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Question
Prove that the function f given by f(x) = log cos x is strictly decreasing on `(0, pi/2)` and strictly increasing on `((3pi)/2, 2pi).`
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Solution
Here f(x) = log cos x
`therefore f'(x) = 1/(cos x) (- sin x) = - tan x`
(i) In the interval `(0, pi/2)`, tan x = + ve
∴ f' (x) = - ve
Hence, f is a decreasing function.
(ii) In the interval `(pi/2, pi)`, tan x = - ve
∴ f' (x) = - tan x = - ve
Hence, f is an increasing function.
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