Question

The function $\frac{1}{1+x^2}$ is decreasing in the interval

• A. $(-\infty ,1]$
• B. $(-\infty ,0]$
• C. $[1,\infty )$
• D. $(0,\infty )$

Answer 1

The correct option is D

$(0,\infty )$

The explanation of the correct option

The given function, $f\left(x\right)=\frac{1}{1+x^2}$.

Differentiate the given function with respect to $x$.

$$\begin{gathered} \frac{\mathrm{d}}{dx}\left[f\left(x\right)\right]=\frac{d}{dx}\left(\frac{1}{1+x^2}\right) \\ \Rightarrow f\text{'}\left(x\right)=\frac{\left(1+x^2\right){\displaystyle \frac{d}{dx}}\left(1\right)-\left(1\right){\displaystyle \frac{d}{dx}}\left(1+x^2\right)}{{\left(1+x^2\right)}^2}\left[\because \frac{d}{d\mathrm{x}}\left(\frac{\mathrm{u}}{\mathrm{v}}\right)=\frac{\mathrm{v}{\displaystyle \frac{d\mathrm{u}}{d\mathrm{x}}}-\mathrm{u}{\displaystyle \frac{d\mathrm{v}}{d\mathrm{x}}}}{{\mathrm{v}}^2}\right] \\ \Rightarrow f\text{'}\left(x\right)=\frac{\left(1+x^2\right)\times 0-\left(2x\right)}{{\left(1+x^2\right)}^2} \\ \Rightarrow f\text{'}\left(x\right)=\frac{-2x}{{\left(1+x^2\right)}^2} \end{gathered}$$

Now, for decreasing intervals $f\text{'}\left(\mathrm{x}\right)<0$.

$$\begin{gathered} \Rightarrow \frac{-2x}{{\left(1+x^2\right)}^2}<0 \\ \Rightarrow -2x<0\times {\left(1+x^2\right)}^2\left[\because {\left(1+x^2\right)}^2\ge 0\right] \\ \Rightarrow -2x<0 \\ \Rightarrow \frac{-2x}{-2}>\frac{0}{-2} \\ \Rightarrow x>0 \end{gathered}$$

So, the function $\frac{1}{1+x^2}$ is decreasing in the interval $\left(0,\infty \right)$.

Hence, option (D) is correct.