Prove that the function f given by f(x)=logsinx is strictly increasing on (0,2π) and strictly decreasing on ( 2π,π )

We have, f(x)=logsinx∴ f′(x)=sinx1cosx=cotxIn interval (0,2π),f′(x)=cotx>0.∴ f is strictly increasing in (0,2π).In interval ( 2π,π ),f′(x)=cotx<0.∴ f is strictly decreasing in ( 2π,π ).

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Prove that the function f given by f(x) = x is strictly incr

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Prove that the function $f$ given by $f (x) = lo g sin x$ is strictly increasing on $(0, \frac{π}{2})$ and strictly decreasing on $(\frac{π}{2}, π)$

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Prove that the function

f

given by

f (x) = lo g sin x

is strictly increasing on

(0, \frac{π}{2})

and strictly decreasing on

(\frac{π}{2}, π)

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

We have,

f (x) = lo g sin x

∴ f^{'} (x) = \frac{1}{sin x} cos x = co t x

In interval

(0, \frac{π}{2}), f^{'} (x) = cot x > 0

∴

f

is strictly increasing in

(0, \frac{π}{2})

.
In interval

(\frac{π}{2}, π), f^{'} (x) = cot x < 0.

∴

f

is strictly decreasing in

(\frac{π}{2}, π)

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Question Text	Prove that the function $f$ given by $f (x) = lo g sin x$ is strictly increasing on $(0, \frac{π}{2})$ and strictly decreasing on $(\frac{π}{2}, π)$
Answer Type	Text solution:1
Upvotes	150

Prove that the function f given by f(x)=logsinx is strictly increasing on (0,2π​) and strictly decreasing on ( 2π​,π )

Text solutionVerified

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Prove that the function $f$ given by $f (x) = lo g sin x$ is strictly increasing on $(0, \frac{π}{2})$ and strictly decreasing on $(\frac{π}{2}, π)$