SATHEE: Application of Derivatives 4 Question 22

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

(d) is -1

Show Answer

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 $x = 0$ .

$∴$ Minimum value of $f (x)$ is $f (0) = - 1$

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Application of Derivatives 4 Question 22

23. If f(x)=x2−1x2+1, for every real number x, then the minimum value of f

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23. If $f (x) = \frac{x^{2} - 1}{x^{2} + 1}$ , for every real number $x$ , then the minimum value of $f$