Question

Prove that the function f given by f(x) = log cos x is strictly decreasing on and strictly increasing on

Answer 1

question is ----> Prove that the function f given by f(x) = log cos x is strictly decreasing (0,π/2) on and strictly increasing on (π/2 , π).

solution :- given f(x) = log cosx
differentiate f(x) with respect to x
f'(x) = $\frac{1}{cosx}(-sinx)$
= -sinx/cosx = - tanx
e.g., f'(x) = -tanx

we know, any function is strictly increasing only when f'(x) > 0 and function is strictly decreasing only when f'(x) < 0 .

$\underline{ for\:strictly\:increasing,}$
f'(x) = - tanx > 0
tanx < 0
and we know, tanx will be negative when x belongs to (π/2 , π).

$\underline{for\:strictly\:decreasing ,}$
f'(x) = - tanx < 0
tanx > 0
and we know , tanx will be positive when x belongs to (0, π/2).


hence, it is clear that f(x) = log cosx is strictly decreasing on (0,π/2) and strictly increasing on (π/2, π).