2.2. CLASS 1 QUIZ 7


Think about the n vectors t, u, v and w, and the scalars a,b,c,d. Click thecorrect statements

1. t0 (u + v) = t0u + t0v for all t, u, v and w

2. (u + v)0

t6=t0 (u + v) for some t, u, v and w.

3. (t + w)0

(u + v) = t0u + w0u + t0v + w0v for all t, u, v and w.

4. (t + w)0 (u + v)6=u0t + u0w + v0t + v0w for some t, u, v and w.

5. (at+bw)0

(cu+dv) = act0u+bcw0u+adt0v+bdw0v for all a,b,c,d, t, u, v and w.

Answer: Statements 1, 3 and 5 are correct. They are implied by thedefinitions of vector addition, inner products and scalar multiplication as

t0 (u + v) =nX

i=1

ti(ui+vi) =nX

i=1

tiui+nX

i=1

tivi = t0u + t0v=

nXi=1

(ui+vi) ti = u0t + v0t

and

(t + w)0 (u + v) =

nXi=1

(ti+wi) (ui+vi) =nX

i=1

(tiui+wiui+tivi+wivi)

= t0u + w0u + t0v + w0v

= u0t + u0w + v0t + v0w.

Similarly

(at+bw)

0

(cu+dv) = (at)

0

(cu) + (bw)

0

(cu) + (at)

0

(dv) + (bw)

0

(dv)= act0u+bcw0u+adt0v+bdw0v.


Ify and xare n vectors and b a scalar, click the correct statements.

1. (ybx)0 (ybx) = y0y + x0xb2.2. (ybx)0 (ybx) = y0y+bx0y+by0x + x0xb2.

3. (ybx)0 (ybx) = y0y x0yby0xb+x0xb2.4. (ybx)0 (ybx) = y0y+2x0yb+x0xb2.

5. (yb

x)

0

(yb

x) =

y0y2

x0yb+

x0xb

2

.

Answer: Statements 3 and 4 are correct because, asx0y= y0x

(ybx)0 (ybx) = y0y x0yby0xb+x0xb2= y0y2x0yb+x0xb2.


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2.3 Length (essential topic, questions 5-7)


What is the length of the vector (x1, x2) in 2 dimensional space?

1. x21+x22

2. x1+x2

3. x1x2

4. (x1+x2)1

2

5.

x21+x22

12

6. none of the above

7. I dont know.

Answer: Pythagoras Theorem implies that answer 5 is correct.


What is the length of the vector (x1, x2, x3) in 3 dimensional space?

1. x21+x22+x

23

2. x1+x2+x3

3. x1x2x3

4. (x1+x2+x3)1

2

5.

x21+x

22+x

23

1

2


7. I dont know

Answer: Pythagoras Theorem implies that answer 5 is correct.


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2.4. THE LEAST SQUARES PROBLEM (USEFUL TOPIC, QUESTIONS 8-10)9


By definitionx0x=Pn

i=1x2i .

As x2i 0for all i, x0x 0. Ifx= 0, then xi = 0 for all i so x

0x= 0.

Thus x0x 0and x =0 implies that x0x= 0. Ifx0x= 0what does this tell you about x?

1. x 0

2. x 0

3. x1 x2 x3 ..... xn4. x= 0


6. I dont know.

Answer: Ifx0x =Pn

i=1x2i = 0 then as xi is real number x

2i 0 for all i

andx2i = 0 if and only ifxi = 0. Thus x0x 0for all x and x0x=0 if and only

ifxi = 0for all i, that is ifx =0. Ifx = 0 so statement 4 holds then statements1, 2, and 3 also hold.

2.4 The Least Squares Problem (useful topic,questions 8-10)


Let

x=

01

and y=

43

Find the value of the scalar b that minimizes kybxk . Note that this is thesame as the value ofb that minimizes kybxk2, which is easier to work with.Show x, y, bx and ybx on an accurate diagram. Which of the followingstatements are true?

1. b= 2

2. b= 3

3. b= 4

4. x0 (ybx) = (ybx)0 x=1.


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2.4. THE LEAST SQUARES PROBLEM (USEFUL TOPIC, QUESTIONS 8-10)11

Find the value of the scalar b that minimizes kybxk . Note that this is thesame as the value of b that minimizes ky

bxk

2, which is easier to work with.

Show x, y, bx and ybx on an accurate diagram. Which of the followingstatements are true?

1. b= 2

2. b= 3

3. b= 4

4. x0 (ybx) = (ybx)0 x=1.

5. x0 (ybx) = (ybx)0 x=0.

6. The vectorsxand ybxare parallel.

7. The vectors xand ybxare at 90 to each other.

Answer:

ybx=

105

b

12

=

10 b5 2b

so

kybxk2 = (10 b)2 + (5 2b)2 = 125 40b+ 5b2.

Completing the square

125 40b+ 5b2 = 5 (b 4)2

+ 45

which is minimized by setting b = 4, and has a minimum value of45.

ybx=

105

4

12

=

63

x0 (ybx) = (ybx)0 x=2X

i=1

(yi bxi) xi= (16) + (2(3)) = 0.

From the diagram the vectors x and ybx are at 90 to each other. Anotherway of saying this is that vectors x and ybx are orthogonal to each other.

This is illustrated in Figure 2.2.


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(10,5)

(4,8)

(1,2)

(6,-3)

0

x = (1,2)

4x = (4,8)

y = (10,5)

y 4x = (6,-3)

Figure 2.2: Question 9


Let

x=

x1x2..

xn

y=

y1y2..

yn

Assume that x6=0. Find the value of the scalar b that minimizes kybxk.Note that the value b that minimizes kybxkis the same as the value ofb thatminimizes kybxk2, which is a quadratic function of b. If possible find theminimum by completing the square.

Note also thatbxis a vector that is a scalar multiple ofx so lies on the sameline as x. The vector ybx joins a point on this line y. Thus bx is the pointon the line through x that is closest to y. Draw a pair of vectors xand y anduse this property to findbx.

Which of the following statements are true?

1. b= (x0x)x0y=Pn

i=1x2i

(Pn

i=1xiyi) .

2. b=

(x0x)x0y=

Pni=1x

2i (Pni=1xiyi) .

3. b= (x0x)

1x0y=

Pni=1x

2i

1

(Pn

i=1xiyi) .

4. x0 (ybx) = (ybx)0 x=1.

5. x0 (ybx) = (ybx)0 x=0.


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2.5. INEQUALITIES FOR VECTORS (USEFUL TOPIC, QUESTIONS 11-12)13

6. The vectorsxand ybxare parallel.

7. The vectorsx

andyb

xare at 90

to each other.Answer b= (x0x)

1x0y=

Pni=1x

2i

1

(Pn

i=1xiyi), x0 (ybx) = (ybx)0 x=0

and the vectors xand ybxare at 90 to each other.

2.5 Inequalities for Vectors (useful topic, ques-

tions 11-12)

Class 1 quiz question 11: the Cauchy-Schwarz Inequality

Click the statements that hold for all vectors xand y.

1. x0y 1kxk kyk

2. x0y (x0x) (y0y)

3. x0y (x0x) 12 (y0y) 12

4. x0y (x0x)1 (y0y)1

5. x0y kxkkyk

6. |x0y| kxk kyk

Answer: Statements 3, 5 and 6 are correct. Statement 6 is the Cauchy-Schwarz inequality.

The Cauchy-Schwarz inequality states that

|x0y| (y0y)1

2 (x0x)1

2 =kxk kyk .

Class 1 quiz question 12: the Triangle Inequality

Click the statements that hold for all vectors x and y.

1. kx + yk< kxk+kyk

2. kx + yk kxk+kyk

3. kx + yk= kxk+kyk

4. kx + yk> kxk+kyk

5. kx yk< kxk kyk6. kx yk kxk+kyk

Answer: statements 2 and 5 are correct. This is the triangle inequality,(see notes on vectors).


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2.6 The Angle Between Two Vectors (essential

topic, questions 13-15)Class 1 quiz question 13

Which of the following statements is true?

1. x0y= cos

kxkkyk

2. x0y= (x0

x) (y0y)cos

3. x0y= (x

0

x)1

2 (y0y)1

2 cos

4. x0y= (x0

x)

1 (y0y)

1 cos

5. x0y= kxk kyk cos

6. dont know.

Answer: The notes on vectors show that

x0y= (x0

x)1

2 (y0y)1

2 cos = kxk kyk cos .

Given this statements 1, 2, and 4 can only hold for particular values ofkxkandkyk .

2.6.1 Class 1 quiz question 14

Ifx0y= 0what is the angle between xand y?

1. 0

2. 45

3. 90

4. 135

5. 180


7. I dont know

Answer: Ifx0y= 0 the angle between xand y is 90.


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2.6. THE ANGLE BETWEEN TWO VECTORS (ESSENTIAL TOPIC, QUESTIONS 13-15)15

45

1

1

2



Ifx0 = (3, 4)and y0 = (

2, 7

2)what is the angle between x and y?

1. 0

2. 45

3. 90

4. 135

5. 180


7. I dont know

Answer:x0 = (3, 4) so k x k= (32 + 42)

1

2 = 5 and y0 = (

2, 7

2) so k y k=

(2 + 98)1

2 = 10. Moreover x0y =3

2 + 28

2 = 25

2. As x0y = k x kky k cos where is the angle between x and y we have

cos = x0y

kx kk y k=

25

2

50 =

2

2 =

12

which implies that = 45.

To see whycos 45 = 12

think about the triangle in Figure 2.3. The vertical

and horizontal sides of this triangle both have length to 1 . From PythagorasTheorem the hypotenuse of this right angled triangle is

12 + 12 =

2 so

cos 45 = 1

2.


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x0

x1

x2

p(x -x0) = 0

a

d

a

c

b

Figure 2.4:

2.7 Vectors, Lines, Planes and Hyperplanes (es-sential topic, questions 16-17)


Figure 2.4 shows the line p0 (x x0) = 0and four vectorsa,b,c andd. Whichof these vectors could not be p?

1. a

2. b

3. c

4. d

Answer The vectorp must be orthogonal (at 90 to) the line p0 (x x0) socould be b or d or indeed any vector of the form b where 6= 0 is a realnumber.


Ifp,x and x0 are elements ofR3 the set of points satisfying p0 (x x0) = 0 is

a:

1. line

2. plane


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2.8. BOUNDARIES, OPEN AND CLOSED SETS (USEFUL TOPIC, QUESTIONS 18-20)17

x10

x2

S

U

T


3. hyperplane


Answer The set is a plane (see notes on vectors section 7.7) and also, bydefinition a hyperplane.

2.8 Boundaries, Open and Closed Sets (usefultopic, questions 18-20)

The questions now move to chapter 8 on "Open Sets, Closed Sets and Contin-uous Functions".


In Figure 2.5 the black lines at the edge of the sets are the parts of the boundaryof the set that lies within the sets. The notation SC means the complementof the set S. Which of the sets are open?

1. S

2. SC

3. T

4. TC

5. U


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6. UC

Answer Set T is open because no points in its boundary lie within the set.Set S is closed because all points in its boundary lie within the set. As Sis closed its complement is open. The set U contains some but not all of itsboundary points so is neither open nor closed. As the boundary ofU is thesame as the boundary ofUC the set UC is neither open nor closed.


Look again at the sets in Figure 2.5 Which of the following sets are closed?

1. S

2. SC

3. T

4. TC

5. U

6. UC

Answer S is closed because it contains its boundary, TC is closed because itis the complement of the open set T.


Consider an infinite collect of open sets A1, A2....... Click the correct state-ments.

1. The union of a finite number of these sets must be open.

2. The union of a finite or infinite number of these sets must be open.

3. The intersection of a finite number of these sets must be open.

4. The intersection of a finite or infinite number of these sets must be open.

Answer See notes. An example of a situation where statement 4 does nothold is the intersection of all intervals of the form

1

n, 1n

, or in open ball

notationB (0, 1/n) where n = 1, 2..... Each of these intervals is an open set,but their intersection{0} is not open.


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2.9. CLOSED SETS, BOUNDED SETS AND CONTINUOUS FUNCTIONS (MORE CHALLENGING TOP

2.9 Closed Sets, Bounded Sets and Continuous

Functions (more challenging topic, questions21-24)


Which of the following intervals are closed? Come to class ready to explainyour answer using the definition of a closed set as one whose complement isopen, and the definition of an open set in terms of open balls.

1. (10, 11)

2. (12,)

3. [10, 11)

4. [12,)

5. [10, 11]

Answer

1. (10, 11) is not closed because 10 is not in the interval, but elements ofB(10,) are in the interval for any > 0. Thus the complement is notopen and the set is not closed.

2. (12,) is not closed because 12 is not in the interval, but elements ofB(12,) are in the interval for any > 0. Thus the complement is notopen and the set is not closed.

3. [10, 11) is not closed because 11 is not in the interval, but elements ofB(11,) are in the interval for any > 0. Thus the complement is notopen and the set is not closed.

4. [12,) is closed because its complement (, 12) is open. To see thisnote that x 11 sothe open ball B

x, 11x2

is a subset of(11,)so (11,)is open. Thus

the complement of[10, 11] is the union of two open sets, so is itself open.


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Which of the following intervals are bounded?

1. (10, 11)

2. (12,)

3. [10, 11)

4. [12,)

5. [10, 11]

Answer Sets 1, 2, and 3 are bounded, because for allx in these sets |x|< 13.

Sets 2 and 4 are not bounded because all suffi

ciently largex are in these sets.


Which of the following intervals are compact?

1. (10, 11)

2. (12,)3. [10, 11)

4. [12,)

5. [10, 11]

Answer Subsets ofR are compact if they are closed and bounded. Only set5 is closed and bounded.


On which of the following intervals does the function f(x) = x2 have a mini-mum?

1. (10, 11)

2. (12,)3. [10, 11)

4. [12,)

5. [10, 11]


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2.10. CONTINUITY FOR CONSUMER THEORY (MORE CHALLENGING TOPIC, QUESTION 25)21

Answer

1. (10, 11) no, the function has a lower bound (100), but does not reach it.In detail the argument is for any >0 however small there is an elementx of(10, 11)withx2 y} are open.

2. For ally the sets{x: x Rn, f(x) y}and {x: x Rn, f(x) y}areopen.

3. For ally the sets{x: x Rn, f(x)< y}and {x: x Rn, f(x)> y}areclosed.

4. For all y the sets {x: x Rn, f(x) y} and {x: x Rn, f(x) y} are closed.

Answer See the proposition in the notes, section 9.3 which gives answer 1.As the complement of an open set is closed, and the sets in answer 4 are thecomplements of the sets in answer 1, answer 4 is also correct.


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2.11 Derivatives and Approximations (more chal-

lenging topic, questions 26-27)The questions now move to chapter 9 , "Introduction to Multivariate Calculus".


Consider the function f :R2 R given by

f(x1, x2) = x1x2

Click the correct statements.

1. Both partial derivatives off(x1, x2)at (0, 0)are zero.

2. The partial derivatives off(x1, x2)exist and are continuous at (0, 0) .

3. The function is well approximated by f(0, 0) + f(0,0)

x1x1+

f(0,0)x2

x2

when (x1, x2)is close to (0, 0) .

Answer The partial derivatives of the functionf(x1, x2) = x1x2are f(x1,x2)

x1=

x2 and f(x1,x2)

x2= x1. These are continuous for all x1 and x2 which implies

that the approximation is a good one. See notes section 3.


Consider the function f :R2 R given by

f(x1, x2) = max[0, min(x1, x2)] .

This function is 0 when one or both ofx1 and x2 is strictly negative or 0. Itis f(x1, x2) = x1 when 0 < x1 x2 and f(x1, x2) = x2 when 0 < x2 < x1.Click the correct statements.

1. Both partial derivatives off(x1, x2)at (0, 0)are zero.

2. The partial derivatives off(x1, x2)exist and are continuous at (0, 0) .

3. The function is well approximated by f(0, 0) + f(0,0)x1

x1+ f(0,0)

x2x2.


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2.11. DERIVATIVES AND APPROXIMATIONS (MORE CHALLENGING TOPIC, QUESTIONS 26-27)23

x1

x2

Figure 2.6:


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Answer Figure 2.6 shows the graph of the function in 3 dimensional space.The functions is 0 when one or both ofx1 and x2 are 0, that is on the axes.

Thus is has partial derivatives at 0 which are

f(0, 0)

x1= 0,

f(0, 0)

x2= 0.

The function does not have partial derivatives when x1 = 0and x2 > 0 or whenx1 > 0 and x2 = 0 or when x1 = x2 > 0. When 0 < x1 < x2 the functionf(x1, x2) = x1 the partial derivatives are

f(x1, x2)

x1= 1,

f(x1, x2)

x2= 0.

When 0 < x2 < x1 then f(x1, x2) = x2 so the partial derivatives are

f(x1, x2)x1

= 0, f(x1, x2)x1

= 1.

Thus the partial derivatives are not continuous at 0 which suggests that theapproximation is not a good one. The approximation argument is impossibleto make rigorously from the information in the notes, but when for example0< x1 < x2 so f(x1, x2) =x1, the best approximation to f(x1, x2) is y = x1,rather than y = 0 that comes from the derivative.

2.12 Level Sets and Vectors of Partial Deriva-tives (essential topic, question 28)


Think about the two good model of consumer choice. Letu (x) be a differen-tiable utility function andx0 a particular value ofx. (xis a2 vector.). Leth

be the vector whose ith component ishi =u (x0)

xi. Assume that h 6=0. Click

the correct statements.

1. The vector h whose ith component is hi = u (x0)

xiis tangent to the

indifference curve at x0.

2.The vector h whose ith componenthi=

u (x0)

xi

is at90 to the tangent to the indifference curve at x0.

3. The set of vectors x satisfying h0x= h0x0 is the tangent to the indifference curve at x0.

4. The set of vectors x satisfying h0x= h0x0 is at 90 to the tangent to the

indifference curve at x0.


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2.12. LEVEL SETS AND VECTORS OF PARTIAL DERIVATIVES (ESSENTIAL TOPIC, QUESTION 28

5. Ifx0 is the utility maximizing choice at price p thenhi =u (x0)

xi=pi.

6.Ifx0 is the utility maximizing choice at price p thenhi =

u (x0)

xi=pi for all i for some >0.

Answer See chapter 9.3.

quiz for class 1 answers

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