Table of Contents:

• Linear Equations
• Quadratic Equations
• Progressions
• Binomial Theorem
• Inequalities
• Permutation and...
• Probability
• Functions
• Set Theory

Linear Equations:

A linear equation is an equation whose graph is a straight line. A linear equation in one variable is an equation that simlpy involves x. A linear equation is any equation that can be written in the form ax + b = 0. There are no terms involving x², x³, x^1/2 etc. Each term has a degree of at most 1. All operations, such as addition or multiplication, involve only x and numerical constants. 3x + 4 = 5 is an example of a linear equation. 2(x+1) = 6(x-4) is also a linear equation. These equations can be solved very easily by performing algebraic operations to the equation to isolate x.

A linear equation in two variables is, as the name suggests, an equation that involves 2 variables. The standard form of this type of equation is Ax + By = C, where A,B and C are real numbers. For example, 3x + y = 7 is a linear equation in two variables. y = 2x + 1/3 is also an example, since it can be rewritten as 2x - y = -1/3 ( or equivalently 6x - 3y = -1 ).

Linear equation in one variable properties.

1. If a = b then a+c = b+c.

2. If a = b then a -c = b-c.

3. If a = b then ac = bc.

4. If a=b then a/b = b/c.

Linear equations in two variables can also be expressed in the slope-intercept form y = mx + b.

The slope of a line, represented by the variable m, is defined as the ratio of change in values of y to change in value of x. The slope is also known as rise over run. For any two points (x₁ ,y₁), (x₂ ,y₂) on a line L, the formula for calculating the slope of L is:

m = (y₂ - y₁)/(x₂ - x₁)

Two lines are parallel if they have equal slopes. Parallel lines never cross each other. The distance between two parallel lines is always the same for every point along the lines.

Two lines are perpendicular, meaning their angle of intersection is 90°, if their slopes are negative reciprocals of each other. For lines L₁ and L₂ with slopes m₁ and m₂, respectively,

m₁m₂ = -1

Example 1: A calculator has been marked up 15% and is being sold for $78.50.  How much did the store pay the manufacturer of the calculator?

Solution: First, let’s define p  to be the cost that the store paid for the calculator.  The stores markup on the calculator is 15%.  This means that 0.15p has been added on to the original price (p) to get the amount the calculator is being sold for.  In other words, we have the following equation p + 0.15p = 78.50 that we need to solve for p.  Doing this gives, 1.15p = 78.50 therefore p = 78.5/1.15.
The store paid $68.26 for the calculator.

Example 2: A shirt is on sale for $15.00 and has been marked down 35%.  How much was the shirt being sold for before the sale?

Solution: Let’s start with defining p to be the price of the shirt before the sale.  It has been marked down by 35%.  This means that 0.35p has been subtracted off from the original price.  Therefore, the equation (and solution) is,

p - 0.35p = 15

0.65p = 15

p = 15/0.65 = 23.07

Algebra: Linear Equations 1

Algebra: Linear Equations 2

Quadratic Equations:

A quadratic equation in one unknown is an equation of the form ax² + bx + c =0,  where  a is not equal to 0. When we solve a linear equation, we may transpose the terms and leave the unknown on one side of the equation. However, this is often not the case for a quadratic equation. There are other methods to solve a quadratic equation, e.g. by factorization, by completing the square and by the quadratic formula. Furthermore, for a linear equation in the form of  mx + n =0, where  m is not 0, there is always a solution  x = -n/m , which is a real number. On the contrary, a quadratic equation may have two real roots, one double root or no real roots.

Solving quadratic equations:

There are several methods to solve a quadratic equation. Some quadratic expressions can be factorized and then the equation is easy to solve.

1. If pq = 0, then p=0 or q=0.

If quadratic equation ax² + bx + c =0 can be factorized into (px+q)(rx+s) =0,

then we have px + q = 0 or rx + s = 0, which gives x =-q/p or x = -s/r respectively.

2. Method of completing square.

The method of completing the square is to change the equation from the form ax² + bx + c =0 to (x+p)² = q. This can be done by dividing the whole equation by a and then we have

x² + b/a x = -c/a

x² + 2.1/2.b/a x + (b/2a)²= -c/a + (b/2a)²

(x + b/2a)² = -c/a + (b/2a)²

and thus p = b/2a and q = -c/a + (b/2a)² . If q > 0 then  x + p = +(-)q or x = -p +(-)q.

If q = 0, then x = -p. If q < 0, equation has no real roots. The result x = -p +(-)q still holds but the roots will be complex.

3. If ax² + bx + c =0 and a is not equal to 0, the roots of the equation is given by

Example 1:

 

Let A and B be the roots of a quadratic equation ax² + bx + c =0 with a not equal to 0  then

A + B = - b/a and

AB = c/a

For a quadratic equation ax² + bx + c =0 with a not equal to 0, the discriminant D = b² - ac.

1. If D > 0, the equation has two distinct real roots.

2. if D = 0, the equation has one double real root.

3. if D < 0, the equation has no real root, it has two distinct unreal roots.

1. Given the quadratic equation x² - (A - 3) x - (A - 2) = 0, for what value of A will the sum of the squares of the roots be zero?

2. A quadratic with integral coefficients has two distinct positive integers as roots, the sum of its coefficients is prime and it takes the value -55 for some integer. The sum of the roots is

3. How many real r are there such that the roots of x² + rx + 6r = 0 are both integers?

 

Arithmetic Progression:

An arithmetic progression is a sequence of numbers such that the difference of any two successive members of the sequence is a constant.
For example, the sequence 3, 5, 7, 9, 11,... is an arithmetic progression with common difference 2.

Arithmetic progression property:
a₁ + a_n = a₂ + a_n-1 = ... = a_k+a_n-k+1

Formulae for the n-th term can be defined as:
a_n = 1/2 x (a_n-1 + a_n+1)

If the initial term of an arithmetic progression is a1 and the common difference of successive members is d, then the n-th term of the sequence is given by
a_n = a₁ + (n - 1)d, n = 1, 2, ...

The sum S of the first n values of a finite sequence is given by the formula:
S = 1/2(a₁ + a_n)n, where a₁ is the first term and a_n the last.

or
S = 1/2(2a1 + d(n-1))n

Example 1: Find the sum of the first 10 numbers from this arithmetic progression 1, 11, 21, 31...
Solution: we can use this formula S = ¹/₂(2a₁ + d(n-1))n
              S = ¹/₂(2.1 + 10(10-1))10 = 5(2 + 90) = 5.92 = 460

Example 2: The sum of the three numbers in A.P is 21 and the product of their extremes is 45. Find the numbers.

Solution: Let the numbers are be a - d, a, a + d
Then a - d + a + a + d = 21
3a = 21
a = 7
and (a - d)(a + d) = 45
a² - d² = 45
d² = 4
d = +2
Hence, the numbers are 5, 7 and 9 when d = 2 and 9, 7 and 5 when d = -2. In both the cases numbers are the same.

A geometric progression is a sequence of numbers such that the quotient of any two successive members of the sequence is a constant called the common ratio of the sequence.

a geometric sequence can be written as:

Formulae for the n-th term can be defined as:

The common ratio then is:

A sequence with a common ratio of 2 and a scale factor of 1 is 1, 2, 4, 8, 16, 32...

A sequence with a common ratio of -1 and a scale factor of 3 is 5, -5, 5, -5, 5, -5,...

If the common ratio is:

Geometric Progression Properties

Formula for the sum of the first n numbers of geometric progression

Infinite geometric series where |q| < 1

 

If |q| < 1 then a_n -> 0, when n -> ∞ So the sum S of such a infinite geometric progression is:

 

Sequence and Series

   

Arithmetic Progressions

Arithmetic Progressions

 

An algebraic expression consisting of two terms with a positive or negative sign between them is called a binomial expression. e.g.  (a+b), ( P / x²)  – (Q / x⁴) etc.

Binomial Theorem: When a binomial expression is raised to a power ‘n’ we would like to be able to expand it. The binomial theorem assists us in doing this. It converts such an expression into a series.

Binomial Theorem for positive integral index:

(x+y)ⁿ = xⁿ + ⁿC₁x^n-1y+ⁿC₂x^n-2y₂+-----+ⁿC_rx^n-ry_r+ -------+---------+ⁿC_n-1xy_n-1   + ⁿc_nyⁿ.

(x+y)ⁿ = ⁿ_r=0∑ⁿC_rx^n-ry^r

therefore Replacing ‘x’ by ‘1’ and ‘y’ by ‘x’, we have :

(1+x)ⁿ = ⁿC_ox^o+ⁿC₁x+ⁿC₂x²+---------+ⁿC_rx^r+------+ⁿC_n-1x^n-1+ⁿC_nxⁿ.

Properties of Binomial – Expansion (x+y)ⁿ :

(i)        There are (n+1) terms in the expansion.

(ii)        In each term, sum of the indices of ‘x’ and ‘y’ is equal to ‘n’.

(iii)        In any term, the lower suffix of ‘c’ is equal to the index of ‘y’, and the index of x = n-(lower suffix of c).

(iv)             Because  ⁿC_r = ⁿC_n-r,

    so we have :

ⁿC_o = ⁿC_n

ⁿC₁=ⁿC_n-1

ⁿC₂=ⁿC_n-2 etc.

It follows that the coefficients of terms equidistant from the beginning and the ends are equal.

General Terms : (r +1) th term from beginning in

      (x+y)ⁿ is called general – term, and

it is denoted by

                        T^r+1 = ⁿC_rx^n-ry^r

Greatest Coefficient : In any binomial expansion middle-term has the greatest.

Coefficient. So

(i)       If ‘n’ is even, then greatest – coefficient  =  ⁿC_n/2

(ii)       If ‘n’ is odd, then greatest – coefficients are ⁿC_{(n+1)/ 2}  and  ⁿC_(n-1)/2

Properties of Binomial coefficients :

(1) The sum of binomial coefficient in (1 + x)ⁿ  is 2n.

(2)  The sum of the coefficients of the odd-terms in (1+x)ⁿ is equal to the sum of coefficients of the even terms and each is equal to 2n-1.

 

The term inequality is applied to any statement involving one of the symbols <, >, , .

Example of inequalities are:

i.       x 1
ii.      x + y + 2z > 16
iii.     p2 + q2 1/2
iv.     a2 + ab > 1

Properties of Inequalities

1.    If a b and c is any real number, then a + c b + c.
e.g. -3 -1 implies -3+4 -1 + 4.

2.    If a b and c is positive, then ac bc.
e.g. 2 3 implies 2(4) 3(4).

3.    If a b and c is negative, then ac bc.
e.g. 3 9 implies 3(-2) 9(-2).

4.    If a b and b c, then a c.
e.g. -1/2 2 and 2 8/3 imply -1/2 8/3. 

Solution of Inequality

By solution of the one variable inequality 2x + 3 7 we mean any number which substituted for x yields a true statement.
e.g. 1 is a solution of 2x + 3 7 since 2(1) + 3 = 5 and 5 is less than and equal to 7.

By a solution of the two variable inequality x - y 5 we mean any ordered pair of numbers which when substituted for x and y, respectively, yields a true statement.
e.g. (2, 1) is a solution of x - y 5 because 2-1 = 1 and 1 5.

Example 1: If -1 < x < 4 the determine a, b in a < 2x + 3 < b.

Solution:  We have -1 < x < 4 so multiply everything by 2. We will get

               -2 < 2x < 8

              Now add 3 to everything. so 1 < 2x + 3< 11.

Therefore a = 1, and b = 11.

1. If |b| > 1 and x = |a|b, then which one of the following is necessarily true?
a) a − xb <= 0 b) a − xb > 0 c) a − xb < 0 d) a − xb >= 0

2. If a, b, c are real numbers such that a < b < c and a + b + c = 6, ab + bc + ca = 9, then which among the following is definitely true?
(a) 0 < a < 1 (b) 1 < b < 3 (c) 3 < c < 4 (d) All of them (e) none of them

3. Let x be a real number such that 1 − 1/n < x <= 3 + 1/n is true for all natural numbers n. Which among the following best describes x?
(a) 1 < x < 3 (b) 1 <= x <= 3 (c) 0 < x < 4 (d) 1 < x <= 3 (e) 0 < x <= 4

4. If a²b³c = 256/27 find min value of a + b + c, given a, b, c are positive real nos.

5. From four positive real nos a, b, c and d, 4 distinct combination of sum of three numbers are formed S1, S2, S3 and S4. IF abcd = 5. FInd the min value of S1S2S3S4.

 

Permutations:
The ways in which a number of given objects can be arranged by taking all of them or a specified number of objects out of them are called PERMUTATIONS.   Thus the number of permutations of three objects, viz.  a, b, and c, taking all of them at a time is 6 i.e., abc, acb, bcd, bac, cab and cba.

The number of ways in which 2 objects can be taken and arranged out of 3 objects a, b and c is 6, viz.  ab, ba, bc, cb, ac and ca.   The number of permutations of r things our of n things is denoted by ⁿp^r.

Formulae:

1. n! = n(n-1)(n-2)....(1)

2. ⁿp^r = n!/(n-r)!

Example: ¹⁰P₄ = 10 x 9 x 8 x 7 = 5040

3. The number of ways in which n objects can be arranged in a circle is (n-1)!

Combinations:
The ways in which a specified number of objects can be taken out of a given number of objects (without regard to their arrangements) are called Combinations.  The symbol ⁿC_r denotes the number of combinations or r things out of n things.  Thus, for example the number of combinations of 2 objects out of three given objects a, b and c is 3, viz.  ab, ca, bc.

Formulae:

1. ⁿC_r = ⁿP_r/r!

2. ⁿC_r = ⁿC_n-r

3. ⁿC_r + ⁿC_r-1 = ⁽ⁿ⁺¹⁾C_r

4. ⁿC₀ + ⁿC₁ + ⁿC₂ + ⁿC₃ + ......ⁿC_n = 2ⁿ

Circular Permutations

There are two types of circular permutations.

(a) If clockwise and anti clock-wise orders are different, then total number of circular-permutations is given by (n-1)!

(b) If clock-wise and anti-clock-wise orders are taken as not different, then total number of circular-permutations is given by (n-1)!/2!

Number of circular-permutations of n different things taken r at a time:-

(a) If clock-wise and anti-clockwise orders are taken as different, then total number of circular-permutations = ^_nP_r /r

(b) If clock-wise and anti-clockwise orders are taken as not different, then total number of circular permutation = ⁿP_r/2r

Example 1: In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together?

Solution: ABACUS is a 6 letter word with 3 of the letters being vowels.

If the 3 vowels have to appear together, then there will 3 other consonants and a set of 3 vowels together.

These 4 elements can be rearranged in 4! Ways.

The 3 vowels can rearrange amongst themselves in 3!/2! ways as "a" appears twice.
Hence, the total number of rearrangements in which the vowels appear together are (4! x 3!)/2!

Example 2: How many different four letter words can be formed (the words need not be meaningful) using the letters of the word "MEDITERRANEAN" such that the first letter is E and the last letter is R?

Solution: The first letter is E and the last one is R.

Therefore, one has to find two more letters from the remaining 11 letters.

Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.

The second and third positions can either have two different letters or have both the letters to be the same.

Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.

Case 2: When the two letters are same. There are 3 options - the three can be either Ns or Es or As. Therefore, 3 ways.

Total number of posssibilities = 56 + 3 = 59.

Example 3: How many different signals can be made by 5 flags from 8-flags of different colours?

Solution:  Number of ways taking 5 flags out of 8-flags  = ⁸P₅

=   8!/(8-5)!      

=  8 x 7 x 6 x 5 x 4 = 6720

Example 4: A child has 3 pocket and 4 coins. In how many ways can he put the coins in his pocket?

Solution: First coin can be put in 3 ways, similarly second, third and forth coins also can be put in 3 ways.
So total number of ways = 3 x 3 x 3 x 3   = 3⁴   = 81

1. A man has nine friends, four boys and five girls. In how many ways can he invite them, if there have to be exactly three girls in the invitees?

2. A company manufactures pencils in boxes of 6, 9, and 20. The boxes are sealed and the pencils cannot  be  sold  loose.  What is  the  largest  number  of  pencils  that  a  wholesaler  cannot  purchase using some combination of these boxes?

3. The number of ways in which 10 candidates A1, A2, ……, A10 can be ranked so that A1 is always above A2 is ?

4. A hosted a party and invited all her friends and asked them to invite their friends.There are n people in the party.Only S is not known to A.Each pair that does not include A or S has exactly 2 common friends.Also,S knows everyone except A.If only 2 friends can dance at a time,how many dance numbers will be there at the party?

5. In a classroom there are 14 students seated in 3 rows of 5 chairs. The place at the centre of the room is unoccupied. A teacher decides to reassign the seats such that each student will occupy a chair  adjacent to his/her present one (i.e. move one desk forward, backward, right or left). In how many ways can this reassignment be done?

6. Consider a 4 digit number. the first 2 digits are equal and last 2 digits are also equal. How many of such digits are perfect square?

7. Suppose u have a currency, named x, in 3 denominations, 1, 10 and 50. In how many ways can 107 x be given in this currency?

An experiment is an act for which the outcome is uncertain. Examples of experiments are rolling a die, tossing a coin, surveying a group of people on their favorite soft drink, etc...

An experiment is said to be a random experiment, if it's out-come can't be predicted with certainty.
Example; If a coin is tossed, we can't say, whether head or tail will appear. So it is a random experiment.

A sample space S for an experiment is the set of all possible outcomes of the experiment such that each outcome corresponds to exactly one element in S.  The elements of S are called sample points.  If there is a finite number of sample points, that number is denoted n(S), and S is said to be a finite sample space.
For example, if our experiment is rolling a single die, the sample space would be S = {1, 2, 3, 4, 5, 6}. If our experiment is tossing a single coin, our sample space would be S = {Heads, Tails}. 

Every subset of a sample space is an event. It is denoted by 'E'. e.g. In throwing a dice S={1,2,3,4,5,6}, the appearance of an event number will be the event E={2,4,6}.

An event, consisting of a single sample point is called a simple event. e.g. In throwing a dice, S={1,2,3,4,5,6}, so each of {1},{2},{3},{4},{5} and {6} are simple events.

Compound event: A subset of the sample space, which has more than on element is called a mixed event.e.g. In throwing a dice, the event of appearing of odd numbers is a compound event, because E={1,3,5} which has '3' elements.

Equally likely events: Events are said to be equally likely, if we have no reason to believe that one is more likely to occur than the other. e.g. When a dice is thrown, all the six faces {1,2,3,4,5,6} are equally likely to come up.

Exhaustive events: When every possible out come of an experiment is considered. e.g. A dice is thrown, cases 1,2,3,4,5,6 form an exhaustive set of events.

Probability of an Event 

If 'S' be the sample space, then the probability of occurrence of an event 'E' is defined as:

P(E) = n(E)/N(S) = (number of elements in 'E'/ (number of elements in sample space 'S')

Empirical Probability

Finding the probability of an empirical event is specifically based on direct observations or experiences.

For example, a survey may have been taken by a group of people.  If the data collected is used to find the probability of an event tied to the survey, it would be an empirical probability.   Or if a scientist did research on a topic and recorded the outcome and the data from this is used to find the probability of an event tied to the research, it would also be an empirical probability.

Equiprobable space

A sample space S is called an equiprobable space if and only if all the simple events are equally likely to occur. e.g. A toss of a fair coin.  It is equally likely for a head to show up as it is for a tail. 

Mutually Exclusive

Events E and F are said to be mutually exclusive if and only if  they have no elements in common.
E.g. if the sample space is rolling a die, where S = {1, 2, 3, 4, 5, 6}, and E is the event of rolling an even number, E = {2, 4, 6} and F is the event of rolling an odd number, F = {1, 3, 5}, E and F are mutually exclusive, because they have NO elements in common. 

Bayes' Theorem The short form of Bayes' Theorem states that if E and F are events, then

P(F|E)

=

P(E|F)P(F) ---------------------- P(E|F)P(F) + P(E|F')P(F')

Properties of Probability

1. 0 <= P(E) <= 1

2. P(not E) = 1 - P(E) So if, P(E) = 1/4 then P(not E) = 3/4.

3. "Or" probabilities with mutually exclusive events P (A or B) = P(A) + P(B)

4. "Or" probabilities with events that are NOT mutually exclusive P (A or B) = P(A) + P(B) - P(A AND B)

5. A and B are Independent Events if an only if P(A AND B) = P(A)P(B)

Example 1: Find the probability of getting a tail in tossing of a coin.

Solution: Sample space S = {H,T}  and n(s) = 2
              Event 'E' = {T}  and n(E) = 1
              therefore P(E) = n(E)/n(S) = 1/2

Example 2: A glass jar contains 6 red, 5 green, 8 blue and 3 yellow marbles. If a single marble is chosen at random from the jar, what is the probability of choosing a red marble? a green marble? a blue marble? a yellow marble?

Solution: Outcomes:         The possible outcomes of this experiment are red, green, blue and yellow.
Probabilities:      
P(red)      =      number of ways to choose red/total number of marbles = 6/22 = 3/11
 
P(green)      =      number of ways to choose green/total number of marbles = 5/22
 
P(blue)      =      number of ways to choose blue/total number of marbles = 8/22 = 4/11
 
P(yellow)      =      number of ways to choose yellow/total number of marbles = 3/22

Example 3: A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target?

Solution: The man will hit the target even if he hits it once or twice or thrice or all four times in the four shots that he takes.
So, the only case where the man will not hit the target is when he fails to hit the target even in one of the four shots that he takes.
The probability that he will not hit the target in one shot = 1 - 1/4 = 3/4
Therefore, the probability that he will not hit the target in all the four shots =3/4 x  3/4 x 3/4 x 3/4 = 81/256
Hence, the probability that he will hit the target at least in one of the four shots = 1 - 81/256
= 175/256 .

Example 4: What is the probability that the position in which the consonants appear remain unchanged when the letters of the word Math are re-arranged?

Solution: The total number of ways in which the word Math can be re-arranged = 4! = 4*3*2*1 = 24 ways.

Now, if the positions in which the consonants appear do not change, the first, third and the fourth positions are reserved for consonants and the vowel A remains at the second position.

The consonants M, T and H can be re-arranged in the first, third and fourth positions in 3! = 6 ways without the positions in which the positions in which the consonants appear changing.

Therefore, the required probability = 3!/4! = 6/24 = 1/4

1. Out of two-thirds of the total number of basket-ball matches, a team has won 17 matches and lost 3 of them. What is the maximum number of matches that the team can lose and still win three-fourths of the total number of matches, if it is true that no match can end in a tie?

2. 4 people played a game of chess, where each one plays every other player. What is the maximum number of points that any player could gather if every win gets him 1 point ?

3. From a pack of 52 cards, all face cards are removed and four cards are drawn. Then the probability that they are of different suit and different denomination is

4. Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five balls. The number of ways in which we can place the balls in the boxes so that no box remains empty is ?

5. How many arrangements can be made of the letters of the word DRAUGHT the vowels never being separated?

6. If the integers m and n are chosen at random from integers from integers  1 to 100 with replacement, then the probability that a no. of the form 7^m + 7ⁿ is divisible by 5 equals?

A relation is a set of ordered pairs where the first components of the ordered pairs are the input values and the second components are the output values.

A function is a relation that assigns to each input number EXACTLY ONE output number.

The domain is the set of all input values to which the rule applies.  These are called your independent variables. These are the values that correspond to the first components of the ordered pairs it is associated with.

The range is the set of all output values.  These are called your dependent variables. These are the values that correspond to the second components of the ordered pairs it is associated with.

Function Notation

f(x) read "f of x"

f is the function name.  Output values are also called functional values. Note that you can use any letter to represent a function name, f is a very common one used.

x is your input variable. 

Think of functional notation as a fancy assignment statement.   When you need to evaluate the function for a given value of x, you simply replace x with that given value and simplify.  For example, if we are looking for f(0), we would plug in 0 as the value of x in our function f.

If the function is constant, that means that the functional value never changes, it is always equal to that constant.

f(x) = c, where c is a constant.

DIRECTIONS for Questions 1 and 2: Answer the questions based on the following information:-
A, S, M and D are functions of x and y, and they are defined as follows:
    A(x, y) = x + y
    S(x, y) = x - y
    M(x, y) = xy
    D(x, y) = x/y, where y in not equal 0.

1. What is the value of M(M(A(M(x, y), S(y,x)), x), A(y, x)) for x = 2, y = 3 ?

2. What is the value of S(M(D(A(a, b), 2), D(A(a, b),2)), M(D(S(a, b), 2), D(S(a, b),2))) ?

3. What is the value of 1.1!+2.2!+3.3!+--------+n.n! ?

4. Let g(x) be a function such that g(x + 1) + g(x − 1) = g(x) for every real x. Then for what value of p is the relation g(x + p) = g(x) necessarily true for every real x?

5. A function y = f(n) is defined, for all natural numbers, as the sum of the digits of n.
if k is a natural number such that f(f(f(f(k)))) = 1 and k > f(f(k)).f (f(f(k))) > 1 what is the least number of digits that k can have?

6. A function y = f(n) is defined, for all natural numbers, as the sum of the digits of n. for a natural number m, what is  the value of f(f(m − f(m)))?

A set is a collection of things. Absolutely anything can be considered a set.
Below you'll see just a sampling of items that could be considered as sets:

• Your favorite clothes
• A coin collection
• The items in a store
• The English alphabet
• Even numbers

A set could have as many entries as you would like.
It could have one entry, 10 entries, 15 entries, or even an infinite number of entries.
On the next page you'll find out that a set could even have no entries at all!
For example, in the above list the English alphabet would have 26 entries, while the set of even numbers would have an infinite number of entries.

Each entry in a set is known as an element. We'll find out more about elements in the next section.

Sets are written using curly brackets ("{" and "}"), with their elements listed in between.
For example the English alphabet could be written as {a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z}
and even numbers could be {0,2,4,6,8,10,...} (Note: the dots at the end indicating that the set goes on infinitely)

A union of two or more sets is another set that contains everything contained in the previous sets.

Union is designated by the symbol ∪.
If A and B are sets then A ∪ B represents the union of A and B

Example:

A={1,2,3,4,5} B={5,7,9,11,13}
A ∪ B = {1,2,3,4,5,7,9,11,13}

The intersection of two (or more) sets is those elements that they have in common.
Intersection is designated by the symbol ∩.
So if A and B are sets then the intersection (the elements they both have in common) is denoted by A ∩ B.

Example:

A={1,3,5,7,9} B={2,3,4,5,6}
The elements they have in common are 3 and 5
A ∩ B = {3,5}

Subset

Let A be the set of objects that you own in your home
Let B be the set of objects that you own which are kept on the second floor of your home
Let C be the set of objects that you own which are kept in your bedroom [Note the bedroom is own the second floor]
Let D be the set of objects that you own which are kept in your bedroom nightstand

Now we could say D is contained within C, which in turn is contained within B, which in turn is contained within A.

This is the notion of a subset.
D is said to be a subset of C since it is completely contained within C (another way to think of this is every element of set D is also an element of set C).
C is said to be a subset of B since it is completely contained within B (another way to think of this is every element of set C is also an element of set A).
B is said to be a subset of A since it is completely contained within A (another way to think of this is every element of set D is also an element of set C).

The symbol for subset is Ì.
So D Ì C and C Ì B and B Ì A.

However if even one element of one set is not contained within the other then thy are not subsets.
If A were defined as {1,2,3,4,5} and B as {3,4,5,6} then B would not be a subset of A since
“6” Î B but 6 Ï A.
The symbol for “not a subset” is Ë.
We would write B Ë A.