Single Choice

If the function $$f : R \rightarrow$$ A given by $$f(x)\, =\, \displaystyle \frac{x^{2}}{x^{2}\, +\, 1}$$ is a surjection, then A is

A$$R$$
B$$[0, 1]$$
C$$(0, 1]$$
D$$[0, 1)$$
Correct Answer

Solution

As 'f' is surjective,
Range of f = co-domain of 'f'
$$\implies$$ A = range of 'f'
Since, $$f(x)=\dfrac{x^2}{x^2+1}$$
$$\implies$$ $$y=\dfrac{x^2}{x^2+1}$$
$$\implies$$ $$y(x^2+1)=x^2$$
$$\implies$$ $$(y-1)x^2+y=0$$
$$\implies$$ $$x^2=\dfrac{-y}{y-1}$$
$$\implies$$ $$x=\sqrt{\dfrac{y}{1-y}}$$
$$\implies$$ $$\dfrac{y}{1-y} \geq 0$$
$$\implies$$ $$y \epsilon [0,1)$$
$$\implies$$ $$A=[0,1)$$

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