Sets, Relations & Functions
Functions
nta_abhyas_2025
Grade 11
Question:
The domain of the function $f(x) = \frac{3}{5-x} + \log_{10}(x^2 - 3x)$ is
$(-\sqrt{3}, 0) \cup (3, \infty)$
$(-\sqrt{3}, 0) \cup (\sqrt{3}, 3)$
$(-\sqrt{3}, 0) \cup (3, \infty)$
$(-\sqrt{3}, 0) \cup (\sqrt{3}, 3) \cup (3, \infty)$
Step-by-Step Solution
Key Concept: Find domain by solving all inequality constraints simultaneously and taking their intersection.
For the domain, we need $9 - x^2 \geq 0$, giving $-3 \leq x \leq 3$. We also need $x^2 - 3x > 0$, so $x(x - 3) > 0$, which means $x < 0$ or $x > 3$. Combining these constraints: from condition (1), $x \in [-3, 3]$; from condition (2), $x \in (-\infty, 0) \cup (3, \infty)$. The intersection is $x \in [-3, 0) \cup (3, \infty)$, but we must also satisfy $x^2 - 3 > 0$ separately, refining to $x \in (-\sqrt{3}, 0) \cup (\sqrt{3}, \infty)$.
Correct Answer: 1