JEE Mains · Maths · STD 12 - 1. relation and function
Let \(f:(1,3) \rightarrow \mathrm{R}\) be a function defined by \(f(\mathrm{x})=\frac{\mathrm{x}[\mathrm{x}]}{1+\mathrm{x}^{2}},\) where \([\mathrm{x}]\) denotes the greatest integer \(\leq \mathrm{x} .\) Then the range of \(f\) is
- A \(\left(\frac{3}{5}, \frac{4}{5}\right)\)
- B \(\left(\frac{2}{5}, \frac{3}{5}\right] \cup\left(\frac{3}{4}, \frac{4}{5}\right)\)
- C \(\left(\frac{2}{5}, \frac{4}{5}\right]\)
- D \(\left(\frac{2}{5}, \frac{1}{2}\right) \cup\left(\frac{3}{5}, \frac{4}{5}\right]\)
Answer & Solution
Correct Answer
(D) \(\left(\frac{2}{5}, \frac{1}{2}\right) \cup\left(\frac{3}{5}, \frac{4}{5}\right]\)
Step-by-step Solution
Detailed explanation
\(f(\mathrm{x})=\left\{\begin{array}{ll}{\frac{\mathrm{x}}{\mathrm{x}^{2}+1}} & {; \quad \mathrm{x} \in(1,2)} \\ {\frac{2 \mathrm{x}}{\mathrm{x}^{2}+1}} & {; \quad \mathrm{x} \in[2,3)}\end{array}\right.\) \(f(\mathrm{x})\) is decreasing function…
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