Differentiation and its Application: If f(x)=dfracx^2-1x^2+1 , for every real number x , then the minimum… | IIT-JEE 1998

IIT-JEE 1998Question 1

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

Differentiation and its Application

does not exist because $f$ is unboundedis not attained even though $f$ is boundedis equal to $1$is equal to $-1$✓ Correct answer

Solution

Concept. Write $f$ as a constant minus a positive fraction. Formulas used. (1) $\dfrac{x^2-1}{x^2+1}=1-\dfrac{2}{x^2+1}$Step 1 - Bound. $x^2+1\ge1$, so $\frac{2}{x^2+1}\le2$ and by (1) $f(x)\ge-1$. Step 2 - Attained. $f(0)=-1$. So the minimum value is $-1$. Answer: (D)

Differentiation and its Application - JEE Mathematics question (IIT-JEE 1998)

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