The domain of the function f(x) = sqrt{frac{1 | vedclass

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Question

(B) For the function $f(x) = \sqrt{\frac{1-|x|}{2-|x|}}$ to be defined,the expression inside the square root must be non-negative and the denominator

Vedclass · Accepted Answer

$[-1, 1] \cup (-\infty, -2) \cup (2, \infty)$

Vedclass · Answer

(B) For the function $f(x) = \sqrt{\frac{1-|x|}{2-|x|}}$ to be defined,the expression inside the square root must be non-negative and the denominator must not be zero.$1$. $\frac{1-|x|}{2-|x|} \geq 0$$2$. $2-|x| 
eq 0 \implies |x| 
eq 2 \implies x 
eq \pm 2$Let $t = |x|$,where $t \geq 0$. The inequality becomes $\frac{1-t}{2-t} \geq 0$.Multiplying by $-1$ in numerator and denominator: $\frac{t-1}{t-2} \geq 0$.Using the wavy curve method for $t \geq 0$,the solution for $t$ is $t \in [0, 1] \cup (2, \infty)$.Now,substitute $t = |x|$ back:Case $I$: $0 \leq |x| \leq 1 \implies x \in [-1, 1]$.Case $II$: $|x| > 2 \implies x \in (-\infty, -2) \cup (2, \infty)$.Combining these,the domain is $x \in (-\infty, -2) \cup [-1, 1] \cup (2, \infty)$.

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

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