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Question

Let $$f=\displaystyle \left\{\left(x,\frac{{x}^{2}}{1+{x}^{2}}  \right):x\in R   \right\} $$ be a function from $$R$$ to $$R$$ then range of $$f$$ is .


A
(0,)
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B
[0,1]
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C
(1,0)
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D
(,0)
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Solution

The correct option is B $$[0,1] $$
$$f(x)=\cfrac { { x }^{ 2 } }{ { x }^{ 2 }+1 } $$
is defined for all real $$x$$
now, say
$$y=\cfrac { { x }^{ 2 } }{ { x }^{ 2 }+1 } \\ \Rightarrow y{ x }^{ 2 }+y={ x }^{ 2 }\\ \Rightarrow { x }^{ 2 }(1-y)=y\\ \Rightarrow x=\sqrt { \cfrac { y }{ 1-y }  } $$
$$x$$ is defined only when,
$$\cfrac { y }{ 1-y } \ge 0\\ \Rightarrow \cfrac { y(y-1) }{ { (1-y) }^{ 2 } } \ge 0\\ \Rightarrow y(1-y)\ge 0\\ \therefore y\quad \epsilon \quad [0,1]$$


Mathematics

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