m343 diagnostic quiz - statistics | statisticsstatistics.open.ac.uk/m343/m343_dq.pdf · m343...

THE OPEN UNIVERSITYFaculty of Mathematics and Computing

M343 Diagnostic Quiz

Prepared by the Course Team[Press ↓ to begin]

c© 2006 The Open UniversityLast Revision Date: March 22, 2006 Version 1.0

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Section 1: Introduction 2

1. Introduction

In embarking on M343, basic mathematical competence is as importantas basic knowledge of probability, such as you could have obtainedfrom M248. This quiz covers a number of key mathematical skillswith which you should be familiar.

Try each question for yourself, using a pencil and paper and yourcalculator where appropriate. Then click on the green section letter(e.g. ‘(a)’) to see the solution. Click on the � symbol at the end ofthe solution to return to the question. Use the ↑ and ↓ keys to movefrom Section to Section.

There is some advice on evaluating your performance at the endof the quiz.

Section 2: Rounding and accuracy 3

2. Rounding and accuracy

Exercise 1.Round 1.24743

(a) to two decimal places, and(b) to four significant figures.

Exercise 2. In part (c) below, 8 is an exact whole number. Theother numbers in the calculations have already been rounded. Useyour calculator to work out the value of each expression and considerhow many decimal places it is reasonable to give in your answer.

(a) 0.163× 0.246(b) 0.163× 7.296(c) 2.05− 0.1237× 8

Section 3: Fractions 4

3. Fractions

Exercise 3. Express each of the following over a suitable commondenominator.

(a)56− 3

8

(b)−2

2x + 5+

1x− 3

(c) 1− 4x2

Section 4: Formulas 5

4. Formulas

Exercise 4.

(a) Evaluate the binomial coefficient(

73

).

(b) Given the values x0 = 0 and y = 0.6, calculate the values of x1

and x2 to four significant figures using the formula

xj+1 =4xj + 7y

4 (1− y),

for j = 0, 1.

Section 5: Powers and logarithms 6

5. Powers and logarithms

Exercise 5. Write each of the following in the form xn, where n issome number.

(a)1x2

(b) x2 × x3

(c)(x2

)3

(d)x2

x3

Exercise 6.

(a) Express the cube root of 0.3 as a power.(b) Use your calculator to evaluate it to four decimal places.

Section 5: Powers and logarithms 7

Exercise 7. In M343, logarithms are always natural logs (to base e).This may be ln on your calculator.

Use the fact that log 3 = 1.0986 and log 4 = 1.3863 to four decimalplaces, to evaluate the following.

(a) log 12(b) log (3/4)(c) log 2

Exercise 8. Simplify

(a) exp (2 log x)(b) exp (−3 log (1− x))

Section 6: Matrices 8

6. Matrices

Exercise 9. A is the 3 × 3 matrix

0.5 0.1 0.40.6 0.0 0.40.7 0.3 0.0

, v is the

column vector

413

and w is the row vector(

4 1 3).

Work out each of the following.

(a) Av(b) wA(c) A2

Section 7: Differentiation 9

7. Differentiation

Exercise 10. Differentiate each of the following with respect to x.

(a) x4 and1x2

(b) e−3x and log(1 + x2

)(c) xe−2x

(d)0.2x

1− 0.8x

Section 8: Integration 10

8. Integration

Exercise 11. Work out each of the following integrals.

(a)∫

x2dx and∫

1x3

dx

(b)∫ 3

1

(1 +

3x2

)dx

(c)∫ 2

0

2x

x2 + 4dx

(d)∫ ∞

0

xe−xdx

Section 9: Differential equations 11

9. Differential equations

Exercise 12. Use the technique of separation of variables to find thegeneral solutions of each of the following differential equations.

(a)dy

dx= 2xy

(b)dy

dx= y + 5

Section 10: Post-mortem 12

10. Post-mortem

You should be familiar with the techniques covered by Exercises 1 to12 before embarking on M343.

If you found these exercises particularly difficult, and have notstudied MST121 and MS221, you would be well advised to considertaking these courses before starting M343.

If you had difficulty only with the later parts of Exercises 10 and11 and with Exercise 12, then this should not delay your registeringfor M343. However, you would be well advised to reinforce yourunderstanding with the help of the relevant units in MST121 andMS221, or any basic textbook on calculus. All the differentiationtechniques required for M343 are covered in Block C of MST121.Basic integration and the technique of separation of variables are alsocovered in Block C of MST121. Integration by parts and integrationby substitution are covered in Block C of MS221.

If you have any queries about your suitability for the course, youshould contact your Regional Office.

Solutions to Exercises 13

Solutions to Exercises

Exercise 1(a) 1.24743 rounded to two decimal places is 1.25.

Since the next digit after the ‘4’ is 7, which is 5 or more, round up;the original number is closer to 1.25 than to 1.24.

�


Exercise 1(b) 1.24743 rounded to four significant figures is 1.247.�


Exercise 2(a) 0.0401. A calculator gives 0.040098 but, since theoriginal numbers are rounded to three significant figures, you cannotexpect more than three significant figures in your answer to be reliable.

�


Exercise 2(b) 1.19. (A calculator gives 1.189248.)�


Exercise 2(c) 1.06. (A calculator gives 1.0604.) You cannot expectanything after the second decimal place to be reliable, because 2.05 isrounded to two decimal places.

�


Exercise 3(a) Here, the lowest common denominator is 24; (6 and8 are both factors of 24).

56− 3

8=

5× 424

− 3× 324

=20− 9

24

=1124

.

�


Exercise 3(b) The common denominator is (2x + 5)(x− 3).

−22x + 5

+1

x− 3=

−2 (x− 3)(2x + 5) (x− 3)

+2x + 5

(2x + 5) (x− 3)

=−2x + 6 + 2x + 5(2x + 5) (x− 3)

=11

(2x + 5) (x− 3).

�


Exercise 3(c) 1− 4x2

=x2 − 4

x2.

�


Exercise 4(a)(

73

)= 7×6×5

1×2×3 = 35.

�


Exercise 4(b) x1 = 4.21.6 = 2.625 exactly; x2 = 14.7

1.6 = 9.188, to foursignificant figures.

�


Exercise 5(a) Using the rule 1xa = x−a gives

1x2

= x−2.

�


Exercise 5(b) Using the rule xa × xb = xa+b gives

x2 × x3 = x5.

�


Exercise 5(c) Using the rule (xa)b = xab gives(x2

)3= x6.

�


Exercise 5(d) Using the rule xa

xb = xa−b gives

x2

x3= x−1.

�


Exercise 6(a) The cube root of 0.3 is 0.313 .

�


Exercise 6(b) 0.313 = 0.6694, to four decimal places.

You can either raise 0.3 to the power (1/3) on your calculator using thexy (or yx) button, or you can use the x

√y button, (if your calculator

has one).�


Exercise 7(a) Using the rule log (a× b) = log a + log b,

log 12 = log (3× 4)= log 3 + log 4= 1.0986 + 1.3863= 2.485.

(Note that we have expressed the solution to three decimal places.Had we expressed it to four decimal places, there may have been anerror of ±1 in the fourth decimal place due to rounding errors.)

�


Exercise 7(b) Using the rule log (a/b) = log a− log b,

log 3/4 = log 3− log 4= 1.0986− 1.3863= −0.288.

Recall that the logarithm of a number between 0 and 1 is negative(and log 1 = 0).


�


Exercise 7(c) Using the rule log ab = b log a,

log 2 = log√

4= log 41/2

=12

log 4

=12× 1.3863

= 0.693.


�


Exercise 8(a) Since 2 log x = log x2,

exp (2 log x) = exp(log x2

)= x2.

This uses the fact that the logarithm function and the exponentialfunction are inverse functions. So exp (log x) = x, for any x > 0.(Similarly, log (expx) = x for all x.)

�


Exercise 8(b)

exp (−3 log (1− x)) = exp(log (1− x)−3

)= (1− x)−3

=1

(1− x)3.

�


Exercise 9(a)

Av =

3.33.63.1

The first entry is (0.5× 4) + (0.1× 1) + (0.4× 3), using the first rowof A. The other two entries are obtained using the second and thirdrows of A.

�


Exercise 9(b)wA =

(4.7 1.3 2.0

)The second entry, for example, is (4× 0.1) + (1× 0.0) + (3× 0.3),using w and the second column of A.

Observe that the answers to parts (a) and (b) are not the same. Theydo not even have the same shape.

�


Exercise 9(c)

A2 =

0.59 0.17 0.240.58 0.18 0.240.53 0.07 0.40

Matrices and vectors are used only in Unit 6 on Markov chains.

�


Exercise 10(a) The derivative of xn is nxn−1, sod

dx

(x4

)= 4x3.

Since1x2

= x−2,d

dx

(1x2

)=

d

dx

(x−2

)= −2x−3, or

−2x3

.

�


Exercise 10(b) The composite function rule, or chain rule, fordifferentiating the composition of one function with a second one isneeded here. (For example, exp (−3x) = exp (y), where y = −3x.)

The derivatives are −3e−3x and2x

1 + x2.

�


Exercise 10(c) This is the product of two functions. The derivativeis

d

dx(x) e−2x + x

d

dx

(e−2x

)= 1e−2x = x (−2) e−2x

= (1− 2x) e−2x.

�


Exercise 10(d) This is the quotient of two functions. The derivativeis

(1− 0.8x) ddx (0.2x)− 0.2x d

dx (1− 0.8x)

(1− 0.8x)2

=(1− 0.8x) 0.2− 0.2x (−0.8)

(1− 0.8x)2

=0.2− 0.16x + 0.16x

(1− 0.8x)2

=0.2

(1− 0.8x)2.

�


Exercise 11(a) For n 6= −1,∫

xndx =xn+1

n + 1+ c, where c is an

arbitrary constant of integration. So∫x2dx =

x3

3+ c.

Since1x3

= x−3,

∫1x3

dx =∫

x−3dx

=x−2

−2+ c

= − 12x2

+ c.

�


Exercise 11(b)∫ 3

1

(1 +

3x2

)dx =

[x− 3

x

]3

1

= (3− 1)− (1− 3)= 4.

�


Exercise 11(c) In general,∫

f ′ (x)f (x)

= log f (x) + c.

Observe that the numerator of the integrand, (2x), is the derivativeof the denominator,

(x2 + 4

). So

∫ 2

0

2x

x2 + 4dx =

[log

(x2 + 4

)]20

= log 8− log 4

= log(

84

)= log 2= 0.6931,

to four decimal places.�


Exercise 11(d) Here, integration by parts is required. First, integratee−x to get −e−x. Then

∫ ∞

0

xe−xdx =[x

(−e−x

)]∞0−

∫ ∞

0

d

dx(x)

(−e−x

)dx

= 0 +∫ ∞

0

e−xdx

=[−e−x

]∞0

= − (0− 1)= 1.

�


Exercise 12(a) Separating the variables, the equation is written inthe form

dy

y= 2xdx.

Observe that x does not appear on the left-hand-side and y is not onthe right-hand-side. Now integrate both sides to obtain

log y = x2 + C

or, taking exponentials,

y = K expx2,

where C is an arbitrary constant and K = eC .

Other techniques for solving differential equations are taught in thecourse where they are needed (mainly in Units 7 and 8).

�


Exercise 12(b) Separating the variables leads to

dy

y + 5= dx.

Integrating both sides gives

log (y + 5) = x + C,

or, taking exponentials,

y = Kex − 5.

As before, C is an arbitrary constant and K = eC .�

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