For the function
f(x)=sinx+3x−π2(x2+x),x∈[0,2π],
consider the following two statements :
(I) f is increasing in (0,2π).
(II) f′ is decreasing in (0,2π).
Between the above two statements,
Solutions · 1
Official solution
f(x)=sinx+3x−π2(x2+x),x∈[0,2π]f′(x)=cosx+3−π2(2x+1)=cosx−π4x−π2+3x∈[0,2π]π4x∈[0,2]⇒3−π2−π4x>0
and also cosx>0 when x∈[0,2π]⇒f′(x)>0⇒f(x) is increasing
Now, f′′(x)=−sinx−π4Hence,{f}\^{'}(x)isdecreasing