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

Given that \( [x] \) represents the greatest integer less than or equal to \( x \), the domain of the function \( f(x) = \sec^{-1}(2[x] + 1) \) is to be determined.

Step 1: Function Definition Analysis

The inverse secant function, \( \sec^{-1}(y) \), is defined for \( |y| \geq 1 \). Consequently, for \( f(x) = \sec^{-1}(2[x] + 1) \), the expression within the inverse secant must satisfy the condition \( |2[x] + 1| \geq 1 \).

Step 2: Inequality Resolution

The inequality \( |2[x] + 1| \geq 1 \) can be split into two cases: \( 2[x] + 1 \geq 1 \) or \( 2[x] + 1 \leq -1 \).
- Case 1: \( 2[x] + 1 \geq 1 \) simplifies to \( 2[x] \geq 0 \), which means \( [x] \geq 0 \).
- Case 2: \( 2[x] + 1 \leq -1 \) simplifies to \( 2[x] \leq -2 \), which means \( [x] \leq -1 \).
Therefore, the inequality holds when \( [x] \geq 0 \) or \( [x] \leq -1 \).

Step 3: Domain Determination

The domain of \( f(x) \) comprises all real numbers \( x \) such that the greatest integer \( [x] \) satisfies \( [x] \geq 0 \) or \( [x] \leq -1 \). This implies that the function is defined for all real numbers except for those where \( 0 \leq [x]<1 \).

Final Answer:

The domain of \( f(x) = \sec^{-1}(2[x] + 1) \) is:

\( (-\infty, 0) \cup [1, \infty) \)