If [x] represents the greatest integer leq x, | vedclass

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(B) Given $f(x) = \frac{1}{\sqrt{[x]^2+[x]-2}}$.For $f(x)$ to be defined, we must have $[x]^2+[x]-2 > 0$.Let $[x] = t$. Then $t^2+t-2 > 0$, which fact

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$(0, \frac{1}{2}]$

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(B) Given $f(x) = \frac{1}{\sqrt{[x]^2+[x]-2}}$.For $f(x)$ to be defined, we must have $[x]^2+[x]-2 > 0$.Let $[x] = t$. Then $t^2+t-2 > 0$, which factors as $(t+2)(t-1) > 0$.This implies $t  1$.Since $t = [x]$ is an integer, $[x] \in \{\dots, -4, -3\} \cup \{2, 3, 4, \dots\}$.Case $1$: If $[x] \geq 2$, then $[x]^2+[x]-2$ takes values $2^2+2-2 = 4, 3^2+3-2 = 10, 4^2+4-2 = 18, \dots$.The function values are $\frac{1}{\sqrt{4}}, \frac{1}{\sqrt{10}}, \dots$, i.e.,$(0, \frac{1}{2}]$.Case $2$: If $[x] \leq -3$, then $[x]^2+[x]-2$ takes values $(-3)^2-3-2 = 4, (-4)^2-4-2 = 10, \dots$.The function values are $\frac{1}{\sqrt{4}}, \frac{1}{\sqrt{10}}, \dots$, i.e.,$(0, \frac{1}{2}]$.Combining both cases, the range is $(0, \frac{1}{2}]$.

If $[x]$ represents the greatest integer $\leq x$, then the range of the real-valued function $f(x) = \frac{1}{\sqrt{[x]^2+[x]-2}}$ is

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