Maharashtra BoardEnglish MediumSTD 11 ScienceMathsFunctions2 Marks
MCQ
The range of the function $f(x)=\frac{1+x^2}{x^2}$ is
A
(0,1)
B
[0, 1]
✓
$(1, \infty)$
D
$[1, \infty)$
✓
Answer
Correct option: C.
$(1, \infty)$
(C) $f (x)$ is defined for all $x \in R -\{0\}$. So, $\operatorname{dom}(f)=R-\{0\}$ Let $y=\frac{1+x^2}{x^2}$ $\Rightarrow x= \pm \sqrt{\frac{1}{y-1}}$ For $x$ to be real, $y-1>0 \Rightarrow y \in(1, \infty)$
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