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**RELATIONS AND FUNCTIONS****Miscellaneous Exercise****Q.1. ****Let f: R â†’ R be defined as f(x) = 10x + 7. Find the function g: R â†’ R such that gof = fog = I _{R}.**

One-one:

Let f(x) = f(y), where x, y âˆˆ

â‡’ 10x + 7 = 10y + 7

â‡’ x = y

âˆ´ f is a one-one function.

Onto:

For y âˆˆ

Therefore, for any y âˆˆ **R**, there exists

âˆ´ f is onto.

Therefore, f is one-one and onto.

Thus, f is an invertible function.

Let us define g: **R** â†’ **R** as f (y) = y-7/10.

Now, we have

gof = I_{R}, fog = I_{R}

Hence, the required function g: R â†’ R is defined as g(y) = y - 7/10**Q.2. Let f: W â†’ W be defined as f(n) = n âˆ’ 1, if is odd and f(n) = n + 1, if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.****Ans. **It is given that:

One-one:

Let f(n) = f(m).

It can be observed that if n is odd and m is even, then we will have n âˆ’ 1 = m + 1.

â‡’ n âˆ’ m = 2

However, this is impossible.

Similarly, the possibility of n being even and m being odd can also be ignored under a similar argument.

âˆ´ Both n and m must be either odd or even.

Now, if both n and m are odd, then we have:

f(n) = f(m) â‡’ n âˆ’ 1 = m âˆ’ 1 â‡’ n = m

Again, if both n and m are even, then we have:

f(n) = f(m) â‡’ n + 1 = m + 1 â‡’ n = m

âˆ´ f is one-one.

It is clear that any odd number 2r + 1 in co-domain **N **is the image of 2r in domain **N **and any even number 2r in co-domain **N **is the image of 2r + 1 in domain **N**.

âˆ´ f is onto.

Hence, f is an invertible function.

Let us define g: W â†’ W as:

Now, when n is odd:

gof(n) = g(f(n)) = g(n - 1) = n - 1 + 1 = n

And, when n is even:

gof(n) = g(f(n)) = g(n + 1) = n + 1 - 1 = n

Similarly, when m is odd:

fog(m) = f(g(m)) = g(m - 1) = m - 1 + 1 = m

When m is even:

fog(m) = f(g(m)) = g(m + 1) = m + 1 - 1 = m

âˆ´ gof = I_{w} and fog = I_{w}

Thus, f is invertible and the inverse of f is given by f^{â€”1} = g, which is the same as f.

Hence, the inverse of f is f itself.**Q.3. If f: R â†’ R is defined by f(x) = x ^{2} âˆ’ 3x + 2, find f(f(x)).**

= (x

= (x

= x

For one â€“ one

Suppose f(x) = f(y), where x, y âˆˆ R.

It can be observed that if x is positive and y is negative,

Then, we have

Since x is positive and y is negative:

x > y â‡’ x âˆ’ y > 0

But, 2xy is negative.

Then, 2xy â‰ x - y.

Thus, the case of x being positive and y being negative can be ruled out.

Under a similar argument, x being negative and y being positive can also be ruled out.

âˆ´ x and y have to be either positive or negative.

When x and y are both positive, we have

When x and y are both negative, we have

âˆ´ f is one â€“ one.

For onto

Now, let y âˆˆ R such that âˆ’1 < y < 1.

âˆ´ f is onto.

Hence, f is one â€“ one and onto.

For one â€“ one

Suppose f(x) = f(y), where x, y âˆˆ R.

â‡’ x

Now, we need to show that x = y.

Suppose x â‰ y, their cubes will also not be equal.

â‡’ x

However, this will be a contradiction to (1).

âˆ´ x = y

Hence, f is injective.

We first show that g is not injective.

It can be observed that:

g(-1) = |-1| = 1

g(1) = |1| = 1

âˆ´ g(âˆ’1) = g(1), but âˆ’1 â‰ 1.

âˆ´ g is not injective.

Now, gof:

Let x, y âˆˆ

â‡’ |x| = |y|

Since x and y âˆˆ

âˆ´ |x| = |y| â‡’x = y

Hence, gof is injective

We first show that g is not onto.

For this, consider element 1 in co-domain **N**. It is clear that this element is not an image of any of the elements in domain **N**.

âˆ´ f is not onto.

Now, gof: **N** â†’ **N** is defined by,

Then, it is clear that for y âˆˆ **N**, there exists x = y âˆˆ **N** such that gof(x) = y.

Hence, gof is onto.**Q.8. Given a non-empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows:****For subsets A, B in P(X), ARB if and only if A âŠ‚ B. Is R an equivalence relation on P(X)? Justify you answer.****Ans. **Since every set is a subset of itself, ARA for all A âˆˆ P(X).

âˆ´ R is reflexive.

Let ARB â‡’ A âŠ‚ B.

This cannot be implied to B âŠ‚ A.

For instance, if A = {1, 2} and B = {1, 2, 3}, then it cannot be implied that B is related to A.

âˆ´ R is not symmetric.

Further, if ARB and BRC, then A âŠ‚ B and B âŠ‚ C.

â‡’ A âŠ‚ C

â‡’ ARC

âˆ´ R is transitive.

Hence, R is not an equivalence relation as it is not symmetric.**Q.9. Given a non-empty set X, consider the binary operation *: P(X) Ã— P(X) â†’ P(X) given by A * B = A âˆ© B âˆ€ A, B in P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation*.****Ans. **It is given the binary operation *:

P(X) Ã— P(X) â†’ P(X) given by A * B = A âˆ© B âˆ€ A, B in P(X)

We know that A âˆ© X = A = X âˆ© A for all A âˆˆ P(X)

â‡’ A * X = A = X * A for all A âˆˆ P(X)

Thus, X is the identity element for the given binary operation *.

Now, an element A âˆˆ P(X) is invertible if there exists B âˆˆ P(X) such that

A * B = X = B * A [As X is the identity element]

or A âˆ© B = X = B âˆ© A

This case is possible only when A = X = B.

Thus, X is the only invertible element in P(X) with respect to the given operation*.

Hence, the given result is proved.**Q.10. Find the number of all onto functions from the set {1, 2, 3, â€¦ , n) to itself. ** **Ans. **Onto functions from the set {1, 2, 3, â€¦ , n} to itself is simply a permutation on n symbols 1, 2, â€¦, n. Thus, the total number of onto maps from {1, 2, â€¦ , n} to itself is the same as the total number of permutations on n symbols 1, 2, â€¦, n, which is n. 9**Q.11. Let S = {a, b, c} and T = {1, 2, 3}. Find F ^{âˆ’1} of the following functions F from S to T, if it exists.**

â‡’ F (a) = 3, F (b) = 2, F(c) = 1

Therefore, F

Since F (b) = F (c) = 1, F is not one â€“ one.

Hence, F is not invertible i.e., F

a * b = |a - b| and a o b = a, "a, b âˆˆ

For a, b âˆˆ

a * b = |a - b|

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

âˆ´ a * b = b * a

âˆ´ The operation * is commutative.

It can be observed that,

(1 * 2) * 3 = (|1 - 2|) * 3 = 1 * 3 = |1 - 3| = 2

and 1 * (2 * 3) = 1 * (|2 - 3|) = 1 * 1 = |1 - 1| = 0

âˆ´ (1 * 2) * 3 * 1 * (2 * 3) where 1, 2, 3 âˆˆ R.

Hence, the operation * is not associative.

Now. consider the operation o

It can be observed that 1 o 2 = 1 and 2 o 1 = 2.

âˆ´ 1 o 2 â‰ 2 o 1 where 1, 2 âˆˆ R.

Hence, the operation o is not commutative.

Let a.b.c âˆˆ R. Then, we have

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

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

âˆ´ a o b) o c = a o (b o c). where a, b, c âˆˆ R

Hence, the operation o is associative.

Now. let a, b, c âˆˆ R. then we have

a * (b o c) = a * b =|a â€” b|

(a * b) o (a * c) = (|a â€” b|) o (|a â€” c|) = |a â€” b|

Hence, a * (b o c) = (a * b)o (a * c).

Now, 1 o (2 * 3) = 1 o (|2 - 3|) = 1 o 1 = 1

(1 o 2) * (1 o 3) = 1 * 1 = |1 - 1| = 0

âˆ´ 1 o (2 * 3) # (1 o 2) * (1 o 3) where 1, 2, 3 âˆˆ R

Hence, the operation o does not distribute over *.

(Hint: (A âˆ’ Î¦) âˆª (Î¦ âˆ’ A) = A and (A âˆ’ A) âˆª (A âˆ’ A) = A * A = Î¦).

Let A âˆˆ P(X). Then, we have

A * Î¦ = (A - Î¦) U (Î¦ - A)= A U Î¦ = A

Î¦ * A = (Î¦ - A) U (A - Î¦) = Î¦ U A = A

âˆ´ A * Î¦ = A = Î¦ * A for all A âˆˆ P(X)

Thus, Î¦ is the identity element for the given operation *.

Now, an elements A âˆˆ P(X) will be invertible if there exists B âˆˆ P(X) such that

A * B = Î¦ = B * A. [As Î¦ is the identity element]

Now. we observed that

A * A = (A - A) U (A - A) = Î¦ U Î¦ = 0 for all A âˆˆ P(X).

Hence, all the elements A of P(X) are invertible with A

**Show that zero is the identity for this operation and each element a â‰ 0 of the set is invertible with 6 âˆ’ a being the inverse of a. ****Ans. **Let X = {0, 1, 2, 3, 4, 5}.

The operation * on X is defined as

An element e âˆˆ X is the identity element for the operation *. if

a * e = a = e * a for all a âˆˆ X

For a âˆˆ X, we have

a * 0 = a + 0 = a [a âˆˆ X â‡’ a + 0 < 6 ]

0 * a = 0 + a = a [a âˆˆ X â‡’ 0 + a < 6 ]

âˆ´ a * 0 = a = 0 * a for all a âˆˆ X

Thus, 0 is the identity element for the given operation *.

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

â‡’ a = - b or b = 6 - a

But, X = {0, 1, 2, 3, 4, 5} and a, b âˆˆ X. Then, a â‰ -b.

âˆ´ b = 6 - a is the inverse o f a for all a E X .

Hence, the inverse o f an element a âˆˆ X. a â‰ 0 is 6 - a i.e., a^{-1} = 6 - a.**Q.15. Let A = {âˆ’1, 0, 1, 2}, B = {âˆ’4, âˆ’2, 0, 2} and f, ??: A â†’ B be functions defined by f(x) = x ^{2} âˆ’ x, x âˆˆ A **

Also, it is given that f, g : A â†’ B are defined by

It is observed that

Hence, the functions f and g are equal.

(a) 1

The smallest relation containing (1, 2) and (1, 3) which is reflexive and symmetric, but not transitive is given by:

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

This is because relation R is reflexive as (1, 1), (2, 2), (3, 3) âˆˆ R.

Relation R is symmetric since (1, 2), (2, 1) âˆˆ R and (1, 3), (3, 1) âˆˆ R.

But relation R is not transitive as (3, 1), (1, 2) âˆˆ R, but (3, 2) âˆ‰ R.

Now, if we add any two pairs (3, 2) and (2, 3) (or both) to relation R, then relation R will become transitive.

Hence, the total number of desired relations is one.

(a) 1

The smallest equivalence relation containing (1, 2) is given by,

R

Now, we are left with only four pairs i.e., (2, 3), (3, 2), (1, 3), and (3, 1).

If we odd any one pair [say (2, 3)] to R

Also, for transitivity we are required to add (1, 3) and (3, 1).

Hence, the only equivalence relation (bigger than R

This shows that the total number of equivalence relations containing (1, 2) is two.

It is given that,

f:

Also, g:

Now, let x âˆˆ (0, 1].

Then, we have:

Thus, when x âˆˆ (0, 1), we have fog(x) = 0and gof (x) = 1.

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

Hence, the total number of binary operations on the set {a, b} is 24 i.e., 16.

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