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  • Please solve RD Sharma class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 13 maths textbook solution

Please solve RD Sharma class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 13 maths textbook solution

Answers (1)

Answer:

f(x) is decreasing on (0,\pi)  and increasing on (-\pi,0) and neither increasing nor decreasing on (-\pi,\pi).

Given:

f(x)=cos\: x

To show:

We have to show that f(x) is decreasing on (0,\pi)  and increasing on (-\pi,0) and neither increasing nor decreasing on (-\pi,\pi).

Hint:

  1. For increasing function f'(x)>0 then f(x) is increasing.
  2. For f(x) to be decreasing f'(x)<0

Solution:

Given

f(x)=cos\: x

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

\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}(\cos x) \\ &\Rightarrow f^{\prime}(x)=-\sin x \end{aligned}

Taking different region from -\pi to \pi.

\begin{aligned} &\text { Let } x \in(0, \pi) \\ &\Rightarrow \sin x>0 \\ &\Rightarrow-\sin x<0 \end{aligned}

By applying negative sign change comparison sign.

\begin{aligned} &\Rightarrow f^{\prime}(x)<0\\ \end{aligned}

Thus f(x) is decreasing (0,\pi)

\begin{aligned} &\text { Let } x \in(-\pi, 0)\\ &\Longrightarrow \sin x<0\\ &\Rightarrow-\sin x>0 \end{aligned}

By applying negative sign change comparison sign.

\Rightarrow f'(x)> 0

Thus f(x) is increasing (-\pi,0).

Therefore from above condition we find that

?f(x) is decreasing in (0, \pi) and increasing in (-\pi,0)

Hence, f(x) is neither increasing nor decreasing in  (-\pi, \pi)

Posted by

Gurleen Kaur

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