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

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

If a relation $R$ on the set of natural numbers $N$ is defined by $x + 2y = 8$,then the domain of $R$ is:

The range of the function $f(x) = (\sin x)^{\sin x}$ defined on $(0, \pi)$ is

The domain of the function $y=f(x)$,where $x$ and $y$ are related by $2^x+2^y=2$ is

The domain of the real valued function $f(x) = \log_2 \log_3 \log_5(x^2 - 5x + 11)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module