Question

hello pls answer fast.
show that the function given by f(x) =sinx is
(a) strictly increasing on (0,pi/2)
(b) strictly decreasing on (pi/2 and pi)




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Answer 1

Answer:

The given function is  f(x)=sinx.

 ∴f′(x)=cosx

(a) Since for each x∈(0,2π,),cosx>0⇒f′(x)>0.

Hence, f is strictly increasing in (0,2π).

(b) Since for each x∈(2π,π),cosx<0⇒f′(x)<0.

Hence, f is strictly decreasing in (2π,π).

(c) From the results obtained in (a) and (b), it is clear that f is neither increasing nor

decreasing in (0,π).