f(x) = 5x^2+6x-9 prove that f is bijective.

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

Solution:- To prove f is bijective,it is to prove f is one-one and onto

Here, $f(x) = 5x^2+6x-9$

One-one: Let $x_1,x_2 \in R_+$ , such that

$f(x_1) = f(x_2)$

$\Rightarrow 5x_1^2+6x_1-9=5x_2^2+6x_2-9$

$\Rightarrow 5x_1^2 +6x_1-5x_2^2-6x_2=0$

$\Rightarrow 5(x_1^2-x_2^2)+6(x_1-x_2)=0$

$\Rightarrow 5(x_1+x_2)(x_1-x_2)+6(x_1-x_2)=0$

$\Rightarrow (x_1-x_2)(5x_1+5x_2+6)=0$

$\Rightarrow x_1-x_2 = 0 \text{ and } 5x_1+5x_2+6 \neq 0$

$x_1 = x_2$

i.e. f is one-one function.

Onto:- Let f(x) = y

Since,
$y = 5x^2+6x-9$

$\Rightarrow 5x^2+6x-(9+y) = 0$

$\Rightarrow x = \frac{-6\pm\sqrt{36+4\times 5(9+y)}}{10}$

$\Rightarrow x = \frac{-6\pm\sqrt{216+20y}}{10}$

$\Rightarrow x = \frac{\pm\sqrt{54+5y}-3}{5}$

$\Rightarrow x = \frac{\sqrt{54+5y}-3}{5}$

Obiviously, $y\in [-9, \infty)$ the value of $x\in R_+$

Every value of codomain have pre-image in domain

$\Rightarrow$ f is onto function.

Hence, f is one-one onto function i.e., bijection

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 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.