Question

F(x)=log sinx is strictly increasing is it possible interval in square braket

Answer 1

$\bf\large\underline\pink{||Solution ||}$

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]

Hence proved

Answer 2

Answer:

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].....