1.relations and functions assignment solutions

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RELATIONS AND FUNCTIONS

Revision Assignment

Class 12 Chapter 1 QUESTION 1 (i).In the given mapping , is f a

SOLUTION: In the given mapping f is not a bijection because

mapping in N

. TheNumber of binary operations on the s

The number of binary operations on a s

The number of

QUESTION 1

binary ope

:

:

ra

(i

tions f m

i)

SOLUTION

QUESTION 2(i). If f(x) = x + 7 and g(x) = x

SOLUTION: Here, f(x) = x + 7 and g(x) = x

fog(x) = f(g(x)) = f(x – 7) = x – 7 + 7 = x.

Now, fog(7) = 7

SOLUTIO

QUESTIO .Find the number of all onto functions f

: Here , A = { 1,2,3,....,10}

For f to be an onto function , Range of

The number o

N 2 (i

f way

.

i)

N

s in which n elements can be arranged taken

The number of ways in which 10 elements ∴ =

QUESTION 3Show that the Relation R in the set

relation. Where the relation R is given by

( ){ }, : is a multiple of 4 a b a b−

SOLUTION: The set [A x x= ∈ ≤ ≤ =

Here, ( ){ , : s a multiple of 4 R a b a b= −

4 or 4a b k b a k− = = +

( ) ( ) ( ) ( ) (( ) ( ) ( ) ( )1,5 , 1,9 2,6 2,10 , 3,7 3,11 4,8 4,12 5,9 6,10 7,11

0,0 , 1,1 , 2, 2 , 4,4 ,................... 12,12R

=

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RELATIONS AND FUNCTIONS

.In the given mapping , is f a not a bijection, give reason for your answer .

In the given mapping f is not a bijection because it is a one-one but into

. TheNumber of binary operations on the set {1, 2} are_________

The number of binary operations on a set of n elements n elementss is n

tions f mro22 4 a set of 2 elements 2 elementss is 2 2 16= =

If f(x) = x + 7 and g(x) = x – 7, x ∈ R, find (fog)(7).

f(x) = x + 7 and g(x) = x – 7, x ∈ R.

7 + 7 = x.

.Find the number of all onto functions from the set {1, 2, 3, ... , 10} to itsel

: Here , A = { 1,2,3,....,10}

For f to be an onto function , Range of f = A

n which n elements can be arranged taken , n at a time = !

The number of ways in which 10 elements can be arranged taken , 10 at a time = 1

P n

∴ =

Show that the Relation R in the set { } : 0 12A x Z x= ∈ ≤ ≤ is an equivalence

relation. Where the relation R is given by

}, : is a multiple of 4 .

] [ ]2 : 0 12 0,1,2,......12A x x= ∈ ≤ ≤ =

}, : s a multiple of 4

)( )( ) ( )( )( ) ( ) ( )( )

1,5 , 1,9 2,6 2,10 , 3,7 3,11 4,8 4,12 5,9 6,10 7,11 8,12 ,

0,0 , 1,1 , 2, 2 , 4, 4 ,................... 12,12

bijection, give reason for your answer .

one but into

2n

et {1, 2} are_________

et of n elements n elementss is n

2 4 a set of 2 elements 2 elementss is 2 2 16ways= =

R, find (fog)(7).

rom the set {1, 2, 3, ... , 10} to itself.

n

10

10

, n at a time = !

can be arranged taken , 10 at a time = 10!

nP n

P

=

∴ =

is an equivalence

8,12 ,

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A relation R is an equivalence relation if it is ,

1. reflexive,

2.symmetric and

3.transitive.

(a) 0 4a a k− = = ⇒R is reflexive

(b) If 4 a b k− = then 4b a k− =

a c a b b c− = − + − ⇒ R is symmetric

(c) When ( a c− ) and ( b c− ) both are multiples of 4 the ( a c− ) is also multiple of 4

⇒R is Transitive

From a, b and c : R is an equivalence relation

QUESTION 4:Prove that the Greatest Integer Function f : R → R, given by f (x) =

[x], is neither one-one nor onto, where [x] denotes the greatest integer less than or

equal to x.

SOLUTION: From any integer x to the next integer (x +1), the function value

f (x) = [x] = x

i.e at infinitely many values of x , [x] = x

i.e f(x) is a many one and not a one-one function.

The function[x] = x where x is an integer only and not any real number

⇒Rf = Z, Z being the set of integers.

∴f(x) is an into function rather than an onto function as R → R.

QUESTION 5:

Show that the Signum Function f : R → R, given by

is neither one-one nor onto.

SOLUTION

f(x) =1 , when, x∈R and x>0

just like f(x) =-1 , when, x∈R and x<0

i.e for infinitely many values of x , f(x) takes the same value

i.e f(x) is a many one and not a one-one function

The function only takes the values : -1, 0 , 1

⇒Rf = {-1,0,1}

∴f(x) is an into function rather than an onto function as R → R.

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QUESTION 6:

Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function

from A to B. Show that f is one-one.

SOLUTION

f = {(1, 4), (2, 5), (3, 6)}⇒f(1) = 4, f(2) = 5, f(3) = 6

i.e points of the domain give rise to a different point in the co-domain

In other words , the second entry of no ordered pair is repeated .

⇒ f is a one-one function.

QUESTION 7:

Let A = R – {3} and B = R – {1}. Consider the function f : A→B defined by

2( )

3

xf x

x

−=

−

Is f one-one and onto? Justify your answer.

SOLUTION

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{ }

{ }

( )( ) ( ) ( )

1 2 1 2

1

1 2

1 2

1 2 2 1

1 2 1 2 1 2 1 2

1 2 1 2

1 2

2( ) ( ) ; 3

3

; , 3

( ) ( )

2 2

3 3

2 3 2 3

3 2 6 2 3 6

3 2 2 3

=> f is a one-one function.

2(ii)Let y= ( )

xi f x x A R

x

Let x x andx x x A R

andf x f x

x x

x x

x x x x

x x x x x x x x

x x x x

x x

xf x

x

−= ∈ = −

−

= ∈ = −

=

− −⇒ =

− −

⇒ − − = − −

⇒ − − + = − − +

⇒ − − = − −

⇒ =

−= { }; 1

3

2y=

3

3 2

3 2

( 1) 3 2

3 2

( 1)

3 2Now note that , we may write

( 1)

3 3 1 3( 1) 1 13

( 1) ( 1) ( 1)

3 a Real number 3

3 23

( 1)

3 2

( 1)

, to each y

y B R

x

x

xy y x

xy x y

x y y

yx

y

y

y

y yas

y y y

y

y

yA

y

Thus

∈ = −−

−

−⇒ − = −

⇒ − = −

⇒ − = −

−⇒ =

−

−

−

− + − += = +

− − −

+ ≠

−∴ ≠

−

−⇒ ∈

−

∈ B , there corresponds an x A

Thus , f is an onto function

∈

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QUESTION 8:

Let f, g and h be functions from R to R. Show that

(f + g)oh = foh + goh

(f . g)oh = (foh) . (goh)

SOLUTION

Here x ∈R

(i) (f + g)oh =(f + g)[h (x)]= f[h (x)] + g[h (x)]= foh(x) + goh(x)= foh + goh

⇒(f + g)oh = foh + goh

(ii)(f . g)oh =(f . g) [h (x)] =(f[h (x)] . g[h (x)]) =(foh(x)) . (goh(x))= (foh) . (goh)

QUESTION 9:

State with reason whether following functions have inverse

(i) f : {1, 2, 3, 4} → {10} with

f = {(1, 10), (2, 10), (3, 10), (4, 10)}

(ii) g : {5, 6, 7, 8} → {1, 2, 3, 4} with

g = {(5, 4), (6, 3), (7, 4), (8, 2)}

(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} with

h = {(2, 7), (3, 9), (4, 11), (5, 13)}

SOLUTION

(i) f = {(1, 10), (2, 10), (3, 10), (4, 10)}

⇒f(1) =10 , f(2) = 10 ,f(3) =10 , f(4) = 10

⇒For more than one value x there is the same image in the range.

⇒ f is a not a one-one function.

∴Even though the function is onto , it is not invertible

(ii) g = {(5, 4), (6, 3), (7, 4), (8, 2)}

⇒g(5) = 4 and g(7) = 4

⇒For more than one value x there is the same image in the range.

⇒ f is a not a one-one function.

R={2,3,4}⊂{1, 2, 3, 4}

⇒ f is not an onto function

⇒ f is neither one-one nor onto

⇒ function does not have an inverse

(iii) h = {(2, 7), (3, 9), (4, 11), (5, 13)}

⇒h(2) = 7 , h(3) =9 , h(4) =11, h(5) = 13

⇒Different points in the domain have different images in the range

Also, range ={7, 9, 11, 13}= codomain

⇒ f is an onto function

⇒ f is both one-one and onto

⇒ function has an inverse

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QUESTION 10:

Show that f : [–1, 1] → R, given by f (x) = 2

x

x + is one-one. Find the inverse

of the function f : [–1, 1] → Range f.

SOLUTION

f (x) =

( ) ( )

1 2 1 2

1 2

1 2

1 2 2 1

1 2 1 1 2 2

1 2

1 2

x(i)f : [-1, 1] R, given by f (x) =

x+2

x and x [-1, 1] , such that f (x ) = f (x )

x x x +2 x +2

x x +2 = x x +2

x x 2x x x 2x

2x 2x

x x

f (x) is a one-one function.

(ii)Range of f

Let ∈

⇒ =

⇒

⇒ + = +

⇒ =

⇒ =

⇒

(x)

xlet y =

x+2

( 2)

2

2

( 1) 2

2

1

, 1

{1}f

y x x

xy y x

xy x y

x y y

yx

y

Obviously y

Range R

⇒ + =

⇒ + =

⇒ − =

⇒ − =

⇒ =−

≠

∴ = −

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QUESTION 11:

Consider f : R+ → [4, ∞) given by f (x) = x

2 + 4. Show that f is invertible with the

inverse f –1

of f given by f –1

(y) =√( y − 4) , where R+ is the set of all non-negative

real numbers.

SOLUTION

f (x) = x2 + 4

Let x1 , x2 ∈ R+, such that f (x1) =f(x2)

⇒ x12 + 4= x2

2 + 4

⇒ x12 = x2

2

But, x1 , x2 ∈ R+

⇒ x1 = x2

f is a one-one function.

Now let y∈ [4, ∞)and

y = x2 + 4

⇒x=√( y − 4)

y∈ [4, ∞) and x>0, ie x∈ R+

⇒ f is an onto function

⇒ f is both one-one and onto

⇒ function has an inverse

Its inverse can be obtained as follows

y = x2 + 4

⇒ x2

= y -4

⇒x=√( y − 4)

. QUESTION 12: Let A = Q x Q , Q being the set of rationals . Let ‘*’ be a binary operation on

A , defined by (a, b) * (c , d) = ( ac , ad + b) . Show that

(i) ‘*’ is not commutative (ii) ‘*’ is associative

(iii The Identity element w.r.t ‘*’ is ( 1 , 0)

SOLUTION:

(i) (a, b) * (c , d) = ( ac , ad + b)

(c , d) *(a, b) = ( ca , c b + d) , ( ac , ad + b) ≠( ca , c b + d)

So, ‘*’ is not commutative [1 Mark]

(ii) Let (a,b ) ( c, d ) , ( e, f ) ∈ A, Then

(( a, b )* (c, d)) ( e*f) = (ac , ad + b) * (e,f) = ( (ac) e , (ac) f + (ad+b))

= ( ace , acf +ad +b)

(a,b) ((c,d)* (e,f)) = (a,b) * ( ce , cf +d) = ( a (ce) , a ( cf+d) +b) = ( ace , acf +ad +b)

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(( a, b )* (c, d)) ( e*f) = (a,b) ((c,d)* (e,f)) [1 Mark]

(iii) Let ( x , y) ∈A. Then ( x , y) is an identity element, if and only if

(x, y ) * (a, b) = (a , b) = (a,b) * (x, y ), for every (a,b) ∈A

(x, y ) * (a, b) = ( xa , xb + y)

(a,b) * (x, y )= ( ax, ay +b )

( xa , xb + y) = (a, b) = ( ax, ay +b ) [1 Mark]

ax = x a = a ⇒ x = 1

bx + y = b = ay +b ⇒ b + y = b = ay +b ⇒ y = 0 = ay ⇒ y = 0

Therefore , ( 1, 0) is the identity element [1 Mark]

:Let '*'be a binary operation on the set { 0,1,2,3,4,5} and

a+b if a+b<6 a*b =

a+b-6

QUESTI

if a+b 6

ON 13

≥

SOLUTION:

* 0 1 2 3 4 5

0 0 1 2 3 4 5

1 1 2 3 4 5 0

2 2 3 4 5 0 1

3 3 4 5 0 1 2

4 4 5 0 1 2 3

5 5 0 1 2 3 4

From the table , the second row and second column are the same as the original set .

⇒0*0 = 0 , 1*) = 0*1 = 1 , 2*0= 0*2 = 2 , 3*0= 0*3 = 3 , 4*0 = 0*4 =4, 0*5 = 5*0 = 5

⇒’0’is the identity element of the operation ‘*’

Now, the element ‘0’ appears in the cell 1*5 = 5*1 =0, 2*4 = 4 *2= 0, 3*3 =0, 0* 0 = 0

⇒Inverse element of 0 is 0, Inverse element of 1 is 5, Inverse element of 2 is 4, Inverse

element of 3 is 3, Inverse element of 4 is 2, Inverse element of 5 is 1.

QUESTION 14 A relation R on the set of complex numbers is defined by

1 21 2

1 2

z zz Rz

z z

−=

+

Show that R is an equivalence relation.

SOLUTION:

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1 21 2

1 2

1 2 1 11 2 1 1

1 2 1 1

1 2 2 11 2 2 1

1 2 2 1

( ) is reflexive : 0,0

( ) is symmetric: ,

( ) is transi

z zz Rz isreal

z z

z z z zi R z Rz R z Rz R

z z z z

z z z zii R z Rz R R z R z R

z z z z

iii R

−=

+

− −= ∈ ⇒ = = ∈

+ +

− −= ∈ ⇒ ∈ ⇒ ∈

+ +

( ) ( )( ) ( )

( ) ( )( ) ( )

( )

1 1 1 2 2 2 3 3 3

2 31 21 2 2 3

1 2 2 3

1 21 2

1 2

1 1 2 2 1 2 1 2

1 1 2 2 1 2 1 2

1 2 1

tive: Let , , z ,

such that

and

z a ib z a ib a ib C

z zz zz Rz R z Rz R

z z z z

z zz Rz R

z z

a ib a ib a a i b bR R

a ib a ib a a i b b

a a i b b

= + = + = + ∈

−−= ∈ = ∈

+ +

−= ∈

+

+ − + − + −⇒ ∈ ⇒ ∈

+ + + + + +

− + −⇒

( )( ) ( )

( ) ( )( ) ( )

( ) ( ) ( ) ( )

[ ] [ ]

2 1 2 1 2

1 2 1 2 1 2 1 2

1 2 1 2 1 2 1 2

1 2 2 1 1 2 2 1 1 2 2 1

2 32 3 2 3 3 2

2 3

he imaginary part = 0

2 0 ...( )

, ....( )

, ( )

a a i b bR

a a i b b a a i b b

T a a b b b b a a

a b a b a b a b a b a b i

z zSimilary z Rz R a b a b ii

z z

Multiplying i and

+ + +× ∈

+ + + + + +

⇒ − × + + − × + =

⇒ − = ⇒ − ⇒ =

−= ∈ ⇒ =

+

1 2 2 3 2 1 3 2

2 2

1 2 2 3 2 1 3 2 1 3 1 3 2

2 2 2

( ),

: When 0

: When 0 0

R is transitive

ii

a b a b a b a b

CaseI b a

a b a b a b a b a b b a z R

CaseII b a z R

=

≠

= ⇒ = ⇒ ∈

= ⇒ = ∈

⇒

The given relation R is (i) Reflexive (ii) Symmetric (iii)Transitive

The given relation R is an Equivalence relation ⇒

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3

1 2 1 2

3 3 3

1 2 1 2 1

f: R Is the function f(x) = 9x injective?

To check whether the function is injective.i.e1-1,

we need to check whether f(x ) = f(x ) x =x

Her

QUESTION

e , f(x ) = f(x ) 9

:

SOLUT

x =9x

15

IO

x

:

N

9 x -

R→

⇒

⇒ ⇒ ( )( ) ( )

3 3 3

2 1 2

2 2

1 2 1 2 1 2

1 2

0 x - x 0

x -x x +x x x 0

x =x

the function is injective

= ⇒ =

⇒ + =

⇒

∴

1.relations and functions assignment solutions

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