Question

If \( f(x) = \frac{3^x}{3^x + \sqrt{3}} \), then \( f(x) + f(1-x) \) is equal to

• A. \( \sqrt{3} \)
• B. \( \frac{1}{\sqrt{3}} \)
• C. \( 2\sqrt{3} \)
• D. \( 1 \)
• E. \( 0 \)

Hint:

Check symmetry \( f(x) + f(1-x) \) — often simplifies to constant.

Answer 1

The Correct Option is D

Note: There appears to be a typo in the question. A common form for this type of problem is \(f(x) = \frac{a^x}{a^x + \sqrt{a}}\). Here, \(a=3\), so the function is likely intended to be \(f(x) = \frac{3^x}{3^x + \sqrt{3}}\). The image confirms this.
Step 1: Understanding the Concept:
This problem tests a specific functional property. We are asked to evaluate the sum \(f(x) + f(1-x)\). The key is to find an expression for \(f(1-x)\) and simplify the sum algebraically.
Step 2: Key Formula or Approach:
1. Write down the expression for \(f(x)\). 2. Find the expression for \(f(1-x)\) by replacing every `x` in the formula for \(f(x)\) with `(1-x)`. 3. Simplify the expression for \(f(1-x)\). A key trick is to multiply the numerator and denominator by \(3^x\). 4. Add the simplified \(f(1-x)\) to \(f(x)\).
Step 3: Detailed Explanation:
We are given: \[ f(x) = \frac{3^x}{3^x + \sqrt{3}} \] Now, find \(f(1-x)\): \[ f(1-x) = \frac{3^{1-x}}{3^{1-x} + \sqrt{3}} = \frac{3/3^x}{3/3^x + \sqrt{3}} \] To simplify this, multiply the numerator and the denominator by \(3^x\): \[ f(1-x) = \frac{(3/3^x) \cdot 3^x}{(3/3^x + \sqrt{3}) \cdot 3^x} = \frac{3}{3 + \sqrt{3} \cdot 3^x} \] We can factor out \(\sqrt{3}\) from the denominator: \[ f(1-x) = \frac{3}{\sqrt{3}(\sqrt{3} + 3^x)} = \frac{\sqrt{3}}{\sqrt{3} + 3^x} \] Now, add \(f(x)\) and \(f(1-x)\): \[ f(x) + f(1-x) = \frac{3^x}{3^x + \sqrt{3}} + \frac{\sqrt{3}}{\sqrt{3} + 3^x} \] Since the denominators are the same, we can add the numerators: \[ = \frac{3^x + \sqrt{3}}{3^x + \sqrt{3}} \] \[ = 1 \] Step 4: Final Answer:
The value of \(f(x) + f(1-x)\) is 1.