Which of the following functions is decreasing on .
A. sin2x
B. tan x
C. cos x
D. cos 3x
(i) Let f(x)=sin 2x
Apply first derivative and get
f’(x)=2cos 2x
Put f’(x)=0, and get
2cos 2x =0
⇒ cos 2x=0
It is possible when
0≤x≤2π
Thus, sin 2x does not decrease or increase on
(ii) Let f(x)=tan x
Apply first derivative and get
f’(x)=
Now. square of every number is always positive,
So, tan x is increasing function in
(iii) Let f(x)=cos x
Apply first derivative and get
f’(x)=-sin x
But, sin x>0 for
And -sin x<0 for
Hence f’(x)<0 for
⇒ cos x is strictly decreasing on
(iv) Let f(x)=cos 3x
Apply first derivative and get
f’(x)=-3sin 3x
Put f’(x)=0, we get
-3sin 3x=0
⇒ sin 3x=0
Because sin θ=0 if θ=0, π, 2π, 3π
⇒ 3x=0,π, 2π, 3π
so we write it on number line as
Now, this point into 2 disjoint intervals.
i.e.
case 1 : for
So wher
Also,
From equation (a), we get
sin 3x <0 for
case 2: for
Now
Also,
Equation (b) gives
Hence, cos 3 x does not decrease or increase on
So, the correct answer is option C i.e., is decreasing in