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

Answers (1)

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

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}} 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)<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 on [0,\infty ).

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].

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.

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