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
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.
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Solutions to Exercises 15
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.
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Solutions to Exercises 17
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.
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Solutions to Exercises 18
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
.
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Solutions to Exercises 19
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).
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Solutions to Exercises 22
Exercise 4(b) x1 = 4.21.6 = 2.625 exactly; x2 = 14.7
1.6 = 9.188, to foursignificant figures.
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Solutions to Exercises 28
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).�
Solutions to Exercises 29
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.)
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Solutions to Exercises 30
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).
(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.)
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Solutions to Exercises 31
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.
(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.)
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Solutions to Exercises 32
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.)
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Solutions to Exercises 33
Exercise 8(b)
exp (−3 log (1− x)) = exp(log (1− x)−3
)= (1− x)−3
=1
(1− x)3.
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Solutions to Exercises 34
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.
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Solutions to Exercises 35
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.
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Solutions to Exercises 36
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.
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Solutions to Exercises 37
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
.
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Solutions to Exercises 38
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.
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Solutions to Exercises 39
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.
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Solutions to Exercises 40
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.
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Solutions to Exercises 41
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.
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Solutions to Exercises 43
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.�
Solutions to Exercises 44
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.
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Solutions to Exercises 45
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).
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