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Solution :

Here, `f(x) = x/(1+x^2)`<br>
`:. f'(x) = ((1+x^2) - x(2x))/(1+x^2)^2`<br>
`=> f'(x) = (1-x^2)/(1+x^2)^2`<br>
So, `f'(x)` is not always greater than `0` and not always less than `0.`<br>
It means `f(x)` is not one - one or injective function.<br>
Now, let `y` is the range of `f(x).`<br>
Then,<br>
`y = x/(1+x^2)`<br>
`=> y+x^2y - x = 0`<br>
`=>x^2y - x+ y = 0`<br>
Now, this equation should have real roots.<br>
`:. D ge 0`<br>
`=> (-1)^2 - 4(y)(y) ge 0`<br>
`=> 1-4y^2 ge 0`<br>
`=> 4y^2 le 1`<br>
`=> y^2 le 1/4`<br>
So, `y` should be between ` -1/2` and `1/2`, which is the co-domain of `f(x).`<br>
As, codomain and range of `f(x)` are equal, `f(x)` is surjective or onto.<br>
**meaning of function**

**Modulus; signum; greatest integer and fractional part function**

**representation of function**

**Physical interpretation**

**Domain codomain and range of function**

**Graph of function: Vertical line test**

**Identity function**

**Explain Constant function with graph**

**Modulus function and properties**

**Greatest integer function and properties**