If f(x) = (x^2 - 1)/(x^2 + 1) for every real number x, then… — JEE Main 1998 Mathematics PYQ

1998

f(x) = \dfrac{x^2 - 1}{x^2 + 1}

for every real number

x

, then the minimum value of

f

A.
does not exist because $f$ is unbounded
B.
is not attained even though $f$ is bounded
C.
is 1
D.
is $-1$

Show answer and solution

Correct option: D

Solution

f(x) = (x²-1)/(x²+1) = 1 - 2/(x²+1). This is minimized when 2/(x²+1) is maximized, which occurs at x = 0, giving f(0) = (0-1)/(0+1) = -1. The minimum value -1 is actually attained at x = 0, so it exists and is achieved.

[1998]

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JEE Main 1998 Mathematics question on Application of Derivatives