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

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

If $f:[-3,2] \rightarrow [0, \sqrt[3]{x}]$ is an onto function defined by $f(n) = \begin{cases} 2+\sqrt[3]{n}, & -3 \leq n \leq -1 \\ n^{2/3}, & -1 < n \leq 2 \end{cases}$,then $x=$

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

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

The least value of $2^{(x^2 - 3)^3+27}$ is-

Vedclass Test Series

Exam Paper Generator

Online Exam Module