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Question

The range of the function $$f(x)=\dfrac{x^{2}}{1+x^{2}}$$ is


Solution

$$\begin{array}{l} Here \\ { x^{ 2 } }\ge 0 \\ { x^{ 2 } }+1>0 \end{array}$$
Since, the denominator is always greater than the numerator for all values other than 0 and at 
$$\begin{array}{l} 0,f\left( x \right) =0 \\ \therefore Range=\left[ { 0,1 } \right)  \end{array}$$

Mathematics

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