Question

Let \( \phi(x) = f(x) + f(2a - x) \), \( x \in [0, 2a] \) and \( f'(x)>0 \) for all \( x \in [0, a] \). Then \( \phi(x) \) is:

• A. increasing on \( [0, a] \)
• B. decreasing on \( [0, a] \)
• C. increasing on \( [0, 2a] \)
• D. decreasing on \( [0, 2a] \)

Hint:

To determine if a function is increasing or decreasing, compute its derivative and analyze its sign over the interval. Use symmetry and test with simple functions like quadratics to understand behavior.

Answer 1

The Correct Option is B

Step 1: Define \( \phi(x) \) and compute its derivative.
We are given \( \phi(x) = f(x) + f(2a - x) \), and we need to determine the behavior of \( \phi(x) \). To check if \( \phi(x) \) is increasing or decreasing, compute its derivative: \[ \phi'(x) = \frac{d}{dx} \left[ f(x) + f(2a - x) \right]. \] Using the chain rule for \( f(2a - x) \), let \( u = 2a - x \), so \( \frac{du}{dx} = -1 \): \[ \phi'(x) = f'(x) + f'(2a - x) \cdot \frac{d}{dx}(2a - x) = f'(x) - f'(2a - x). \]
Step 2: Analyze \( \phi'(x) \) on \( [0, a] \).
For \( x \in [0, a] \), the argument \( 2a - x \in [a, 2a] \). We know \( f'(x)>0 \) for \( x \in [0, a] \), so \( f'(x)>0 \). However, \( 2a - x \in [a, 2a] \), and we have no direct information about \( f'(x) \) on \( [a, 2a] \). To proceed, consider the symmetry of \( \phi(x) \): \[ \phi(2a - x) = f(2a - x) + f(2a - (2a - x)) = f(2a - x) + f(x) = \phi(x). \] This shows \( \phi(x) \) is symmetric about \( x = a \), but we focus on \( [0, a] \).
Step 3: Determine the sign of \( \phi'(x) \) on \( [0, a] \).
Since \( \phi'(x) = f'(x) - f'(2a - x) \), and \( x \in [0, a] \), we need to compare \( f'(x) \) and \( f'(2a - x) \). Since \( f'(x)>0 \) on \( [0, a] \), \( f(x) \) is increasing on \( [0, a] \). For \( 2a - x \in [a, 2a] \), assume \( f'(x) \) may decrease or change behavior beyond \( x = a \). If \( f'(x) \) is large near \( x = 0 \) and \( f'(2a - x) \) is smaller near \( x = 2a \), then \( f'(x)>f'(2a - x) \), making \( \phi'(x)<0 \).
Step 4: Hypothesize \( f(x) \) and test.
Assume \( f(x) \) is linear, say \( f(x) = kx + b \), with \( k>0 \) since \( f'(x) = k>0 \). Then: \[ \phi(x) = f(x) + f(2a - x) = (kx + b) + (k(2a - x) + b) = kx + b + 2ka - kx + b = 2b + 2ka. \] \[ \phi'(x) = 0. \] This implies \( \phi(x) \) is constant, not decreasing, so a linear \( f(x) \) doesn’t work. Try a concave function, say \( f(x) = x^2 \) on \( [0, a] \), with \( a = 1 \) for simplicity: \[ f'(x) = 2x>0 \text{ on } [0, 1], \quad \phi(x) = x^2 + (2 - x)^2 = x^2 + 4 - 4x + x^2 = 2x^2 - 4x + 4. \] \[ \phi'(x) = 4x - 4 = 4(x - 1). \] On \( [0, 1] \), \( \phi'(x) \leq 0 \), and \( \phi'(x) = 0 \) at \( x = 1 \), so \( \phi(x) \) is decreasing on \( [0, 1] \). This matches option (B).
Step 5: Generalize the behavior.
Since \( f'(x)>0 \), \( f(x) \) is increasing on \( [0, a] \). If \( f(x) \) is concave (\( f''(x)<0 \)), then \( f'(x) \) is decreasing. For \( x \in [0, a] \), \( 2a - x \geq x \), so \( f'(2a - x) \leq f'(x) \), making \( \phi'(x) \leq 0 \), confirming \( \phi(x) \) is decreasing on \( [0, a] \).