ExamPlay Dark Logo
Войти

JEE Advance - Mathematics (1998 - No. 39)

If $$f\left( x \right) = {{{x^2} - 1} \over {{x^2} + 1}},$$ for every real number $$x$$, then the minimum value of $$f$$
does not exist because $$f$$ is unbounded
is not attained even though $$f$$ is bounded
is equal to 1
is equal to -1

Комментарии (0)

Войдите, чтобы оставить комментарий
Реклама
BrainBehindX Inc Logo
©2026; Питаться от BrainBehindX Inc