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The range of $f (x)=\large\frac{1+x^2}{x^2}$ is

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

1 Answer

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

 

answered Apr 28, 2013 by rvidyagovindarajan_1
 
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