#### please solve rd sharma class 12 chapter 16 Increasing and Decreasing Functions exercise 16.1 , question 7 maths textbook solution

$f(x)$ is neither increasing nor decreasing on R.

Hint:

1. $f(x)$ is increasing on$(a,b)$, if the values of$f(x)$ increase with the increase in the values of x.
2. $f(x)$ is decreasing on $(a,b)$, if the values of$f(x)$ decrease with the increase in the values of x.

Given:

$f(x)=\frac{1}{1+x^{2}}$

To prove:

$f(x)=\frac{1}{1+x^{2}}$ is neither increasing nor decreasing on R

Solution:

Here we have two cases:

Case I: if $x \in [0,\infty)$

Let $x_{1},x_{2} \in [0,\infty)$ Such that $x_{1}>x_{2}$

Then,

$x_{1}>x_{2}$

$\Rightarrow x_{1}^{2}>x_{2}^{2}$     [Squaring on both sides]

$\Rightarrow 1+x_{1}^{2}>1+x_{2}^{2}$    [Adding 1 on both sides]
\begin{aligned} &\Rightarrow \quad \frac{1}{1+x_{1}^{2}}<\frac{1}{1+x_{2}^{2}} \\ &\Rightarrow \quad f\left(x_{1}\right)

Thus, $x_{1}>x_{2} \Rightarrow f\left(x_{1}\right) for all $x_{1},x_{2} \in [0,\infty)$.

So, $f(x)$is a decreasing function on $[0,\infty )$.

Case II: if $x \in (-\infty ,0]$

Let $x_{1},x_{2} \in (-\infty ,0]$ Such that $x_{1}

Then

$x_{1}

$\Rightarrow x_{1}^{2}>x_{2}^{2}$     [Squaring on both sides]

$\Rightarrow 1+x_{1}^{2}>1+x_{2}^{2}$ [Adding 1 on both sides]
\begin{aligned} &\Rightarrow \quad \frac{1}{1+x_{1}^{2}}>\frac{1}{1+x_{2}^{2}} \\ &\Rightarrow \quad f\left(x_{1}\right)>f\left(x_{2}\right) \end{aligned}

Thus, $x_{1}>x_{2} \Rightarrow f\left(x_{1}\right)>f\left(x_{2}\right)$ for all $x_{1},x_{2} \in (-\infty ,0]$.

So, $f(x)$is increasing function on$(-\infty ,0]$.

In both cases we get, $f(x)$is a decreasing function on$[0,\infty )$and increasing function on$(-\infty ,0]$.

Thus, $f(x)$is neither increasing nor decreasing on R.