Question

answer with explanation​

Answer 1

Answer:

The given function is f(x)=sinx.

∴f

′

(x)=cosx

(a) Since for each x∈(0,

2

π

,),cosx>0⇒f

′

(x)>0.

Hence, f is strictly increasing in (0,

2

π

).

(b) Since for each x∈(

2

π

,π),cosx<0⇒f

′

(x)<0.

Hence, f is strictly decreasing in (

2

π

,π).

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

decreasing in (0,π).

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