Mathematics Solution Chapter 1
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EXERCISE 1.2 Relations and Functions Questions solutions

7. Explanation of Function Properties

  1. f : ℝ → ℝ defined by f(x) = 3 - 4x
    • One-to-one (Injective): To determine if f(x) = 3 - 4x is one-to-one, we check if different inputs produce different outputs. Since the coefficient of x is negative (-4), the function is decreasing. Therefore, for every different input x1 and x2, f(x1) ≠ f(x2), making f(x) = 3 - 4x one-to-one.
    • Onto (Surjective): To determine if f(x) = 3 - 4x is onto, we check if every real number y has at least one pre-image in ℝ. Since the function is decreasing, there are real numbers that do not have pre-images (e.g., y > 3), making f(x) = 3 - 4x not onto.
    • Bijective: Since f(x) = 3 - 4x is one-to-one but not onto, it is not bijective.
  2. f : ℝ → ℝ defined by f(x) = 1 + x2
    • One-to-one (Injective): To determine if f(x) = 1 + x2 is one-to-one, we check if different inputs produce different outputs. Since x2 is always non-negative, adding 1 ensures that every input x1 and x2 produces a different output, making f(x) = 1 + x2 one-to-one.
    • Onto (Surjective): To determine if f(x) = 1 + x2 is onto, we check if every real number y has at least one pre-image in ℝ. Since x2 is always non-negative, adding 1 ensures that every real number y has at least one pre-image, making f(x) = 1 + x2 onto.
    • Bijective: Since f(x) = 1 + x2 is both one-to-one and onto, it is bijective.

8. Proof that the function f : A × B → B × A, defined by f(a, b) = (b, a), is bijective:

To prove that the function f(a, b) = (b, a) from the Cartesian product of sets A and B to the Cartesian product of sets B and A is bijective:

  1. Injectivity (One-to-one): Let (a1, b1) and (a2, b2) be two distinct ordered pairs in A × B such that f(a1, b1) = f(a2, b2). Then, (b1, a1) = (b2, a2). Since the order of elements in an ordered pair matters, we must have b1 = b2 and a1 = a2, which implies that (a1, b1) = (a2, b2). Therefore, f is injective.
  2. Surjectivity (Onto): Let (b, a) be an arbitrary ordered pair in B × A. Since the function f(a, b) = (b, a), for every (b, a) in B × A, there exists an ordered pair (a, b) in A × B such that f(a, b) = (b, a). Therefore, f is surjective.

Since the function f is both injective and surjective, it is bijective.

9. Checking if the function f : ℕ → ℕ is bijective:

To determine if the function f(n) = { n + 1/2 if n is odd, n/2 if n is even } for all n ∈ ℕ is bijective:

  1. Injectivity (One-to-one): Let m and n be two distinct natural numbers such that f(m) = f(n). If m and n are odd, then (m + 1/2) = (n + 1/2) implies m = n. If m and n are even, then (m/2) = (n/2) implies m = n. If one of m and n is odd and the other is even, then their images under f will be different, ensuring that f is injective.
  2. Surjectivity (Onto): For every natural number n, there exists a natural number m such that f(m) = n. If n is odd, we can choose m = 2n - 1. If n is even, we can choose m = 2n. Therefore, every natural number has a pre-image under f, ensuring that f is surjective.

Since the function f is both injective and surjective, it is bijective.

10. Checking if the function f : A → B is one-to-one and onto:

To determine if the function f(x) = (2/3)x - 3 is one-to-one and onto:

Let's analyze the function f(x) = (2/3)x - 3:

  • Domain A: A = ℝ - {3}
  • Codomain B: B = ℝ - {1}
  1. One-to-one (Injective): If f(x1) = f(x2), then (2/3)x1 - 3 = (2/3)x2 - 3. Simplifying, we get x1 = x2. Therefore, f is one-to-one.
  2. Onto (Surjective): Every element in the codomain B can be achieved as an output of the function. Since the function is a linear function, its range is all real numbers, and it never reaches the excluded value 1. Therefore, f is onto.

Since the function f is both one-to-one and onto, it is bijective.

11. Choosing the correct answer for the function f(x) = x4:

To determine the properties of the function f(x) = x4:

  • One-to-one: To check if f is one-to-one, we need to verify if different inputs produce different outputs. Since every real number has a unique fourth power, f(x) = x4 is one-to-one.
  • Onto: To check if f is onto, we need to verify if every real number is covered by the function. Since the function's range is all non-negative real numbers (including 0), it does not cover negative real numbers. Therefore, f is not onto.

So, the correct answer is (C) f is one-one but not onto.

12. Choosing the correct answer for the function f(x) = 3x:

To determine the properties of the function f(x) = 3x:

  • One-to-one: To check if f is one-to-one, we need to verify if different inputs produce different outputs. Since every real number has a unique scalar multiple by 3, f(x) = 3x is one-to-one.
  • Onto: To check if f is onto, we need to verify if every real number is covered by the function. Since the function's range is all real numbers, it covers the entire codomain. Therefore, f is onto.

So, the correct answer is (A) f is one-one onto.