Set, Relation and FunctionHard
Question
Let f(x) be a non-constant twice differentiable function defined on (- ∞, ∞) such that f(x) = f(1 - x) and f′ 
= 0. Then
= 0. ThenOptions
A.f′(x) vanishes at least twice on [0, 1]
B.f′
= 0
= 0C.
sin x dx 0
sin x dx 0D.
f(t)esin πt dt = 
f(1 - t)esin πt dt
f(t)esin πt dt = f(1 - t)esin πt dtSolution
f(x) = f(1 - x)
Put x = 1/2 + x
Hence f(x + 1/2) is an even function or f(x + 1/2) sin x is an odd function.
Also, f′ (x) = - f′(1 - x) and for x = 1/2, we have f′(1/ 2) = 0.
Alos,
f(1 - t)esin πt dt = - 
f(y)esin πy dy (obtained by putting, 1 - t = y).
Since f′(1/4) = 0, f′(3/4) = 0. Also f′(1/2) = 0
⇒ f′′(x) = 0 atleast twice in [0, 1] (Rolle′s Theorem)
Put x = 1/2 + x
Hence f(x + 1/2) is an even function or f(x + 1/2) sin x is an odd function.
Also, f′ (x) = - f′(1 - x) and for x = 1/2, we have f′(1/ 2) = 0.
Alos,
f(1 - t)esin πt dt = - f(y)esin πy dy (obtained by putting, 1 - t = y).Since f′(1/4) = 0, f′(3/4) = 0. Also f′(1/2) = 0
⇒ f′′(x) = 0 atleast twice in [0, 1] (Rolle′s Theorem)
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