1. Given function and domain

We have $f(x)=\frac{[x]}{1+x^2}, \qquad x\in[2,6).$ Here $[x]$ denotes the greatest integer function.

We must find the range of $f(x)$ on $[2,6)$.

2. Break the domain according to the value of $[x]$

Since $x\in[2,6)$, the possible values of $[x]$ are:

• $2$ for $x\in[2,3)$
• $3$ for $x\in[3,4)$
• $4$ for $x\in[4,5)$
• $5$ for $x\in[5,6)$

So on each interval,

$f(x)=\frac{n}{1+x^2} \quad \text{for } x\in[n,n+1), \; n=2,3,4,5.$

3. Find the range on each sub-interval

For fixed $n$, the function $\frac{n}{1+x^2}$ is decreasing in $x$ for $x>0$, because denominator increases as $x$ increases.

So on each interval $[n,n+1)$, the maximum occurs at $x=n$, and the infimum occurs as $x\to (n+1)^-$.

(i) For $x\in[2,3)$

$f(x)=\frac{2}{1+x^2}$ Hence range is $\left(\frac{2}{1+3^2},\frac{2}{1+2^2}\right]=\left(\frac{2}{10},\frac{2}{5}\right]=\left(\frac15,\frac25\right].$

(ii) For $x\in[3,4)$

$f(x)=\frac{3}{1+x^2}$ Hence range is $\left(\frac{3}{1+4^2},\frac{3}{1+3^2}\right]=\left(\frac{3}{17},\frac{3}{10}\right].$

(iii) For $x\in[4,5)$

$f(x)=\frac{4}{1+x^2}$ Hence range is $\left(\frac{4}{1+5^2},\frac{4}{1+4^2}\right]=\left(\frac{4}{26},\frac{4}{17}\right]=\left(\frac{2}{13},\frac{4}{17}\right].$

(iv) For $x\in[5,6)$

$f(x)=\frac{5}{1+x^2}$ Hence range is $\left(\frac{5}{1+6^2},\frac{5}{1+5^2}\right]=\left(\frac{5}{37},\frac{5}{26}\right].$

4. Take the union of these ranges

So total range is $\left(\frac15,\frac25\right]\;\cup\;\left(\frac{3}{17},\frac{3}{10}\right]\;\cup\;\left(\frac{2}{13},\frac{4}{17}\right]\;\cup\;\left(\frac{5}{37},\frac{5}{26}\right].$

Now compare endpoints numerically:

• $\frac{5}{37}\approx 0.1351$
• $\frac{2}{13}\approx 0.1538$
• $\frac{4}{26}=\frac{2}{13}$
• $\frac{3}{17}\approx 0.1765$
• $\frac15=0.2$
• $\frac{5}{26}\approx 0.1923$
• $\frac{4}{17}\approx 0.2353$
• $\frac{3}{10}=0.3$
• $\frac25=0.4$

These intervals overlap continuously:

• $\left(\frac{5}{37},\frac{5}{26}\right]$
• overlaps with $\left(\frac{2}{13},\frac{4}{17}\right]$ since $\frac{2}{13}<\frac{5}{26}$
• overlaps with $\left(\frac{3}{17},\frac{3}{10}\right]$
• overlaps with $\left(\frac15,\frac25\right]$

Hence the union becomes one continuous interval: $\left(\frac{5}{37},\frac25\right].$

5. Check whether any values inside are missing

Because the four interval-ranges overlap, there are no gaps. So the range is exactly $\boxed{\left(\frac{5}{37},\frac25\right]}.$

6. Match with options

This is Option B.

7. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.