Question

The domain of the function $\log_{10}(x^2 - 5x + 6)$ is

• A. $(-\infty, \infty)$
• B. $(-\infty, 2) \cup (3, \infty)$
• C. $(2, 3)$
• D. None of these

Hint:

For any strict quadratic inequality of the form $(x - \alpha)(x - \beta) > 0$ where $\alpha < \beta$, the valid solution intervals always lie strictly outside the roots, namely $(-\infty, \alpha) \cup (\beta, \infty)$. If it were $< 0$, the solution would lie inside the roots!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the mathematical domain of a logarithmic function containing a quadratic expression in its argument.

Step 2: Key Formula or Approach:
For a logarithmic function $\log_b(f(x))$ to be well-defined over real numbers, its interior argument must be strictly positive:
$$f(x) > 0$$

Step 3: Detailed Explanation:
Set up the domain inequality constraint for our function:
$$x^2 - 5x + 6 > 0$$ Factor the quadratic polynomial by splitting the middle term:
$$x^2 - 2x - 3x + 6 > 0$$ $$x(x - 2) - 3(x - 2) > 0 \implies (x - 2)(x - 3) > 0$$ To solve this product inequality, we analyze the critical points $x = 2$ and $x = 3$ using the standard sign wave method (Wavy Curve method):
For $x > 3$: Both factors are positive $\implies (+)(+) = + > 0$ (Valid)
For $2 < x < 3$: The first factor is positive, second is negative $\implies (+)(-) = - < 0$ (Invalid)
For $x < 2$: Both factors are negative $\implies (-)(-) = + > 0$ (Valid)
Therefore, the inequality holds true when $x < 2$ or $x > 3$. Expressed in formal set-interval notation, this corresponds to:
$$x \in (-\infty, 2) \cup (3, \infty)$$ This matches option (B).

Step 4: Final Answer:
The domain of the function is $(-\infty, 2) \cup (3, \infty)$, which corresponds to option (B).