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  • Provide solution for RD Sharma maths class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 26

Provide solution for RD Sharma maths class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 26

Answers (1)

Answer:

f(x) is increasing in (0,\infty)

and f(x) is decreasing (-1,0)

Given:

f(x)=log(1+x)-\frac{x}{1+x}

To find:

We have to find the intervals in which f(x) is increasing and decreasing.

Hint:

First, we find critical point then use property of increasing and decreasing.

Solution:

We have,

f(x)=log(1+x)-\frac{x}{1+x}

Differentiating w.r.t. x we get,

\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left\{\log (1+x)-\frac{x}{1+x}\right\} \\ &f^{\prime}(x)=\frac{1}{1+x}-\left[\frac{(1+x)-x}{(1+x)^{2}}\right] \end{aligned}

\begin{aligned} &=\frac{1}{1+x}-\frac{1}{(1+x)^{2}},\left[\therefore \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x} \cdot u \frac{d v}{d x}}{v^{2}}\right] \\ &=\frac{x}{(1+x)^{2}} \end{aligned}

For critical points. We must have,

\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow \frac{x}{(1+x)^{2}}=0\\ &\Rightarrow x=0 \text { and domain of }(1+x)^{2} \text { is }(-1, \infty)\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x>0\\ &\text { and } f^{\prime}(x)<0 \text { if }-1<x<0 \end{aligned}

Hence,f(x) is increasing in (0,\infty), decreases in (-1,0).

Posted by

Gurleen Kaur

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