JEE MAIN - Mathematics (2023 - 31st January Morning Shift - No. 7)
If the domain of the function $$f(x)=\frac{[x]}{1+x^{2}}$$, where $$[x]$$ is greatest integer $$\leq x$$, is $$[2,6)$$, then its range is
$$\left(\frac{5}{37}, \frac{2}{5}\right]-\left\{\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}\right\}$$
$$\left(\frac{5}{37}, \frac{2}{5}\right]$$
$$\left(\frac{5}{26}, \frac{2}{5}\right]$$
$$\left(\frac{5}{26}, \frac{2}{5}\right]-\left\{\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}\right\}$$
Explanation
$f(x)=\frac{k}{1+x^{2}}$ is a decreasing function where $k>0$
$$ \begin{gathered} \therefore \quad x \in[2,3) \Rightarrow f(x)=\frac{2}{1+x^{2}} \in\left(\frac{2}{10}, \frac{2}{5}\right]=R_{1} \\\\ x \in[3,4) \Rightarrow f(x)=\frac{3}{1+x^{2}} \in\left(\frac{3}{17}, \frac{3}{10}\right]=R_{2} \\\\ x \in[4,5) \Rightarrow f(x)=\frac{4}{1+x^{2}} \in\left(\frac{4}{26}, \frac{4}{17}\right]=R_{3} \\\\ x \in[5,6) \Rightarrow f(x)=\frac{5}{1+x^{2}} \in\left(\frac{5}{37}, \frac{5}{26}\right]=R_{4} \\\\ \text { Range }=R_{1} \cup R_{2} \cup R_{3} \cup R_{4} \\\\ =\left(\frac{5}{37}, \frac{2}{5}\right] \end{gathered} $$
$$ \begin{gathered} \therefore \quad x \in[2,3) \Rightarrow f(x)=\frac{2}{1+x^{2}} \in\left(\frac{2}{10}, \frac{2}{5}\right]=R_{1} \\\\ x \in[3,4) \Rightarrow f(x)=\frac{3}{1+x^{2}} \in\left(\frac{3}{17}, \frac{3}{10}\right]=R_{2} \\\\ x \in[4,5) \Rightarrow f(x)=\frac{4}{1+x^{2}} \in\left(\frac{4}{26}, \frac{4}{17}\right]=R_{3} \\\\ x \in[5,6) \Rightarrow f(x)=\frac{5}{1+x^{2}} \in\left(\frac{5}{37}, \frac{5}{26}\right]=R_{4} \\\\ \text { Range }=R_{1} \cup R_{2} \cup R_{3} \cup R_{4} \\\\ =\left(\frac{5}{37}, \frac{2}{5}\right] \end{gathered} $$
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