Range of the function f (x) = [ 1 ( x^2 + e) ] , , + , ,1 1 + x^2… - Vidyadip

MCQ

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 $e =$ $\mathop {Limit}\limits_{\alpha \to 0} {(1 + \alpha )^{1/\alpha }}\,$

A
$\left( {0,\,\frac{{e + 1}}{e}} \right)$ $\cup \{2\}$
B
$(0, 1)$
C
$(0, 1] \cup \{2\}$
✓
$(0, 1) \cup \{2\}$

✓

Answer

Correct option: D.

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

d
$\left[ {\frac{1}{{\ln ({x^2} + e)}}} \right]\, - \left[ \begin{gathered} 0\,\,\,\,x\, \ne \,0 \hfill \\ 1\,\,\,\,\,x\, = \,0 \hfill \\ \end{gathered} \right.$

$f(x) - \left[ \begin{gathered} 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\, = \,0 \hfill \\ \frac{1}{{\sqrt {1 + {x^2}} }}\,\,\,\,x \ne 0 \hfill \\ \end{gathered} \right.$
Hence range of $f (x)$ is $(0, 1) \cup {2}$

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