The range of the function f(x) = left[ frac{1 | vedclass

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(D) Let $g(x) = \frac{1}{\ln(x^2 + e)}$. Since $x^2 \ge 0$,we have $x^2 + e \ge e$,so $\ln(x^2 + e) \ge \ln(e) = 1$. Thus,$0 < \frac{1}{\ln(x^2 + e)}

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$(0, 1) \cup \{2\}$

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(D) Let $g(x) = \frac{1}{\ln(x^2 + e)}$. Since $x^2 \ge 0$,we have $x^2 + e \ge e$,so $\ln(x^2 + e) \ge \ln(e) = 1$. Thus,$0 If $x = 0$,then $\frac{1}{\ln(e)} = 1$,so $[g(0)] = [1] = 1$. If $x 
e 0$,then $x^2 > 0$,so $x^2 + e > e$,which means $\ln(x^2 + e) > 1$. Thus,$0 Now,$f(x) = [g(x)] + \frac{1}{\sqrt{1 + x^2}}$. If $x = 0$,$f(0) = [g(0)] + \frac{1}{\sqrt{1 + 0}} = 1 + 1 = 2$. If $x 
e 0$,$f(x) = 0 + \frac{1}{\sqrt{1 + x^2}}$. Since $x^2 > 0$,$1 + x^2 > 1$,so $0 As $x 	o \pm \infty$,$f(x) 	o 0$,and as $x 	o 0$,$f(x) 	o 1$. Thus,for $x 
e 0$,the range is $(0, 1)$. Combining these,the range of $f(x)$ is $(0, 1) \cup \{2\}$.

The range of the function $f(x) = \left[ \frac{1}{\ln(x^2 + e)} \right] + \frac{1}{\sqrt{1 + x^2}}$ is,where $[*]$ denotes the greatest integer function and $e = \lim_{\alpha \to 0} (1 + \alpha)^{1/\alpha}$.

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