The domain of definition of f(x) = sqrt{frac{ | vedclass

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(B) For $f(x) = \sqrt{\frac{1-|x|}{2-|x|}}$ to be defined,the expression inside the square root must be non-negative: $\frac{1-|x|}{2-|x|} \geq 0$. Mu

Vedclass · Accepted Answer

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

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(B) For $f(x) = \sqrt{\frac{1-|x|}{2-|x|}}$ to be defined,the expression inside the square root must be non-negative: $\frac{1-|x|}{2-|x|} \geq 0$. Multiplying by $-1$ on both sides,we get $\frac{|x|-1}{|x|-2} \leq 0$. Let $t = |x|$. Then $\frac{t-1}{t-2} \leq 0$. Since $t = |x| \geq 0$,the critical points are $t=1$ and $t=2$. The inequality holds for $1 \leq t Thus,$1 \leq |x| This implies $x \in (-2, -1] \cup [1, 2)$ is incorrect based on the original expression. Re-evaluating $\frac{1-|x|}{2-|x|} \geq 0$: The expression is $\geq 0$ when the numerator and denominator have the same sign. Case $1$: $1-|x| \geq 0$ and $2-|x| > 0 \Rightarrow |x| \leq 1$ and $|x| Case $2$: $1-|x| \leq 0$ and $2-|x|  2$ $\Rightarrow |x| > 2$ $\Rightarrow x \in (-\infty, -2) \cup (2, \infty)$. Combining these,the domain is $(-\infty, -2) \cup [-1, 1] \cup (2, \infty)$.

The domain of definition of $f(x) = \sqrt{\frac{1-|x|}{2-|x|}}$ is: (Here $(a, b) = \{x : a < x < b\}$ and $[a, b] = \{x : a \leq x \leq b\}$)

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