If $$f(x)=x2$$−$$1x2+1$$, for every real number, then minimum value of f

Correct option is 4is equal to -1

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$If f (x) = \frac{x^{2} - 1}{x^{2} + 1}, for every real number, then minimum value of f$

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does not exist

is not attained even though f is bounded

is equal to 1

is equal to -1

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detailed solution

Correct option is D

We have f(x)=1−2x2+1f(x) will be minimum if 2/x2+1 is maximum i.e. if x2+1 is least i.e. when x=0. Thus minimum value of f(x) is f (0) = −1

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If f(x)=x2−1x2+1, for every real number, then minimum value of f

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$If f (x) = \frac{x^{2} - 1}{x^{2} + 1}, for every real number, then minimum value of f$