RELATIONS & FUNCTIONS · Relations & Functions [2] RELATIONS a R b means 'a is R-related to b' i.e. a is related to b under relation R. If (a, b) R; (a, b) is called ordered pair

Relations & Functions [1]

CONTENTS

RELATIONS & FUNCTIONS

KEY CONCEPT ................................................................Page 02 - 32

EXERCISE–I ....................................................................Page 33 - 39

EXERCISE–II ...................................................................Page 39 - 44

EXERCISE–III .................................................................Page 44 - 49

EXERCISE–IV ..................................................................Page 49 - 54

ANSWER KEY ..................................................................Page 55 - 56


RELATIONSa R b means 'a is R-related to b' i.e. a is related to b under relation R. If (a, b) R; (a, b) is calledordered pair in the sense that a and b can't be interchanged as a A and b B.

Ordered Pair :It is a pair of objects written in a particular order. Two members are written in a particular order separatedby a comma and enclosed in parentheses. Hence in ordered pair (a, b) a is called the firstcomponent or the first element or the first co-ordinate and b the second.Ordered pairs (a, b) and (b, a) are different.(a, b) = (c, d) iff a = c and b = di.e. (1, 3) = (1, 3); (1, 3) (1, 2) (2, 3) (3, 1)

CARTESIAN PRODUCT

Cartesian product of two sets A × B : For any two non empty sets A and BA × B = {(a, b) : a A and b B}It is a set of all ordered pairs such that in each ordered pair first element belongs to set A and secondelement belongs to set B.A × B is read as 'A cross B' or 'Product set of A and B'A × B = {(a, b) : a A b B}Thus (a, b) A × B a A and b B.B × A = {(b, a) : b B a A}A × B B × A (not commutative)n(A × B) = n(A) n(B) and n(P(A × B)) = 2n(A) n(B)

A = and B = A × B = Cartesian product of n non empty sets A1, A2, ......An is a set of all n tuples (a1, a2,........an) such thateach aiAi, i = 1, 2.........n.

A1 × A2 ×.......× An =

n

ii

1

A

A × A = A2 : R ×R = R2 is a set of all points lying in the planeR × R × R = R3 represents set of all points in 3-D space.If at least one of A and B is infinite set then A × B is also infinite set,provided that other is non-empty set.

Ex. Let A = {a, b}, B = {c, d}, C = {e, f}then n(A × B × C) = n(A). n(B). n(C) = 8A × B × C = {(a, c, e), (a, c, f), (a, d, e), (a, d, f), (b, c, e), (b, c, f), (b, d, e), (b, d, f)}

Each element of A × B shall be an ordered pair or 2–tupleEach element of A × B × C shall be an ordered triplet or 3–tupleEach element of A1 × A2 × .....An shall be n–tuple

RELATIONS & FUNCTIONS


Ex. A1 × A2 × A3× A4 = {(1, 1, 1, 1), (2, 4, 8, 16), (3, 9, 27, 81), ........}. Find A1, A2, A3 and A4.Sol. Each ordered pair {x1, x2, x3, x4} is of the form {x, x2, x3, x4}

Hence x1 A1 A1 = {x x N} = {1, 2, 3, 4, .....}x2 A2 A2 = {x2 x N} = {12, 22, 32, 42, .....}x3 A3 A3 = {x3 x N} = {13, 23, 33, 43, .....}x4 A4 A4 = {x4 x N} = {14, 24, 34, 44, .....}

Key Results on Cartesian Product :If A, B, C, D are four sets, then.1. A × (B C) = (A × B) (A × C)2. A × (B C) = (A × B) (A × C)3. A × (B – C) = (A × B) –(A × C)4. (A × B) (C × D) = (A C) ×(B D)5. If A B, then (A × C) (B × C)6. If A B, then (A × B) (B × A) = A2

7. If A B, then A × A (A × B) (B × A)8. If A B and C D, then (A × C) (B × D)9. A × B = B × A A = B10. A × (B' C')' = (A × B) (A × C)11. A × (B' C')' = (A × B) (A × C)If A and B are two non-empty sets having n elements in common then (A × B) and (B × A) have n2

elements in common.

Ex. If n(A) = 7, n(B) = 8 and n(A B) = 4, then match the following columns.(i) n(A B) (a) 56(ii) n(A × B) (b) 16(iii) n((B × A) × A) (c) 392(iv) n((A × B) (B × A)) (d) 96(v) n((A × B) (B × A)) (e) 11

Sol. (i) n(A B) = n(A) + n(B) – n(A B) = 7 + 8 – 4 = 11(ii) n(A × B) = n(A) n(B) = 7 × 8 = 56 = n(B × A)(iii) n((B × A)× A) = n(B × A). n(A) = 56 × 7 = 392(iv) n((A × B) (B × A)) = (n((A B))2 = 42 = 16(v) n((A × B) (B × A)) = n(A × B) + n(B × A) – n(A × B) (B × A)

= 56 + 56 – 16 = 96

Ex. If A = {2, 4} and B ={3, 4, 5}, then (A B) × (A B) is(1) {(2, 2), (3, 4), (4, 2), (5, 4)}(2) {(2, 3), (4, 3), (4, 5)}(3) {(2, 4), (3, 4), (4, 4), (4, 5)}(4) {(4, 2), (4, 3), (4, 4), (4, 5)}


Sol. A B = {4} and A B = {2, 3, 4, 5} (A B) × (A B) = {(4, 2), (4, 3), (4, 4), (4, 5)}

Pictorial Representation of Cartesian Product of Two Sets :Arrow diagram :Let A = {1, 2, 3} B = (a, b)

A × B

1

2

3

a

b

Lattice-Diagram :Axis OX represents elements of A and perpendicular axis OY represents set B. Each dot represents anordered pair of A × B.Let A = (1, 2, 3) B = (1, 3)

1

1 2 3

2

3 (1, 3) (2, 3) (3, 3)

(1, 1) (2, 1) (3, 1)

O

y

x

RELATIONSFor any two non-empty sets A and B, every subset of A × B defines a relation from A to B and everyrelation from A to B is a subset of A × B.a R b A × B RIf (a, b) R, then a R b is read as 'a is related to b'

If (a, b) R, then a R b is read as 'a is not related to b'

Domain and Range of Relation :Domain of R = Dom(R) = Set of first components of all the ordered pairs belonging to R.Range of R = Set of second components of all the ordered pairs belonging to R.Co-domain of R = B where R is a relation from A to BRange of R Co-domain of RDom(R) = {a A : (a, b) R for some b B}


Range of R = {b B : (a, b) R for some a A}If R = A × B, then Dom(R) = A and Range of R = BDom () = ; Range of =

Ex. Let A = {1, 3, 4, 5, 7} and B = {1, 4, 6, 7} and R be the relation 'is one less than' from A to B,then list the domain, range and co-domain sets of R.

Sol. R = {(3, 4), (5, 6)}So, Dom(R) = {3, 5}Range of R = {4, 6}Co-domain of R = B = {1, 4, 6, 7}Clearly Range of R co-domain of R.

Representation of a Relation :Let A = {–2, –1, 4} B = {1, 4, 9}A relation from A to B i.e. a R b is defined as a is less than b.This can be represented in the following ways.1. Roster form :

R = {(–2, 1), (–2, 4), (–2, 9), (–1, 1), (–1, 4), (–1, 9), (4, 9)}2. Set builder notation :

R = {(a, b): a A and b B, a is less than b}3. Arrow - diagram :

–2

–1

4

1

4

9

A B

4. Lattice-diagram :

–2 –1 0

y

x41 2 3

4

9

1


5. Tabular form :

R 1 4 9 – 2 1 1 1 – 1 1 1 1 4 0 0 1

Ex. Let A = {1, 2, 3, 4}, B = {1, 2, 3,.......10}

R1 = {(1, 4), (2, 5), (3, 6), (4, 7)}R2 = {(2, 5), (3, 6), (4, 7), (5, 8)}R3 = {(1, 1), (2, 4), (3, 9)}Among R1, R2, R3, choose those, that represent a relation from A to B, and represent the relations in set-builder form.

Sol. R1 A × B; R A × B, R A × BHence R2 is not a relation as (5, 8) A × BR1 = {(a, b) : a A , b B and a + 3 = b}R3 = {(a, b) : a A , b B and a2 = b}Any subset of A × A is a relation on A. If n(A) = p and n(B) = q then n(A × B) = pqTotal number of subsets of (A × B) = 2pq

Hence 2pq different relations are possible from A to B.

Inverse Relation :Let R A × B be a relation from A to B.The inverse relation of R (denoted by R–1) is a relation from B to A defined as R–1 = {(b, a) : (a, b) R}If (a, b) R, then (b, a) R–1, a A, b B.domain of R–1 = Range of RRange of R–1 = domain of R

(R–1)–1 = R

Identity Relation :The identity relation on a set A is the set of ordered pairs belonging to A × A and is denoted by IA.

IA = {(a, a) : a A}i.e. every element of A is related to only itself.R is an identity relation if (a, b) R iff a = b, a A, b A.

IA–1 = IA

Domain of IA= Range of IA = A

Universal Relation :If A be a set and R is the set A × A, then R is called universal relation in A.If R = A × A, then R is universal relation in A.


Void Relation : is called the empty or void relation if A × A

Types of Relations on a Set :If A is a non-empty set, then a relation R on A is said to be

1. Reflexive :If (a, a) R, a A.

i.e. a R a, a A

“is equal to”, “is a friend of”, “is parallel to”, are some of reflexive relations.

2. Symmetric :If a R b b R a, a, b A

i.e. if (a, b) R (b, a) R, a, b A

“is a friend of”, “is parallel to”, “is equal to”, are some of symmetric relations.

3. Anti-Symmetric :If a R b and b R a a = b, a, b A (If R R–1 = Identity, then R is anti-symmetric)

“is divisible by” is an anti symmetric relation.

4. Transitive :If a R b and b R c a R c, a, b, c A

i.e. If (a, b) R and (b, c) R (a, c) R, a, b, c A

“is parallel to”, “is equal to”, “is congruent to” are some of the transitive relation.

Equivalence Relation :A relation R on a non-empty set A is called an equivalence relation if and only if it is Reflexive, Symmetricas well as Transitive.

"is parallel to", "is equal to", "is congruent to" "Identity relation" are some of the equivalence relations.

Every identity relation is an equivalence relation but every equivalence relation need not to be identityrelation.

Ex. Check the following relations for being reflexive, symmetric, transitive and thus choose the equivalencerelations if any.

(i) a R b if a b; a, b set of real numbers.

(ii) a R b iff a < b; a, b N.

(iii) a R b iff ba > 21 ; a, b R.

(iv) a R b iff a divides b; a, b N.

(v) a R b iff (a – b) is divisble by n; a, b I, n is a fixed positive integer.


Sol. (i) Not reflexive, not symmetric but transitive

Let a = –2 and b = 3; (–2, 3) R. Since 2 3 is true

Since 2 2 –2 hence relation is not Reflexive

Since 3 –2 is wrong hence relation is not symmetric

Now Let a, b, c be three real numbers such that a b and b c

a b b 0, so b c b c

Hence a c is true so the given relation is transitive.

(ii) Not reflexive, not symmetric but transitive.Since no natural number is less than itself hence not reflexive,If a < b then b < a is false. Hence not symmetric.If a < b then b < c clearly a < c. Hence transitive

(iii) Not reflexive, symmetric, not transitive.

210aa hence it is not reflexive.

|x|x2 hence symmetric.

Let a = 1, b = – 1 and c = 23

, 21

25cb)b,a

212ba R; ( so so (b, c) R

But R)c,aso( 21

21

231ca . Hence R is not a transitive relation.

(iv) Reflexive, not symmetric, transitive

Since aa

= 1 i.e. every number divides itself, hence R is reflexive.

If a divides b then b does not divide a (unless (a = b) hence the relation is not symmetric (but anti-symmetric).If a divides b and b divides c then it is clear that a will divide c. Hence transitive.

(v) Relfexive, symmetric as well as transitive, hence it is an equivalence relation.

Since 0 is divisible by n

0

n0

so given relation is reflexive

If a – b is divisible by n, then (b – a) will also be divisible by n. Hence, symmetric.If a – b = nI1 and b – c = nI2, where I1, I2 are integer.Then, a – c = (a – b) + (b – c) = n(I1 + I2) so a – c is also divisible by n, hence transitive.

Ordered Relation :R is an ordered relation if it is transitive but not equivalence relation. e.g. a R b iff a < b, a, b Nis an ordered relation.


e.g. R = {(1, 1), (1, 3), (1, 2), (2, 1), (2, 2) (2, 3)} is not reflexive, not symmetric and transitive, hencenot an equivalence relation.

so, R is an ordered relation.

Partial Order Relation :

R is an partial order relation if it is reflexive, transitive and antisymmetric at the same time.

Ex. a R b iff a divides b; a, b N is partial order relation since it is reflexive, transitive and anti-symmetric.

If R is reflexive R–1 is reflexive

If R is symmetric R–1 is symmetric

If R is transitive R–1 is transitive

Hence if R is an equivalence relation R–1 is equivalence relation

FUNCTION

Let A and B are two non empty sets. A function f from set A to set B is a rule which associated eachelement of A to a unique element of B, denoted by f : A B

set A is called domain of function f

set B is called co-domain of function f

If element x of A corresponds to y(B) under the function f, then we say that y is the image of x andwrite f(x) = y.

Ex. Which of the following given below is/are a function, from R to R?

(i) f(x) = x2

(ii) f(x) = x(iii) f(x) = 3x + 4.

Sol. (i) Yes, because all element of domain (which is R) have images in co-domain (R).

(ii) No, this is not a function because all negative number in a domain (R), do not have images inco-domain.

i.e. f(–1) = 1 (imaginary no.)

f(–2) = 2 (imaginary no.)

(iii) Yes, because all real numbers in domain have images in co-domain.

Note : (i) Range of f co-domain of f

(ii) f : A B is not a function, if there is atleast one element in A which does not have af-image in B or if there is an element in A which has more than one f-images in B.

(iii) A function can also be represented as a set of ordered pairs e.g. f = {(1, 2), (2, 3),(3, 4), (4,4)} is a function from {1, 2, 3, 4} to {2, 3, 4}. Clearly f = {(1,2), (1, –1),(2, 2), (3, 3)} is not a function as 1 2 and 1 – 1.


Ex. If 2x1x

2x1xf 2

2

, then find the value of f(2).

Sol.2x1x

2x1xf 2

2

....(1)

Put y2x1x

1yy21x

Put these value in equation (1)

So f(y) = 2

1yy21

11yy21

2

2

f(y) = 22

22

)1y(2y21)1y(y21

So value of f(2) = 118

)12(2)41()12()41(

22

22

.

Ex. If f(x) =

4x1x2–61x03x4

then find f (1/2) and f(2).

Sol. f(1/2) = 4. (1/2) + 3 = 5, f (2) = 6 – 2.2 = 2

Ex. Which of the following is a function?(A) {(2,1), (2,2), (2,3), (2,4)} (B) {(1,4), (2,5), (1,6) , (3,9)}(C) { (1,2), (3,3), (2,3), (1,4)} (D) { (1,2), (2,2), (3,2), (4,2)}

Sol. We know that for a relation to be function, every element of first set should be associated with one andonly one element of second set but elements of first set can have same f-image in second set which is givenin (D). Ans.[D]

Ex. If f (x) = 1–xx

, then

x1–xf equals

(A) x (B) x – 1 (C) x + 1 (D) 1 – x

Sol. .x–1x–1–x

1–x

1–x

1–xx/)1–x(

x1–xf

Ans.[D]

Ex. If f(x) = cos (log x), then f(x) f(y) – 12 [f (x/y) + f(xy)] is equal to

(A) –1 (B) 1/2 (C) –2 (D) 0


Sol. cos (log x) cos (log y) – 12 [ cos (log x/y) + cos (log xy)]

= 12 [cos (log x + log y) + cos (log x – log y)]

– 12 [cos (log x – log y) + cos (log x + log y)] = 0 Ans.[D]

Ex. If f(x) = 2 2

2

x x

, then f (x + y) . f (x – y) is equal to

(A) 12 [f (x+ y) + f(x – y)] (B) 1

2 [f (2x) + f (2y)]

(C) 12

[f(x+ y) . f(x – y)] (D) None of these

Sol. f(x + y). f(x – y) = 222.

222 yx–y–xy–x–yx

= 4

2222 y2–x2–y2x2

222

222

21 y2–y2x2–x2

= 21

[f(2x) + f(2y) ] Ans.[B]

The Graph of a Function :The graph of a function y = f (x) consists of all points (x, f(x)) in the Cartesian plane since by definitionof a function, there is exactly one value of y for each x, it follows that no vertical line can intersect thegraph of a function of x for twice or more.

y

0 x

not a function of x a function of x a function of x

y

x x

y

Domain of the Function :Domain of the function is the set of all those real numbers x for which f(x) exists or f(x) is meaningful.

f(x) or any imaginary no.) f(x) = x1

, f(x) exist, if x 0, so domain is R0. f(x) = 2x , f(x) exist,if x 2 so domain [2, ).

Range of Function :Set of all the images of elements in domain is called the range.Range = {f(x) : x domain}


Algebraic Operation on Functions :1. Given functions f and g, their sum f + g, difference f –g, and fg are defined on dom f dom g as:

(f + g) (x) = f (x) + g (x), (f – g) (x) = f (x) – g (x) and (fg) (x) = f(x) g (x). Moreover f/g is definedon dom f {x dom g: g(x) 0} by (f/g) (x) = f(x)/g(x).

2. If k is any real number and f is a function then kf is defined on the domain of f by (k f) (x) = k f(x).We have the following formulae for domains of functions(i) dom (f g) = dom f dom g(ii) dom (f g) = dom f dom g(iii) dom (f /g) = dom f {x dom g: g (x) 0}

(iv) dom f = {x dom f; f (x) 0}

Ex. If f : R+ R+, f(x) = x2 + 2 and g : R+R+, g(x) = 1x then (f + g) (x) equals

(A) 3x2 (B) x + 3 (C) )1x(2x2 (D) )1x(2x2

Sol. (f + g) (x) = f(x) + g(x) = x2 + 2 + 1x Ans. [D]

IDENTICAL FUNCTION Two functions f : AB and g : CD are called equal functions if and only if(a) domain of f = domain of g(b) co-domain of f = co-domain of g

(c) f(x) = g(x) , x domain

Ex. If A = {0,2}, B = {0,4}, then find whether f : A B, f(x) = 2x and g : A B, g (x) = x2 are equalfunctions or not.

Sol. f = {(0,0), (2,4) } and g = {(0,0),(2,4)}Now domain of f = domain of gco-domain of f = co-domain of gf (0) = 0, f (2) = 4 and g(0) = 0, g(2) = 4

f(x) = g(x), x A.

f = g Ans.

Ex. Find the domain and range of the function, f(x) = 3xx6 C

.

Sol. f(x) = 3xx6 C

f(x) exist if 6 – x > 0 and (6 – x) Nx – 3 0 and x – 3 W6 – x x – 3


so x < 6 x {...–2, –1, 0, 1, 2, 3, 4, 5,} ....(1)x 3 x {3, 4 5, 6, 7......} ....(2)

6 – x x – 3 2x 9 x 4.5so x {...–2, –1, 0, 1, 2, 3, 4} ....(3)so final value from (1), (2) and (3) is x {3, 4} so domain is {3, 4}

Range, f(3) = 3336

C = 3C0 = 1, f(4) = 2C1 = 2 so range is {1, 2}.

SOME STANDARD FUNCTIONS AND THEIR GRAPHS

Constant Function :A function denoted by f(x) = C (where C R) is known as constant functionDomain = R, Range = C

C

f x( ) y

xx' x

y'

O

Identity Function :A function which is associated to itself is known as identity function and denoted by f(x) = xSince x can take any value so domain of this function is R, corresponding value of f(x) is also R, so rangeis RDomain = R, Range = R

y

y'

x'x

x

I x( )

Modulus Function :This is also known as absolute value function and denoted by f(x) = |x|

i.e. f(x) =

0x , x0x , x

Domain of this function is set of all real numbers because f(x) existsfor all x R but |x| 0 so range is all non-negative real numbers.Domain = R , Range = [0, ] or R+ {0}

f x( )y

y'

x' xO x

Properties of modulus function :

(a) |x|x2 (d) |x| a – a x a

(b) | x y | = | x | | y | (e) |x| a x a or x – a

(c)yx

yx , (y 0) (f) a |x| b a x b or – b x – a


Ex. If

0x ; x0x;1x

)x(f and

1x;2x

1x; 1x)x(g .

Then find (f + g) (x) and draw its graph.

Sol.

0x ; x0x1 ; 1x

1x ; 1x)x(f

2x ; 2x2x1 ; 2x1x0 ; 1x

0x ; 1x

)x(g

12

3

– 1 0 1 2

2x ; 2 2x1 ; 2x21x0 ; 1x20x1 ; 2

1x ; x2

2x ; 2xx2x1 ; 2xx1x0 ; 1xx0x1 ; 1x1x

1x ; 1x1x

)x(g)x(f

Ex. If f(x) = 2|x – 2| – 3|x – 3|, then the value of f(x) when 2 < x < 3 is(A) 5 – x (B) x – 5 (C) 5x – 13 (D) None of these

Sol. 2< x< 3 |x – 2| = x – 2 |x – 3| = 3 – x

f(x) = 2 (x – 2) – 3 (3 – x) = 5x – 13. Ans. [C]

Ex. If f : RR. f (x) = 2x + |x| , then f (3x) – f (–x) – 4x equals-(A) f(x) (B) – f(x) (C) f (–x) (D) 2f(x)

Sol. f (3x) – f (–x) – 4x= 6x + |3x| – {–2x + | –x|} – 4x = 6x + 3 |x| + 2x – |x| – 4x= 4x + 2 |x| = 2 f(x) . Ans.[D]

Signum Function :

The function f(x), defined as f(x) =

0x ; 0

0x ; xx

is called signum function. This signum function may also

defined as f(x) =

0x ; 10x ; 0 0x ; 1

x

–1

1f x( )

O

y

Domain = R, Range = {–1, 0, 1}


Greatest Integer Function :This function is also known as step function or floor function denoted by f(x) = [x]. By [x] we meangreatest integer less then or equal to x. If n is an integer and x is any real number between n and n + 1i.e. n x < n + 1, then [x] = nThus [3.4] = 3, [3.99] = 3 , [–4.99] = –5, [–4.001] = –5, [0.3] = 0, [–0.2] = –1Domain of [x] is set of all real numbers because [x] exist x RBut [x] is always integral number so range is set of all integers Z.

Some Properties of [x] :(a) [x + k] = [x] + k, if k Z(b) [–x] = –[x] – 1(c) [x] + [–x] = 0, x Z(d) [x] + [–x] = –1, x Z

1 2 3 4–1–2–3–4

3

2

1

–1

–2

–3

x

f x( )

(e) [x] – [–x] = 2x, x Z(f) [x] – [–x] = 2[x] + 1, x Z(g) x – 1 < [x] x(h) [x] n x n, n z(i) [x] > n x n + 1, n z(j) [x] n x < n + 1, n z(k) [x] < n x < n, n z

Ex. Find the range of f(x) = 3 + x – [x+2]

Sol. f(x) = 3 + x – [x + 2]

= 1 + 2 + x – [x + 2] {where {.} is fractional part function}

= 1 + {2 + x} 0 {2 + x} < 1 0 +1 {2 + x} + 1 < 1 + 1 1 f(x) < 2 Range of f(x) is [1, 2)

Ex. The range of f(x) = cos 2]x[

is -

(A) {0,1} (B) {–1,1} (C) {–1,0,1} (D) [–1,1]

Sol. [x] is an integer, cos (–x) = cos x and cos

2 = 0, cos 2

2 = –1.

cos 0

2 = 1, cos 3

2 = 0,..... Hence range = {–1,0,1} Ans.[C]


Ex. Find the range of the function f(x) = 5)]1xx([tan])1x[sin(

221

. (where [] denotes step-function)

Sol. Here denominator 0 x R and [x + 1] = Z (due to step. function)sin [x + 1] = 0 (because sin of integer multiple of is always zero)

so f(x) = 5]1xx[[tan]1x[xsin

221

= 0 (for x R)

f(x) = 0 x R , Range of the function = {0}

Fractional Part Function :y = {x}Any number is sum of integral & fractional part i.e. x = [x] + {x} y = {x} = x – [x] = x – k, k x < k + 1

y = {x} =

..............3x2,2x2x1,1x1x0,x

.......1xy = 1

Domain = R , Range = [0, 1)From graph 0 {x} < 1This function is a periodic function with period 1. This is also many-one function discontinuous atx .

Ex. Find the value of

10099

31..........

1003

31

1002

31

1001

31

Where [] denote greatest integer function.Sol. Using properties of step function.

3310099

31..........

1003

31

1002

31

1001

31

.

Ex. Write the equivalent function of the function f(x) = |x + 2| + |x – 3|.Sol. First we find the critical values (values of x where modulus function vanish) which is x = – 2, 3.

If x < – 2, then f(x) = – (x + 2) – (x – 3) = –2x + 1If –2 x < 3, then f(x) = x + 2 – (x – 3) = 5If x 3, then f(x) = x + 2 + x – 3 = 2x – 1

so f(x) =

3x ; 1x23x2 ; 5

2x ; 1x2.


LOGARITHMIC FUNCTIONIf f : R+ R, f(x) = loga x, then f(x) is known as logarithmic functionHere f(x) exist if x > 0 and 0 < a < 1 or a > 1 (a 1)

f x( )

0 x

a > 1

loga x

y

Properties of logarithmic function :(i) loga m.n = log m + log n

(ii) loga nm

= log m – log n f x( )

0 xx'

y'

y

Domain = RRange = R

+

log a x

0 < < 1a(iii) loga mn = nloga m

(iv) pa bqlog = q

ploga

b

(v) logab

= alogblog

x

x = log xb.log ax

(vi) 1log.log ab

ba

(vii)If loga f(x) = y f(x) = (a)y

(viii) If loga f(x) loga g(x)

1a0 if )x(g)x(f1a if )x(g)x(f

(ix) If loga f(x) y

1a0 if )a()x(f

1 a)a()x(fy

y

(x) If loga f(x) y

1a0 )a()x(f

1a )a()x(fy

y

Ex. Find the domain of the function f(x) = logx (x2 – 3x + 2).

Sol. Logax exist if x > 0 and 0 < a < 1, a > 1Now f(x) = logx (x

2 – 3x + 2)x2 – 3x + 2 > 0 (x – 1) (x – 2) > 0, x (– 1) (2, )But x > 1 or 0 < x < 1So common value of x is (0, 1) (2, ).

Domain of f(x) (0, 1) (2, )

Ex. Find the domain of the function f(x), if f(x) = xlog 5.0 .


Sol. f(x) = xlog 5.0

Now we now that f(x) exist if log0.5 x 0x > 0 (because log x not defined for zero and negative numbers)log0.5 x 0 x (0.5)° x 1 x (– , 1]But x > 0 sox (0, 1].

Exponential Functionf(x) = ax is known as exponential function (a > 0)

0 xx

f x( )

x

(0, 1)

ax

If > 1 a

0 xx

f x( )ax

If 0 < 1 a <

Domain = RRange = R+ (0, 1)

Ex. How many solutions are there for equation log4 (x – 1) = log2 (x – 3)?

Sol. log4 (x – 1) = log2 (x – 3)

log22 (x – 1) = log2 (x – 3) 2

1 log2 (x – 1) = log2 (x – 3)

log2 (x – 1)1/2 = log2 (x – 3) (x – 1)1/2 = (x – 3) x – 1 = x2 – 6x + 9 (x – 2) (x – 5) = 0 x = 2, 5But x – 1 > 0 and x – 3 > 0 , x > 1 and x > 3 So only one solution x = 5

Ex. The domain of function f(x) = xx 3–2 is -

(A) (– , 0] (B) R (C) [0,) (D) No value of xSol. Domain = {x ; 2x – 3x 0} = {x ; (2/3)x 1}

= x (– ,0] Ans.[A]

Inverse Trigonometric Function :y = sin–1 x y = cos–1x

/2

1

– /2

–1O

1–1 O

/2

Domain = [–1, 1] Domain = [–1, 1]

Range =

2,

2 Odd function Range = [0, p] Neither even nor odd


y = tan–1x y = cot–1 x

xO

/2

y

–/2 O

/2

Domain = R Domain = R

Range =

2,

2 Odd function Range = (0, p)Neither even nor odd

y = sec–1x y = cosec–1x

O

/2

1–1

y

y

x -1 1

- /2

0

/2

• •

•

•

Domain = (–, –1] [1, ) Domain = (–, –1] [1, )

Range = [0, p] –

2Range =

2

,2

– {0}

Neither even nor odd Odd functionAll inverse trigonometric functions are monotonic.

Ex. Find the domain of the function f(x) =

52x

cos 1.

Sol. f(x) exist if –1 15

2x

– 5 < | x | – 2 5 – 3 | x | 7 | x | – 3 true x Ror | x | 7 x [–7, 7].

Ex. Find the domain of the function,

41x

cos5

3xsin)x(f 11 .

Sol. Domain of sin–1

5

3x say D1 and Domain of cos–1

4

1x say D2


So Domain of f(x) is D1 D2

For D1 53x

1

for D2 – 1 4

1x

–5 |x| –3 5 –4 |x| + 1 4–2 |x| 8 –5 |x| 3|x| – 2 and |x| 8 |x| – 5, and |x| 3sox [–8, 8] so x [–3, 3]

Domain of f(x) is [–8, 8] [–3, 3] = [–3, 3].

Ex. Find the domain of the function f(x) =

6x–x5log

2

.

Sol. For a given function to be defined

6x–x5log

2

0 6x–x5 2

1

5x – x2 6 x2 – 5x + 6 0(x – 3)(x – 2) 0

x

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

2xlog

2

2 is -

(A) [–2,2] – (–1,1) (B) [–1,2] – {0} (C) [1,2] (D) [–2,2] – {0}Sol. We know that the domain of sin–1x is [–1,1]. So for f(x) to be meaningful , we must have

–1 2xlog

2

2 1

2–1 2x2

2 , x 0 1 x2 4, x 0

x[–2,–1] [1,2] x[–2,2] – (–1,1) Ans.[A]

Polynomial Function :The function f(x) = a0 + a1x + a2x2 + ........ + anxn where a0, a1, a2, ....... an R and n N is called a polynomial function of degree n.

Rational Function :A function defined by the quotient of two polynomial function is called rational function.

1xx1x

3

2

is a rational function.


Ex. Find the domain and range of the function f(x) = 1xx1xx

2

2

.

Sol. x2 – x + 1 0 for any value of x ( b2 – 4ac < 0) so domain of f(x) is RRange Let f(x) = y

y1xx1xx

2

2

x2 (1 – y) + x(1 + y) + (1 – y) = 0But x is real so b2 – 4ac 0

(1 + y)2 – 4 (1 – y)2 0 3y2 – 10y + 3 0

(y – 3) (3y – 1) 0 y

3,31

. So range of f(x)

3,31 .

Irrational Function :A function involving one or more radicals of polynomial is called a irrational function

5xx3x2x ,xxx 3

222

3

etc.

Ex. The domain of f(x) = x–x

13 is-

(A) R – {–1,0,1} (B) R (C) R – {0,1} (D) None of theseSol. Domain = {x; x R; x3 – x 0}= R – {–1, 0,1} Ans.[A]

Ex. The range of function f(x) = 2

2

x1x

is-

(A) R – {1} (B) R+ {0} (C) [0,1] (D) None of theseSol. Range is containing those real numbers y for which f(x) = y where x is real number.

Now f(x) = y 2

2

x1x

= y x2 = y + yx2 (y –1) x2 + y = 0

D 0 0 – 4.y (y – 1) 0 y (y – 1) 0 y [0, 1)

Algebraic Function :An algebraic function is one which consist of a finite number of terms involving power and roots of thevariable x and simple operation, addition, subtraction, multiplication and division i.e. all rational, andirrational functions are algebraic functions.

Transcendental Function :All function which are not algebric are called transcendental function.(a) All trigometric function i.e. sin x, cos x etc.(b) All exponential function, ex, log x, ax etc.(c) Inverse trigonometric function sin–1 x, cos–1 x, etc.

Note : A transcendental function is not expressed in a finite number of algebraic terms.


Explicit Function :A function in which dependent variable (y) is expressed directly in terms of independent variable (say x)

i.e. y = x3 + x2 + 1, y = 2x

5x3x 2

, etc.

Implicit Function :A function in which we can't express dependent variable in terms of independent variable.

x3 + y3 + 3xy = 0, note that we can't write y or x in terms of x, or y separately.

Even or Odd Function :(a) Even function : If f(–x) = f(x) then f(x) is said to be even function.

f(x) = cos x is a even function [ f(–x) = cos (–x) = cos x = f(x)]

(b) Odd function : If f(–x) = –f(x) then f(x) is said to odd function.

If f(x) = x3 + tan3 x is a odd function

because f(–x) = (–x)3 + [tan (–x)]3

= – x3 – tan3 x = – [x3 + tan3 x] = – f(x)

so f(–x) = – f(x)

Note : (a) Even function is symmetrical about y-axis while odd function is symmetrical about origin(i.e. in opposite quadrant)

(b) Addition and subtraction of two even function is always even function.

(c) Sum of even and odd function is neither even nor odd function.

(d) Any function 'f' can be represented as the sum of an even and an odd function.

)]x(f)x(f[21)x(f + )]x(f)n(f[

21

where )]x(f)x(f[21

is an even and )]x(f)x(f[21

is an odd function

(e) f(x) = 0 is the only function which is both odd and even.

Ex. Is a function f(x) = xx

xx

eeee.x

even ?

Sol. Yes, f(x) is even, because f(–x) = (–x). xx

xx

xx–

xx–

eeeex

eeee

= f(x)

so f(–x) = f(x).

Ex. Show that f(x) is a odd function if f(x) = log )x1x( 63 .

Sol. f(x) = log )x1x( 63

f(–x) = log

63 )x(1)x( = log [–x3 + 6x1 ]


= log

63

6363

x1x

x1xx1x = log

6363

66

x1x

1logx1x

x1x

f(–x) =

63

63x1xlog

x1x

1log

so f(–x) = –f(x) so f(x) is odd function.

Ex. If f(x) = 1–a1a

x

x then find whether it is an even or odd function .

Sol. f(–x) = 1–a1a

x–

x– = –

1–a1a

x

x

f(–x) = – f(x)

f(x) is an odd function. Ans.

Periodic Function :A function 'f' defined on its domain is said to be periodic function if their exist a positive number T suchthat f(x + T) = f(x) x D. Also both x + T and x – T should belong to D.The least value of T, if exists is called, the period of the function.f(x) = sin x , f(x) = sin (x + 2) = sin (x + 4) = sin (x + 6) = ..................Here T = 2, 4, 6 ......................Least value of T is 2, so period of sin x is 2

Some Standard Functions and their Period :Function Periodsin x 2cos x 2tan x

{x} 1

Some Special Point about Periodic Function :If period of f(x) is 'T' then

(a) (i) Period of f(ax) and f(ax + b) is aT

.

(ii) Period of

axf is |a|.T..

(b) If period f(x) is T1 and g(x) is T2. Then period of f(x) ± g(x) is given by L.C.M. of T1 and T2

(same for )x(g)x(f

)


Note : (i) LCM of f,d,b of HCFe,c,a of LCM

fe,

dc,

ba

.

(ii) Sin x and sin x2 is not a periodic function because these can't be written in the form off(x + T) = f(x)

(iii) L.C.M. of rational with irrational is not possible, e.g., L.C.M. of (, 2, 2) is not possibleas , 2 irrational and 2 rational

Ex. Calculate the period of f(x) = sin 3x + cos 2x.

Sol. Period of sin 3x = 32

, Period of cos 2x =

22

So, Period of f(x) is L.C.M. of 1,

32

= 2

Ex. If f(x) = x2sin1 is a periodic function, then find its period.

Sol. f(x) = x2sin1

= 222 )xcosx(sinxcosxsin2xcosxsin

f(x) = |sin x + cos x|

xxRemember 2

Now period of sin x + cos x is 2

So, period of |sin x + cos x| is 2

2

Ex. If f(x) = xcose xsin 3 is a periodic function. If yes, then calculate its period.

Sol. For periodic function f(x + T) = f(x)Now f(x + T) = )Txcos(e T)(xsin 3

If T = 2

f(x + 2) = )2xcos(e )2π(xsin 3 = )x(fxcose xsin 3

f(x + 2) = f(x); f(x) is periodic function having period 2.

Ex. Show that xsin is not periodic

Sol. Suppose that f(x) = xsin is periodic with period T. Then,

0x ; xsin)x(fTxsin)Tx(f

02

xTxsin2

TTxcos2

02

TTxcos or 0

2xTxsin


In ,)1n2(2

)1n2(2xTx

or In ,n2xTx

The above equalities gives T as function of x. But for f(x) to be periodic T should be constant i.e.independent of x.Hence f(x) can not be periodic.

Ex. Find the period of n1–n32 2xcot

2xcos.......

2xcot

2xcos

2xcotxcos

Sol. Since the period of cos ax(a > 0) is 2/a and the period of cot ax (a > 0) is /a, the periods of

cosx, cosx/22, ....., cos x/2n–1 are 2, 22(2) ......., 2n–1(2) and the period of n2xcot,......,

2xcot are

2, ...., 2n. Hence the period of the given function is L.C.M of (2, 23, ..., 2n) = 2n.

Bounded and Unbounded Function :f(x) is said to be bounded above, if there exists a fixed number say M such that f(x) is never greater thenM for all value of x. Similarly it's bounded below if there exists a fixed number m (say) that f(x) is neverless then m i.e. M f(x) m for all value of x.f(x) is said to be unbounded if one or both of the upper and lower (M and m) bounds of the functionare infinite.f(x) = 3 + sin x is a bounded function because maximum and minimum value of sin x are +1 and –1So, 2 f(x) 4 for all value of x.

TYPES OF MAPPINGS OR FUNCTIONS

One-one Function or Injective Function :A function is said to be one-one function if different element in a domain have different images inco-domain.

123

abcd

(domain) (co-domain)

if ( ) = ( ) then = ( ) is one - one function

f x f x x xf x

1 2 1 2

Set A Set B

Note : (i) Linear polynomial function (ax + b), x, ex, log x, are always one-one functions.(ii) If 0

dxdy

or 0dxdy

domain x , then y = f(x) is said to be one-one function.

Number of one-one function : If A and B are finite sets having m and n elements respectively,then number of one-one function from A to B = nPm, if n m = 0, if n < m.

Ex. If A = {–2, 0, 2} and B = {1, 10, 17} and a function f from A to B is defined by the rule f (x) =x4 + 1 then find whether the function is one-one or not.

Sol. f (–2) = (–2)4 + 1 = 17,f (0) = (0)4 + 1 = 1, f (2) = (2)4 + 1 = 17

Here – 2 2 but f (–2) = f (2) so function is many-one. Ans.


Ex. Which of the following functions defined from R to R are one-one.(A) f(x) = |x| (B) f(x) = cos x (C) f(x) = ex (D) f(x) = x2

Sol. x1 x2 ex1 ex2

f(x1) f(x2)

f(x) = ex is one-one. Ans. [C]

Many–one Function :A function f : A B is said to be many one if more than one element in set A have same image inSet B.

1234

abcde

A B

Note : (i) All even function, modulus function, periodic function are always many-one function.(ii) Square function, Trigonometric function are also many–one function in their domain.

Into Function :A function f: A B is said to be into function if there exist at least one element in set B having nopre-image in set A.

123

abcd

A B

In fig set B (co-domain) there is no pre-image, for element d, in set A, so function is into function.

Onto Function :f : A B, said to be onto function if every element in set B has a pre image in set A.Range of f = co-domain of f. log x, linear polynomials, are always onto function.Possible mappings are

(i) One-one and onto (bijective function)(ii) Many one and onto(iii) One-one and into(iv) Many one-into

Ex. If f : R R, f(x) = x2 + 3x + 2 then show that f(x) is many one function.

Sol. f(x) = x2 + 3x + 2 = 41

23x

2


f(–2) = 041

232

2

f(–1) = 041

231

2

So image of –2 and –1 are same f(x) is many one.

Ex. Prove that f(x) = 2x + sin x, is one-one function.Sol. f (x) = 2 + cos x, minimum value of cos x is –1.

f (x) > 0 for all x Domain = R

Ex. If f : [–1, 1] B and f(x) = sin–1 x, if f(x) is one-one onto function then find value of B.Sol. In one-one onto function range of f = co-domain of f (which is B)

So range of function for x [–1, 1] is

2,

2 so B

2,

2 .

Ex. A function f : R R, is defined by : 2

2

x8x68x6x)x(f

Find the interval of values of for which f is onto. Is the function one-one for = 3 ?

Sol. Let 0)m8(x)m1(6x)m8(x8x6

8x6xm 22

2

Since x is real, D 036 (1 – m)2 + 4( + 8m)(8 +m) 0

(9 + 8) m2 + (46 + 2) m + (9 + 8) 0f is onto if and only if the above relation hold for all m R. This will happen if

9 + 8 > 0 and (46 + 2)2 –4 (9 + 8)2 0 9 + 8 > 0 and (2 + 16 + 64) (2 – 16 +28) 0 9 + 8 > 0 and ( + 8)2 ( – 2) ( –14) 0 > –9/8 and 2 14. Thus 2 14

when = 3, 2

2

x8x638x6x3m

For m = 0, we get 3x2 + 6x – 8 = 0

33331

696366x

Hence f is not one–one when = 3

Ex. The function f : R R, f(x) = x2 is-(A) one-one but not onto (B) onto but not one-one(C) one-one onto (D) None of these


Sol. 4 – 4, but f (4) = f(–4) = 16

f is many one function.

Again f (R) = R+ {0} R, therefore f is into. Ans. [D]

Ex. If f : I0 N, f(x) = |x|, then f is-

01 2 3–3 – 2 – 1

3

2

1

(A) one-one (B) onto (C) one-one onto (D) none of these

Sol. Observing the graph of this function, we find that every line parallel to x-axis meets its graph at more than onepoint so it is not one-one.

Now range of f = N = Co-domain, so it is onto. Ans. [B]

Ex. If f : R – {3} R – {1}, f (x) = 3–x2–x

then function f(x) is -

(A) Only one-one (B) one-one into (C) Many one onto (D) one-one onto

Sol. f(x) = 3–x2–x

f (x) = 23)–(x1–

23)–(x2).1–(x–3).1–(x

f (x) < 0 x R – {3}

f(x) is monotonocally decreasing function

f is one-one function.

for onto/ into : Let y R – {1} ( co-domain)

Then one element x R – {3} in domain is such that

f(x) = y xx

23

= y x – 2 = xy – 3y

x =

1–y2–y3

Þ y R – {1}

the pre-image of each element of co-domain R – {1} exists in domain R – {3}.

f is onto. Ans. [D]

Ex. Function f : N N , f(x) = 2x + 3 is

(A) one-one onto (B) one-one into (C) many one onto (D) many one into


Sol. f is one- one because for any x1, x2Nx1 x2 2x1 + 3 2x2 + 3

f (x1) f (x2)

Further f –1(x) = x 32

N (domain ) when

x = 1, 2, 3 etc. f is into which shows that f is one- one into. Ans. [B]

Ex. Function f : R R, f(x) = x3 – x is -(A) one-one onto (B) one-one into (C) many-one onto (D) many-one into

Sol. Since –1 1, but f (–1) = f(1), therefore f is many-one. Also let, f(x) = x3 – x = x3 – x –= 0. This is a cubic equation in x which has at least one real root because complex roots alwaysoccur in pairs. Therefore each element of co-domain R has pre-image in R. Thus function f is onto function f is many-one onto. Ans. [C]

Composition of Function :Let f : A B and g : B C then the composition of g and fis denoted by gof and is defined as gof : A C given bygof (x) = g(f(x))Similarly fog is defined. Note that, gof is defined only if Rangef dom g and fog is defined only if Range g dom f.Let f(x) = x2 + 3 and g(x) = x . Since dom g = [0, ),dom f = R

we have fog (x) = f(g(x)) = 3x3)x()x(f 2

So dom fog = {x [0, ) : g(x) R} = [0, )Let us now find gof, we have (gof) (x) = g(f(x)) = g(x2 +3)= 3x 2 , then dom gof = {x R : f(x) [0, )} = R.

Ex. If f(sin 2x) = cos2x (sec2 x + 2 tanx) then find domain and range of f.Sol. f(sin2x) = 1 + 2sin x cosx = 1 + sin2x

Let u = sin2x so f (u) = 1 + usince –1 sin 2x 1 so dom f = [–1, 1]Now – 1 u 1 0 1 + u 2. Thus the range f = [0, 2]

Ex. If f : R R, f(x) = cos x, g: R R , g(x) = x2 then find fog(x) and gof(x) and analyse the result.Sol. fog (x) = f[g(x)] gof (x) = g[f(x)]

= f[x2] = g[cosx] = cos x2 = (cosx)2 = cos2x

So, we can see fog gof.

x

A B C

f x( ) g f x( ( ))f g

gof


Ex. If f(x) = xx

31 , then f [f {f (x)}] equals-

(A) x (B) 1/x (C) –x (D) –1/x

Sol. Here f {f(x) } =

1x3–xf =

11x3–x

3–1x3–x

= x–1

3x

f[f{f(x)}] = x4x4

1x–13x

3–x–13x

Ans. [A]

Ex. If f : R R, f(x) = 2x – 1 and g : R R, g(x) = x2 + 2, then (gof) (x) equals-(A) 2x2 – 1 (B) (2x – 1)2 (C) 2x2 + 3 (D) 4x2 – 4x + 3

Sol. Here (gof) (x) = g [f(x)] = g (2x – 1) = (2x – 1)2 + 2 = 4x2 – 4x + 3. Ans. [D]

Ex. f(x) = |1–x| and g (x) = sin x then (fog ) (x) equals-

(A) sin |1–x| (B) |sin x/2 – cos x/2| (C) |sin x – cos x| (D) None of these

Sol. (fog) (x) = f [g(x) ] = f [sin x]

= |1–xsin| = |xsin–1| = 2/xcos2/xsin2–2/xcos2/xsin 22

= |)2/xcos–2/xsin(| 2 = |)2/xcos–2/x(sin| 2 = |sin x/2 – cos x/2| Ans.[B]

Ex. If g(x) = x2 + x – 2 and 12

(gof) (x) = 2x2 – 5x + 2, then f(x) is equal to

(A) 2x – 3 (B) 2x + 3 (C) 2x2 + 3x + 1 (D) 2x2 – 3x – 1Sol. g(x) = x2 + x – 2

(gof) (x) = g[f(x)] = [f(x)]2 + f(x) – 2

Given, 12

(gof) (x) = 2x2 – 5x + 2

12

[f(x)]2 + 12

f(x) – 1 = 2x2 – 5x + 2 [f(x)]2 + f(x) = 4 x2 – 10x + 6

f(x) [f(x) + 1] = (2x – 3) [(2x – 3) + 1] f(x) = 2x – 3 Ans.[A]

Ex. If f (x) = |x| and g(x) = [x], then value of fog

41– + gof

41– is -

(A) 0 (B) 1 (C) –1 (D) 1/4


Sol. fog =

41–gf = f (–1) = 1 and gof

41– = g

41–fg

= g

41

= [1/4] = 0

Required value = 1 + 0 = 1. Ans.[B]

Ex. If f(x) = –1 + |x – 1|, –1 x 3, g(x) = 2 – |x + 1|, –2 x 2 then for x (0,1), (fog) (x) equals-

(A) x – 1 (B) 1 – x (C) x + 1 (D) –(x + 1)

Sol. 0< x < 1 f(x) = –1 – (x – 1) = –x and g(x) = 2 – (x + 1) = 1 – x

x (0,1) (fog) (x) = f[g(x) ] = f(1 – x) = –(1 – x) = x – 1 Ans.[A]

Inverse Function :

Two functions f and g are inverse of each other if f (g(x)) = x for x dom g and g(f(x)) = x for x domf . To find the inverse of f, write down the equation y = f(x) and then solve x as a function of y. Theresulting equation is x = f–1 (y).

Ex. Find the inverse of 2ee)x(f

xx

Sol. We write x

x2xx

e1ey2

2eey

2

4y4y2e01ye2e

2xxx2

1yye 2x since ex 0 so 1yye 2x

1yylogx 2

e . Thus f–1(x) = loge

1xx 2

The graph of f and f–1 are related to each other in the

y = ex

y = x

y = xlog

following way :

If the point (x, y) lies on the graph of f then the point (y, x) lies on the graph of f–1 and vice versa. Thusthe graph of f–1 is the reflection of the graph of f in the line y = x as below (since we know that y = logx and y = ex are inverse of each other).

Existence of inverse function :

A function need not have an inverse. e.g. the function f(x) = x2 has no inverse (where dom f = R). Tohave an inverse, a function must be both one-one and onto, i.e. bijective.


Ex. If f : (3, 4) (2, 4), f(x) :

2xx where [×] denote step function then find the f–1(x).

Sol. f(x) =

2xx , Domain of this function is (3, 4) in this domain

2x

= 1

So function is f(x) = x – 1or y = x – 1 x = y + 1On interchanging x and y we gety = x + 1 f–1(x) = x + 1.

Ex. If f : R R, f(x) = x3+2 then find f –1 (x).Sol. f(x) = x3 + 2, x R.

Since this is a one– one onto function therefore inverse of this function (f –1) exists.Let f –1(x) = y

x = f (y ) x = y3 + 2 y = (x – 2)1/3

f –1 (x) = (x – 2)1/3. Ans.

Ex. If f : R R, f(x) = 4x3 + 3, then f –1(x) equals-

(A) 3/1

43–x

(B)

43–x 3/1

(C) 41

(x – 3)1/3 (D) None of these

Sol. Since f is a bijection, therefore f –1 exists.But f(x) = 4x3 + 3 y = 4x3 + 3

x = 3/1

43–y

f –1(x) = 3/1

43–x

Ans. [A]

Ex. If f : R R, f(x) = 2x + 1 and g : R R, g(x) = x3,then (gof)–1(27) equals-(A) –1 (B) 0 (C) 1 (D) 2

Sol. Here f(x) = 2x + 1 f –1(x) = x 12

and g(x) = x3g–1(x) = x1/3

(gof)–1(27) = ( f –1og–1) (27)

= f –1 [g–1(27) ] = f –1[(27)1/3] = f –1(3) = 21–3

= 1 Ans.[C]

RELATIONS & FUNCTIONS · Relations & Functions [2] RELATIONS a R b means 'a is R-related to b' i.e. a is related to b under relation R. If (a, b) R; (a, b) is called ordered pair

Documents