Question

prove that the function f given by f(x) =log|cosx| is decreasing on(0,pi/2) and increasing on (3pi/2, 2pi)


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Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = log(cosx)$

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

$\rm :\longmapsto\:\dfrac{d}{dx}f(x) = \dfrac{d}{dx} log(cosx)$

$\rm :\longmapsto\:f'(x) =\dfrac{1}{cosx} \dfrac{d}{dx} cosx$

$\rm :\longmapsto\:f'(x) =\dfrac{1}{cosx} ( - sinx)$

$\rm :\longmapsto\:f'(x) = - \: tanx$

$\begin{gathered}\boxed{\begin{array}{c|c} \bf interval & \bf sign \: of \: f'(x) \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf (0, \dfrac{\pi}{2}) & \sf - ve \\ \\ \sf ( \dfrac{3\pi}{2},2\pi) & \sf + ve \end{array}} \\ \end{gathered}$

$\rm :\implies\:f(x) \: is \: increasing \: on \: \bigg(\dfrac{3\pi}{2}, 2\pi \bigg)$

and

$\rm :\implies\:f(x) \: is \: deccreasing \: on \: \bigg(0,\dfrac{\pi}{2} \bigg)$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\green{ \boxed{ \bf \: \dfrac{d}{dx}cosx = - \: sinx}}$

$\green{ \boxed{ \bf \: \dfrac{d}{dx}logx = \dfrac{1}{x} }}$

$\green{ \boxed{ \bf \: tanx > 0 \: if \: x \in \: \bigg(0,\dfrac{\pi}{2} \bigg) \cup \: \bigg(\pi,\dfrac{3\pi}{2} \bigg) }}$

$\green{ \boxed{ \bf \: tanx < 0 \: if \: x \in \: \bigg(\dfrac{\pi}{2},\pi \bigg) \cup \: \bigg(\dfrac{3\pi}{2},2\pi \bigg) }}$

Additional Information :-

1. If the functions f(x) and g(x) are increasing (or decreasing) on the interval (a,b), then the sum of the functions f(x) + g(x) is also increasing (or decreasing) on this interval.

2. If the function f(x)  is increasing (or decreasing) on the interval (a,b), then the opposite function − f(x) is decreasing (or increasing) on this interval.

3. If the function f(x) is increasing (or decreasing) on the interval (a,b), then the inverse function of f(x)  is decreasing (or increasing) on this interval.