Q. 505.0( 1 Vote )

Which of the foll A. sin2x

B. tan x

C. cos x

D. cos 3x

Answer :

(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 3rd 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

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