Question

The function $ f(x)=\frac{x}{1+{{x}^{2}}} $ decreases in the interval

• A. $ (-\infty ,\,\,-1]\cup [1,\infty ) $
• B. $ (-1,1) $
• C. $ (-\infty ,\infty ) $
• D. $None\, of\, the\, above$

Answer 1

Answer: (A)

Given, $ f(x)=\frac{x}{1+{{x}^{2}}} $
On differentiating w. r. t. x, we get
$ f'(x)=\frac{(1+{{x}^{2}})\,(1)\,-x\,(2x)}{{{(1+{{x}^{2}})}^{2}}} $
$ =\frac{1-{{x}^{2}}}{{{(1+{{x}^{2}})}^{2}}} $
For $ f(x) $ to be decreasing
$ f'(x)\le 0 $
$ \Rightarrow $ $ (1-{{x}^{2}})\le 0 $
$ \Rightarrow $ $ {{x}^{2}}\ge 1 $
$ \Rightarrow $ $ x\in (-\infty ,\,\,-1]\,\cup [1,\,\infty ) $