If $f(x)=\frac{x^2-1}{x^2+1}$, for every

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Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Relations and Functions

Question:

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

Options:

-1

$-\frac{1}{2}$

$\frac{1}{2}$

Correct Answer:

-1

Explanation:

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 $[x^2+1]>1$ also $(\frac{2}{[x^2+1]})≤2$]

So $1-[\frac{2}{[x^2+1]}]≥1-2$

$-1≤f(x)≤1$

Thus, f (x) has the minimum value equal to –1.