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1.Relation and Function
medium
If $f(x) = \frac{{{x^2} - 1}}{{{x^2} + 1}}$, for every real numbers. then the minimum value of $f$
A
Does not exist because $f$ is bounded
B
Is not attained even through $f$ is bounded
C
Is equal to $+1$
D
Is equal to $-1$
Solution
(d) Let $f(x) = \frac{{{x^2} – 1}}{{{x^2} + 1}} $
$= \frac{{{x^2} + 1 – 2}}{{{x^2} + 1}} = 1 – \frac{2}{{{x^2} + 1}}$
$\because {{x^2} + 1} > 1; \,\, \therefore \frac{2}{{{x^2} + 1}} \le 2$
$\therefore 1 – \frac{2}{{{x^2} + 1}} \ge 1 – 2$;
$\therefore – 1 \le f(x) < 1$
Thus $f(x)$ has the minimum value equal to $-1$.
Standard 12
Mathematics
