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Exercises of Mathematics for Business

Claudio Mattalia

October 2010


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ii


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Chapter 1

Inequalities

1.1 Definitions

An inequality is an expressione of the kind:

A(x) B(x) x R

Solutions of the inequality are those values of the unknown that satisfy thisexpression.

Two inequalities are said equivalent if they admit the same solutions; holdin this sense the two basic principles:

1. Adding or substracting to both members of the inequality the same (con-stant or variable) quantity, an inequality equivalent to the given one isobtained:

A(x)> B(x) A(x) +C(x)> B(x) +C(x)

2. Multiplying or dividing both members of an inequality for the same (con-stant) positive quantity, an inequality equivalent to the given one is ob-

tained, multiplying or dividing them for the same (constant) negativequantity an inequality equivalent to the given one is obtained by reversingthe direction of the inequality:

A(x)> B(x)

c A(x)> c B(x) if c >0

c A(x)< c B(x) if c


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4 CHAPTER 1. INEQUALITIES

Example 1 Solve the inequality:

x 5 0

Adding the (constant) quantity+5 to both members we get (appling the1st

principle of equivalence):

x 5 + 5 0 + 5

i.e.:

x

5

this is the solution.


2x+ 3 > x

Adding the (variable) quantityxto both members we get (appling the1stprinciple of equivalence):

2x+ 3 x > x x

i.e:

x+ 3 > 0

and then adding the (constant) quantity

3 to both members we get (stillapplying the 1st principle of equivalence):

x+ 3 3> 0 3

i.e.:

x >3

that is the solution.


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1.1. DEFINITIONS 5


4x >5

Multiplying both members for the (constant and positive) quantity1

4we get

(applying the2nd principle of equivalence):

1

4 4x > 1

4 5

i.e.:

x >5

4that is the solution.


4x >5

Multiplying both members for the same (constant and negative) quantity 14

we get (applying the2

nd

principle of equivalence and reversing the inequality):1

4

(4x)0

(ifa 0).The solution is then easily given by:

x ba


4x 12> 0

In this case the inequality is already written in the canonic form, applyingthe equivalence principles seen above we get:

4x >12

and then:

x >3

that is the solution.


4x 12> 0

In this case the inequality is not written in the canonic form (as a < 0),multiplying both members for1(and reversing the inequality) we get first ofall:

4x+ 12< 0

that is the inequality written in the canonic form (as a > 0). We then easilyget (applying the equivalence principles):

4x


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1.3. RATIONAL INTEGER INEQUALITIES OF 2ND DEGREE 7

1.3 Rational integer inequalities of 2nd degree

The rational integer inequalities of 2nd degree can always be reduced to thecanonic form:

ax2 +bx+c 0 with a >0

(ifa 0).To solve an inequality of this type, first of all it is necessary to consider theassociated2nd degree equation:

ax2 +bx+c= 0

and to compute its roots x1 andx2 (where it is assumed x1 < x2 in the case ofdistinct roots).

We now have the following rule:

For values ofx external to the interval having as extremes the roots of theequation (that is for x < x1 and for x > x2) the trinomial ax2 +bx+chas the same sign of the coefficient a.

For values ofx internal to the interval having as extremes the roots of the

equation (that is for x1 < x < x2) the trinomial ax2 +bx + c has theopposite sign of the coefficient a.

For values ofx equal to the roots of the equation (that is for x= x1 andfor x = x2) the trinomial ax

2 +bx+c is null.

More precisely, keeping in mind that an inequality of2nd degree can have2real distinct roots, 2 real coincident roots or no real roots, the following threecases can be distinguished:

1. =b2 4ac >0. In this case the associated equationax2 +bx+c = 0has2 dinstict real roots x1, x2 (where it is assumed thatx1 < x2) and forthe trinomial ax2 +bx+c we have:

ax2 +bx+c >0 for x < x1 x > x2

ax2 +bx+c


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2. =b2 4ac= 0. In this case the associated equation ax2 +bx+c = 0has 2 real coincident roots x1 =x2 and for the trinomil ax

2

+bx+c wehave:ax2 +bx+c >0 for x =x1, x2

ax2 +bx+c= 0 for x= x1, x2

3. =b2 4ac 0 xR

The rule stated above then turns out to be valid in general, keeping in mindthat if = 0 all the values of x different from the roots x1 = x2 are to be

considered external to the interval having as extremes the same roots (and thenthere are no values ofxinternal to such interval), while if< 0 all the valuesofx are to be considered external to the interval having as extremes the roots(interval that in reality turns out to be empty, since these roots do not exist, sothat also in this case there are no values ofx internal to it).


x2 2x 8> 0

In this case the inequality is already written in the canonic form, and the

associated equation:x2 2x 8 = 0

has two distinct real roots x1 =2and x2 = 4. Applying the rule seen beforewe have that the solution of the inequality is given by:

x 4


x2 + 2x+ 8> 0

In this case the inequality is not written in the canonic form, hence multi-pliying both members for1(and reversing the inequality) we get:

x2 2x 8< 0

whose associated equation (the same of the previous excercise) has roots x1 =2and x2= 4. Applying the rule seen before we have that the solution of theinequality is given by:

2< x < 4


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1.3. RATIONAL INTEGER INEQUALITIES OF 2ND DEGREE 9


6x2 + 36x 0

whose associated equation:

6x2

36x

= 0

has roots x1= 0andx2= 6. The solution of the inequality is then given by:

x 6


x2 9

First of all the inequality can be written in the canonic form:

x2 9 0

then it is possible to observe that the associated equation:

x2 9 = 0

has rootsx1=3and x2 = 3,so that the solution of the inequality is given by:

x 3 x 3


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x2 2x+ 1 < 0

In this case the associated equation:

x2 2x+ 1 = 0

has two real coincident roots x1 = x2 = 1, and applying the rule seen beforewe have that the inequality is never satisfied (the same result can be obtainedobserving that the first member of the inequality is simply (x 1)2 that, beinga square, can never be 0

In this case the associated equation:

6x2 + 5 = 0

has no real roots, applying the rule seen before we then have that the inequalityis satisfiedx R (the same result can be obtained observing that the firstmember of the inequality is the sum of a non-negative term, 6x2, and of apositive term, 5, therefore it is strictly positive for every value of x, and theinequality is always satisfied).

1.4 Rational fractional inequalities

The rational fractional inequalities are those in which the unknown appears in

the denominator of a fraction, and they can always be reduced to the canonicform:

N(x)

D(x) 0

In this case first of all it is necessary to exclude the values ofx that makethe denominator of the fraction D(x) equal to 0 (since a fraction with nulldenominator has no meaning), then it is possible to study separately the signofN(x) and that ofD(x) and, combining them through the rule of signs, todetermine the sign of the fraction and to solve the inequality.


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1.4. RATIONAL FRACTIONAL INEQUALITIES 11


x+ 1

4x 8 >0

First of all it must be 4x 8= 0, from which x= 2 (condition of realityof the fraction). Studing separately the sign of the numerator and that of thedenominator of the fraction we then have:

N(x)> 0 x+ 1 > 0 x >1

D(x)> 0 4x 8> 0 x > 2

The sign of N(x) and of D(x), together with that of the fraction, can be

represented graphically in the following way (where the continous line indicatesthe intervals in which the sign is positive and the dashed line the intervals inwhich the sign is negative, while the cross indicates the value excluded from theexistence range):

From the analysis of this graphic we have that the solution of the inequalityis given by:

x 2


x+ 4x2 10

First of all it must be x2 1= 0, from which x=1 (condition of realityof the fraction). Studying the sign of the numerator and of the denominator ofthe fraction we then have:

N(x) 0 x+ 40 x 4

D(x)> 0 x2 1> 0 x < 1 x >1


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and graphically:

and the solution of the inequality is given by:

1< x 0

Studying separately the sign of the two factors we get:

1st factor> 0 x+ 1 > 0 x > 1

2nd factor> 0 x2 9> 0 x 3

and graphically:


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1.5. SYISTEMS OF INEQUALITIES 13

so that the solution of the inequality is given by:

3< x 3


(x 2)(x2 x+ 1) 0

Studing separately the sign of each of the two factors we get:

1st factor 0 x 2 0 x 2

2nd factor0 x2 x+ 10 xR

and combining them graphically:

so that the solution of the inequality is given by:

x 2

1.5 Syistems of inequalities

A system of inequalities is a set of two or more inequalities that must be satisfiedsimultaneously. The solution of the system is given by the intersection of thesolutions of the single inequalities, and to solve a system of inequalities it isnecessary to solve each of the inequalities that compose it and then to consideronly the solutions that satisfy at the same time all the inequalities of the system.To this end it is possible to use a graphic representation, in which the solution ofeach inequality is represented with a continous line; the system is then satisfiedin the intervals in corrispondence of which all the lines (as many as the numberof inequalities that compose the system) are continous.


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Example 17 Solve the system of inequalities:

x+ 1

4x 8 >0

x2 + 2x+ 8> 0

Each of the two inequalities that form the system has already been solvedpreviously; in particular, the fractional inequality has solution:

x 2while the 2nd degree inequality has solution:

2< x < 4

At this point it is possible to represent graphically these sets of solutions,getting:

from which it is possible to deduce that the initial system has solution:

2< x < 1 2< x 0x+ 4x2 1 0

Each of the two inequalities that form the system has already been solvedpreviously; in particular, the first inequality has solution:

x


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1.6. INEQUALITIES WITH ABSOLUTE VALUE 15

while the second has solution:

1< x


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|x+ 3|


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Using this method, to solve the previous inequality we proceed in the

following way:

a) Ifx + 30, i.e. x 3, then the inequality is:

x+ 3 < 1

from which:

x


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|x+ 3 ||x 4 |


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and then:

x


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The initial inequality can then be broken in the two systems:

x+ 3

x 40

x+ 3

x 4


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1.7. IRRATIONAL INEQUALITIES 21

1.7 Irrational inequalities

The irrational inequalities are those in which the unknown appears under thesign of a root. In order to solve them it is necessary to distinguish the case inwhich the index of the root is odd and the case in which the index is even.

If the root in the inequality has an odd index n, an inequality equivalent tothe initial one is obtained raising both members to the power n; no other con-ditions are required because a root of odd index can have the radical quantityof any sign and can itself assume whatever sign. If the roots are more than one,eventually with different indexes (always odd), both members of the inequal-ity must be raised to the appropriate power in such a way that the roots areeliminated.

Example 23 Sove the inequality:

3

x2 9 2

In this case raising both members to the cube we get:

x2 9 8

and then:

x2 1 0

from which:

1 x 1

that is the solution of the inequality.


3

x+ 2 < 9

x3 + 6x2

In this case raising both members to the power nine we get:

(x+ 2)3 < x3 + 6x2

and then:

x3 + 6x2 + 12x+ 8 < x3 + 6x2

from which:

x


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If the root in the inequality, on the contrary, has an even index n, it is

possible to solve the inequality using the following procedure:

1. The existence field of the root is found out (imposing that the radicalquantity is non-negative), then the signs of the two members are discussed.

2. If the two members are of opposite sign, then the values of the unknownfor which the inequality is satisfied can be determined immediately.

3. If the two members are of the same sign (in particular non-negative, oth-erwise it is possible to multiply them by1changing the direction of theinequality) it is possible to raise them to an opportune power so that theroots are eliminated, and then to solve the inequality.

4. The union of the solutions found at points2) and 3) is considered.


x 4 x

Applying the procedure described before we have:

a) First of all it must be x 0 (condition of reality of the root).

b) If x 4 < 0, that is x < 4, the two members are of opposite sing (thefirst negative, the second positive or null) and the inequality is always satisfied(since a negative quantity is always less than or equal to a positive or nullquantity), provided thatx 0 (that is the condition of reality) and x


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provided that x 0 (that is the condition of reality) and x 4 (thatcharacterizes the interval considered). This part of the inequality has thereforeas a solution:

4 x9 +

17

2

d) Combining the results found at points b) and c) (i.e. considering theirunion), finally, it turns out that the solution of the initial inequality is:

0 x9 +

17

2


2x+ 6 6x 1

a) First of all it must be 6x 10, that is x 16

(condition of reality of

the root).

b) If2x+ 6 < 0, that is x


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x2 4x 12< x+ 2

First of all we must have, for the condition of reality of the roots:

x2 4x 120

x+ 2 0

x 2 x 6

x 2 x=2 x 6

At this point it is possible to observe that surely both members of the in-equality have the same sign (non-negative), since they are two roots with evenindex, therefore they can be squared getting:

x2

4x 12< x+ 2that is:

x2 5x 14< 0with solution:

2< x < 7Combining this result with the condition of reality it turns out that the

solution of the initial inequality is:

6 x


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and then:

x4 +15

8 x3 +x2 0and finally:

x2

x2 +15

8x+ 1

0

that is verified only for x = 0 (compatible with the condition of reality x0and the conditionx 1 x 0that characterizes the interval considered).

d) Combining the results found at points b) and c), finally, it turns out thatthe solution of the initial inequality is:

x= 0

In the case of inequalities with roots with even index it is also possibleto transform such inequalities in systems equivalent to them. In particular,considering an inequality of the type:

A(x)

B(x)

first of all it must be B(x) 0 (condition of reality of the root) and alsoA(x)0 (since the root in the right-hand side is surely non-negative, and forthe inequality to be satisfied also the left-hand side must be non-negative). Atthis point it is possible to square both members (that are surely non-negative)in order to eliminate the root, and therefore the initial inequality is equivalentto the system:

B(x) 0

A(x) 0

[A(x)]2 B(x)Considering then an inequality of the type:

A(x)

B(x)

first of all it must beB(x) 0(condition of reality of the root), then ifA(x) 0it is possible to square both members (that are surely non-negative) in order toeliminate the root, and the initial inequality is satisfied by the values ofx thatare solutions of the system:

B(x) 0

A(x) 0

[A(x)]2 B(x)


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where the first condition is superfluous since it is implicit in the third one

(in fact ifB(x) [A(x)]2

then surely B (x) 0). In this case, furthermore, theinitial inequality is satisfied also when A(x)< 0 (because a negative quantity isalways less than or equal to a non-negative quantity like

B(x)), therefore it

is satisfied also by the values ofx that are solutions of the system:

B(x) 0

A(x)< 0

and, in conclusion, the initial inequality is equivalent to the union of the twosystems:

A(x) 0[A(x)]2 B(x)

B(x) 0A(x)< 0

In these cases, therefore, given the initial inequality first of all it is possibleto write the system (or the systems) equivalent to it, then the solution of thissystem corrisponds to the solution of the initial inequality.


2x+ 6 6x 1

This inequality (already solved in Example 1.26) is written in the formA(x) B(x) and therefore it is equivalent to the system:

6x 1 0

2x+ 6 0

(2x+ 6)2 6x 1

from which we get:

6x 1 0

2x+ 6 0

4x2 + 18x+ 37 0


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that is:

x 16x 3

xand finally:

x 16

that is the solution of the initial inequality, and coincide with the one foundpreviously.


x 4 x

This inequality (already solved in Example 1.25) is written in the formA(x)B(x)and therefore it is equivalent to the union of the two systems:

x 40

(x 4)2

x

x 0

x 4< 0from which we get:

x 40

x2 9x+ 16 0

x 0

x 4< 0

that is:

x 4

9 172

x

9 +

17

2

x 0

x < 4

and then:

4 x9 +

17

2 0 x


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1.8 Logarithmic inequalities

Given two numbers a,b >0 (witha = 1) the logarithm in basis a ofb is definedas the number c at which a must be raised in order to obtain b, that is:

logab= c ac =b

By the definition of logarithm we have therefore:

logaab =b alogab =b

so that a number b can be expressed through the logarithm in a basis a > 0(and different from1) using one of these two relations (the second can be usedonly when b >0). The logarithms then satisfy the following properties:

(i) loga1 = 0 a >0, a= 1

(ii) if0 < a logax with x, x >0

(iii) ifa >1 then x < x logax 0

(iv) logax+ logay = loga(xy) with x, y >0

(v) logax logay = logax

y with x, y >0

(vi) loga

xp =p loga

x with x > 0

(vii)logab= logcb

logca with a >0, a= 1, b >0, c >0, c= 1

The logarithmic inequalities are those in which the unknown appears in theargument of a logarithm. To solve them first of all it is necessary to impose thatthe argument of the logarithm is strictly positive (condition of reality), then theproperties listed above are exploited in order to obtain the solution.


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1.8. LOGARITHMIC INEQUALITIES 29


log 12

x < 3

First of all it must be x > 0 (condition of reality of the logarithm), then(simply applying the definition of logarithm) it is possible to write:

log 12

x 8

that is the solution of the inequality (being compatible with the condition ofreality x >0).

The same result can be obtained applying the properties of the exponentials(presented in the next paragraph); in this case first of all it is possible to write

(applying to both members of the initial inequality the exponential of basis 1

2,

that requires to reverse the inequality since the basis of the exponential is lowerthan 1):

12log 12 x

>

123

and then (simply applying in the first member the definition of logarithm):

x > 8



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log2x >4

First of all it must be x > 0 (condition of reality of the logarithm), then(applying the definition of logarithm) it is possible to write:

log2x >log2(2)4

that is:

log2x >log216

and finally (applying the property(iii)seen above as the basis of the logarithm

in this case is larger than 1, so that passing from the inequality between thelogarithms to the inequality between the corresponding arguments the directionof the inequality is preserved ):

x >16


The same result can be obtained applying the properties of the exponentials,in this case first of all it is possible to write (applying to both members of theinequality the exponential of basis 2, that requires to preserve the direction ofthe inequality since the basis of the exponential is larger than 1):

2log2x >24

and then (applying to the left-hand side the definition of logarithm):

x >16



log 13

x2

x+ 3 0

x > 3 x = 0


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1.8. LOGARITHMIC INEQUALITIES 31

then the initial inequality can be written in the form:

log 13

x2

x+ 3 1

(the same expression can be obtained if, in the initial inequality, we apply to the

two members the exponential of basis1

3

). This is a rational fractional inequality,

solving it as shown above (after taking it back to the canonic form x2

x+ 31> 0

and doing the computations) we get:

3< x < 1

13

2 x > 1 +

13

2

that is compatible with the condition of reality determined initially (x > 3 x= 0), so that the solution found is also the solution of the initial inequality.


log

x+ 42

First of all we must have, for the conditions of reality of the root and of thelogarithm:

x+ 4 0

x+ 4 > 0

x+ 40

x+ 4 > 0 x >4

At this point (keeping in mind that when the basis is not specified then thelogarithm is considered with basis e = 2.7182..., therefore larger than 1) theinitial inequality can be written as:

log

x+ 4 log e2

from which:

x+ 4 e2

(the same expression can be obtained if, in the initial inequality, we apply toboth members the exponential of basis e).


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This is an irrational inequality that can be solved as seen previously (in

particular, since both members are non-negative they can be squared) getting:

xe4 4

Combining this solution with the condition of reality found at the beginning(x > 4) it is possible to obtain the solution of the initial inequality, that is:

4< x e4 4

1.9 Exponential inequalities

Given a numbera >0, a power ofa with real exponent x is denoted by ax.Thisis called exponential, and the exponentials satisfy the following properties:

(i) ax >0 x with a >0

(ii) if0 < a ax

(iii) ifa >1 then x < x ax < ax

The exponential inequalities are those in which the unknown appears in theexponential of a certain expression, and to solve them it is possible to use the

proprierties listed above.


1

2

x5



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1.9. EXPONENTIAL INEQUALITIES 33

The same results can be obtained using the logarithms (indeed the exponen-

tial inequalities and the logarithmic inequalities are strictly linked, since expo-nentials and logarithms are inverse functions one with respect to the other). Inparticular, starting from the initial inequality and applying to both members

the logarithm in basis1

2(and reversing the direction of the inequality, being the

basis lower than 1) we get, first of all:

log 12

1

2

x>log 1

2

32

and then (simply applying the definition of logarithm):

x >5that is the solution of the inequality.


4x >16

First of all the inequality can be written in the form:

4x >42

and then (applying the property(iii)seen before as the basis of the exponen-tial in this case is greater than 1, so that passing from the inequality betweenthe exponentials to the inequality between the exponents the direction of theinequality is maintained):

x > 2


The same result can be obtained applying to both members of the initial

inequality the logarithm in basis4 (maintaining the direction of the inequality,since the basis in this case is larger than 1), so that we have:

log44x >log416

and then (applying the definition of logarithm):

x > 2



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32x+1 + 3x 1< 0First of all this inequality can be written as:

3 32x + 3x 1< 0and then, putting3x =z, we have:

3z2 +z 1< 0that is an inequality of2nd degree whose solution is:

1

13

6 < z (x+ 1) log 4and with some simple computations we get:

x(log 3 log 4)> log3 + log 4from which (keeping in mind that the quantity(log3 log4)is negative, so thatwhen we divide both members for this quantity the direction of the inequalitymust be reversed):

x 0

4) 3x2 15x0

5) x2 6x+ 9 > 0

6) 2x2 15x+ 30 < 0

7) 2x+ 1

x 3 0

8) 4x 5

5x 0

3x2 15x 0

14)

x2

6x+ 9 > 0

2x2 15x+ 30< 0

15) |x2 2x+ 5| 3

16)x2 4 > x 2

17) |x 2|


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19)

x2 4< x

20) x+ 5< x 1

21)

x2 4< x 4

22)

x+ 1> 3

x 1

23) log3(x+ 2)< 2

24) log 13

(x+ 2)< 2

25) log 14

(x2 + 7)< 2

26) log 13

x 1 1

27)

2

3

x24x1

28) 4x + 2x 2< 0

29) 4 3x 2 4x

30) 2x+1 >3x1