ExamPlay Dark Logo
Đăng nhập

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

Bình luận (0)

Đăng nhập để bình luận
Quảng cáo
BrainBehindX Inc Logo
©2026; Được cung cấp bởi BrainBehindX Inc