The domain of the real-valued function f(x) = | vedclass

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

Question

(D) Given,$f(x) = \sqrt{\frac{2-|x|}{3-|x|}}$.For $f(x)$ to be defined,we must have $\frac{2-|x|}{3-|x|} \geq 0$.Let $t = |x|$,where $t \geq 0$. The i

Vedclass · Accepted Answer

$(-\infty, -3) \cup [-2, 2] \cup (3, \infty)$

Vedclass · Answer

(D) Given,$f(x) = \sqrt{\frac{2-|x|}{3-|x|}}$.For $f(x)$ to be defined,we must have $\frac{2-|x|}{3-|x|} \geq 0$.Let $t = |x|$,where $t \geq 0$. The inequality becomes $\frac{2-t}{3-t} \geq 0$,which is equivalent to $\frac{t-2}{t-3} \geq 0$.Using the wavy curve method for $t \geq 0$,the solution for $t$ is $t \in [0, 2] \cup (3, \infty)$.Now,substitute $t = |x|$ back:Case $I$: $0 \leq |x| \leq 2 \Rightarrow x \in [-2, 2]$.Case $II$: $|x| > 3 \Rightarrow x \in (-\infty, -3) \cup (3, \infty)$.Combining these,the domain is $x \in (-\infty, -3) \cup [-2, 2] \cup (3, \infty)$.

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

Similar Questions

If $[x]$ denotes the greatest integer $\leq x$, then the range of the real-valued function $f(x) = \frac{1}{\sqrt{x-[x]}}$ is

The domain of the function $f(x) = \frac{\sqrt{1 + x} - \sqrt{1 - x}}{x}$ is

The number of integral values of $\lambda$ for which the function $f(x) = \sqrt{\ln(2\lambda \cos x + 5)}$ is defined for all $x \in R$ is:

The domain of the definition of the function $y(x)$ given by the equation $2^x+2^y=2$ is

Let $a > 1$ be a constant. If $f: A \rightarrow A$ and $(x, y) \in f$ satisfy $a^x + a^y = a$, then $A =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module