If f:[2, infty) rightarrow B defined by f(x)= | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the function $f(x) = x^2 - 4x + 5$ defined on the domain $[2, \infty)$.To find the range $B$ for the function to be a bijection, we need to

Vedclass · Accepted Answer

$[1, \infty)$

Vedclass · Answer

(B) Given the function $f(x) = x^2 - 4x + 5$ defined on the domain $[2, \infty)$.To find the range $B$ for the function to be a bijection, we need to determine the set of all possible values of $f(x)$.First, we rewrite the function by completing the square:$f(x) = (x^2 - 4x + 4) + 1 = (x - 2)^2 + 1$.Since the domain is $x \in [2, \infty)$, we have $x - 2 \geq 0$.Therefore, $(x - 2)^2 \geq 0$.Adding $1$ to both sides, we get $(x - 2)^2 + 1 \geq 1$.Thus, $f(x) \geq 1$.The range of the function is $[1, \infty)$.Since the function is strictly increasing on $[2, \infty)$, it is injective. For it to be a bijection, the codomain $B$ must be equal to the range.Therefore, $B = [1, \infty)$.

If $f:[2, \infty) \rightarrow B$ defined by $f(x)=x^2-4x+5$ is a bijection, then $B$ is equal to

Similar Questions

If the domain of the function $f(x) = \sqrt{\log_{0.6} (\left| \frac{2x-5}{x^2-4} \right|)}$ is $(-\infty, a] \cup \{b\} \cup [c, d) \cup (e, \infty)$, then the value of $a+b+c+d+e$ is ————

The range of the function $f(x) = \sqrt{9 - x^2}$ is

The domain of the real valued function $f(x) = \frac{\sqrt{\log _{0.5}(x-3)}}{\sqrt{x-1}}$ is

Domain of the real valued function $f(x) = \frac{x+2}{9-x^{2}}$ is

Range of the function $f(x) = 9 - 7 \sin x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module