Question:
Let \(f:(1,3)\to\mathbb{R}\) be a function defined by \(f(x)=\dfrac{x[x]}{1+x^2}\), where \([x]\) denotes the greatest integer \(\le x\). Then the range of \(f\) is
Let \(f:(1,3)\to\mathbb{R}\) be a function defined by \(f(x)=\dfrac{x[x]}{1+x^2}\), where \([x]\) denotes the greatest integer \(\le x\). Then the range of \(f\) is
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Split the domain: \([x]=1\) on \((1,2)\) and \([x]=2\) on \([2,3)\); on each piece the function is decreasing, so read the range off the end points.
Updated On: Nov 5, 2026
- \(\left(\dfrac25,\dfrac12\right)\cup\left(\dfrac34,\dfrac45\right]\)
- \(\left(\dfrac35,\dfrac45\right)\)
- \(\left(\dfrac25,\dfrac45\right]\)
- \(\left(\dfrac25,\dfrac35\right]\cup\left(\dfrac34,\dfrac45\right)\)
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The Correct Option is A
Solution and Explanation
For \(1\lt x\lt2\): \([x]=1\), \(f(x)=\dfrac{x}{1+x^2}\). Its derivative \(\dfrac{1-x^2}{(1+x^2)^2}\lt0\), so \(f\) is decreasing. Values go from \(f(1)=\dfrac12\) (not attained) down to \(f(2)=\dfrac25\) (not attained): range \(\left(\dfrac25,\dfrac12\right)\).
For \(2\le x\lt3\): \([x]=2\), \(f(x)=\dfrac{2x}{1+x^2}\), again decreasing. \(f(2)=\dfrac45\) (attained), \(f(3)=\dfrac{6}{10}=\dfrac35\) (not attained): range \(\left(\dfrac35,\dfrac45\right]\).
Total range \(=\left(\dfrac25,\dfrac12\right)\cup\left(\dfrac35,\dfrac45\right]\).
The printed option (1) has this shape (open interval \(\left(\tfrac25,\tfrac12\right)\) joined to an interval closed at \(\tfrac45\)); the paper prints its middle end point as \(\tfrac34\) where the computation gives \(\tfrac35\). It is the only option consistent with the result.
Answer: (1) \(\left(\dfrac25,\dfrac12\right)\cup\left(\dfrac34,\dfrac45\right]\) (as printed in the paper)
For \(2\le x\lt3\): \([x]=2\), \(f(x)=\dfrac{2x}{1+x^2}\), again decreasing. \(f(2)=\dfrac45\) (attained), \(f(3)=\dfrac{6}{10}=\dfrac35\) (not attained): range \(\left(\dfrac35,\dfrac45\right]\).
Total range \(=\left(\dfrac25,\dfrac12\right)\cup\left(\dfrac35,\dfrac45\right]\).
The printed option (1) has this shape (open interval \(\left(\tfrac25,\tfrac12\right)\) joined to an interval closed at \(\tfrac45\)); the paper prints its middle end point as \(\tfrac34\) where the computation gives \(\tfrac35\). It is the only option consistent with the result.
Answer: (1) \(\left(\dfrac25,\dfrac12\right)\cup\left(\dfrac34,\dfrac45\right]\) (as printed in the paper)
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