Paper 1 maths question 10.f(x)=f(1-x),f(x) --- a nonconstant differentiable fn. defined on R and f'(1/4)=0.If we differentiate the given condition then we get f'(x)=-f'(1-x) and now we put x=1/4 then ,f'(1/4)=-f'(3/4)===>besides 1/4 we also have 3/4 as our maxima or minima.Next ,we put -x in place of x in the given condition and differentiate it again.f(-x)=f(1+x)===>-f'(-x)=f'(1+x) and put x=-1/2 and that gives -f'(1/2)=f'(1/2)===>2f'(1/2)=0 or f'(1/2)=0 which is the second option.So,now we have 1/4,1/2 and 3/4 as points of maxima or minima.Thus,f''(x) at these points should vanish.So,between [0,1] f''(x) should vanish at least thrice .However, option A is included in correct answers.Somebody please help.
The inference which states that f"(x) at these points should vanish is not clear. If \(f'(\frac{1}{4})=f'(\frac{1}{2})=f'(\frac{3}{4})=0\), then by Rolle's theorem f"(x) = 0 at least once in \((\frac{1}{4},\frac{1}{2})\) and at least once in \((\frac{1}{2},\frac{3}{4})\). So, ovearll it should be zero at least 2 times in \((\frac{1}{4},\frac{3}{4})\).
Yes,Sir I understood your point.I thought that at 1/4,1/2 and 3/4 function attains zero height and lhd and rhd at these points won't match as these are points of maxima and minima for f(x) so,f''(x) won't exist.