question_answer

Let \[f:\left( 1,\text{ }3 \right)\to R\] be a function defined by\[f\left( x \right)=\frac{x\left[ x \right]}{1+{{x}^{2}}}\], where [x] denotes the greatest integer\[\le x\]. Then the range of f is             [JEE MAIN Held on 08-01-2020 Evening]

A) \[\left( \frac{2}{5},\left. \frac{3}{5} \right] \right.\cup \left( \frac{3}{4},\frac{4}{5} \right)\]

B) \[\left( \frac{2}{5},\frac{1}{2} \right)\cup \left( \frac{3}{5},\frac{4}{5} \right]\]

C) \[\left( \frac{2}{5},\frac{4}{5} \right]\]    

D) \[\left( \frac{3}{5},\frac{4}{5} \right)\]

Correct Answer: B

Solution :

\[f\left( x \right)=\frac{x\left[ x \right]}{{{x}^{2}}+1};1<x<3\] \[\Rightarrow \,\,\,\,\,\,\,f\left( x \right)=\left\{ \frac{\frac{1}{\left( x+\frac{1}{x} \right)}\,\,\,\,\,\,\,1<x<2}{\frac{2}{\left( x+\frac{1}{x} \right)}\,\,\,\,\,2\le x<3} \right.\] \[\therefore \,\,\,\,\,\,f\left( x \right)\]is decreasing function. \[\therefore \,\,\,\,\,\,\]Range is \[\left( \frac{2}{5},\,\,\frac{1}{2} \right)\bigcup \left( \frac{3}{5},\,\,\frac{4}{5} \right]\]