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

The function \( f(x) = \frac{1}{2} [ g(x) + g(2 - x) ] \) exhibits symmetry about \( x = 1 \), indicating the significance of its behavior around this point.

The derivative \( f'(x) \) is calculated as:
\( f'(x) = \frac{1}{2} \left[ g'(x) + g'(2 - x) \right] \)

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

The second derivative \( f''(x) \) is:  
\( f''(x) = \frac{1}{2} \left[ g''(x) - g''(2 - x) \right] \)

Since \( g \) is a non-constant and twice-differentiable function, the Intermediate Value Theorem implies that \( f''(x) = 0 \) must have at least two occurrences within the interval \( (0, 2) \).