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 \[\left( \frac{2}{5},\,\,\frac{1}{2} \right)\bigcup \left( \frac{3}{5},\,\,\frac{4}{5} \right]\]You need to login to perform this action.
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