Question

If x is a real number, then \(\sqrt{log_e\frac{4x-x^2}{3}}\)  is a real number if and only if

• A. 1≤x≤2
• B. -3≤x≤3
• C. 1≤x≤3
• D. -1≤x≤3

Answer 1

The Correct Option is C

To determine when the expression \[ \sqrt{\log_e\left(\frac{4x - x^2}{3}\right)} \] is a real number, we must ensure the value inside the square root is non-negative.

Step 1: Inside the Square Root Must Be Non-Negative

\[ \log_e\left(\frac{4x - x^2}{3}\right) \geq 0 \] This implies: \[ \frac{4x - x^2}{3} \geq 1 \] Multiplying both sides by 3: \[ 4x - x^2 \geq 3 \] Rearranging: \[ x^2 - 4x + 3 \leq 0 \]

Step 2: Solving the Quadratic Inequality

Factor the quadratic: \[ x^2 - 4x + 3 = (x - 1)(x - 3) \] We now solve: \[ (x - 1)(x - 3) \leq 0 \] This inequality holds when: \[ 1 \leq x \leq 3 \]

Step 3: Verifying the Intervals

• For \( x = 0 \): \( (0 - 1)(0 - 3) = 3 > 0 \) → Invalid
• For \( x = 2 \): \( (2 - 1)(2 - 3) = -1 < 0 \) → Valid
• For \( x = 4 \): \( (4 - 1)(4 - 3) = 3 > 0 \) → Invalid

✅ Final Answer:

The domain of the expression is: \[ \boxed{1 \leq x \leq 3} \]