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  • Explain solution RD Sharma class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 31 maths

Explain solution RD Sharma class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 31 maths

Answers (1)

Answer:

f(x) \text { is strictly increasing on }(-\frac{\pi }{2},0) \text { and strictly decreasing on }(0,\frac{\pi }{2})

Given:

f(x)=log\: cos\: x

To prove:

\text { We have to prove that } f(x) \text { is strictly increasing on }(-\frac{\pi }{2},0) \text { and strictly decreasing on }(0,\frac{\pi }{2})

Hint:

  1. for f(x) to be increasing we must have f'(x)>0
  2. for f(x) to be decreasing we must have f'(x)<0

Solution:

Given

f(x)=log\: cos\: x

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

\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}[\log \cos x] \\ &\Rightarrow f^{\prime}(x)=\frac{1}{\cos x} \times(-\sin x) \\ &\Rightarrow f^{\prime}(x)=-\tan x \\ &\text { In interval }\left(0, \frac{\pi}{2}\right), \tan x>0 \\ &\Rightarrow-\tan x<0 \end{aligned}

By applying negative sign change comparison sign

\therefore f^{\prime}(x)<0 \text { on }\left(0, \frac{\pi}{2}\right)

\text { Hence } f(x) \text { is strictly decreasing on }(0,\frac{\pi }{2})

\begin{aligned} &\text { In interval }\left(\frac{\pi}{2}, \pi\right), \tan x<0 \\ &\Rightarrow-\tan x>0 \end{aligned}

By applying negative sign change comparison sign

\therefore f^{\prime}(x)>0 \text { on }\left(\frac{\pi}{2}, \pi\right)

\text { Hence } f(x) \text { is strictly increasing on }(\frac{\pi }{2},\pi )

Posted by

Gurleen Kaur

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