If f : A → B be defined by f(x) = {x – 2}/{x – 3}

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Question 6:- Let A = R – {3}, B = R – {1}. If f : A → B be defined by f(x) = {x – 2}/{x – 3}, ∀x ∈ A. Then, show that f is bijective.

Solution:- Given that, A = R – {3}, B = R – {1}.

f : A → B be defined by f(x) = \dfrac{x - 2}{x - 3}, ∀x ∈ A

For injectivity

Let  f(x) = f(y) ⇒ \dfrac{x - 2}{x - 3} = \dfrac{y - 2}{y - 3}

⇒ (x – 2 )(y -3) = (y – 2)(x – 3)

⇒ x y – 3x – 2y + 6 = x y – 3y – 2x + 6

⇒ – 3 x – 2y = -3y – 2x

⇒ – x = -y ⇒ x = y

So, f(x) is an injective function.

For surjectivity

Let, y =f(x) = \dfrac{x - 2}{x - 3}

⇒ x – 2 = xy – 3y

⇒ x(1-y) = 2 – 3y

⇒ x = \dfrac{2 - 3y}{1 - y}

⇒ x = \dfrac{3y - 2}{y - 1} \in A, ∀ y ∈ B [codomain]

Since, every value of codomain have preimage in domain

So, f(x) is surjective function.

Hence, f(x) is a bijective function.

Other question:-

Question 1:-  Show that the relation R on the set R of real numbers, defined as R= {(a, b): a ≤ b²} is neither reflexive nor symmetric nor transitive.

Solution: See full solution

Question 2: Check whether the relation R in R defined as R={(a, b): a ≤  b³} is reflexive, symmetric or transitive.

Solution: See full solution

Question 3:-    Consider \large f:R_+ \rightarrow [-9, \infty) given by \large f(x) = 5x^2+6x-9 prove that f is bijective.

Solution: See full solution

Question 4:- Consider f:R_+ → [4, ∞) given by f(x) = x² + 4. Show that f is invertible.         [CBSE(AI) 2013]

Solution: See full solution

Question 5:-  Let f : R – {-4/3} → R be a function defined as f(x) = 4x/(3x+4). Show that, f:R – {-4/3} → Range of f, f is one-one and onto.                      [CBSE  2017(C)]

Solution:- See full solution

Question 7:- Let f : W → W, be defined as f(x) = x – 1, if x is odd and f(x) = x + 1, Show that f is bijective.

Solution: See full solution

 

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