The function f: ℝ → ℝ defined by f(x) = x³ is a function from the real numbers to the real numbers.

Analyzing the function:

1. One-to-one (Injective):
A function is one-to-one if different inputs map to different outputs.
Suppose f(x₁) = f(x₂), i.e., x₁³ = x₂³.
Then, x₁ = x₂.
This demonstrates that f(x) = x³ is one-to-one (injective).

2. Onto (Surjective): A function is onto if every element in the target set, here ℝ, has a pre-image in the domain. For any real number y ∈ ℝ, we can find an x ∈ ℝ such that f(x) = x³ = y.
Specifically, x = ∛y. This demonstrates that the function f(x) = x³ is onto (surjective).

Conclusion: The function f(x) = x³ is one-to-one and onto, so the correct answer is: – One-one and onto.

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We have the relation R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} on the set A = {1, 2, 3, 4}.

Definitions:
1. Function: A relation R is a function if each element of A occurs as the first component in at most one pair of R.
2. Transitive: A relation R is transitive if for all a, b, c ∈ A, whenever (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.
3. Symmetric: A relation R is symmetric if for all a, b ∈ A, whenever (a, b) ∈ R, then (b, a) ∈ R.
4. Reflexive: A relation R is reflexive if for all a ∈ A, (a, a) ∈ R.

Analysis:
1. Is R a function?
– The element 2 occurs as the first element in both (2, 4) and (2, 3), which goes against the definition of a function. Thus, R is not a function.

2. Is R transitive?
– Since (1, 3) ∈ R and (3, 1) ∈ R, we would need (1, 1) ∈ R for it to be transitive, but it isn’t there.
– For (2, 4) ∈ R and (4, 2) ∈ R, we would need (2, 2) ∈ R, which is absent too.
– Thus, R is not transitive.

3. Is R symmetric?
– Do we have (a, b) ∈ R implies (b, a) ∈ R?
– (1, 3) ∈ R, and (3, 1) ∈ R (symmetric pair).
– (4, 2) ∈ R, but (2, 4) ∈ R (symmetric pair).
– All needed symmetric pairs are there. So, R is symmetric.

4. Is R reflexive?
– See if (a, a) ∈ R for every a ∈ A:
– (1, 1), (2, 2), (3, 3), and (4, 4) are not in R.
– Hence, R is not reflexive.

Conclusion:
The relation R is:
– Not a function
– Not transitive
– Symmetric
– Not reflexive

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We are given the relation R on the set N, defined as:
R = {(a, b) : a = b – 2, b > 6}
This is to say, for every pair (a, b), a should be b – 2, and b must be greater than 6.
 
(6, 8) ∈ R
– a = 6 and b = 8.
– a = b – 2 gives 6 = 8 – 2, which is true.
– b = 8 > 6, so the condition is satisfied.
– Therefore, (6, 8) ∈ R.

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