function theory

7/17/2019 Function Theory

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FUNCTIONS

. ORDERED PAIR

An ordered pair can be def ined as (a , b ) = {{a},{a,b}}.

2. CARTESIAN PRODUCT

For sets A and B, it is the set containing all ordered pairs. A B = {(a , b ) : a A and b B}

Resul t s : 1) Two ordered pairs (a ,b ) and (c ,d ) are equal iff a = c and b = d 2) A B = C D iff A = C and B = D.

3. RELATION

Definition : A relation from A to B is a non-empty subset of A B.

i .e. A relation is a set of ordered pairs. Any set of ordered pairs is, therefore, arelation. Since a relation is a set, therefore any statement about the sets is appro-

priately applied to the relations. One can thus speak of the intersection or union of two relations, or speak of one relation being a subset of another. A relation, whichis a subset of A B, may be obtained by just choosing the ordered pairs randomly,

wi thout any specif ic cond it ions in two variables a & b of an ordered pair (a , b ).

There are various ways in which a relation can be expressed.

i ) Rost er f or m : By exhibiting a set of pairs e.g.{(4, 2), (9, 3), (16,4), (25,5)}

i i ) Set bui lder form : By standard description, using a rule or a formula{(x , y ) : }, the blank to be replaced by a rule or

a condition.e.g. {(x , y ): y = x } , {(a , b ) : 3a –b > 4}

iii ) By arrow diagrams : In constructing the arrow diagrams the arrow headshou ld

indicate the direction from A to B. e.g.



2

Set A

Dollar

Pound

Mark

Rupee

Set B

Scotland

Canada

England

U.S.A.

Germany

India

is the

currency of

iv) As a Venn diagram :

Sometimes the relations are not between two different sets but between the mem-bers of thesame set. For example, let us consider the relation “ is a factor of ” on theset A, A ={2, 3, 6, 8}.

As a Venn diag ram the rela ti on looks li ke the fol lowing

6

2

4

8

A lot more can be studied about Relations but our main focus is towards learningth efunctions. So let us move towards functions.

4. FUNCTIONS

The word ‘function’ is used in casual conversation like the word ‘relation’ is used.The math-ematical concept of a relation is fairly close to the idea of a relation asused outside mathematics. But the mathematical concept of a function is entirelydifferent from its meaning in ordinary usage. The mathematical concept of a func-tion is an outgrowth of scientific study of interrelated phenomena. For example,the atmospheric pressure depends upon the altitude. This example suggests a cer-tain kind of association between two measurable quantities, one of which is re-garded as dependent on the other. We then say that the first is a function of thesecond. A function can be defined in several ways.

4. First definition of a function (Function as an association)

If with each element of a set A, exactly one element of a set B is associated, then

t h i s a ssoc ia ti on is cal led a f un ct ion fr om A t o B. Th e set A i s c all ed th edomain of the functionand the set C of all those elements of B, which are assigned to the elements to Aby the function, is called the range of the function. The set B is cal led the co-

domain.

4. 2 Second definition of a function (function as a rule)

Let A & B be two given sets. A function from A to B is a rule which assigns to eachx in A, a unique y in B.



3

It is customary to use letters to denote the functions.The letter f is mostly used. Other popular

notation are g , h , , u , v etc.x

( )x

If f is a function and x is any element of the domainof f , we use the symbol f (x ) to denote theobject which f associates with x .

f (x ) is the functional value at x . The symbol f (x ) is read as “ the value of f at x or “f at x . f (x ) is also called the image of x . e.g.

A

2

b

3

4

5

B

Some important facts regarding a function f : A B are

i ) Every x A is assigned some y B. i.e. a function is defined only when the domainis entirely used up.

ii ) The set B, co-domain may or may not be entirely used up by the function.

ii i) The function may assign the same y B to more than one x A.

i v) For no x in A, do there exist two choices for f (x ). i.e. to every x A, f should assignone and only one y B.

Note that (i) & (iv) are essential requirements of the definition of a function.

The word ‘rule’ covers di fferent kinds of schemes for making assignments andconsequently we have different methods of describing a function. A function isexpressed by means of

i ) an arrow diagram (as shown in relation)

ii)a table :

e.g.

x

3

4

4

5

5

6

6

7

represents a function.

i i i ) a graph :

e.g.

O

x

represents a function.



4

iv) a formula (as an algebraic expression or an equation)e.g. the equation y = 2x – 3 represents a function.

4. 3 Fourth definition (as a special class of relations)

We have described the concept of function in terms of words such as relate, assign,assoc iat e etc. We would now give a more abstract definition of this concept.

Consider a function

f

: A

B

A

2 b

3

4

B

1 a 2 b 3 b 4 c

d

(i) ( i i )

1 a = f (1) 1, a = f (1)2 b = f (2) 2, b = f (2)3 b = f (3) 3, b = f (3)4 c = f (4) 4, c = f (4)

( ii i) (iv)

(1, a ) = (1, f (1))

(2, b ) = (2, f (2))(3, b ) = (3, f (3))(4, c ) = (4, f (4)) f = {(1, a ), (2, b ), (3, b ), (4, c )}

(v) (vi)(i) to (iv) are several ways of showing a function. The final diagram shows thefunction as a set of ordered pairs (x , f (x )). The first element of each pair is theelement of the domain and the second element is i ts image under thecorrespondance. Since every set of ordered pairs is a relation, therefore every func-tion is a relation. Clearly all relations are not functions so a function is also de-fined asA function f : A B is a subset of A B such that

i) for every

a

A there is

b

B such that (

a

,

b

)

f

ii) (a, b) f and (a , b ) f b = b

We can test a graph of a relation for it to be function. A vertical line can be foundthat would cut the graph of a relation that is not a function, at atleast two points.On the other hand, any vertical l ine would cut the graph of a relation that is afunction, at atmost one point.



5

O

O

O

(Graphs of non-function relation)

O

O O

(Graphs of function relations)

5 . D OMA IN , CO -D OMA IN & R AN GE OF A F UN CT IO N

Let f: A B, then the set A is known as the domain of f & the set B is knownas co-domain of f. If a member 'a' of A is associated to the member 'b' of B,

then 'b' is called the f -image of 'a' and we write b = f (a). Further 'a' is calleda pre-image of 'b'. The set {f(a): a A} is called the range of f and is denotedby f(A). Clearly f(A) B.Sometimes i f only defini t ion of f (x) is given (domain and codomain are notmentioned), then domain is set of those values of ' x' for which f (x) is defined,

while codom ain is consider ed to be (– , ). A function whose domain andrange both are sets of real numbers is cal led a real funct ion . Conventional lythe word "FUNCTION” is used only as the meaning of real function.

l l lus t ra t ion

A = {a , b , c } , B = {x , y , z }Consider f 1 = {(a , x ), (b , y ), (c , z )}f 2 = {(a , x ), (a , y ), (b , z ), (c , z )}f 3 = {(a , y ), (b , y ),(c , y )}f 4 = {(a , z ), (b , x )}

Which of f 1 , f 2 , f 3 , f 4 is (are) function(s)?Sol . f 1 , f 3 are functions as for every element of A, there is unique image f 2 is not a

function as element a of set A does not have a unique image. f 4 is not a function asevery element of set A does not have image.

l l lus t ra t ion 2

Let f : R R be defined by

f (x ) = 2x + 3 x < – 2 =x 2 – 2 – 2 x < 3 = 3x – 1 x 3

Find f (2) , f (4) , f (–1) , f (– 3).

Sol . 2 [– 2, 3) , f (2) = 22 – 2= 2

4 [3,) , f (3) = 3 (3) – 1= 8



6

– 1 [– 2, 3) , f (–1)=(–1)2 – 2= – 1

– 3 (– , – 2) , f (– 3) = 2 (– 3) + 3 = – 3l l lus t ra t ion 3

Find the domain and range of the function f given by f (x ) = x 2.Sol . f (x ) = x 2

Here x can take all real values, therefore the domain of f is R. As x varies, f (x )can take all non-negative real values. Therefore the range is R+ {0}


Find the domain and range of the function f given by f (x ) = x .

Sol . f (x )= x

Here x can take all non-negative real values, therefore the domain of f is R+ {0}.

As x varies, f (x ) can take all non-negative real values. Therefore the range of f isa l so

R+ {0}

6 . C LASSI FI CAT ION OF FU NC TI ONS

6. One-One Funct ion (Inject ive mapping):

A fu nc tion f : A B is said to be a one-one function or injective mapping if different elements of A have different f images in B. Thus for x1, x2 A &f(x1), f(x2) B,f(x1) = f(x2) x1 = x2 or x1 x2 f(x1) f(x2).

Example s : R R f (x) = x3 + 1 ; f (x) = e – x ; f (x) = l n xDiagramatical ly an injective mapping can be shown as

O R

No te :

( i ) A con t inuous funct ion wh ich i s a lways increas ing or decreas ing in who led o ma i nis one-one.

( i i ) A function is one to one i f and only if no horizontal l ine intersects i ts graphmore than once.

6 . 2 Many-one func ti on : (not i nj ec t iv e)

A func tion f : A B is said to be a many one function if two or moreelements of A have the same f image in B. Thus f : A B is many one if for

at leas t onex1, x2 A , f(x1) = f(x2) but x1 x2.Examples : f : R R , f (x) = | x | ; f (x) = ax2 + bx + c ; f (x) = sin x

Diagramatical ly a many one mapping can be shown as

OR



7

Note:

( i ) Any cont inuous function which has at leas t one local maximum or loca l minimumis manyone. In other words, if any line parallel to x-axis cuts the graph of the functionatleast at two points, then f is many-one.

(i i ) If a function is one-one, it cannot be many-one and vice versa.Number of One One + Number of Many One mappings = Total number of

mappings.6 . 3 O nto func t io n (S ur je ct iv e mapp ing) :

If the function f : A B is such that each element in B (co-domain) is the f image of atleast one element in A, then we say that f is a function of A 'onto' B.Thus f : A B is surjective iff b B, some a A such that f (a) = b.f : R R f (x) = 2x +1; f : R R+ f (x) = ex; f : R+ R f (x) =l n x

Diagramatically surjective mapping can be shown as

O R

Note that :

if range = co-domain, then f(x) is onto .

6 .4 Into function:

If f : A B is such that there exists atleast one element in co-domain which isnot the image of any element in domain, then f(x) is into.e.g. f : R R f (x) =| x |, sgn x, f (x) = ax2 + bx + c

Diagramatically into function can be shown as

O R

Note that :

If a function is onto, it cannot be into and vice versa . A polynomial of degreeeven define from R R will always be into & a polynomial of degree odd definedfrom R R will always be onto..Thus a function can be one of these four types :(a) one-one onto (injective & surjective) (I S)

(b) one-one into (injective but not surjective) (I S )

( c ) many-one onto (surjective but not injective) (S I )



8

(d) many-one into (neither surjective nor injective) ( I S )

Note :

( i ) I f f is both injective & surjective, then i t is cal led a Bi ject ive mapping. Thebijective functions are also named as invertible, non singular or biuniformfunct ions.

7 . I DENT IT Y / CONSTA NT FUN CT ION

7. Identity function :

The function f : A A defined by f(x) = x x A is called the identity of Aand is denoted by I A. It is easy to observe that identity function defined on R is abi ject ion.

7 . 2 Const ant fu nc tion :

A function f : A B is said to be a constant function if every element of A hasthe same f image in B. Thus f : A B ; f(x) = c , x A , c B is a constantfunction. Note that the range of a constant function is singleton.

8 . B ASI C ELEMENTARY F UNC TI ONS

The bas i c e lementary func i ton are the fo l lowing funct ions wi th ana ly t i crepresent t ion.

8. Power funct ion

ny x , n is Rat ional ( No te )

o

1

1 1

1X

Y4

y x2y x

1,1

1,1

Domain : R

Range : R 0

Nature : Many one into

3

y x

o1

1

1

1

X

Y5y x

1, 1

1,1 Domain : R

Range : R

Nature : one one onto



9

Y

X

4y 1/ x2y 1/ x

1,12

y 1/ x2y 1/ x

4y 1/ x

4y 1/ x2y 1/ x

4

y 1/ xO

1,1

Domain : R 0

Range : R

Nature: Many one into

Y

XO

3y 1/ x

3y 1/ x

3y 1/ x

y 1/ x

Domain : R 0

Range : R 0

Nature : one one into

8 .2 Ex pon en ti al Fu nc ti on

xy a , a 0, a 1

X O

Y

0 < a < 1

a > 1

y = 2x

y = 4

x

y = 10x y = 10

x –

y = 4

–x

y = 2

–x Domain : R

Range : R +

Nature : one-one into

8 .3 Lo gar it hmic F un ct ion

ay log x, a 0, a 1



0

2y log x

4y log x

10y log x

1/10y log x

1/ 4y log x

1/ 2y log x

O

Y

X

+Domain : R

Range : R Nature : one one onto

8 .4 Tr i gonomet r ic func ti on or c i rcu lar funct i on

y sin x, y cos x, y tan x

y secx, y cot x, y cosecx

Note :

The graph of y sin x is symmetric about origin i .e. symmetric in oppositequadrants .

y cos x

Clearly cos x is an even function therefore it is symmetrical about axis of y.

y tan x

x

Y

2 3 / 2 / 2

o/ 2

3 / 2

2

2n 1Domain : R

2

Range : ,

Nature : Many one onto



Principal value : / 2, / 2

y cot x

x

Y

2

3 / 2 / 2

o/ 2

3 / 2

2

Domain : R n

Range : ,

Nature : Many one onto

Principal value : 0,

y sec x

y 1

X

y 1 3π/2,0

2π,1

π, 1

π/2,0 0,1

O π, 1 π/2,0

12π,

Y

Domain : R 2n 1 / 2

Range : R 1,1 or ( ,1] [1, )


Principal value : 0, / 2

y cosec x

O

1

1

y cosec x

y sin x

2X

Y

Domain : R n

Range : R 1,1 or ( ,1] [1, )


Principal value : / 2, / 2 0

8.5 Inverse c i r cu lar funct ion or Inverse Tr igonometr i c Funct ion

1 1 1

1 1 1

y sin x, y cos x, y tan x

y cot x, y sec x, y cosec x



2

x

y

/ 2

O x 1

x 1

/ 2

1y sin x

x

y

/ 2

O x 1x 1

1y cos x

1y tan x and 1cot x

0 1 2 3 4 5 6

/ 21y tan x

1y cot x/ 2

X

Y

1y sec x

1 2 3 41234 0

/ 2

/ 2

y

x

1y cos x

1y cosec x

1y sin x

0

/ 2

/ 2

y

x11

Function Dom ai n Ran ge

sin–1x [ –1, 1]

2,

2



3

cos –1 x [–1, 1] [0, ]

t an –1x R

2,

2

cot –1 x R (0, )sec –1 x R – (–1, 1). [0, ] – { /2}

cosec–1 x R – (–1, 1) ,2 2

– {0}

9 . S OM E S PEC IA L F UN CT IO NS AN D T HEI R G RA PH S

9. Linear funct ion

f x ax b,a 0 and x R

Where a and b are constant

OX

Y

o, b

b / a, 0

Domain : R, Range : R

9. 2 M odu lu s f unc tion

f(x) = |x| =

0x,x

0x,x

Domain : R, Range : [0, )

y

y x

Ox

I t is many one functionGraph is symmetrical with respect to y-axis.

9. 3 Si gnu m Fun ct i on

f (x) =

| x |, x 0

x0, x 0

, or f(x) =

0x,00x,1

0x,1



4

0,1 1,1

0, 1

X

Y

O

Domain ; R, Range; {–1, 0, 1}It is a many one and discontinuous function

9 . 4 G re at es t I nt eg er Fu nc t io n

A function is said to be greatestinteger function i f i t is defined asf(x) = [x] where [x] = greatest integer less than or equal to x.

e.g.

4.2 4, 3.6 3, 4.4 5, 5.8 6

The graph of this function is as fol lows

1

1

2

2

3

3

4

41

12

2

3

3

4

4

0

Y

X

f x y x

0 x 1 y 0

1 x 2 y 1

2 x 3 y 2

and so on

Note : Important Ident i t ies

( i ) x – 1 < [x] x

( i i ) [x] + 1 > x



5

( i i i ) I f f (x ) = [x + n], where n I and [.] denotes the greatest integer functionthe n

f (x) = n + [x]

( iv) x = [x] + {x} where [ . ] and {.} denotes the integral and fract ional par tof x respect ive l y

(v) x 1 x x

[– x] = – [x], if x I

[– x] = –[x] – 1, if x I

[x] – [– x] = 2n, if x = n, n I ; [x ] – [–x] = 2n + 1, i f x =

n+{x}, n I , x 0

[x] n x n, n I ; [x] n x < n + 1, n I[x] > n x n + 1, n I ; [x] < n x < n, n I

9 . 5 Fr act ional Par t of X

1

O12 1 2 3 4 X

Y

f(x) = x – [x], x Ri.e., f(x) = {x}

x 1,x [ 1,0)

x, x [0,1)

x 1, x [1,2)

0, x W

Domain : R, Range : [0, 1), Nature : Many one intoThis is a many one function with period 1. It is discontinuous at every integer

9 .6 R ec ta ng ul ar Hy pe rb ol a

1

f xx

X

Y

O



6

Domain : R 0

Range : R 0

I t is a discontinuous and one one into function.

0 . T RANS FORMATION

If Graph of y f x be known then to find graph of

0. = f x a

or = f x +a

To find y f x a (Let a 2 )

y f x 2 y f x y f x 2

S h i f t e d

y f x

2 u n it s

t ow a r d s le ft

Shifted y f x

horizontally 2 unit

distance rightwise

Y

X22O

0. 2 = f x +a

or = f x a

(Let a 2 )

1

2

3

4

5

2

2

2

2

2

2

1

23

y f x 2

y f x 2

y f xX

Y

Shifted vertically up

the previous graph

y f ( x) by 2 units

Shifted vertically down

the previous graph

y f ( x) by 2 units

0. 3

x

y = f

a

or = f ax

(Let1

a 2,2

)



7

y f 2x

y f x y f x / 2

X

Y

O

stretched the graph 2 times ho rizontally

compressed the graph 2 times horizontally

See more examples about the same :

1x 12x1x 1

4x

y f x

y f x / 2 stretched double

y f 2 x

compressed

made half

0. 4 = f x

Y

XO

y f x y f x

Reflection of y f x w.r.t. axis of y is y f x

0. 5 = k f x

Rule - Strech previous graph k t imes vert ical lye .g . see below y 2 sin x , y 3 sin x



8

1

1

1

y 2sin x

y sin x

y 3sin x

X

Y

O/ 2

y f x :

X

Y

y f x

y f x

Reflection of y f x w.r.t axis of x is y f x

0. 6

y = f x

y f xX

Y

Remove the graph ly ing in I I and I I I quadrant and take the image of graphlying in I & IV quadrant w.r.t. axis of y.

The new graph including i ts image is cal led y f x .

Here we took the image of the portion of the lying in first quadrant about axisof y and left the portion which was lying in second quadrant.

0. 7 = f x



9

Y

XO

y f x

Y

XO

y f x

Rule : Take the image of the portion lying below axis of x about axis of x andkeep the remaining portion as it is above the axis of x.


Draw the graph of y sin x

So l .

2 O

Y

X

Note : You can draw y sin x just by taking mirror image of portion line in Iand IV quadrant w.r.t. axis of y.


Draw the graph of y cos x

So l .

O

Y

X5 / 2

3 / 2 / 2 / 2

3 / 25 / 2


Draw xxy e , y e .

So l .

xy e

Y

OX

1




20

Draw e e ey log x, y log x , y log x

So l . Y

X1O

ey log x

ey log x

1

ey log x

O X

Y

Y

X11 O

ey log x

. A LGE BR AIC O PE RATIONS O N FUNCTIO NS

I f f & g are real valued functions of x with domain set A and B respectively,then both f & g are defined in A B. Now we define f + g, f - g, (f . g) & (f /g) asfo l lows :

domain in each case is A B

(i) (f ± g) (x) = f(x) ± g(x)

(ii) (f.g) (x) = f(x). g(x)

(iii)

f

g

(x) =

f(x)

g(x) domain is {x,

x A B such that g(x) 0}.Note : For domain of (x) = {f(x)}g(x) , conventionally, the conditions are f(x)

> 0 and g(x) must be defined.

For domain of (x) = f(x)Cg(x) or (x) = f(x)Pg(x) conditions of domain areef(x) g(x) and f(x) N and g(x) W

I l lus trat ion 8

Find the domain of following functions :

( i ) f(x) = 5x2

( i i ) sin–1

(2x – 1)Sol .

( i ) f(x) = 5x2 is real iff x2 – 5 0 |x| 5 x – 5 or x 5

the domain of f is (– , – 5 ] [ 5 , )

( i i ) –1 2x – 1 + 1 domain is x [0, 1]

I l lus t ra t ion 9

Find the domain of 10f x log sin x



2

Sol . For existence of f x ,sin x 0 and from the graph of y sin x it is clear that sinx is

positive when 2n x 2n 1 . Hence Dom f x 2n , 2n 1 , where n I .


Find the domain of definition of 1

f xx x

Sol .

x x 0 x x and already we know x x.

this contradiction Dom f x

I l lus t ra t ion

Find the domain of xf x sin x cos x e tan x

Sol. Dom sin x R

Dom cos x R

2n 1

Dom tan x R 2

xDom e R

xDom f Dom sin x Dom cos x Dom e tan x

2n 1R R R

2

2n 1

R 2


Find the domain of fol lowing functions :

( i ) f(x) = 2sinx 16 x ( i i ) f(x) =2

3

4 x log(x3 x)

So l .

( i ) sinx is real iff sin x 0 x [2n , 2n + ], n I.

216 x is real iff 16 x2 0 4 x 4.

Thus the domain of the given function is{x : x [2n , 2n + ], n I } [ 4, 4] = [ 4, ] [0, ].

( i i ) Domain of 24 x is [ 2, 2] but 24 x = 0 for x = ± 2

x ( 2, 2)log(x3 x) is defined for x3 - x > 0 i.e. x(x 1)(x + 1) > 0.

domain of log(x3 - x) is ( 1, 0 ) (1, ).Hence the domain of the given function is {( 1, 0 ) (1, )} ( 2, 2) =

(-1, 0 ) (1, 2).

IN CHAPTER EXERCISE -

1 Which of the following relation is a function?

(a) {(1, 4), (2, 6), (1, 5), (3, 9)} (b) {(3, 3), (2, 1), (1, 2), (2, 3)}

(c) {(1, 2), (2, 2), (3, 2), (4, 2)} (d) {(3, 1), (3, 2), (3, 3), (3, 4)}



22

2 If y f x , f : R R then which of the following rules is not a function

(a) y = 9 – x2 (b) y = 2x2 (c) |x|xy (d) y = x2 + 1

3 Domain of the function f(x) =2

1

x is

(a) R (b) (-2, ¥) (c) [2, ¥] (d) [0, ¥]

4 The domain of the function2

3log

xis

(a) (3, ¥) (b) (-¥, 3) (c) (0, 3) (d) (-3, 3)

5 Domain of the function cos–1 (4x – 1) is(a) (0, 1/2) (b) [0, 1/2] (c) [1/2, 2] (d) None of these

6 Domain of the function log |x2 – 9| is(a) R (b) R – [-3, 3] (c) R – {-3, 3} (d) None of these

7 The domain of the function x x x f 61)(

(a) (1, 6) (b) [1, 6] (c) [1, ¥) (d) (-¥, 6)

8 The domain of the function )22()(2

x x x f is

(a) 33 x (b) 3131 x

(c) 22 x (d) 3232 x

9 Domain of a function f(x) = sin–1 5 x is

(a)

5

1,

5

1(b)

5

1,

5

1(c) R (d)

5

1,0

10 Domain of f(x) =1

log(2 x) + x 1 , is

(a) [ 1, 2) (b) [ 1, 1) (1, 2](c) ( 1, 1) (1, 2) (d) ( 1, 2)

11. If graph of y = f(x) is as shown, then which of the fol lowing is the graph of y = f(|x|)



23

(a) (b)

(c) (d)

12. Which of the fol lowing is graph of |y| = f(x) if graph of y = f(x) ia as shown?

(a) (b)

(c) (d) None of these

13 . If domain of f(x) is [1, 2], then the domain of 2f log x is

(a) (2, 4) (b) 2,4 (c) [1, 4] (d) [1, 2]

1 4. If domain of 2

2f x x

9

, then domain of 1f cos x is



24

(a)1

0,2

(b)1 1

,2 2

(c)1 1

,2 2

(d) 1,1

1 5 . N um ber of sol ut ion s o f 2sin x ln x is

(a) 2 (b) 4 (c) 6 (d) 0

16 . domain of the function f(x) = 1 x – sin–1

2x 1

3

is

(a) [ , ] (b) ( , ) (c) [ , ) (d) ( , ]

17 . domain of the function

2

10

5x xy log

4

is

(a) [ , 4] (b) [ , 4 ] (c) [ 4, ] (d) ( , 4)

18 . domain of the function 1

3y cosec x sin x is

(a) 1 2n 2n π < x < π,n I(b) nπ < x < n + 1 π,n I

(c) 2nπ < x < 2n +1 π,n I(d) 1 n n π < x < π,n I

9 . d omai n of t he f un ct ion

x 2 1 xy

x 2 1 x

is

(a) ( , 0) (b) R (c) ( , 0] (d)

20. Number o f i nteger s i n the domain of

11

2 xy cos n 3 x

4

is

(a) 7 (b) 8 (c) 9 (d) 0

ANSWER KEY

. c 2 . c 3 . b 4 . b 5 . b 6 . c 7 . b

8 . b 9 . b 0 . b . b 2 . c 3 . b 4 . C

5. a 16. a 17. b 18. c 19. d 20. b

2 . METHODS OF DETERMINING RANGE :

2 . Representing x in terms of y

Definition of the function is usually represented as y (i.e. f(x) which is dependent var iabl e) in te rms of an express ion of x (which is independent va riab le ). To find

range rewrite given definition so as to represent x in terms of an expression of yand thus obtain range (possible values of y). If y = f(x) x = g(y), then domainof g(y) represents possible values of y, i.e. range of f(x).

I l lus trat ion 1 3



25

Find the range of f(x) =2

2

x x 1

x x 1

Sol . f(x) =2

2

x x 1

x x 1

{x2 + x + 1 and x2 + x – 1 have no common factor}

y =

2

2

x x 1

x x 1

yx2

+ yx – y = x2

+ x + 1 (y – 1) x2

+ (y – 1) x – y – 1= 0

If y = 1, then the above equation reduces to –2 = 0. Which is not true.Further if y 1, then (y – 1) x2 + (y – 1) x – y – 1 = 0 is a quadratic and has real

roots if (y – 1)2 – 4 (y – 1) (–y – 1) 0 i.e. if y –3/5 or y 1 but y 1Thus the range is (– , –3/5] (1, )

2 .2 Graph ical Method

Values covered on y-axis by the graph of function is range

I l lus trat ion 1 4

Find the range of f(x) =2x 4

x 2

Sol . f(x) =2x 4

x 2

= x + 2; x 2

graph of f(x) would be

Thus the range of f(x ) is R – {4 }

2.3 Using Monotonocity/Maxima-Minima

(a ) Contin uou s f un ction

If y = f(x) is continuous in its domain then range of f(x) is y [min f(x), max. f(x)]( b ) S ec ti onal ly con ti nuous f unct io n

In case of sectionally continuous functions,range will be union of [min f(x), max. f(x)]over all those intervals where f(x) is continuous,



26

as shown by following example.Let graph of function y = f(x) be

Then range of above sectionally continuous function is[y2, y3] (y4, y5] (y6, y7]

Note : In case of monotonic functions minimum and maximum values lie at end

points of interval.

I l lus trat ion 1 5Find the range of following functions :

( i ) y = n (2x – x2) (i i ) y = sec–1 (x2 + 3x + 1)Sol .

( i ) Step – : Using maxima-minima, we have 2x – x2 (– , 1]Step – 2 : For log to be defined accepted values are 2x – x2 (0, 1]

{i.e. domain (0, 1]}Now, using monotonocity

n (2x – x2

) (– , 0] range is (– , 0] Ans .

(i i ) y = sec–1 (x2 + 3x + 1)Le t t = x2 + 3x + 1 for x R

then t 5

,4

but y = sec–1 (t) t 5

, 14

[1, )

from graph range is y 0,2 1

5sec ,

4

3 . EQUAL OR IDENTICAL FUNCTION :

Two func tions f & g are said to be equal if ( i ) The domain of f = the domain of g.( i i ) The range of f = the range of g and( i i i ) f(x) = g(x) , for every x belonging to their common domain.

e.g . f (x) =1

x & g(x) = 2

x

x are identical functions.


Find whether f(x) = x + cos x is one-one.Sol . The domain of f(x) is R. f ' (x) = 1 sin x.

f ' (x) 0 x complete domain and equality holds at discrete pointsonly f(x) is strictly increasing on R. Hence f(x) is one-one.


Identify whether the function f(x) = x3 + 3x

2 2x + 4 ; R R is ONTO or



27

INTOSol . As codomain range, therefore given function is ONTO


f : R R f(x) = 2x +|sin x|, find whether f(x) is bijective.So l . f i s cont inuous

note that f ' (x) > 0 x R

x ; y and x – ; y –& f(0) = 0 one–one onto bijective


f : R R f(x) =2

2

x 4x 30x 8x 18

is many one

So l . Assume f (x) = 5/3 x = 0 or 26 f(x) is many oneNote that the technique is True if f (x) is defined for x = 0 and x = 26


State f : R R f(x) =

2

2x 4x 30x 8x 18 is Onto/Into?

So l .

22

2

x 4 x 30y y 1 x 4 2y 1 x 18y 30 0

x 8x 18

2

For x R, 16 2y 1 4 y 1 18y 30

2or y 16y 13 0

Hence 8 51 y 8 51 .

As ran ge is not R, f(x) is Into.


State whether f(x) = x + sin x, f : R R, is Onto or Into & one - one or

many - one?

S o l . x 1 x s in x x 1 , hence range is al l real numbers.f(x) is Onto.Further f’(x) = 1 + cos x.

As f’(x) can not change sign and f(x) is a continuous function hen ce f(x) is aOne - one function.

4 . P ER MU TAT IO N B AS ED P RO BL EM S

f : A B

Case-I: When both the sets A and B contain an equal number of elements( i ) Total number of funct ions = 44 = 256(i i) Number of functions one-one = 4!

(ii i) Number of functions many-one = 44 – 4!

(iv) Number of onto functions 4! number of bijective mapping = 4!



28

(v) Number of into funct ions = 44 – 4! Alternate : 4C1 [34 – {3C2 + 3C1(2

4 – 2)}] + 4C2(24 – 2) + 4C3 · 1 = 44 – 4! ]

Case-II: When number of elements in A(domain) is more than B(i ) Total number of funct ions = 45 = 1024( ii ) One-one ( in jec tive) = Nil( ii i) Many-one = 1024

(5 books in 4 groups 1, 1, 1, 2)(v) Number of into functions = 1024 – 240 = 784

Case-III: Number of elements in codomain (B) is more than A( i) Total functions = 54 = 625(i i ) Number of injective mapping = 5C4 · 4! = 120(iii) Number of many-one = 625 – 120 = 505(iv) Number of onto functions = 0(v) Number of into funct ions = 625

(iv) Number of onto functions =

!3·!2·!1·!1

!5 4! = 240

IN CHAPTER EXERCISE - 2

1. domain of the function f(x) =cosx

x x

1

2

6 35 6 2

is

(a)4

1 π π, , 6

6 3 3(b)

2

1 π π, , 6

6 3 3

(c)

1 π 5π, , 6

6 3 3

(d)

2

1 π 5π, , 6

6 3 3

2. Range of the function f(x) = cos–1[x] , where [.] is G.I.F.

(a) 0, ,2

(b)

0,2

(c)

,0,2 2

(d) 0,

3. Range of the function f(x) = 4 tan x cos x(a) [ 4, 4] (b) ( 4, 4) (c) [ 4, 4) (d) ( , )

4. Range of the function f(x) = cos4 x

5 – sin4 x

5(a) [ , ] (b) ( , ) (c) [ , ) (d) ( , ]

5. Range of the function f(x) = sin x

(a) [ , ] (b) ( , ) (c) [ , ) (d) ( , ]

6. Range of the function f(x) = cos(2sin x)(a) [cos 2, ] (b) [ cos 2, cos 2]

(c) [ , ] (d) [ cos2, ]



29

7. Range of the function f(x) = sin(log2x)(a) [ , ] (b) ( , ) (c) [ , ) (d) ( , ]

8. Range of the function f(x) = 3 – 2x

(a) ( , ] (b) ( , 3) (c) [ , 3] (d) ( , 3]

9. Range of the function f(x) =x

|x|x3

(a) R (b) ( , 4] (c) {2, 4} (d) ( , 2]

10 . Range of the function f(x) = cot2 x4

(a) [1, ) (b) (1, ) (c) [0, ) (d) [0, 1]

11 . Range of the function f(x) = 1 2 1 21sin x tan x 1

2

, is

(a) ,2 2

(b)

,8 2

(c) 0, (d)

,2 2

12 . Which of the following pair of functions are indentical

(a) f(x) = l n x2 & g(x) = 2l n x (b) f (x) = cosecx & g (x) =1

sinx

( c ) f (x) = tan x & g (x) =1

cot x(d) f(x) = l n ex & g(x) = el n x

13 . Which of the following pair of functions are indentical

(a) f(x) = cot2x.cos2x & g(x) = cot2x – cos2x(b) f (x ) = tanx · cotx & g(x) = s inx · cosecx

(d) f(x) =1 2

2

cos x & g(x) = sin x

(e) f(x) = x2 & g(x) = ( x )2

14. Let A = {1, 2, 3, 4}, then number of one - one funct ions f : A A , such

that f i i , is

(a) 81 (b) 256 (c) 24 (d) 9

1 5 . The numb er of O nto f unct io ns , f : a, b, c,d p,q, r , is

(a) 81 (b) 36 (c) 24 (d) 33

16 . f : R R f(x) = 2|x| – 2–x is(a) many one & into (b) many one & onto(c) one one & into (d) one one & onto

17 . f: [–1, 1] [–1, 1] f(x) = sin2x is



30

(a) many one & into (b) many one & onto(c) one one & into (d) one one & onto

18 . f (x) = x3 – 2x2 + 5x + 3, is(a) many one & into (b) many one & onto(c) one one & into (d) one one & onto

1 9 . L et f: {x, y, z}

{a, b, c} be a one–one function. It is known that onlyone of these statements is true and the remaining two are false.(i ) f(x) b; ( i i) f(y)=b;( i i i ) f (z ) a, then f -1 (a) =(a) x (b) y (c) z (d) x or y

20 . If f x 3 sin x 5 cos x a, f : R 0,10 is an Onto function, then which of

the fo l lowing i s correct?(a) a = 5 (b) a < 5 (c) a > 5 (d) None of these

ANSWER KEY

1. c 2. a 3. b 4. a 5. a 6. a 7. a

8. b 9. c 10. c 11. b 12. b 13. a 14. d

15. b 16. a 17. b 18. d 19. b 20. d

5 . C OMP OSI TE F UN CTI ON

Let f: X Y1

and g: Y2 Z be two functions and the set D = {x X: f(x) Y

2

}.If D , then the function h defined on D by h(x) = g{f(x)} is called compos-i te function of g and f and is denoted by gof. I t is also cal led function of afunc t i on .

Note : Domain of gof is D which is a subset of X (the domain of f ). Range of gof is a subset of the range of g. If D = X, then f(x) Y2.

Properties of Composite Functions :

(a) In general gof fog (i .e. not commutative)

(b) The composi te of funct ions are associa tive i.e . if three funct ions f , g , hare such that fo (goh) & (fog) oh are defined, then fo (goh) = (fog) oh .(c ) I f f and g both are one-one, then gof and fog would also be one-one.(d) If f and g both are onto, then gof or fog may or may not be onto.(e) The composi te of two bi jections i s a bi ject ion i ff f & g are two bi ject ions

such that gof is defined, then gof is also a bijection only when co-domain

of f is equal to the domain of g .( f ) I f g is a func t ion such that go f i s def i ned on the domain o f f and f is



3

periodic with T, then gof is also periodic with T as one of i ts periods.Further i f # g is one-one, then T is t he per iod o f g of # g is also periodic with T' as the period and the range of f is a

sub-set of [0, T' ], then T is the period of gof

l l lustration 22

Describe fog and gof wherever is possible for the fol lowing functions( i ) f(x) = x 3 , g(x) = 1 + x2

( i i ) f(x) = x , g(x) = x2 - 1.Sol .

( i) Domain of f is [-3, ), range of f is [0, ).Domain of g is R, range of g is [1, ).Since range of f is a subset of domain of g,domain of gof is [-3, ) {equal to the domain of f }

gof (x) = g{f(x)} = g ( x 3 ) = 1 + (x+3) = x + 4. Range of gof is [1, ).Further since range of g is a subset of domain of f,

domain of fog is R {equal to the domain of g}fog (x) = f{g(x)}= f(1+ x2 ) = 2x 4 Range of fog is [2, ).

( i i ) f(x) = x , g(x) = x2 1.Domain of f is [0, ), range of f is [0, ).Domain of g is R, range of g is [ 1, ).Since range of f is a subset of the domain of g,

domain of gof is [0, ) and g{f(x)}= g( x ) = x 1. Range of gof is [-1, )Further since range of g is not a subset of the domain of f i.e. [ 1, ) [0, ) fog is not defined on whole of the domain of g.

Domain of fog is {xR, the domain of g : g(x) [0, ), the domain of f}.Thus the domain of fog is D = {xR: 0 g(x) }i.e. D = { xR: 0 x2 1}= { xR: x 1 or x 1 } = ( , 1] [1, )

fog (x) = f{g(x)} = f(x2 1) = 2x 1 Its range is [0, ).

l l lustration 23

Let f(x) = ex ; R+ R and g(x) = sin–1 x; [ 1, 1] ,2 2

. Find domain

and range of fog (x).

Sol . Domain of f(x) : (0, ) Range of g(x) :,

2 2

values in range of g(x) which are accepted by f(x) are 0,2

0 < g(x) 2

0 < sin–1x 2

0 < x 1

Hence domain of fog(x) is x (0, 1]



32

Therefore Domain : (0, 1]Range : (1, ep/2]

Example of composite function of non-uniformly defined functions :

l l lustration 24

I f f ( x) = | |x – 3| – 2 | 0 x 4g(x) = 4 – |2 – x| – 1 x 3

then find fog(x) and draw rough sketch of fog(x).S o l f(x) = | | x 3| 2| 0 x 4

=|x 1| 0 x 3|x 5| 3 x 4

=

1 x 0 x 1

x 1 1 x 35 x 3 x 4

g(x) = 4 |2 x| 1 x 3

=4 (2 x) 1 x 2

4 (x 2) 2 x 3

=

2 x 1 x 2

6 x 2 x 3

fog (x) =

1 g(x) 0 g(x) 1

g(x) 1 1 g(x) 35 g(x) 3 g(x) 4

1 (2 x) 0 2 x 1 and 1 x 22 x 1 1 2 x 3 and 1 x 2

5 (2 x) 3 2 x 4 and 1 x 2

1 6 x 0 6 x 1 and 2 x 36 x 1 1 6 x 3 and 2 x 35 6 x 3 6 x 4 and 2 x 3

1 x 2 x 1 and 1 x 21 x 1 x 1 and 1 x x3 x 1 x 2 and 1 x 2

x 5 6 x 5 and 2 x 35 x 5 x 3 and 2 x 3x 1 3 x 2 and 2 x 3



33

1 x 2 x 1 and 1 x 21 x 1 x 1 and 1 x 2

3 x 1 x 2 and 1 x 2x 5 5 x 6 and 2 x 3

5 x 3 x 5 and 2 x 3

x 1 2 x 3 and 2 x 3

=

1 x 1 x 13 x 1 x 2x 1 2 x 3

Alternate method for f inding fog

g(x) =

2 x 1 x 2

6 x 2 x 3

graph of g(x) is

fog(x) =

1 g(x) 0 g(x) 1g(x) 1 1 g(x) 35 g(x) 3 g(x) 4

=

1 g(x) for no valueg(x) 1 1 x 15 g(x) 1 x 3

=

2 x 1 1 x 15 (2 x) 1 x 25 (6 x) 2 x 3

=

x 1 1 x 13 x 1 x 2x 1 2 x 3

l l lustration 25

If f (x) = 10100x

2log x 2log

x

and g(x) = {x}. If the function (fog)(x) exists then

find the range of g(x).

Sol . (i) 100 x > 0 & 100 x 1 x 100

1

(ii) x > 0 and log10x + 1 < 0 0 < x < 110

& x 1100

(fog)(x) exists range of g(x) domain of f (x)

l l lustration 26

Let f 1(x) = x, f 2(x) = 1 – x; f 3(x) =x

1, f 4(x) =

x1

1

; f 5(x) =

1x

x

; f 6(x) =

x

1x



34

Suppose that )x(m6 f f = f 4(x) and )x(4n f f = f 3(x) then find the value of m &n .

Sol . Given f 6(x) =

x 1x

....(1) 6 mf f (x) =m

m

f (x) 1f (x)

....(2)

b u t 6 mf f (x) = f 4(x) =1

1 x (given)

6 mf f (x) =m

m

f (x) 1f (x)

=

x1

1

p u t f m(x) = k,k 1

k

=1

1 x

k – kx – 1 + x = k k =x 1

x

f m(x) =x 1

x

= f 6(x) m = 6

aga in n 4f f (x) = f 3(x) =1x

f n1

1 x

=x

1; l e t

11 x

= t

t – tx = 1 x =t 1

t

f n(t) =t

t 1 f n(x) =x

x 1 = f 5(x)

h e n c e n = 5 Ans.

l l lustration 27

If f (x) =3x

7x2

find a function g such that g [f (x)] = x for all x in the domain

of f and find its domain and range.

Sol . f : R – {– 3} R – {+2}

let2x 7x 3

= t x =3t 72 t

;

Hence g (t) =3t 72 t

; g (x) =3x 72 x

]

l l lustration 28

Let f : R R and g : R R be two functions given byf (x ) = x 3 and g (x )= 3x + 2. Find gof and fog .

Sol . ( gof ) (x ) = g(f (x )) = g (x 3) = 3x 3 + 2

(fog ) x = f ( g (x )) = f (3x + 2) = (3x + 2)3

gof and fog are both defined but are different from each other.

l l lustration 29



35

Let f (x ) =||1 x

x

, x R, g (x )= ||1 x

x

for –1 < x < 1, Find gof and

fog .Sol . ( gof ) x = g (f (x ))

= g

||1 x

x =

||11

||1

x

x

x

x

= ||||1 x x

x

(Observe that 1+|x | > 0 |1+ |x || = 1+|x |)

=x

(fog ) x = f ( g (x )) , – 1 < x < 1

=f

||1 x

x =

||11

||1

x

x

x

x

=

||||1 x x

x

(Observe that for – 1< x < 1, |x |<1 1 – |x | > 0 |1–|x || = 1–|x |) (fog )x = x

6. PERIODIC F UNCTIONS

A function f(x) is called per iod ic with a per iod T if there exists a rea l numberT > 0 such that for each x in the domain of f the numbers x – T and x + T arealso in the domain of f and f(x) = f(x + T) for a l l x in the domain of ' f ' .Domain o f a per iod ic funct ion i s a lways unbounded. Graph o f a per iod ic

function with period T is repeated after every interval of 'T'.e.g. The function sin x & cos x both are periodic over 2 & tan x is periodicover .The least posi t ive period is cal led the principal or fundamental period of f orsimply the period of f .

Note : f (T) = f (0) = f ( T), where ‘ T ’ is the period.

Inverse of a periodic function does not exist.

Every constant function is always periodic, with no fundamental period.Properties of Periodic Function

( a) I f f (x) has a period T, then

1

f(x) and f(x) also have a period T..

(b ) I f f (x ) has a per iod T then f (ax + b) has a periodT

|a| .

( c ) If f (x) has a period T1 & g (x) also has a period T2 then period of f(x) ± g(x) orf(x) . g(x)

orf(x)g(x) is L.C.M. of T1 & T2 provided their L.C.M. exists. However that L.C.M.



36

(if exists) need not to be fundamental period. If L.C.M. does not exists f(x) ±

g(x) or f(x) . g(x) orf(x)g(x) is aperiodic. (not periodic)

e.g. |sinx| has the period , | cosx | also has the period |sinx| + |cosx| also has a period .

But the fundamental period of |sinx| + |cosx| is 2

.

Note : Is 2cos x is periodic function ? (check your self)


f(x) = sinx2

+ cosx3

Sol . Period of sinx2

is 4 while period of cosx3

is 6 .

Hence period of sin x2

+ cos x3

is 12 {L.C.M. of 4 & 6 is 12}


f(x) = {x} + sin xSol . Period of sin x = 2

Period of {x} = 1but L.C.M. of 2 & 1 is not possible it is aperiodic


f(x) = cos x . cos 3xSo l . f(x) = cos x . cos 3x

period of f(x) is L.C.M. of2

2 ,3

=2 , but 2 may or may not be the funda-

mental period but fundamental period =2n

, where n N. Hence cross-check-

ing forn = 1, 2, 3,...we find to be fundamental period f( + x)=(– cos x) (– cos 3x)= f(x)


f(x) = sin3x2

– cosx3

– tan2x3

So l . Period of f(x) is L.C.M. of2

3 / 2

,2

1/ 3

, 3 / 2



37

= L.C.M. of43

, 6 ,23

= 12

NOTE :

L.C .M .( , , )

L .C .M . o f , ,

b q m H.C .F .( b , q , m )


If f (x) = (a + 3)x + 5a, x R is periodic. [Ans. a = – 3]So l . f (x) = mx + c is periodic only if m = 0]


Find the period of f x x x m ,m I , where [] denotes the greatest integer

func t i on .

So l . Given f x x x m x m x m m x m m

x m i s per iodic

m x m wil l also be periodi c and period wi ll be 1.

Period of x Period of x a 1

l l lustration 35

Draw the graph of y sin x and also find the period if possible.

So l .

sinx 0, 0 x2

1, x2

0, x2

31, x , ...............

2

Clearly f(x) is periodic and its period is 2 .

l l lustration 36

xf x sin is i t periodic? If periodic f ind i ts period.

So l .

x x0 0

xx 2 1

1

O

1 π/2

25π/2

X

Y



38

x2 x 3 2

x3 x 4 3

, Clearly f (x) is not periodic.


1 . If g(x) = 2x + 1 and fog(x) = 4x2 + 4x + 7, then range of f(x) is

(a) R (b) (6, ) (c) (1, ) (d) 6,

2 . f(x) =

1 x if 0 x 1x 2 if 1 x 24 x if 2 x 4

, then f(f(x)), for 3 x 4 , is

(a) 6 x (b) x 3 (c) 2 x (d) x

3 . f (x) =2

1 x x 0if

x x 0if

and g (x) =

x x 1if

1 x x 1if

, then the range of gof(x) is

(a) ,0 (b) ,0 (c) ,1 (d) ,1

4 . If f(x) 1 x 2 , 0 x 4 & g(x) 2 x , 1 x 3 , then fog(x), in

appropr iate domain, i s(a) 1 |x| (b) |x| (c) |x| 1 (d) |x| + 1

5 . f : R R be the function defined by f (x) = ax2 2 for some positive a.

If (fof) ( 2 ) = 2 then the value of 'a' is

(a) 2 (b) 2 (c)12

(d)12

6. Let P and Q be polynomials such that P(x) and )x(QPQ have the same numberof roots. If the degree of P is 7, then the degree of Q, is(a) 0 (b) 1 (c) 2 (d) 7

7 . Fundamental period of the function f(x) = sin x + | sin x |

(a) 2 (b ) ( c ) 3 (d) 2

8 . Fundamental period of the function f(x) = 3 cos x – sin3

x

(a) 2 (b ) 6 ( c ) 3 (d)

9 . Fundamental period of the function sin5

x2 – cos

7

x3



39

(a) 35 (b) 140 (c) 70 (d) None of these

10 . Fundamental period of the function f(x) = sin2x + cos4x

(a)4

(b ) ( c ) 4 (d)2

11 . Fundamental period of the function f(x) = cos 23x – sin 4

5x

(a) 35 (b) 15 (c) 30 (d) None of these

12 . Fundamental period of the function f(x) = cos(sin x)

(a)4

(b ) ( c ) 2 (d)2

13. Period of the funct ion, f (x) = [x] + [2x] + [3x] + ....... + [nx] –n(n 1)

2

x

where n Î N an d [ ] den otes the greatest integ er funct io n, is

(A) 1 (B) n (C) 1n

(D) non periodic

14. Consider those funct ions f that satisfy f (x + 4) + f (x – 4) = f (x) for all realx . Any such function is periodic, and there is a least common posi t ive period

p for all of them. The value of p , is(A) 8 (B) 12 (C) 16 (D) 24

15 . f : R R i s an odd funct ion such that f x 1 f x 2 f x 1 f x , then

f(2) is(a) 0 (b) 1 (c) 2 (d) can’t be determined

ANSWER KEY

. d 2. b 3. a 4. c 5. d 6. b 7 . a

8 . b 9 . c 0 . b . b 2 . b 3 . a 4 . d

5. a

7 . O DD & EVEN FUNCTIONS

( i ) I f f ( -x) = f (x) for a l l x in the domain of ‘ f ’ then f is said to be an even

funct ion.If f (x) - f (-x) = 0 f (x) is even.e.g. f (x) = cos x; g (x) = x² + 3.

(i i ) I f f (-x) = -f (x) for al l x in the domain of ‘ f’ then f is said to be an oddfunct ion.

If f (x) + f (-x) = 0 f (x) is odd.e.g. f (x) = sin x; g (x) = x3 + x.

Note : A function may neither be odd nor even. (e.g. f(x) = ex , cos–1x)



40

If an odd function is defined at x = 0, then f(0) = 0

Properties of Even/Odd Function

(a) Every even function i s symmetr ic about the y-axis & every odd functionis symmetr ic about the or ig in.For example graph of y = x 2 is symmetric about y-axis, while graph of y= x3 is symmetric about origin

(b ) A l l f unct i on s (who se d oma in i s s y mme tr i ca l abo u t or i gi n ) can b eexpressed as the sum of an even & an odd function, as follows

f(x) =

(c ) The o n ly f unc t io n wh i ch i s de f ined o n t he en t ir e numbe r li ne and i seven & odd at the same time is f(x) = 0.

(d ) I f f and g both a re even or both a re odd then the func ti on f . g wi l l beeven but if any one of them is odd then f.g will be odd.

f (x) g (x) f (x) + g (x) f (x) - g (x) f (x) . g (x) f(x) / g(x) ( gof ) (x) (f o g) (x)

odd odd odd odd even even odd odd

even even even even even even even evenodd even neither odd nor even neither odd nor even odd odd even even

even odd neither odd nor even neither odd nor even odd odd even even

( e) I f f( x) is even t hen f ' (x) is odd.

Even extension / Odd extension

Let the definition of the function f(x) is given only for x 0. Even extensionof this funct ion impl ies to def ine the funct ion for x < 0 assuming i t to beeven. In order to get even extension replace x by –x in the given defini t ionSimilar ly, odd extension implies to define the function for x < 0 assuming i t

to be odd. In order to ge t odd ex tens ion, mul t ip ly the de f in i t ion o f evenextension by –1

l l lustration 37

Show that log 2x x 1 is an odd function.



4

So l . Let f(x) = log 2x x 1 . Then f(–x) = log 2x ( x) 1

= log 2 2

2

x 1 x x 1 x

x 1 x

= log

2

1

x 1 x – log 2x x 1 = –f(x)

Hence f(x) is an odd function.

l l lustration 38

Show that ax +a–x is an even function.So l . Let f(x) = ax + a–x

Then f(–x) = a–x + a–(–x) = a–x +ax = f(x). Hence f(x) is an even function

l l lustration 39

Show that cos–1 x is neither odd nor even.So l . Let f(x) = cos–1x. Then f(–x) = cos–1 (–x) = – cos–1 x which is neither equal to

f(x) norequal to f(–x). Hence cos–1 x is neither odd nor even

l l lustration 40

What is even and odd extension of f(x) = x3 – 6x2 + 5x – 11 ; x > 0So l . Even extension f(x) = –x3 – 6x2 – 5x – 11 ; x < 0

Odd extension f(x) = x3 + 6x2 + 5x + 11 ; x < 0

8 . INV ERSE OF A FUNCTION

Let f : A B be a one-one & onto function , then their exists a uniquefunct iong : B A such that f(x) = y g(y) = x, x A & y B. Then g is said tobe inverse of f. Thus g = f -1 : B A = {(f(x), x) ½ (x, f(x)) f}.

Note: To find the inverseSt ep -1: write y = f (x), f : A B , and interchange x & yStep-2: solve th is equation for y in terms of x (i f poss ib le)Step-3: Express y = f –1 (x), 1f : B A .

Properties Of Inverse Function

(i ) The inverse of a bi jection is unique.Proof :

Let f : A B be a bijection. If possible let g : B A and h : B A be two

inverse function of f . Also let a1, a2 A and b B such that g (b) = a1 and h(b) = a2 theng(b) = a1 f (a1) = bh(b) = a2 f (a2) = b.But since f is one-one, so f (a1) = f (a2) a1 = a2 g(b) = h(b), b B

( i i ) If f : A B is a bijection & g : B A is the inverse of f, then fog = IB andgof = I A, where I A & IB are identity functions on the sets A & B respectively.



42

Note that the graphs of f & g are the mirror images of each other in the liney = x.

As shown in th e figure gi ven be low a point (x ', y' ) cor re sponding to y = x2(x

> 0) changes to (y', x') corresponding to xy , the changed form of x = y .

( i i i ) The inverse of a bi jection is also a bi jection.

Proof :

Let f : A B be a bijection and g : B A be its inverse.We have to show that g is one-one and onto.One-one : Let g(b1) = a1 and g(b2) = a2 : a1, a2 A , b1, b2 B

Then g(b1) = g(b2) a1 = a2 f(a1) = f(a2) [ f is a bijection] b1 = b2 [ g(b1) = a1 b1 = f(a1)

g(b2) = a2 b2 = f(a2)] which proves that g is one- oneOnto : Again, if a A, then

a A b B s.t. f (a) = b (by defintion of f) b B s.t. g(b) = a [ f(a) = b a = g(b)]

which proves that g is onto.Hence g is also a bijection.

(iv) If f & g are two bijections f : A B, g : B C then the inverse of gof existsand(gof)–1 = f –1 og –1 .Proof :

Since f : A B and g : B C are two bijections, gof : A C is also a bijection.

[by theorem the compos i te o f two b i jec t ion i s ab i j e c t i on]

As su ch gof has an in ver se function (gof)–1 : C A. We have to show that(gof)–1 = f –1 og –1 .

Now let a A, b B, c C such thatf (a) = b and g(b) = c

s o (gof) (a) = g[f( a)] = g(b) = c,n ow f(a) = b a = f –1(b) . . . . ( i)

g(b) = c b = g–1(c) . . . . ( i i )

(gof)(a) = c a = (gof)–1

(c) . . . . ( i i i ) A l s o (f –1og–1) (c ) = f –1 [g–1 (c)] [by defintion]= f –1(b) [by ( ii )]= a [by (i)]= (gof)–1 (c) [by (i i i)]

(gof)–1 = f –1 og–1 , which proves the theorem.



43


Determine whether f(x) =2x 3

4

; R R, is invertible or not? If so find it.

So l . As given function is one-one and onto, there fore i t is inver tible.

y =2x 3

4

x =4y 3

2

f –1

(x) =4x 3

2

.


Let f(x) = x2 + 2x; x –1. Draw graph of f –1(x) also find the number of so lut ions o f the equation, f(x) = f –1(x)

Sol. f(x) = f –1(x) is equilavent to solving

y = f(x) and y = x

x2 + 2x = x x(x + 1) = 0 x = 0, –1Hence two solution for f(x) = f –1(x )


If y = f(x) = x2 – 3x + 1, x 2. Find the value of g(1) where g is inverse of f So l . y = 1 x2 – 3x + 1 = 1 x (x – 3) = 0 x = 0, 3

B u t x 2 x = 3Now g(f(x) ) = xDifferentiating both sides w.r.t . x

g ' (f(x)). f ' (x) = 1 g ' (f(x)) =1

f (x)

g ' (f(3)) =1

f (3) g ' (1)=

1

6 3(As f ' (x) = 2x – 3) =

1

3

l t e r n a t e M e t h o d

y = x2 – 3x + 1x2 – 3x +1 – y = 0

x =3 9 4(1 y)

2

=

3 5 4y

2

x 2

x =3 5 4y

2

g(x) = 3 5 4y2

g ' (x) = 0 +1

x 5 4x x

g ' (1) =1

5 4 =

1

9 =

1

3



44


1. Consider the fo llowing functions in thei r defau lt domains

(i)

x

x

2 12 1

(ii)2x 1

x s i n x(iii)

1 xln

1 x(iv) |x| c os xx e

Which of these is/are odd(a) (i) & (iii) (b) (i) & (iv) (c) all four (d) (i), (iii) & (iv)

2. f(x) is an even function and g(x) is an odd function, then which of the following,if defined appropriately, must be an even function(a) f(x) + g(x) (b) f(x) g(x) (c) f(x).g(x) (d) f(g(x))

3. Odd extension of 2f x x s in x ln x, f : 0, R , is

(a) 2x s in x ln x , x , 0 (b) 2x s in x ln x , x ,0

(c) 2

x s in x ln x , x , 0 (d) Can’t be defined4. Even extension of 2f x x s in x ln x, f : 0, R , is

(a) 2x s in x ln x , x , 0 (b) 2x s in x ln x , x ,0

(c) 2x s in x ln x , x , 0 (d) Can’t be defined

5. f(x) is a real valued even function, then the number of real values of x

necessaryly satisfying

x 1f x f

x 2, is

(a) 0 (b) 2 (c) 4 (d) infinite

6. A real valued function satisf ies f(x)f(y) = f(x + y) and f x 0 .

Let

f x 1g x

f x 1 , then g(x) is

(a) Odd (b) Even

(c) nether even nor odd (d) can’t be determined

7. f(x) is an even function and equation of tangent to the graph of y = f(x) at (2, 0)is y = 2x 4, then equation of tangent at ( 2, 0) is(a) y = 2x + 4 (b) 2x + y = 4 (c) 2x + y = 4 (d) can’ t be determined

8. Inverse of the function f : (– , –1) (– , –2), 2f(x) x 2x 3 , is

(a) – + 2

(b) – – 2

(c) – 2

(d) doesn’t exist



45

9. Inverse of the function f :

6

7,

6 [–1, 1], f(x) = sin

3x is

(a) sin

–

x (b) 1s in x

3(c)

2π

3

– sin

–

x (d) 12 s in x

6

10 . Inverse of the function

f : R R

+

, f(x) = 10

x + 1

is(a) log

10 x - 1 (b) log

10 x + 1 (c) log

10 (x - 1) (d) log

10 (x + 1)

11 . Inverse of the function f : (– 2, ) R, f(x) = 1 + ln (x + 2) is

(a) ex 1 2 (b) ex 1 2 (c) ex 1 2 (d) ex 1 2

12 . Inverse of the function f : R (0, 1), f(x) =2

1 2

x

x is

(a)

2

xlog

1 x

(b)

2

xlog

x 1

(c)

2

1 xlog

x

(d)

2

x 1log

x

13 . For the functions f : R R, f(x) = ex and g : R R, g(x) = 3x – 2, domain of (gof)–1 (x) is(a) (2, ) (b) ( – , ) (c) (–2, ) (d) [–2, )

14 . If f : R Rf (x) = x3 + (a + 2)x2 + 3ax + 5 is an invertible mapping find‘a’.

(a) , 4 (b) [ ,4]

(c) ( ,4]

(d) )[ ,4

15 . A function f :3

2

7

4

, ,

is defined as f(x) = x2 – 3x + 4. Number of

solutions of the equation f(x) = f –1(x) is(a) 1 (b) 2 (c) 3 (d) 4

ANSWER KEY

1. d 2. d 3. a 4. c 5. c 6. a 7. c

8. b 9. c 10. a 11. d 12. a 13. c 14. b

15. a

9 . IMPORTANT TYPES OF FUNCTIONS

9. Polynomial Function

If a function f is defined by f (x) = a0 xn + a1 xn-1 + a2 xn-2 + ... + an-1 x + an where n is a no n negat ive int eger an d a0, a1, a2, ..., an are real numbers and

a0 0, then f is called a polynomial function of degree n .NOTE: (a) A pol ynomial of deg ree one with no constant term is cal led an odd

linear function . i.e. f(x) = ax , a 0(b) There are two polynomial functions , satisfying the relation ;



46

f(x).f(1/x) = f(x) + f(1/x) . They are :(i ) f(x) = xn + 1 & (ii) f(x) = 1 - xn , where n is a positive integer.

Proof : Consider f (x) = a0 + a1x + a2x2 + ..... + anxn. Equate f (x). f(1/x) = f (x) + f (1/x) & equate coefficients.

9 . 2 A lg eb ra ic Funct io n

A funct ion f is cal led an algebraic funct ion i f i t can be constructed us ingalgebraic operat ions such as addi t ion, substract ion, mul t ipl icat ion, d iv is ionand taking roots , s tar ted wi th polynomials .

e . g . f (x) = 1x2 ; g (x) =xx

x16x 24

+ (x – 2) × 3 1x

9.3 Fract iona l Rat ional Function

A ra tional func tion is a func tion of the form. y = f (x) =g(x)h(x) , where g (x) &

h (x) are polynomials & h (x) 0. The domain of f (x) is set of real x such thath (x) 0.

e.g . f (x) =4 2

22x x 1x 4

; D = {x | x ± 2}

9 .4 Homogeneous Func t ions

A funct ion is said to be homogeneous wi th respect to any set of var iables when each of its ter ms is of the same deg ree wi th re spec t to those variabl es .

For example 5x2 + 3y2 – xy is homogeneous in x & y. Symbolically if,f (tx, ty) = tn . f (x , y) then f (x , y) is homogeneous function of degree n.


f (x, y) =x y x

y x x

cos

sin is not a homogeneous function and f(x, y) =

x

yn

y

x

y

xn

x

yl l ;

x y2 2 + x; x + y cos

y

x are homogeneous functions of degree one.

9 . 5 B o und ed Func ti o n

A function is sa id to be bounde d if |f(x) | M , where M is a finite quantity..


Which of the following function(s) is(are) bounded on the intervals as indicated

(A) f(x) = 2

1

1x on (0, 1) (B) g(x) = x cos1

x on (– , )

(C) h(x) = xe–x on (0, ) (D) l (x) = arc tan2x on (– , )

Sol. (A)x 0

Limit

f (x) =1

h 1

h 0

1Limit 2

2

;

x 1Limit

f (x) =

1h

h 0Limit 2 0

1

f (x) 0,2

bounded



47

(C)h 0

Limit

x e–x =h 0

Limit

h e –h = 0 ;x

Limit

x e–x =x

Limit x

x0

e

Also x

xy

e

x x

2x

e xey '

e

e x (1 – x) 1

h(x) 0,e

]

9 .6 Impli c it & Exp l ic i t Func ti on

A fu ncti on def ined by an eq uat io n not so lv ed for the depen dent variable is

called an I

MPLICIT

F

UNCTION. For eg. the equation x3

+ y3

= 1 defines y as animplicit function. If y has been expressed in terms of x alone then it is calledan E

XPLICIT

F

UNCTION

.


( 1) Implicit x y1 + y x1 = 0; explicit y =x

1 x

or y = x (rejected)

(2) y2 = x represents two separate branches. Implicit form

y x

y x

(3) x3 + y3 – 3xy = 0

folium of descartes , Implicit

(4) x = 2y – y2 , Implicit

9 . 7 S OME GE NE RA L FUNCTIO NS

If x, y are independent variables, then:

( i) f (xy) = f (x) + f (y) f (x) = k ln x or f (x) = 0.

( i i) f (xy) = f (x). f (y) f (x) = xn, n R

( i i i ) f (x + y) = f (x). f (y) f (x) = akx.

( i v) f (x + y) = f (x) + f (y) f(x) = kx, where k is a constant.

(v) f(x) . f1x

= f(x) + f1x

f(x) = 1 ± xn where n N



48


If f(x) is a polynomial function satisfying f(x) . f1x

= f(x) + f1x

x R –

{0} and f(2) = 9, then find f (3)So l . f(x) = 1 ± xn

As f(2) = 9 f(x) = 1 + x3

Hence f(3) = 1 + 33 = 28


Let f be a real valued function of real and positive argument such that

f (x) + 3x f

x

1 = 2(x + 1) for all real x > 0. Find the value of f (10099).

S o l replace x 1/x and solve to get f (x) =2

1x

f (10099) =10099 1

2

=10100

2 = 5050.


Let 1

f x f x2

. If f (2) = 5 and f

4

9 = 2. Find f (–3) and f

4

1.

So l . f

4

9 = 2 f

4

12 = 2 f

4

1 = 2

ag ai n f (–3) = f (4 – 3) = f (1) = f (2) = 5 Ans. ]


Let f : R R be a function such that

x3

f(x 1) 5 , x (0, ) then find

the value of

273 y

3

27 yf

y

for y (0, ) .

So l .

27y

3

27f 1

y

=

3 33 y3

f 1y

=

33

3 y3f 1

y

;

let

3

y = x, then

3x3

f x 1

= 53


1. I f f (x)f (y) = f(x + y) for al l real x , y & f(1) =12

, then

r 0

f r =

(a) 2 (b) 4 (c) 1 (d)

2 . f(x) + f(y) = f(x + y) + 1 for all real x, y & f(2) = 5, then f(100) =



(a) 104 + 1 (b) 299 (c) 201 (d) 1/10

3. Number of so lut ions of 5{x} = [x], where {x} denotes fract ional part of x and[x] denotes greates t integer funct ion, i s(a) 6 (b) 5 (c) 4 (d) None of these

4 . Rang e of f( x) = 3[s in x] + 4[ cos x] is

(a) { 7, 4, 3, 0, 3, 4} (b) {7, 4, 3, 0, 3, 4, 7}(c) {7, 4, 3, 1, 0, 1, 3, 4} ( d) Non e of th es e

5 . W hi ch o f t he f ol lo wi ng r ep re se nt s a homog en io us e xp re ss io n?

(a) x2 + y2 + 1 (b) 2 2x x y (c)

2

2

x 1y 1 (d) None of these

6. Which of the fo l lowing implic it relat ions can represent y explic it ly as a functionof x?

(i) x2 + y2 = 1 (ii ) ln x + ln y = 1 ( ii i)

2

2x 1y 1 = 2 (iv)

x y

x ye 3 2e 1

(a) (i) & (iii) (b) (i) & (iv) (c) all four (d) (ii) & (iv)

7 . If f(xy) = f(x)f(y), f x 0 and f(2) = 4, then f(10) is

(a) 20 (b) 28 (c) 40 (d) 100

8. I f 2f(cos x) + 3f(sin x) = x, then f (x) may be

(a) 1cos x

5(b)

1s in x5

(c) 15 s in x (d) 15 cos x

9. I f f(x) i s a polynomial function sat is fy ing f (x) f 1x

= f(x) + f 1x

x R – {0}

and f(3) = 10, then f(4) =(a) 13 (b) 14 (c) 17 (d) None of these

1 0. If 2 f (x) – 3 f 1x

= x2, (x 0), then f (2) =

(a) –4

7(b) –

5

4(c) –

4

7(d) –

4

5

ANSWER KEY

1. a 2. c 3. b 4. a 5. d 6. d 7. d

function theory

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