R= {(a, b): a ≤ b²} is neither reflexive nor symmetric nor transitive

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, R=\{(a,b): a\leq b^2\} where a, b ∈ R

Reflexivity:- Obiviously, \frac{1}{2} is a real number and \frac{1}{2}\leq (\frac{1}{2})^2 is not true.

\Rightarrow \left(\frac{1}{2},\frac{1}{2}\right)\notin R

Therefore, R is not reflexive.

Symmetric:- Let the real numbers 1 and 2

such that (1,2)\in R \Rightarrow 1\leq 2^2

\Rightarrow 2 \leq 1^2 is not true

Thus (2,1)\notin R

Hence, R is not symmetric.

Transitive:- Let a = 2, b = -2, c = 1

(2,-2)\in R and (-2, 1)\in R \quad \Rightarrow 2 \leq (-2)^2 and -2\leq (1)^2

\quad \Rightarrow 2 \leq (1)^2 is not true

\Rightarrow (2,1)\notin R

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 \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 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

Class 12 ncert solution math exercise 1.1

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