Question

Let \(g(x) = f(x)+f(1-x)\) and \(f^{''}(x) < 0,0\leq x\leq 1\), then \(\ldots\)

• A. \(g(x)\) increases on \([\frac{1}{2},1]\) and \(g(x)\) decreases on \([0,\frac{1}{2}]\)
• B. \(g(x)\) decreases on \([0,1]\)
• C. \(g(x)\) increases on \([0,1]\)
• D. \(g(x)\) decreases on \([\frac{1}{2},1]\) and \(g(x)\) increases on \([0,\frac{1}{2}]\)

Hint:

g'(x) = f'(x) - f'(1-x). With f'' negative, f' is decreasing, so compare x with 1-x.

Answer 1

The Correct Option is D

Step 1: Use symmetry:
$g(x) = g(1-x)$, so the graph of $g$ is symmetric about $x = \tfrac12$. Whatever happens on the left half is mirrored on the right.

Step 2: Use concavity:
$g''(x) = f''(x) + f''(1-x) < 0$, since both terms are negative. So $g$ is concave on $[0,1]$.

Step 3: Combine:
A concave function symmetric about $x=\tfrac12$ has its maximum at $x = \tfrac12$. So it must rise before $\tfrac12$ and fall after it.

Step 4: Match:
That is option (D): increasing on $[0,\tfrac12]$ and decreasing on $[\tfrac12,1]$.

Final Answer:
Option (D). \[ \boxed{\text{(D)}} \]