Question

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

Answer 1

The given function f is defined as,

f( x )=logsinx

The derivative of the function f is given as,

f ′ ( x )= df( x ) dx = d( logsinx ) dx = 1 sinx ×cosx =cotx

For the given interval ( 0, π 2 ),

cotx>0

Hence, f ′ ( x )>0.

As f ′ ( x )>0 in the interval ( 0, π 2 ), therefore f is strictly increasing in the interval ( 0, π 2 ).

For the given interval ( π 2 ,π ),

cotx<0

Hence, f ′ ( x )<0.

As f ′ ( x )<0 in the interval ( π 2 ,π ), therefore f is strictly decreasing in the interval ( π 2 ,π ).

Hence, it is proved that the function f is strictly increasing in interval ( 0, π 2 ) and strictly decreasing in the interval ( π 2 ,π ).