Sets, Relations & Functions
Domain of Function with Floor Function
nta_pyq_2025_apr
Grade 11
Question:
Let $[x]$ denote the greatest integer less than or equal to $x$. Then the domain of $f(x) = \sec^{-1}(2[x] + 1)$ is:
$(-\infty,-1] \cup [0,\infty)$
$(-\infty,-1] \cup [1,\infty)$
$(-\infty, \infty)$
$(-\infty, \infty) - \{0\}$
Step-by-Step Solution
Key Concept: For $\sec^{-1}(t)$ to be defined, need $t \leq -1$ or $t \geq 1$, i.e., $2[x]+1 \leq -1$ or $2[x]+1 \geq 1$, giving $[x] \leq -1$ or $[x] \geq 0$.
Need $2[x]+1 \geq 1$ or $2[x]+1 \leq -1$: i.e., $[x] \geq 0$ ($\Leftrightarrow x \geq 0$) or $[x] \leq -1$ ($\Leftrightarrow x < 0$). Union = $\mathbb{R}$. Domain $= (-\infty, \infty)$.
Correct Answer: $(-\infty, \infty)$