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: We have, where a, b ∈ R
Reflexivity:- Obiviously, is a real number and is not true.
Therefore, R is not reflexive.
Symmetric:- Let the real numbers 1 and 2
such that
is not true
Thus
Hence, R is not symmetric.
Transitive:- Let
and and
is not true
Hence, R is not transitive.
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 given by prove that f is bijective.
Solution: See full solution
Question 4:- Consider → [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 6:- Let A = R – {3}, B = R – {1}. If f : A → B be defined by , ∀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
Class 12 ncert solution math exercise 1.1