If f: A rightarrow B is an onto function such | vedclass

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

Question

(B) Given the function $f(x) = \sqrt{|x|-x} + \frac{1}{\sqrt{|x|-x}}$.For $f(x)$ to be defined,the expression inside the square root must be strictly

Vedclass · Accepted Answer

$(-\infty, 0), [2, \infty)$

Vedclass · Answer

(B) Given the function $f(x) = \sqrt{|x|-x} + \frac{1}{\sqrt{|x|-x}}$.For $f(x)$ to be defined,the expression inside the square root must be strictly positive:$|x| - x > 0 \Rightarrow |x| > x$.This inequality holds true for all $x Now,for $x \in (-\infty, 0)$,we have $|x| = -x$.Substituting this into the function:$f(x) = \sqrt{-x - x} + \frac{1}{\sqrt{-x - x}} = \sqrt{-2x} + \frac{1}{\sqrt{-2x}}$.Let $t = \sqrt{-2x}$. Since $x  0$,so $t > 0$.The function becomes $f(t) = t + \frac{1}{t}$.Using the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality for $t > 0$:$\frac{t + \frac{1}{t}}{2} \geq \sqrt{t \cdot \frac{1}{t}} = 1 \Rightarrow t + \frac{1}{t} \geq 2$.Since $f(x)$ is an onto function,the codomain $B$ must be equal to the range of $f(x)$,which is $[2, \infty)$.Therefore,$A = (-\infty, 0)$ and $B = [2, \infty)$.

If $f: A \rightarrow B$ is an onto function such that $f(x)=\sqrt{|x|-x}+\frac{1}{\sqrt{|x|-x}}$,then $A$ and $B$ are respectively.

Similar Questions

Find the domain of the function $f(x) = \frac{x^{2}+3x+5}{x^{2}-5x+4}$.

Let $f: R - \{2, 6\} \rightarrow R$ be a real-valued function defined as $f(x) = \frac{x^2+2x+1}{x^2-8x+12}$. Then the range of $f$ is

The range of the function $f(x) = \frac{x^2+x+1}{x^2-x+1}$ is

Find the range of the following function:
$f(x) = 2 - 3x$,where $x \in R$ and $x > 0$.

The domain of the real valued function $f(x) = \sqrt{2+x} + \sqrt{3-x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module