Question

Let f(x) be a twice differentiable function defined on (-∞, ∞) such that f(x) = f(2 – x) and $f'(\frac{1}{2})=f'(\frac{1}{4})=0$. Then The minimum number of values where f"(x) vanishes on [0, 2] is :

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (C)

4

Answer 2

The condition $f(x) = f(2-x)$ implies symmetry about $x=1$. Differentiating twice, we get $f'(x) = -f'(2-x)$ and $f''(x) = f''(2-x)$.

Given $f'(\frac{1}{2}) = 0$. Using $f'(x) = -f'(2-x)$, we get $f'(\frac{1}{2}) = -f'(2-\frac{1}{2}) \implies 0 = -f'(\frac{3}{2})$, so $f'(\frac{3}{2}) = 0$.

Given $f'(\frac{1}{4}) = 0$. Using $f'(x) = -f'(2-x)$, we get $f'(\frac{1}{4}) = -f'(2-\frac{1}{4}) \implies 0 = -f'(\frac{7}{4})$, so $f'(\frac{7}{4}) = 0$.

Using $f'(x) = -f'(2-x)$ with $x=1$, we get $f'(1) = -f'(1) \implies 2f'(1) = 0 \implies f'(1) = 0$.

Thus, $f'(x)$ has at least five distinct roots in $[0, 2]$: $\frac{1}{4}, \frac{1}{2}, 1, \frac{3}{2}, \frac{7}{4}$.

Since $f(x)$ is twice differentiable, $f'(x)$ is differentiable. Applying Rolle's Theorem to $f'(x)$ on the intervals defined by consecutive roots: \begin{itemize} \item On $[\frac{1}{4}, \frac{1}{2}]$, there exists $c_1 \in (\frac{1}{4}, \frac{1}{2})$ such that $f''(c_1) = 0$. \item On $[\frac{1}{2}, 1]$, there exists $c_2 \in (\frac{1}{2}, 1)$ such that $f''(c_2) = 0$. \item On $[1, \frac{3}{2}]$, there exists $c_3 \in (1, \frac{3}{2})$ such that $f''(c_3) = 0$. \item On $[\frac{3}{2}, \frac{7}{4}]$, there exists $c_4 \in (\frac{3}{2}, \frac{7}{4})$ such that $f''(c_4) = 0$. \end{itemize} These four roots $c_1, c_2, c_3, c_4$ are distinct because they lie in disjoint open intervals. We also know $f''(x) = f''(2-x)$. If $c_1 \in (\frac{1}{4}, \frac{1}{2})$, then $2-c_1 \in (\frac{3}{2}, \frac{7}{4})$. This implies $c_4 = 2-c_1$. If $c_2 \in (\frac{1}{2}, 1)$, then $2-c_2 \in (1, \frac{3}{2})$. This implies $c_3 = 2-c_2$. The four roots are $c_1, c_2, 2-c_2, 2-c_1$. These are guaranteed to be distinct and lie in $(0, 2)$. Rolle's Theorem guarantees at least 4 roots for $f''(x)$ in $(0, 2)$. It is possible to construct a function where $f''(1) \neq 0$ and these are the only roots. Therefore, the minimum number of values where $f''(x)$ vanishes on $[0, 2]$ is 4.