Question

The range of $\frac{1 + x^{2}}{x^{2}}$

• A. $(0,1)$
• B. $(1,\infty)$
• C. [0, 1]
• D. $\lbrack 1,\infty)$

Answer 1

Answer: (B)

$(1,\infty)$

Answer 2

Let $y = \frac{1 + x^{2}}{x^{2}} \Rightarrow x^{2}y = 1 + x^{2}$ $\Rightarrow x^{2}(y - 1) = 1$

$\Rightarrow x^{2} = \frac{1}{y - 1}$

Now since, $x^{2} > 0 \Rightarrow \frac{1}{y - 1} > 0 \Rightarrow (y - 1) > 0$ $\Rightarrow y > 1$

$\Rightarrow y \in (1,\infty)$

Trick : $y = \frac{1 + x^{2}}{x^{2}} = 1 + \frac{1}{x^{2}}$. Now since, $\frac{1}{x^{2}}$ is always > 0 $\Rightarrow y > 1 \Rightarrow y \in (1,\infty)$.