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please solve rd sharma class 12 chapter 16 Increasing and Decreasing Functions exercise 16 .1 , question 6 maths textbook solution

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               f(x) is a decreasing function on (0,\infty ].

f(x) is increasing on  [-\infty , 0)

Hint:

  1. f(x) is increasing on(a,b), if the values off(x) increase with the increase in the values of x.
  2. f(x) is decreasing on (a,b), if the values off(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}} decreases in the interval  (0,\infty ] and increases in the interval [-\infty , 0).

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)>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 [0,\infty).

So, f(x)is a decreasing function.

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

Let x_{1},x_{2} \in (-\infty ,0] 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)<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].

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