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

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

Answers (1)

Answer:

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)

Posted by

Gurleen Kaur

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