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Home  >>  CBSE XI  >>  Math  >>  Relations and Functions
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The range of $f (x)=\large\frac{1+x^2}{x^2}$ is

$\begin{array}{1 1} [0,1] \\ (0,1] \\ (1,\infty ) \\ [1, \infty ) \end{array}$

Can you answer this question?

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Ans  (D)  $[1,\infty)$
Both numerator and denominator of $f(x)$ is positive.
$1+x^2$ is > $x^2$


answered Apr 28, 2013 by rvidyagovindarajan_1
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