Let F:R→R be a thrice differentiable function. Suppose that F(1)=0,F(3)=−4 and F′(x)<0 for all x∈(1/2,3). Let f(x)=xF(x) for all x∈R
The correct statement is
f′(1)<0
f(2)>0
f′(x)≠0 for any x∈(1,3)
f′(x)>0 for some x∈(1,3)
f'x=xF'x+Fx
⇒f′(1)=f′(1)+f(1)=f′(1)<0⇒(A)f(2)=2f(2)<0⇒(B)
For x∈(1,3),f′(x)=xf′(x)+f(x)<0