In order to determine which of the relations listed is reflexive, we must recall that a relation R on a set A is reflexive if for every element x ∈ A, the pair (x, x) is in R. This means that x must be related to itself.

Let’s examine each of the relations listed:

(a) R = {(x, y) : x > y, x, y ∈ ℕ}

For this relation to be reflexive, we must have x > x for all x ∈ ℕ. However, this is never the case since x is never greater than itself. So this relation is not reflexive.

(b) R = {(x, y) : x + y = 10, x, y ∈ ℕ}

For this relation to be reflexive, we require that x + x = 10 for all x ∈ ℕ. This yields 2x = 10, or x = 5. So, the only element that satisfies this is x = 5. Hence, the relation is **not reflexive** for all elements of ℕ, but only for x = 5.

(c) R = {(x, y) : xy is a square number, x, y ∈ ℕ}

For this relation to be reflexive, we must have x * x to be a square number for all x ∈ ℕ. Since x * x = x² is always a square number for all natural numbers x, this relation is reflexive.

(d) R = {(x, y) : x + 4y = 10, x, y ∈ ℕ}

For this relation to be reflexive, we require x + 4x = 10 for all x ∈ ℕ. This simplifies to 5x = 10, or x = 2. Hence, the only element satisfying this condition is x = 2, so this relation is not reflexive for all elements of ℕ, only for x = 2.

Conclusion:
The only reflexive relation is:
– (c) R = {(x, y) : xy is a square number, x, y ∈ ℕ}.

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We are given the relation R on ℤ, defined by:

aRb if and only if a² – 7ab + 6b² = 0

We need to find out the properties of this relation: whether it is reflexive, symmetric, and/or transitive.

Step 1: Check if the relation is reflexive.

A relation R is reflexive if for every element a ∈ ℤ, we have aRa. That is, we must check whether the equation a² – 7a ⋅ a + 6a² = 0 holds for every integer a.

Substitute a = b into the relation:

a² – 7a ⋅ a + 6a² = a² – 7a² + 6a² = 0

This reduces to:

0 = 0

Thus, the relation is reflexive because the equation holds for all a ∈ ℤ.

Step 2: Check if the relation is symmetric.

A relation R is symmetric if for every pair (a, b) ∈ ℤ, whenever aRb, we also have bRa. That is, if a² – 7ab + 6b² = 0 we need to see if b² – 7ba + 6a² = 0 also holds.

Since the relation aRb is defined by the equation a² – 7ab + 6b² = 0, we simply reverse a and b in the equation:

b² – 7ba + 6a² = 0

This is exactly the same as a² – 7ab + 6b² = 0 (just switching a and b does not affect the equation because it is symmetric in a and b).

Thus, the relation is symmetric.

Step 3: Determine if the relation is transitive.

A relation R is transitive if for every a, b, c ∈ ℤ, whenever aRb and bRc, we must have aRc. To check transitivity, we would need to check the equation a² – 7ab + 6b² = 0 and b² – 7bc + 6c² = 0, and then determine if a² – 7ac + 6c² = 0 holds. This is however complicated and does not ensure the transitivity simply by observing the structure of the equation (an algebraic check would be necessary).

For now, let us focus on the properties that have been given, namely, the reflexive and symmetric.
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We have a function f: ℝ → ℝ defined as

f(x) = 4 + 3 cos(x)

Whether the given function is bijective, one-one, onto, or neither can be determined.
Step 1 Check if the given function is one-one (injective).

A function is said to be one-one (injective) if the different elements in the domain of a function go to different elements in the range i.e. f(x₁) = f(x₂) => x₁ = x₂.

Let’s see if the function is injective. Assume that f(x₁) = f(x₂), that is:

4 + 3 cos(x₁) = 4 + 3 cos(x₂)

We simplify this to obtain:

3 cos(x₁) = 3 cos(x₂)
cos(x₁) = cos(x₂)

The cosine function is periodic and not injective. That means that there may be different values of x₁ and x₂, like x₁ = 0 and x₂ = 2π, for which cos(x₁) = cos(x₂). In other words, the function is not injective.

Step 2: Determine whether the function is onto (surjective).

A function is onto (surjective) if for every element y ∈ ℝ, there exists an x ∈ ℝ such that f(x) = y.

For f(x) = 4 + 3 cos(x), the range of the cosine function is between -1 and 1, so the range of f(x) will be:

f(x) = 4 + 3 cos(x)

Since cos(x) lies between -1 and 1, the values of f(x) will lie between:

f(x) = 4 + 3(-1) = 1 and f(x) = 4 + 3(1) = 7

So, f(x) can only be taken in the interval [1, 7]. Since f(x) cannot take any value outside that interval, it is not onto because no values of x map to values of y that are outside that interval.

Conclusion:
The function f(x) = 4 + 3 cos(x) is neither one-one nor onto. So the correct option is
(d) neither one-one nor onto.

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