Question

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

Answer 1

The given function f is defined as,

f( x )=logcosx

The derivative of the function f is given as,

f ′ ( x )= df( x ) dx = d( logcosdx ) dx = 1 cosx ×( −sinx ) =−tanx

For the given interval ( 0, π 2 ),

tanx>0 −tanx<0

Hence, f ′ ( x )<0.

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

For the given interval ( π 2 ,π ),

tanx<0 −tanx>0

Hence, f ′ ( x )>0.

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

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