(i) Let f(x)=sin 2x

Applying the first derivative we get

f’(x)=2cos 2x

Putting f’(x)=0, we get

2cos 2x =0

⇒ cos 2x=0

Is possible when

0≤x≤2π

Thus sin 2x is neither decreasing nor increasing on

(ii) Let f(x)=tan x

Applying the first derivative we get

f’(x)=sec² x

As square of any number is always positive,

So f’(x)>0 ∀

Hence tan x is increasing function in

(iii) Let f(x)=cos x

Applying the first derivative we get

f’(x)=-sin x

But we know, 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

Applying the first derivative we get

f’(x)=-3sin 3x

Putting f’(x)=0, we get

-3sin 3x=0

⇒ sin 3x=0

As sin θ=0 if θ=0, π, 2π, 3π

⇒ 3x=0,π, 2π, 3π

Now

Since so we write it on number line as

Now the point divide the interval into 2 disjoint interval.

i.e.,

case 1: for

Now

0<3x<π

So when

We know that

sin θ>0 for θ∈ (0,π)

sin 3x>0 for 3x∈ (0,π)

From equation (a), we get

sin 3x>0 for

-sin 3x<0 for

⇒ f’(x)<0 for

⇒ f(x) is strictly decreasing on

case 2: for

Now

So when

We know that

Sin θ<0 in 3^rd quadrant

sin θ<0 for θ∈ (0,π)

sin θ <0 for

sin 3x<0 for

From equation (b), we get

sin 3x<0 for

-sin 3x>0 for

⇒ f’(x)<0 for

⇒ f(x) is strictly increasing on

Thus cos 3x is neither decreasing nor increasing on

So the correct option is option C, i.e., cos x is decreasing in