Given by f(x) = x² + 4. Show that f is invertible

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

Solution:- One-one: Let x, y ∈ $R_+$ (Domain)

such that f(x) = f(y) ⇒ x² + 4 = y² + 4

⇒ x² = y²

⇒ x = y [∴ x, y are +ve real number]

Hence, f is one-one function.

Onto:- Let y ∈ [4, ∞) such that

y = f(x) ∀ x ∈ $R_+$ [set of non-negative reals]

⇒ y = x² + 4

⇒ x = √(y – 4) [∴ x is + ve real number]

Obiviously, ∀ y ∈ [4, ∞), x is real number ∈ $R_+$ (domain)

i.e. all elements of co-domain have pre image in domain.

⇒ f is onto

Hence,f is invertible being one-one onto.

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 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 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.