Application of Derivatives 4 Question 22
####23. If $f(x)=\frac{x^{2}-1}{x^{2}+1}$, for every real number $x$, then the minimum value of $f$
(1998, 2M)
(a) does not exist because $f$ is unbounded
(b) is not attained even though $f$ is bounded
(c) is 1
(d) is -1
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Answer:
Correct Answer: 23. (b)
Solution:
- Given, $f(x)=\frac{x^{2}-1}{x^{2}+1}=1-\frac{2}{x^{2}+1}$
$f(x)$ will be minimum, when $\frac{2}{x^{2}+1}$ is maximum,
i.e. when $x^{2}+1$ is minimum,
i.e. at $x=0$.
$\therefore \quad$ Minimum value of $f(x)$ is $f(0)=-1$
