Given that f is twice differentiable, the only true statement in the options provided is that if f'' is a polynomial, then f is also a polynomial. So, the correct option is D.
Given that f is twice differentiable, let's examine the provided options:
a) If f is bounded then f′ is bounded: This is not necessarily true. A counterexample would be
, which is bounded, but its derivative,
, is not bounded.
b) If f and f' are bounded then f'' is bounded: This is not necessarily true as well. Even if f and f' are bounded, it does not ensure that the second derivative f'' will also be bounded. There is no direct implication from the boundedness of the first derivative to the boundedness of the second derivative.
c) If f(x)>0 and f'(x)>0 , ∀x∈(0,1) then f''(x)>0, ∀x∈(0,1): This statement is not necessarily true. The second derivative being greater than zero would imply that the function is concave upward, but it's possible for a function to be increasing while its concavity is switching.
d) If f′′ is a polynomial then f is a polynomial: This is true. If the second derivative f'' is a polynomial, then integrating twice will result in f being a polynomial as well.
Therefore, the correct statement is d) If f′′ is a polynomial then f is a polynomial.
The probable question may be:
"Suppose f is twice differentiable (i.e., f is differentiable and f ' is also differentiable). Which of the following is true?
a). If f is bounded then f′ is bounded.
b). If f and f′ are bounded then f′′ is bounded.
c). If f(x)>0 and f′(x)>0, ∀x∈(0,1) then f′′(x)>0, ∀x∈(0,1)
d). If f′′ is a polynomial then f is a polynomial.."