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

One-one:

Let f(x) = f(y), where x, y ∈

⇒ 10x + 7 = 10y + 7

⇒ x = y

∴ f is a one-one function.

Onto:

For y ∈

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

∴ f is onto.

Therefore, f is one-one and onto.

Thus, f is an invertible function.

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

Now, we have

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

Hence, the required function g: R → R is defined as g(y) = y - 7/10**Q.2. Let f: W → W be defined as f(n) = n − 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 − 1 = m + 1.

⇒ n − 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.

∴ 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) ⇒ n − 1 = m − 1 ⇒ n = m

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

f(n) = f(m) ⇒ n + 1 = m + 1 ⇒ n = m

∴ 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**.

∴ f is onto.

Hence, f is an invertible function.

Let us define g: W → 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

∴ gof = I_{w} and fog = I_{w}

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

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

= (x

= (x

= x

For one – one

Suppose f(x) = f(y), where x, y ∈ 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 ⇒ x − y > 0

But, 2xy is negative.

Then, 2xy ≠ 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.

∴ 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

∴ f is one – one.

For onto

Now, let y ∈ R such that −1 < y < 1.

∴ f is onto.

Hence, f is one – one and onto.

For one – one

Suppose f(x) = f(y), where x, y ∈ R.

⇒ x

Now, we need to show that x = y.

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

⇒ x

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

∴ 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

∴ g(−1) = g(1), but −1 ≠ 1.

∴ g is not injective.

Now, gof:

Let x, y ∈

⇒ |x| = |y|

Since x and y ∈

∴ |x| = |y| ⇒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**.

∴ f is not onto.

Now, gof: **N** → **N** is defined by,

Then, it is clear that for y ∈ **N**, there exists x = y ∈ **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 ⊂ 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 ∈ P(X).

∴ R is reflexive.

Let ARB ⇒ A ⊂ B.

This cannot be implied to B ⊂ A.

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

∴ R is not symmetric.

Further, if ARB and BRC, then A ⊂ B and B ⊂ C.

⇒ A ⊂ C

⇒ ARC

∴ 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) × P(X) → P(X) given by A * B = A ∩ B ∀ 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) × P(X) → P(X) given by A * B = A ∩ B ∀ A, B in P(X)

We know that A ∩ X = A = X ∩ A for all A ∈ P(X)

⇒ A * X = A = X * A for all A ∈ P(X)

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

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

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

or A ∩ B = X = B ∩ 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, … , n) to itself. ** **Ans. **Onto functions from the set {1, 2, 3, … , n} to itself is simply a permutation on n symbols 1, 2, …, n. Thus, the total number of onto maps from {1, 2, … , n} to itself is the same as the total number of permutations on n symbols 1, 2, …, n, which is n. 9**Q.11. Let S = {a, b, c} and T = {1, 2, 3}. Find F ^{−1} of the following functions F from S to T, if it exists.**

⇒ F (a) = 3, F (b) = 2, F(c) = 1

Therefore, F

Since F (b) = F (c) = 1, F is not one – one.

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

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

For a, b ∈

a * b = |a - b|

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

∴ a * b = b * a

∴ 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

∴ (1 * 2) * 3 * 1 * (2 * 3) where 1, 2, 3 ∈ 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.

∴ 1 o 2 ≠ 2 o 1 where 1, 2 ∈ R.

Hence, the operation o is not commutative.

Let a.b.c ∈ R. Then, we have

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

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

∴ a o b) o c = a o (b o c). where a, b, c ∈ R

Hence, the operation o is associative.

Now. let a, b, c ∈ R. then we have

a * (b o c) = a * b =|a — b|

(a * b) o (a * c) = (|a — b|) o (|a — c|) = |a — 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

∴ 1 o (2 * 3) # (1 o 2) * (1 o 3) where 1, 2, 3 ∈ R

Hence, the operation o does not distribute over *.

(Hint: (A − Φ) ∪ (Φ − A) = A and (A − A) ∪ (A − A) = A * A = Φ).

Let A ∈ P(X). Then, we have

A * Φ = (A - Φ) U (Φ - A)= A U Φ = A

Φ * A = (Φ - A) U (A - Φ) = Φ U A = A

∴ A * Φ = A = Φ * A for all A ∈ P(X)

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

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

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

Now. we observed that

A * A = (A - A) U (A - A) = Φ U Φ = 0 for all A ∈ 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 ≠ 0 of the set is invertible with 6 − 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 ∈ X is the identity element for the operation *. if

a * e = a = e * a for all a ∈ X

For a ∈ X, we have

a * 0 = a + 0 = a [a ∈ X ⇒ a + 0 < 6 ]

0 * a = 0 + a = a [a ∈ X ⇒ 0 + a < 6 ]

∴ a * 0 = a = 0 * a for all a ∈ X

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

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

⇒ a = - b or b = 6 - a

But, X = {0, 1, 2, 3, 4, 5} and a, b ∈ X. Then, a ≠ -b.

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

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

Also, it is given that f, g : A → 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) ∈ R.

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

But relation R is not transitive as (3, 1), (1, 2) ∈ R, but (3, 2) ∉ 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 ∈ (0, 1].

Then, we have:

Thus, when x ∈ (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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