# Class 12 NCERT Solutions- Mathematics Part I – Chapter 1 Relations And Functions -Miscellaneous Exercise on Chapter 1 | Set 2

### (i) F = {(a, 3), (b, 2), (c, 1)}

Solution:

As, F = {(a, 3), (b, 2), (c, 1)} and S = {a,b,c} and T={1,2,3}

F: Sâ†’T is defined as

F(a) = 3, F(b) = 2 and F(c) = 1

F is one-one and onto.

Taking F-1, so F-1: Tâ†’S

a = F-1(3), b = F-1(2) and c = F-1(1)

F-1 = {(3,a),(2,b),(1,c)}

### (ii) F = {(a, 2), (b, 1), (c, 1)}

Solution:

As, F = {(a, 2), (b, 1), (c, 1)}

F: Sâ†’T is defined as

F(a) = 2, F(b) = 1 and F(c) = 1

Here, F(b) = F(c) but b â‰  c

Hence, F is not one-one.

So, F is not invertible and F-1 doesn’t exists.

### Question 12. Consider the binary operations âˆ— : R Ã— R â†’ R and o : R Ã— R â†’ R defined as a âˆ—b = |a â€“ b| and a o b = a, âˆ€ a, b âˆˆ R. Show that âˆ— is commutative but not associative, o is associative but not commutative. Further, show that âˆ€ a, b, c âˆˆ R, a âˆ— (b o c) = (a âˆ— b) o (a âˆ— c). [If it is so, we say that the operation âˆ— distributes over the operation o]. Does o distribute over âˆ—? Justify your answer.

Solution:

Binary operations âˆ— : R Ã— R â†’ R defined as a âˆ—b = |a â€“ b|

a*b = |a-b|

b*a = |b-a| = |-(a-b)| = |a-b|

a*b = b*a

Hence, âˆ— is commutative.

Now, let’s take a=1, b=2 and c=3 for better understanding

a*(b*c) = a*|b-c| = |a-|b-c|| = |1-|2-3|| = 0

(a*b)*c = |a-b|*c = ||a-b|-c| = ||1-2|-3| = 2

a*(b*c) â‰  (a*b)*c

Hence, âˆ— is not associative.

Binary operations o : R Ã— R â†’ R defined as a o b = a, âˆ€ a, b âˆˆ R

a o b = a

b o a = b

a o b â‰  b o a

Hence, o is not commutative.

a o (b o c) = a o b = a

(a o b) o c = a o c = a

a o (b o c) â‰  (a o b) o c

Hence, o is associative.

Let’s check for a âˆ— (b o c) = (a âˆ— b) o (a âˆ— c) a, b, c âˆˆ R

a âˆ— (b o c) = a * b = |a-b|

(a âˆ— b) o (a âˆ— c) = |a-b| o |a-c| = |a-b|

Hence, a âˆ— (b o c) = (a âˆ— b) o (a âˆ— c)

Now, let’s check for a o (b * c) = (a o b) * (a o c)

a o (b * c) = a

(a o b) * (a o c) = a * a = |a-a| = 0

Hence, a o (b * c) â‰  (a o b) * (a o c)

o does not distribute over âˆ—

### (Hint : (A â€“ Ï†) âˆª (Ï† â€“ A) = A and (A â€“ A) âˆª (A â€“ A) = A âˆ— A = Ï†).

Solution:

Set X, such that P(X) Ã— P(X) â†’ P(X) be defined as A * B = (A â€“ B) âˆª (B â€“ A), âˆ€ A, B âˆˆ P(X)

Ï†*A = (Ï†-A) U (A-Ï†) = Ï† U A = A

A*Ï† = (A-Ï†) U (Ï†-A) = A U Ï† = A

Hence, Ï† is the identity element for the operation * on P(X)

A*A = (A-A) U (A-A) = Ï† U Ï† = Ï†

â‡’ A = A-1

Hence, all the elements A of P(X) are invertible with Aâ€“1 = A.

### Show that zero is the identity for this operation and each element a â‰  0 of the set is invertible with 6 â€“ a being the inverse of a.

Solution:

Let the set x = {0, 1, 2, 3, 4, 5}

Let’s take i as identity element, where a*i = a = i*a âˆ€ a âˆˆ x

a*0 = a

0*a = a, when (a+0<6)

Hence, zero is the identity for this operation

An element a âˆˆ x is invertible if there exists b âˆˆ x such that a*b = b*a = 0

From above equations, we have

a = -b or b = 6-a

But, as x = {0, 1, 2, 3, 4, 5} and a,bâˆˆ x. Then aâ‰ -b

Hence, b = 6-a is the inverse of an element aâˆˆ x

aâ‰ 0

a-1 = 6-a

### (Hint: One may note that two functions f : A â†’ B and g : A â†’ B such that f(a) = g (a) âˆ€ a âˆˆ A, are called equal functions).

Solution:

Given, f, g : A â†’ B be functions defined by f(x) = x2 â€“ x, x âˆˆ A and g(x) =   x âˆˆ A

At x = -1

f(0) = (-1)2 – (-1) = 2

g(0) =  = 2

Here, f(-1) = g(-1) and 2=2

At x = 0

f(0) = 02 – 0 = 0

g(0) =  = 0

Here, f(0) = g(0) and 0=0

At x = 1

f(1) = 12 – 1 = 0

g(1) =  = 0

Here, f(1) = g(1) and 1=1

At x = 2

f(1) = 22 – 2 = 2

g(1) =  = 2

Here, f(2) = g(2) and 2=2

For, every câˆˆ A, f(c) = g(c)

Hence, f and g are equal functions.

### (D) 4

Solution:

R = {(1,1),(1,2),(1,3),(2,1),(2,2),(3,1),(3,3)}

Reflexive : (1,1), (2,2), (3,3) âˆˆ R

Symmetric: (1,2), (2,1)âˆˆ R and (1,3), (3,1) âˆˆ R

R is not Transitive because, (1,2), (1,3) âˆˆ R but (3,2) âˆ‰R

So, if we will add (3,2) and (2,3) or both, then R will become transitive.

New, R = {(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)}

Hence, A is the correct option.

### Question 17. Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is

(A) 1

(B) 2

(C) 3

(D) 4

Solution:

Smallest equivalence relations containing (1, 2):

R = {(1,1),(2,2),(1,2),(2,1),(3,3)}

or R = {(1,1),(2,2),(1,2),(2,1),(3,2),(2,3)(3,3)}

Hence, B is the correct option.

### and g : R â†’ R be the Greatest Integer Function given by g (x) = [x], where [x] is greatest integer less than or equal to x. Then, does fog and gof coincide in (0, 1]?

Solution:

Given, f : R â†’ R and g : R â†’ R

when x âˆˆ (0,1]

[x] = 1, when x=1

[x] = 0, when 0<x<1

Now, fog(x)=f(g(x)) = f([x])

And, Now gof(x) = g(f(x))

g(1) = [1] = 1

g(0) = [0] = 0

g(-1) = [-1] = -1

When x âˆˆ (0,1), fog = 0 and gof = 1. fog(1) â‰  gof(1)

Hence, fog and gof do not coincide in (0, 1].

### (D ) 8

Solution:

Let A = {a,b}

A x A = {a,b} x {a,b}

R = {(a,a),(a,b),(b,a),(b,b)}

Number of elements are 4.

Hence, the number of binary operations on the set will be 24 = 16

Hence, B is the correct option.

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