1. 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 thatRead more

    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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  2. In order to find the number of one-one and onto mappings or bijective functions from set A to set B, consider the following: 1. One-one mapping: Each element of A should map to a unique element of B. 2. Onto mapping: Every element of B should be associated with at least one element of A. ObservationRead more

    In order to find the number of one-one and onto mappings or bijective functions from set A to set B, consider the following:

    1. One-one mapping: Each element of A should map to a unique element of B.
    2. Onto mapping: Every element of B should be associated with at least one element of A.

    Observations:
    Set A has 5 elements (|A| = 5), and set B has 6 elements (|B| = 6).
    – For a bijection (one-one and onto), the no. of elements in both the sets must be equal, that is, |A| = |B|.
    Now, |A| ≠ |B| hence it is not possible to have a one-one and onto mapping.
    Conclusion:
    The number of one-one and onto mappings from A to B is: 
    0.

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  3. 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, wheneverRead more

    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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  4. Let A = {1, 2, 3} and B = {4, 5, 6, 7}. Let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. 1. Injective (One-to-one): Each element of A goes to a distinct element in B. There are no two elements of A going to the same element in B. Therefore, f is injective. 2. Surjective (Onto): Not all elRead more

    Let A = {1, 2, 3} and B = {4, 5, 6, 7}.
    Let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B.

    1. Injective (One-to-one): Each element of A goes to a distinct element in B. There are no two elements of A going to the same element in B.
    Therefore, f is injective.

    2. Surjective (Onto):
    Not all elements of B are covered by f. For instance, 7 is not mapped by any element of A.
    Therefore, f is not surjective.

    3. Bijective (One-to-one and Onto):
    Since f is not surjective, it cannot be bijective.

    4. Function:
    f is a function since every element of A is related to exactly one element of B.

    Therefore, the correct answer is: Injective function.

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  5. 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.Read more

    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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