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 \( (0, 2) \).