Class 9 – Functions
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Introduction: Let A and B be two nonempty sets. Then a function ‘f ’ from set A to set B is a rule or method or correspondence which associates elements of set A to elements of set B such that. i) All elements of set A are associated to elements in set B. ii) An element of set A is associated to a unique element in set B. In other words, a function ‘ f ’ from a set A to set B associates each element of set Terms such as “map” (or “mapping”), “correspondence” are used as synonyms for “Function”. If f is a function from a set A to a set B, then we write or, which is read as f is a function from A to B or f maps A to B. If an element a ∈ A is associated to an element , then b is called ‘the fimage of a (or) image of a under f (or) the value of the function f at a. Also a is called the preimage of b under the function f. We write it as: b = f (a). 
Example: Let and be two sets and let be rules associating elements A to elements of B as shown in the following figures. 

We observe that
i) f1 is not a function from set A to set B,since there is an element whichis not associated to any element of B.
ii) f2 is not a function from A to B because an element 4 ∈ A is associated to two elements cand e in B.
iii) f3 and f4 are functions from A to B, because under f3 and f4 each element in A is associated to a unique element in B.
Domain,co domain and range of a function:
Let f : A⟶B.Then, the set A is known as the domain of f and the set B is known as the
Codomain of f. The set of all fimagesof elements of A is known as the range of f or image set of A under f and is denoted by f (A).
Thus, = Range of f.
Clearly,
Example: Let A = {2, 1, 0, 1, 2} and B = { 0, 1,2, 3, 4, 5, 6 }. Consider a rule.
Then, Clearly, each element of A is associated to unique element of B. So, f : A⟶Bgiven by is a function.
Clearly, domain of f = A = {2,1, 0, 1, 2} and range of f = {0, 1, 4}.
Example: Consider a function f : N ®N defined by the rule f(x) = 2x – 3. This rule does not define a function from N to itself, because i.e., is not associated to any element of N.
Number of Functions: Let A and B be two finite sets having m and n elements respectively. Then each element of set A can be associated to any one of n elements of set B. So, total number of functions from set A to set B is equal to the number of ways of doing m jobs where each job can be done in n ways. The total number of such ways is
Example: Given A = {1, 0, 2, 5, 6, 11}, B = {2,1, 0, 18, 28, 108} and Find. Is
Solution:We have:
Weobserve that. So,.
Example: Let be given by .
Find (i) (ii)The preimages of 39 and 2 under f.
Solution: (i) We have
(ii)Let x be the preimage of 39. Then,
So,preimages of 39 are 6 and 6.
Letx be the preimage of 2. Then,
Since,no real value of x satisfies the equation ,so, 2 does not have any preimage in f.
The total number of functions from A to B is
The total number of relations from a set A having m elements to a set Bhaving n elements is .So, the number of relations from A to B which are not functions is
i.e.,
Example: The total number of functions from a set A = {a, b, c, d} to set B = {1, 2, 3} is
Function as a relation: Let A and B be two nonempty sets. A relation f from A to B i.e. a subset of , is called a function (or a mapping ora map) from A to B if
(i) for each there exists a unique such that .
(ii)
Thus, a nonvoid subset of is a function from A to B if each element of A appears in some ordered pair in f and no two ordered pairs in f have the same first element.
If , then b is called the image of a under f and a is called the preimage of b under f .
Example: Let A = {1,2,3}, B = {2,3,4} and be three subsets of as given below.
, ,
Then
i) is a function from A to B.
ii) is not a function from A to B, because has two images .
iii) is not a function from A to B because has no image in B.
If a function f is expressed as the set of ordered pair, the domain of f is the set of all first components of the ordered pairs of f and the range is the set of second components of ordered pairs of f i.e.,
Domain of
Range of
Example: Expressthe following functions as sets of ordered pairs and determine their ranges:
(i)
(ii)
Solution:
i)
So, . Range of.
ii) We have A = {1, 2,3, ……. 10}. So,
Equal Functions: Two functions f and g are said to be equal if
i)the domain of f = domain of g.
ii)the codomain of =the codomain of g, and
iii) for every x belonging to their common domain.
If two function f and g are equal, then we write .
Example: Let given by and given by . Then, we observe that are having same domain and codomain i.e., Hence,
Algebra of real valued functions: If f and are real valued functions with domains A and B respectively, then both f and are defined on when
(i) Let and suppose that , we define and on as and .
Let and c be aconstant and a function is defined on A then from the above definition
and
Thefunction isdenoted by
(ii) Let , we define on E by
If ,then is not defined.
(iii) Let and . We define and and A by and
for all
(iv) If , then we define on E by for all .
Example: If and then find
(i) (ii) (iii) (iv) (v ) (vi)
(vii) (viii) (ix) (x)
Solutions: Domain of f = A = {4, 5, 6}, Domain of g = B = {4, 6, 8}
Domainof
(i)
(ii)
(iii) Domain of ,Domain of
(iv)Domain of f + 4 = A = {4,5,6}
(v) Domain of
(vi)Domain of
(vii) Domain of
(viii) Domainof
(ix) Domain of
(x) Domain of
Functional Equations: Functional equations are equations for unknown functions instead of unknown numbers. In this Synopsis, we shall try to explore how we can find the unknown function when we know that the condition is satisfied.
Functional equations in one variable: Functional equations in one variable are usually easierto solve. Although there is no definite method to solve functional equations,there are some tips.
Example: If , then find
Solution:Let . On direct substitution of y,we get. Thus
Example: If , find
Solution: Let , then . On substitution weget
Thus
Example: If then find
Solution: It is equivalent to.Simplifying it,
wehave