Answer :

(A) Let f1(x) = cosx


In interval,


Therefore, f1(x) = cosx is strictly decreasing in interval.


(B) Let f2(x) = cos2x



Now, 0 < x <


0 < 2x < π


sin2x > 0


-2sin2x < 0



Therefore, f2(x) = cos2x is strictly decreasing in interval.


(C) Let f3(x) = cos3x



Now,


sin3x = 0


3x = π, as xϵ


x =


The point x = divides the interval into two distinct intervals.


i.e. and


Now, in interval, ,


f3'(x) = -3sin3x < 0 as (0 < x < => 0 < 3x < π)


Therefore, f3 is strictly decreasing in interval.


Now, in interval


f3'(x)=-3sin3x > 0 as


Therefore, f3 is strictly increasing in interval.


(D) Let f4 = tanx



In interval,



Therefore, f4 is strictly increasing in interval .


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