f(x)=sinx+(x3−3x2+4x−2)cosx x∈(0,1) f(0)=−2>0 f(1)=sin1<0 becausef(0).f(1)<0⇒f(x) has a zero in (0,1) Now, f(x)=sinx+[(x−1)3+(x−1)]cosx ⇒f′(x)=(3(x−1)2+2)cosx−sinx[(x−1)3+(x.−1)] =[3(x−1)2+2]cosx+[(1−x)3+(1−x)]sinx >0∀x∈(0,1) ⇒f(x) is monotone in (0,1)