Let g:R→R be a non constant twice differentiable such that g′(12)=g′(32). If a real valued function f is defined as f(x)=12[g(x)+g(2−x)], thenf′′(x)=0 for atleast two x in (0,2)f′′(x)=0 for exactly one x in (0,1)f′′(x)=0 for no x in (0,1)f′(32)+f′(12)=1

The question seems to be asking for the properties of the function f(x) = 12[g(x) + g(2-x)], given that g is a twice differentiable function with g′(12)=g′(32).

Let's break it down:

1. f′′(x)=0 for at least two x in (0,2)

To find the second derivative of f, we first need to find the first derivative. Using the chain rule, we get:

f′(x) = 12[g′(x) - g′(2-x)]

Then, the second derivative is:

f′′(x) = 12[g′′(x) + g′′(2-x)]

Since g is twice differentiable and non-constant, there must exist at least two x in (0,2) such that f′′(x) = 0.

2. f′′(x)=0 for exactly one x in (0,1)

This is not necessarily true. The number of x in (0,1) for which f′′(x) = 0 depends on the specific function g.

3. f′′(x)=0 for no x in (0,1)

This is also not necessarily true. As mentioned above, the number of x in (0,1) for which f′′(x) = 0 depends on the specific function g.

4. f′(32)+f′(12)=1

Substituting x = 3/2 and x = 1/2 into the first derivative, we get:

f′(32) + f′(12) = 12[g′(32) - g′(2-32)] + 12[g′(12) - g′(2-12)]

Given that g′(12) = g′(32), the above equation simplifies to 0, not 1. So this statement is false.