Question

Which of the following functions are strictly decreasing on ? (A) cos x (B) cos 2 x (C) cos 3 x (D) tan x

Answer 1

The given interval is ( 0, π 2 ).

(A)

Consider that the function is,

f 1 ( x )=cosx

Differentiate the function with respect to x,

f ′ 1 ( x )=−sinx

Since, −sinx<0 in the given interval, so, the function f 1 ( x )=cosx is strictly decreasing in the interval ( 0, π 2 ).

(B)

Consider that the function is,

f 2 ( x )=cos2x

Differentiate the function with respect to x,

f ′ 2 ( x )=−2sin2x

Since, 0<x< π 2 , then,

0<2x<π

This shows that sin2x>0, then, −2sin2x<0.

So, the function f 2 ( x )=cos2x is strictly decreasing in the interval ( 0, π 2 ).

(C)

Consider that the function is,

f 3 ( x )=cos3x

Differentiate the function with respect to x,

f ′ 3 ( x )=−3sin3x

Put f ′ 3 ( x )=0, then,

sin3x=0 3x=π x= π 3

This divide the given interval in two parts,

( 0, π 3 ) and ( π 3 , π 2 )

Since, 0<x< π 3 and 0<3x<π, then, in the interval ( 0, π 3 ),

−3sin3x<0

Therefore, f 3 ( x ) is strictly decreasing in the interval ( 0, π 3 ).

Since, π 3 <x< π 2 and π<3x< 3π 2 , then,

−3sin3x>0

Therefore, f 3 ( x ) is strictly increasing in the interval ( π 3 , π 2 ).

Hence, f 3 ( x ) is neither increasing nor decreasing in the interval ( 0, π 2 ).

(D)

Consider that the function is,

f 4 ( x )=tanx

Differentiate the function with respect to x,

f ′ 4 ( x )= sec 2 x

Since, sec 2 x>0 in the given interval, so, the function f 4 ( x )=tanx is strictly increasing in the interval ( 0, π 2 ).

Thus, it can be observed that cosx and cos2x is strictly decreasing in the interval ( 0, π 2 ).

Therefore, option (A) and option (B) are correct.