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

We are given that:

Let \( [x] \) denote the greatest integer less than or equal to \( x \). Then the domain of \( f(x) = \sec^{-1}(2[x] + 1) \) is:

Step 1: Understanding the Function 

The function involves the inverse secant, \( \sec^{-1}(y) \), which is defined for \( |y| \geq 1 \). Therefore, for \( f(x) = \sec^{-1}(2[x] + 1) \), we need to ensure that the expression inside the inverse secant is valid. \[ |2[x] + 1| \geq 1 \] This inequality must hold for the domain of the function.

Step 2: Solving the Inequality

Consider the inequality \( |2[x] + 1| \geq 1 \): \[ 2[x] + 1 \geq 1 \quad \text{or} \quad 2[x] + 1 \leq -1 \] Solving each case separately: - Case 1: \( 2[x] + 1 \geq 1 \) gives \( 2[x] \geq 0 \) or \( [x] \geq 0 \). - Case 2: \( 2[x] + 1 \leq -1 \) gives \( 2[x] \leq -2 \) or \( [x] \leq -1 \). Therefore, the solution is \( [x] \geq 0 \) or \( [x] \leq -1 \).

Step 3: Domain of \( f(x) \)

The domain of \( f(x) \) is all real values of \( x \) for which the greatest integer \( [x] \) satisfies one of the conditions above: \( [x] \geq 0 \) or \( [x] \leq -1 \). This means the function is defined for all real numbers except those between 0 and 1, exclusive.

Final Answer:

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

\( (-\infty, -\infty) \)

Answer 2

Step 1: Understand the function.
We are given: \[ f(x) = \sec^{-1}(2[x] + 1) \] where \([x]\) denotes the greatest integer less than or equal to \(x\).

Step 2: Recall the domain condition for \( \sec^{-1}(y) \).
The inverse secant function is defined for: \[ y \leq -1 \quad \text{or} \quad y \geq 1 \] Hence, for \( f(x) \) to be defined: \[ 2[x] + 1 \leq -1 \quad \text{or} \quad 2[x] + 1 \geq 1 \]

Step 3: Solve the inequalities.
From \( 2[x] + 1 \leq -1 \): \[ 2[x] \leq -2 \Rightarrow [x] \leq -1 \] From \( 2[x] + 1 \geq 1 \): \[ 2[x] \geq 0 \Rightarrow [x] \geq 0 \]

Step 4: Combine the results.
Thus, the function is defined when: \[ [x] \leq -1 \quad \text{or} \quad [x] \geq 0 \] This means: \[ x \in (-\infty, 0) \cup [0, \infty) \] But since \([x] = -1\) for \(x \in [-1, 0)\), this condition fits perfectly.

Step 5: Write the final domain.
\[ \boxed{(-\infty, 0) \cup [0, \infty)} \] or equivalently, \[ \boxed{(-\infty, \infty)} \] since all real numbers are covered except the non-integral transition points already included.

Final Answer:
\[ \boxed{(-\infty, \infty)} \]