Let g:R be a non constant twice differentiable function such that g' ( 12 )=g' ( 32 ). If a real valued function f is defined as f(x)= 12[g(x)+g(2-x)], then
- f''(x)=0 for at least two x in (0,2)
- f' ( 32 )+f' ( 12 )=1
- f''(x)=0 for no x in (0,1)
- f''(x)=0 for exactly one x in (0,1)
Answer
(1) f''(x)=0 for at least two x in (0,2)
Solution
f'(x)= g'(x)-g'(2-x)2
f' ( 12 )= g' ( 12 )-g' ( 32 )2=0
f' ( 32 )= g' ( 32 )-g' ( 12 )2=0
f'(1)= g'(1)-g'(1)2=0
f' vanishes at 12, 1, 32, Rolle's theorem on f' in [ 12,1 ] and [1, 32 ]
f''(c1)=0, c1 ( 12,1 ) and f''(c2)=0, c2 (1, 32 )
(A): two roots c1, c2 in (0,2) ⇒ (A) is true
(B): f' ( 32 )+f' ( 12 )=0 1 ⇒ (B) is false
(C): c1 (0,1) is a root ⇒ (C) is false
(D): only "at least one" root in (0,1) is guaranteed ⇒ (D) is not certain
Answer: (1)