Let f : R – {-4/3} → R be a function defined as f(x) = 4x/(3x+4).

by

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:- One-one: Let x, y ∈ R -{-4/3}

Now, f(x) = f(y) ⇒  \dfrac{4x}{3x + 4} = \dfrac{4y}{3y + 4}

⇒ 12 x y + 16 x = 12x y + 16 y

⇒ 16 x = 16 y

⇒ x = y

Hence f is one-one function.

Onto:- Let, y = f(x) = \dfrac{4x}{3x + 4}

⇒ 3xy + 4y = 4x

⇒ 4y = x(4 – 3y)

⇒ x = \dfrac{4y}{4 - 3y}

Since, y ∈ Range of f ⇒ x ∈ R – {-4/3}

Since, every value of codmain have one and only one pre image in domain

Thus, f is onto function

So, f : R – {-4/3} → R is one-one onto 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 6:- Let A = R – {3}, B = R – {1}. If f : A → B be defined by f(x) = \dfrac{x - 2}{x - 3}, ∀x ∈ A. Then, show that f is bijective.

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

 

Leave a Comment