#### Need solution for RD Sharma maths class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 19

f(x) is neither decreasing nor increasing on (0,1)

Given:

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

To prove:

We have to show that f(x) is neither decreasing nor increasing on (0,1).

Hint:

1. condition for f(x) to be increasing is f'(x)>0
2. condition for f(x) to be decreasing is f'(x)<0

Solution:

Given

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

On differentiating both sides w.r.t x we get

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

Taking different region from (0,1)

\begin{aligned} &\text { Let } x \in\left(0, \frac{1}{2}\right) \\ &\Rightarrow 2 x-1<0 \\ &\Rightarrow f^{\prime}(x)<0 \end{aligned}

\begin{aligned} &\text { Thus } f(x) \text { is decreasing }\left(0, \frac{1}{2}\right) \\ \end{aligned}

\begin{aligned} &\text { Let } x \in\left(\frac{1}{2}, 1\right) \\ &\Rightarrow 2 x-1>0 \\ &\Rightarrow f^{\prime}(x)>0 \end{aligned}

\begin{aligned} &\text { Thus } f(x) \text { is increasing }\left(\frac{1}{2},1\right) \\ \end{aligned}

Therefor from above condition we find that

$\Rightarrow f(x) \text { is decreasing in } (0,\frac{1}{2}) \text { and increasing in } (\frac{1}{2},1)$

Hence, f(x) is neither increasing nor decreasing in (0,1)