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Joint Entrance Examination
Mathematics
Functions
Question 141
Question 141 of 162
Medium
MCQ
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
A
\[\left(\frac{5}{37}, \frac{2}{5}\right]-\left\{\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}\right\}\]
B
\[\left(\frac{5}{37}, \frac{2}{5}\right]\]
C
\[\left(\frac{5}{26}, \frac{2}{5}\right]\]
D
\[\left(\frac{5}{26}, \frac{2}{5}\right]-\left\{\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}\right\}\]
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Official solution
f
(
x
)
=
k
1
+
x
2
f(x)=\frac{k}{1+x^{2}}
f
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x
)
=
1
+
x
2
k
is a decreasing function where
k
>
0
k>0
k
>
0
\[ \begin{gathered} \therefore \quad x \in[2,3) \Rightarrow f(x)=\frac{2}{1+x^{2}} \in\left(\frac{2}{10}, \frac{2}{5}\right]=R_{1} \\\\ x \in[3,4) \Rightarrow f(x)=\frac{3}{1+x^{2}} \in\left(\frac{3}{17}, \frac{3}{10}\right]=R_{2} \\\\ x \in[4,5) \Rightarrow f(x)=\frac{4}{1+x^{2}} \in\left(\frac{4}{26}, \frac{4}{17}\right]=R_{3} \\\\ x \in[5,6) \Rightarrow f(x)=\frac{5}{1+x^{2}} \in\left(\frac{5}{37}, \frac{5}{26}\right]=R_{4} \\\\ \text { Range }=R_{1} \cup R_{2} \cup R_{3} \cup R_{4} \\\\ =\left(\frac{5}{37}, \frac{2}{5}\right] \end{gathered} \]
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