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:

• A. \( (-\infty, -1] \cup [0, \infty) \)
• B. \( (-\infty, -\infty) \)
• C. \( (-\infty, -1] \cup [1, \infty) \)
• D. \( (-\infty, \infty) - \{ 0 \} \)

Hint:

The secant function is defined for values where \( |x| \geq 1 \), so always check for values that lie within the function's range.

Answer 1

The Correct Option is B

Step 1: Analyze the Given Inequality

We are given the inequality: \[ 2[x] + 1 \leq -1 \quad \text{or} \quad 2[x] + 1 \geq 1 \]

Step 2: Solve Each Condition

Solving the first inequality: \[ 2[x] + 1 \leq -1 \] \[ 2[x] \leq -2 \quad \Rightarrow \quad [x] \leq -1 \] Solving the second inequality: \[ 2[x] + 1 \geq 1 \] \[ 2[x] \geq 0 \quad \Rightarrow \quad [x] \geq 0 \]

Step 3: Combine the Solutions

Combining these results, the solution for the domain is: \[ x \in (-\infty, 0) \cup [0, \infty) \]

Final Answer: \( x \in (-\infty, 0) \cup [0, \infty) \)