JEE MATHEMATICS MCQ: MAXIMA & MINIMA

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Analyzing +4 | -1

f\left( x \right) = \frac{{{x^2} - 1}}{{{x^2} + 1}}

, for every real number

x

, then minimum value of

f\left( x \right)

View answer and explanation

Correct option: D

I

equal to

- 1

Option D (

I

equal to

- 1

) is correct because it is the minimum value of the function, found by analyzing its derivative.

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