Question

Prove that the function f given by f(x) = log sin x is strictly increasing on [0,π/2] and strictly decreasing on[π/2,π]

Answer 1

given, f(x) = logsinx
differentiate f(x) with respect to x ,
f'(x) = d{logsinx}/dx
= 1/sinx × d(sinx)/dx
= 1/sinx × cosx
= cosx/sinx = cotx
hence, f'(x) = cotx

in interval [0, π/2], cotx > 0
e.g., cotx = f'(x) > 0
therefore, function f(x) is strictly increasing on [0, π/2]

in interval [π/2, π] , cotx < 0
e.g., cotx = f'(x) < 0
therefore, function f(x) is strictly decreasing on [π/2, π]

$\textbf{\underline{hence proved}}$

Answer 2

Answer:

Step-by-step explanation:

Hope you understood.