Question

Let \( g : \mathbb{R} \rightarrow \mathbb{R} \) be a non-constant twice differentiable function such that \( g'\left(\frac{1}{2}\right) = g'\left(\frac{3}{2}\right) \). If a real-valued function \( f \) is defined as \[ f(x) = \frac{1}{2} \left[ g(x) + g(2 - x) \right], \] then

• A. \( f''(x) = 0 \) for at least two \( x \) in \( (0, 2) \)
• B. \( f''(x) = 0 \) for exactly one \( x \) in \( (0, 1) \)
• C. \( f''(x) = 0 \) for no \( x \) in \( (0, 1) \)
• D. \( f'\left(\frac{3}{2}\right) + f'\left(\frac{1}{2}\right) = 1 \)

Answer 1

The Correct Option is A

Since \( f(x) = \frac{1}{2} [ g(x) + g(2 - x) ] \), we observe that \( f(x) \) is symmetric about \( x = 1 \), suggesting that the behavior around \( x = 1 \) is crucial.

Calculate \( f'(x) \):  
\( f'(x) = \frac{1}{2} \left[ g'(x) + g'(2 - x) \right] \)

Given \( g'\left( \frac{1}{2} \right) = g'\left( \frac{3}{2} \right) \), we find: 
\( f'\left( \frac{1}{2} \right) = \frac{1}{2} \left[ g'\left( \frac{1}{2} \right) + g'\left( \frac{3}{2} \right) \right] = 0 \)  
and similarly,  
\( f'\left( \frac{3}{2} \right) = 0 \)

Calculate \( f''(x) \):  
\( f''(x) = \frac{1}{2} \left[ g''(x) - g''(2 - x) \right] \)

Since \( g \) is non-constant and twice differentiable, by the Intermediate Value Theorem, \( f''(x) = 0 \) must occur at least twice in \( (0, 2) \).

Answer 2

Given the function \( f(x) = \frac{1}{2} [g(x) + g(2-x)] \), we are asked to analyze the behavior of its second derivative, \( f''(x) \), and determine when it equals zero within the interval \( (0, 2) \).

Let's first compute the first derivative \( f'(x) \):

\(f'(x) = \frac{1}{2} \left[ g'(x) - g'(2-x) \cdot (-1) \right] = \frac{1}{2} \left[ g'(x) + g'(2-x) \right]\)

Next, we compute the second derivative \( f''(x) \):

\(f''(x) = \frac{1}{2} \left[ g''(x) - g''(2-x) \right]\)

We know from the problem statement that \( g'\left(\frac{1}{2}\right) = g'\left(\frac{3}{2}\right) \). This implies that the behavior of \( g \) at these symmetric points suggests possible symmetry in its derivatives or other peculiar aspects to exploit.

To satisfy \( f''(x) = 0 \), we require:

\(g''(x) = g''(2-x)\)

Given that \( g \) is a non-constant twice differentiable function, and based on the similarity provided by the equality of derivatives at specific symmetric points, this condition is likely satisfied at least at two values of \( x \) in the interval \( (0, 2) \). This reflects a potential of behavior of \( g \)'s symmetry or periodicity leading to cancellation in its second derivatives.

Therefore, amongst the options provided, the first is correct:

\( f''(x) = 0 \) for at least two \( x \) in \( (0, 2) \)