Types of Functions JEE Notes | EduRev

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FUNCTIONS
A function is a relation that maps each element x of a set A with one and only one element y of another set B. In other words, it is a relation between a set of inputs and a set of outputs in which each input is related with a unique output. A function is a rule that relates an input to exactly one output.

Types of Functions JEE Notes | EduRevIt is a special type of relation. A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B and no two distinct elements of B have the same mapped first element.  A and B are the non-empty sets. The whole set A is the domain and the whole set B is codomain.

Definition: A function is a rule (or a set of rules) which relates or associates each and every element of a non empty set A with the unique element of the non empty set B.

REPRESENTATION
A function f: A → B is represented as f(a) = b such that for a ∈ A there is a unique element b ∈ B such that (a, b) ∈ f.

For any function f, the notation f(a) is read as “f of a” and represents the value of b when a is replaced by the number or expression inside the parenthesis. The element b is the image of a under f and a is the pre-image of b under f.

The set ‘A’ is called the domain of ‘f’.

The set ‘B’ is called the co-domain of ‘f’.

Set of images (outputs) of different elements of the set A is called the range of ‘f’. It is obvious that range could be a subset of the co-domain as we may have some elements in the co-domain which are not the images of any element of A (of course, these elements of the co-domain will not be included in the range). Range is also called domain of variation.

If R is the set of real numbers and A and B are subsets of R, then the function f (A) is called a real−valued function or a real function.

Domain of a function ‘f’ is normally represented as Domain (f). Range is represented as Range (f). Note that some times domain of the function is not explicitly defined. In these cases domain would mean the set of values of ‘x’ for which f (x) assumes real values that is if y = f (x) then called Domain (f) = {x : f (x) is a real number}.

In other words, domain is defined as a set of all those values of x for which the given function is defined.


Note: Every function is a relation but every relation need not be a function.
For example: Let A = {1, 2, 3}
R1 = {(1, 1), (2, 3), (3, 3)}
R2 = {(1, 1)(1, 2), (3, 2)(2, 1)}
Here R1 is a function but R2 is not a function because here element 1 has two images as 1 and 2.

Q. Which of the following is a function?

Types of Functions JEE Notes | EduRev

Types of Functions JEE Notes | EduRev

Types of Functions JEE Notes | EduRevAns. Figure 3 is an example of function since every element of A is mapped to a unique element of B and no two distinct elements of B have the same pre-image in A.


Q. Find the domain and range of the function 
Types of Functions JEE Notes | EduRev
Ans. The function
Types of Functions JEE Notes | EduRev
is defined for x ≥ 5
⇒ The domain is [5, ∞] .
Also, for any
Types of Functions JEE Notes | EduRev
⇒ The range of the function is [0, ∞] .

Q. Find the domain and the range of the function y = f (x), where f (x) is given by
(i) x2 − 2x − 3

Types of Functions JEE Notes | EduRev
Types of Functions JEE Notes | EduRev
(iv) tan x
(v) tan−1x
(vi) log10 (x).
(vii) sin x

Ans. 
(i) Here y = (x − 3) (x + 1).
The function is defined for all real values of x
⇒ Its domain is R.
Also x2 − 2x − 3 − y = 0 for real x
⇒ 4 + 4 (3 + y) ≥ 0 ⇒ − 4 ≤ y < ∞.
Hence the range of the given function is [- 4, ∞].
(ii) Here
Types of Functions JEE Notes | EduRev

⇒ (x − 3) (x + 1) ≥ 0 so that x ≥ 3 or x ≤ − 1.
Hence the domain is R − (− 1, 3) or (− ∞, - 1] ∪ [3, ∞].

Since f(x) is non−negative in the domain, the range of f (x) is the interval [0, ∞].
(iii) The given function is defined for all x ≥ 0
⇒ The domain is [0, ∞].
Moreover - 1 ≤ f (x) < ∞
⇒ The range is [- 1, ∞].
(iv) The function f (x) = tan x = sin x/cos x is not defined when
cos x = 0, or x = (2n + 1) π/2, n = 0, ± 1, ± 2, …….
Hence domain of tan x is R -{(2n + 1)π/2, n = 0, + 1, + 2,....}, and its range is R. 

(v) The function f (x) = tan−1 x, is defined for all real values of x and -π/2 < tan-1 < π/2.
Hence its domain is R and the range is (-π/2, π/2).
(vi) The function f (x) = log10 x is defined for all x > 0. Hence its domain is (0, ∞) and range is R. 

(vii) The function f (x) = sin x, (x in radians) is defined for all real values of x
⇒ domain of f (x) is R. Also − 1 ≤ sin x ≤ 1, for all x,
so that the range of f (x) is [− 1, 1].

Different types of functions and their graphical representation

Identity function

A function f:R → R is said to be an identity function if f(x) = x, ∀ x ϵ R denoted by IR.

Let A = {1, 2, 3}

The function f: A → A defined by f(x) = x is an identity function.

f = {(1,1), (2,2), (3,3)}.

Types of Functions JEE Notes | EduRev


The graph of an identity function is a straight line passing through the origin.Types of Functions JEE Notes | EduRev

Each point on this line is equidistant from the coordinate axes.

The straight line makes an angle of 45° with the coordinate axes.
Example:
Let A = {1, 2, 3, 4, 5, 6}
Then Identity function on set A will be defined as
IA : A → A , IA = x , x ∈ A
for x = 1 , IA(1) = x = 1
for x = 2 , IA(1) = x = 2
for x = 3 , IA(1) = x = 3
for x = 4 , IA(1) = x = 4
for x = 5 , IA(1) = x = 5
Domain, Range and co-domain will be Set A

Constant function
A function f:R → R is said to be a constant function, if f(x) = c , ∀ x ϵ R, where c is a constant.

Let f:R → R be a constant function defined by f(x) = 4 , ∀ x ϵ R.

The ordered pairs satisfying the linear function are: (0, 4), (-1, 4), (2, 4).

If the range of a function is a singleton set, then it is known as a constant function.

On plotting these points on the Cartesian plane and then joining them, we get the graph of the constant function f of x = 4,∀ x ϵ R as shown.

Types of Functions JEE Notes | EduRev

Also, from the graph, we can conclude that the graph of a constant function, f(x) = c, is always a straight line parallel to the X-axis, intersecting the Y-axis at (0, c).

Q. Which of the graph represent a constant function?
Types of Functions JEE Notes | EduRevAns.
The graph should be parallel to x-axis for the function to be constant function
So C and D are constant function.

Polynomial function

A polynomial function is defined by y = a+ a1x + a2x2 + … + anxn, where n is a non-negative integer and a0, a1, a2,…, n ∈ R. The highest power in the expression is the degree of the polynomial function. Polynomial functions are further classified based on their degrees:
If the degree is zero, the polynomial function is a constant function (explained above).

Examples of polynomial functions

f(x) = x2 + 5x + 6 ,∀ x ϵ R (Degree is 2)

and f (x) = x3 + 4x + 2 ,∀ x ϵ R (Degree is 3)

Note: In a polynomial function, the powers of the variables should be non-negative integers.For example, f(x) = √x + 2 (∀ x ϵ R) is not a polynomial function because the power of x is a rational number.

Consider the polynomial function, f(x) = 3x2 +2x -3 ,∀ x ϵ RThe ordered pairs satisfying the polynomial function are (0, -3), (-1, -2), (1, 2), (2, 13), (-2, 5).

On plotting these points on the Cartesian plane and then joining them, we get the graph of the polynomial function f of x is equal to 3 x2 + 2 x - 3 ,∀ x ϵ R as shown.

Types of Functions JEE Notes | EduRev


Linear function
A function f:R → R is said to be a linear function if f (x) = ax + b, where a ≠ 0, a and b are real constants, and x is a real variable. It is a polynomial function with degree one.

Consider the linear function, f(x) = 3x + 7, ∀ x ϵ R

The ordered pairs satisfying the linear function are (0, 7), (-1, 4), (-2, 1).

On plotting these points on the Cartesian plane and then joining them, we get the graph of the linear function f of x is equal to 3x + 7, ∀ x ϵ R as shown in the figure.

Types of Functions JEE Notes | EduRev

Quadratic Function: If the degree of the polynomial function is two, then it is a quadratic function. It is expressed as f(x) = ax2 + bx + c, where a ≠ 0 and a, b, c are constant & x is a variable. The domain and the range are R. The graphical representation of a quadratic function say, f(x) = x2 – 4 is

Types of Functions JEE Notes | EduRev

Cubic Function: A cubic polynomial function is a polynomial of degree three and can be denoted by f(x) = ax3 + bx2 + cx +d, where a ≠ 0 and a, b, c, and d are constant & x is a variable. Graph for f(x) = y = x3 – 5. The domain and the range are R.

Types of Functions JEE Notes | EduRev

Rational function
If f(x) and g(x) be two polynomial functions, then f(x)/g(x) such that g(x) ≠ 0 and ∀ x ϵ R, is known as a rational function.

Let us consider the function f ( x) = 2x -5/3x - 2 (x ≠ ⅔).

The ordered pairs satisfying the polynomial function are: (0, 5), (2, -¼), (1, -3).

On plotting these points on the Cartesian plane and then joining them, we get the graph of the given rational function as shown.
Types of Functions JEE Notes | EduRev


Modulus function
A function f: R → R defined by f(x) = |x| (∀ x ϵ R) is known as a modulus function.

If x is negative, then the value of the function is minus x, and if x is non-negative, then the value of the function is x. i.e. f(x) = x if x ≥ 0 = - x if x < 0.

The ordered pairs satisfying the polynomial function are (0, 0), (-1, 1), (1, 1), (-3, 3), (3, 3).

On plotting these points on the Cartesian plane and then joining them, we get the graph of modulus function f of x is equal to mod of x.
Types of Functions JEE Notes | EduRev

Greatest integer function
A function f: R → R defined by f(x) = [x],∀ x ϵ R assumes the value of the greatest integer, less than or equal to x.

From the definition of [x], we can see that

[x] = -1 for -1 £ x < 0

[x] = 0 for 0 £ x < 1

[x] = 1 for 1 £ x < 2

[x] = 2 for 2 £ x < 3, and so on.

⇒  f(2.5) will give the value 2 and f(1.2) will give the value 1, and so on…

Hence, the graph of the greatest integer function is as shown.
Types of Functions JEE Notes | EduRev

Signum Function

A function f: R→ R defined by

f(x) = { 1, if x > 0; 0, if x = 0; -1, if x < 0

Signum or the sign function extracts the sign of the real number and is also known as step function.

Types of Functions JEE Notes | EduRev

Trigonometric function
(i) Function: f(x) = sin x
Domain: x ∈ R
Range: y ∈ [–1, 1]
Curve:
 Types of Functions JEE Notes | EduRev
(ii) Function: f(x) = cos x
Domain: x ∈ R
Range: y ∈ [–1, 1]
Curve:
Types of Functions JEE Notes | EduRev
(iii) Function: f(x) = tan x
Domain: x ∈ R
Range: y ∈ R
Curve:
Types of Functions JEE Notes | EduRev
(iv) Function: f(x) = cot x
Domain: x ∈ R – (2n + 1) π / 2, n ∈ I
Range: y ∈ R
Curve:
Types of Functions JEE Notes | EduRev
(v) Function: f(x) = cosec x
Domain: x ∈ R – nπ, n ∈ I
Range: y ∈ (-∞, –1] ∪ [1, ∞)
Curve:
Types of Functions JEE Notes | EduRev
(vi) Function: f(x) = sec x
Domain: x ∈ R – (2n + 1) π / 2 , n ∈ I
Range: y ∈ (-∞, –1] ∪ [1, ∞)
Curve:
Types of Functions JEE Notes | EduRev

Algebraic Function
A function is called an algebraic function. If it can be constructed using algebraic operations such as additions, subtractions, multiplication, division taking roots etc.
All polynomial functions are algebraic but converse is not true
Example: f(x) =Types of Functions JEE Notes | EduRev+ x + (x3 + 5)3 / 5, f(x) = x3 + 3x2 + x + 5
Remark : Function which are not algebraic are called as transcendental function.
Example: Types of Functions JEE Notes | EduRev
Example:Types of Functions JEE Notes | EduRev

Logarithmic function
f(x) = logax, where x > 0, a > 0, a ≠ 1
a → base, x → number or argument of log.
Case – I : 0 < a < 1
 f(x) = loga x
Domain : x ∈ (0, ∞)
Range : y ∈ R
Types of Functions JEE Notes | EduRev
Case – II : a > l
  f(x) = ln x
Types of Functions JEE Notes | EduRev

Exponential function
f(x) = ax, where a > 0, a ≠ 1
a → Base x → Exponent
Case – I : 0 < a < 1 ; a = 1 / 2
Types of Functions JEE Notes | EduRev
Domain : x ∈ R
Range : y ∈ (0, ∞)
Types of Functions JEE Notes | EduRev
Case – II : a > 1
Types of Functions JEE Notes | EduRev

Fractional part function
y = f(x) = {x} = x – [x]
Domain : x ∈ R; Range : [0, 1)
Types of Functions JEE Notes | EduRev
Example:
Types of Functions JEE Notes | EduRev
Properties:
(i) Fractional part of any integer is zero.
(ii) {x + n} = {x}, n ∈ I
(iii)Types of Functions JEE Notes | EduRev

Examples
Example.1. Find the range of the following functions:
(a)Types of Functions JEE Notes | EduRev
(b)Types of Functions JEE Notes | EduRev
Ans. (a) We have,
Types of Functions JEE Notes | EduRev
i.e. Types of Functions JEE Notes | EduRev
i.e.Types of Functions JEE Notes | EduRev
since, Types of Functions JEE Notes | EduRev
therefore we haveTypes of Functions JEE Notes | EduRev
i.e.Types of Functions JEE Notes | EduRev
i.e.Types of Functions JEE Notes | EduRev
i.e.Types of Functions JEE Notes | EduRev
Hence, the range isTypes of Functions JEE Notes | EduRev
(b) We have,
Types of Functions JEE Notes | EduRev
Now, we have 2 ≤ x2 + 2 < ∞ i.e.Types of Functions JEE Notes | EduRev
i.e.Types of Functions JEE Notes | EduRev
i.e.Types of Functions JEE Notes | EduRev
i.e.Types of Functions JEE Notes | EduRev
i.e.Types of Functions JEE Notes | EduRev
givesTypes of Functions JEE Notes | EduRev
Hence, the range isTypes of Functions JEE Notes | EduRev

Example.2. Find the range of following functions :  
(i) y = ln (2x – x2)
(ii) y = sec – 1 (x2 + 3x + 1)
Solution. (i)
using maxima–minima, we have (2x – x2) ∈ (–∞, 1]
For log to be defined accepted values are 2x – x2 ∈ (0, 1] {i.e. domain (0, 1]}
ln (2x – x2) ∈ (0, 1] ∴ range is (–∞, 0]
(ii) y = sec–1 (x2 + 3x + 1)
Let t = x2 + 3x + 1 for x ∈ R thenTypes of Functions JEE Notes | EduRev
but y = sec–1 (t)Types of Functions JEE Notes | EduRev
from graph range isTypes of Functions JEE Notes | EduRev
Types of Functions JEE Notes | EduRev

Example.3. Find the range ofTypes of Functions JEE Notes | EduRev
Solution.
We have,
Types of Functions JEE Notes | EduRevwhich is a positive quantity whose minimum value is 3 / 4.
Also, for the functionTypes of Functions JEE Notes | EduRevto be defined, we have x+ x + 1 ≤ 1
Thus, we have
Types of Functions JEE Notes | EduRev
[∴ sin-1 x is an increasing function, the inequality sign remains same]
i.e.Types of Functions JEE Notes | EduRev
i.e.Types of Functions JEE Notes | EduRev
Hence, the range is y ∈ [ln π / 3, ln π / 2]

Example.4.Types of Functions JEE Notes | EduRev.If the range of this function is [– 4, 3) then find the value of (m2 + n2).
Solution.

Types of Functions JEE Notes | EduRevTypes of Functions JEE Notes | EduRev
Types of Functions JEE Notes | EduRevfor  y to lie in [– 4, 3) mx + n – 3 < 0 ∀ x ∈ R
this is possible only if m = 0 when, m = 0 thenTypes of Functions JEE Notes | EduRev
note that n - 3 < 0 (think !) n < 3 if x → ∞, ymax → 3-

now ymin occurs at x = 0 (as 1 + x2 is minimum)

ymin = 3 + n - 3 = n ⇒ n = - 4 so m2 + n2 = 16

Example.5. Find the domain and range of f(x) =Types of Functions JEE Notes | EduRev
Solution.
Types of Functions JEE Notes | EduRevis positive and x2 < 4 ⇒ –2 < x < 2
1 – x should also be positive. ∴ x < 1
Thus the domain ofTypes of Functions JEE Notes | EduRevis –2 < x < 1 sine being defined for all values, the domain of sinTypes of Functions JEE Notes | EduRevis the same as the domain ofTypes of Functions JEE Notes | EduRev
To study the range. Consider the functionTypes of Functions JEE Notes | EduRev
As x varies from –2 to 1,Types of Functions JEE Notes | EduRevvaries in the open interval (0, ∞) and hence

Types of Functions JEE Notes | EduRevvaries from –∞ to + ∞. Therefore the range of sinTypes of Functions JEE Notes | EduRev

Example.6. Find the range of the function f(x) =Types of Functions JEE Notes | EduRev
Solution.
ConsiderTypes of Functions JEE Notes | EduRevAlso g(x) is positive ∀ x ∈ R and g(x) is continuous ∀ x ∈ R and g(0) = 1 andTypes of Functions JEE Notes | EduRev
⇒ g(x) can take all values from (0, 1] ⇒ Range of f(x) = sin–1 (g(x)) is (0, π / 2]

Example.7.Types of Functions JEE Notes | EduRevfind the domain and range of f(x) (where [ * ] denotes the greatest integer function).
Solution.
If cos–1 x = θ, then– 1 ≤ x ≤ 1
Types of Functions JEE Notes | EduRev
Types of Functions JEE Notes | EduRev
Types of Functions JEE Notes | EduRev
Types of Functions JEE Notes | EduRev
1 ≤ [x3] + 1 < 9 0 ≤  [x3] < 8 ∴ 0 ≤  x < 2
∴ Domain of f(x) = Dr in x ∈ [0, 2) Range of f(x) When 0 ≤ x < 2
Then 1 ≤  x3 + 1 < 9 ∴ 1 ≤  [x3 + 1] ≤  8
Types of Functions JEE Notes | EduRev
Case I:
Types of Functions JEE Notes | EduRev
∴ Range in cos–1 {log 1} and cos–1 {log 2}
Case II
Types of Functions JEE Notes | EduRev
∴ Rf is (π / 2, cos–1 (log 2))

Example.8. Find the range of the following functions
(i) f(x) = loge (sinx sinx + 1) where 0 < x < π / 2.
(ii) f(x) = loge (2 sin x + tan x - 3x + 1) where π / 6 ≤ x ≤ π /3
Solution. (i)
 0 < x < π / 2 ⇒ 0 < sin x < 1

Range of loge (sin xsin x + 1) for 0 < x < π / 2 = Range of loge (xx + 1) for 0 < x < 1

Let h(x) = xx + 1 = exlogex + 1

h'(x) = exlogex (1 + loge x) ⇒ h'(x) > 0 for x > 1/e and h'(x) < 0 for x < 1/e

∴ h(x) has a minima at x = 1/e
Types of Functions JEE Notes | EduRev
Types of Functions JEE Notes | EduRev
∴ 0 < x< 1
Types of Functions JEE Notes | EduRev
Types of Functions JEE Notes | EduRev
Types of Functions JEE Notes | EduRev
Types of Functions JEE Notes | EduRev
(ii) Let h(x) = (2 sin x + tan x – 3x +1) ⇒ h'(x) = (2 cos x + sec2 x – 3)
Types of Functions JEE Notes | EduRev
∴ h'(x) > 0 ⇒ 2 cos3 x – 3 cos2 x + 1 > 0
Types of Functions JEE Notes | EduRev
⇒ h(x) is an increasing function of x
⇒ h(π / 6) ≤ h(x) ≤ h(π / 3)
Types of Functions JEE Notes | EduRev
Types of Functions JEE Notes | EduRev

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