Question

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be such that $f(2+x)=f(2-x)\,\,\forall x\in\mathbb{R}$. If $f(x)$ is twice differentiable such that $f^{\prime}(1)=0$, then which one of the following is true?

• A. there exist at least one $c$ in $(0, 1)$ such that $f^{\prime}(c)=0$
• B. there exist at least one $c in (1, 2)$ such that $f^{\prime\prime}(c)=0$
• C. there exist at least one $c$ in $(0, 1)$ such that $f^{\prime\prime}(c)=0$
• D. there exist at least one $c$ in $(1, 2)$ such that $f^{\prime}(c)=0$

Hint:

Differentiating functional symmetry equations like $f(a+x)=f(a-x)$ immediately provides helpful values for the derivative function $f^{\prime}(x)$.

Answer 1

The Correct Option is B

Step 1: Concept
The functional equation $f(2+x) = f(2-x)$ indicates that the graph of $f(x)$ is symmetric about the vertical line $x = 2$. Differentiating this relation using the chain rule gives $f^{\prime}(2+x) = -f^{\prime}(2-x)$.

Step 2: Meaning
We are given $f^{\prime}(1) = 0$. Let's substitute $x = -1$ into the derivative relation: $f^{\prime}(2 + (-1)) = -f^{\prime}(2 - (-1)) \implies f^{\prime}(1) = -f^{\prime}(3)$. Since $f^{\prime}(1) = 0$, we have $0 = -f^{\prime}(3) \implies f^{\prime}(3) = 0$.

Step 3: Analysis
We can also substitute $x = 1$ into the original symmetry equation: $f(2+1) = f(2-1) \implies f(3) = f(1)$. Since $f(x)$ is differentiable on $[1,3]$, Rolle's Theorem states that there must exist at least one value $c \in (1,3)$ such that $f^{\prime}(c) = 0$.

Step 4: Conclusion
By analyzing the symmetric halves across the midpoint interval properties and Rolle's theorem criteria, there must exist at least one value $c \in (1,2)$ such that $f^{\prime}(c) = 0$, which corresponds to option (D).

Final Answer: (D)