Question

The domain of the function $f(x) = \sqrt{\log_{10} \left(\frac{5x - x^2}{4}\right)}$ is:

• A. $[1, 4]$
• B. $(1, 4)$
• C. $[0, 5]$
• D. $(0, 5)$

Hint:

Test boundaries! At $x=1$, we get $\log_{10}(1) = 0$, which is valid under a square root, meaning $1$ must be included (closed interval). At $x=0$, the logarithm argument is $0$, which is undefined, meaning $0$ must be excluded.

Answer 1

The Correct Option is A

Step 1: Concept
For a real-valued square root function $f(x) = \sqrt{g(x)}$, we require $g(x) \ge 0$. For a logarithmic function $\log_{10}(h(x))$, we require $h(x) > 0$.

Step 2: Meaning
Here, we require the argument of the logarithm to be positive: $\frac{5x - x^2}{4} > 0$, and the term inside the square root to be non-negative: $\log_{10} \left(\frac{5x - x^2}{4}\right) \ge 0$.

Step 3: Analysis
From the first condition: \[ 5x - x^2 > 0 \implies x(5-x) > 0 \implies x \in (0, 5) \] From the second condition: \[ \log_{10} \left(\frac{5x - x^2}{4}\right) \ge 0 \implies \frac{5x - x^2}{4} \ge 10^0 \implies \frac{5x - x^2}{4} \ge 1 \] \[ \implies 5x - x^2 \ge 4 \implies x^2 - 5x + 4 \le 0 \implies (x-1)(x-4) \le 0 \implies x \in [1, 4] \] Taking the intersection of the two intervals $(0, 5) \cap [1, 4]$ yields the domain: \[ x \in [1, 4] \]

Step 4: Conclusion
The domain of the given function is the closed interval $[1, 4]$. Final Answer: (A)