Solveeit Logo
Login

Question

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

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

A

(−∞,−1]∪[0,∞)(- \infty, -1] \cup [0, \infty)

B

(−∞,−1]∪[1,∞)(- \infty, -1] \cup [1, \infty)

C

(−∞,0]∪[1,∞)(- \infty, 0] \cup [1, \infty)

D

R\mathbb{R}

Answer

The domain of f(x)=sec⁡−1(2[x]+1)f(x) = \sec^{-1}(2[x] + 1) is R\mathbb{R}.

Explanation

Solution

The domain of the function f(x)=sec⁡−1(y)f(x) = \sec^{-1}(y) is defined for ∣y∣≥1|y| \ge 1, which means y≤−1y \le -1 or y≥1y \ge 1.

In this problem, y=2[x]+1y = 2[x] + 1. Therefore, we must satisfy the condition: 2[x]+1≤−1or2[x]+1≥12[x] + 1 \le -1 \quad \text{or} \quad 2[x] + 1 \ge 1.

Let's solve each inequality:

Case 1: 2[x]+1≤−12[x] + 1 \le -1 Subtract 1 from both sides: 2[x]≤−22[x] \le -2 Divide by 2: [x]≤−1[x] \le -1

This inequality means that the greatest integer less than or equal to xx must be less than or equal to −1-1. This implies that [x][x] can be −1,−2,−3,…-1, -2, -3, \dots. If [x]=−1[x] = -1, then −1≤x<0-1 \le x < 0. If [x]=−2[x] = -2, then −2≤x<−1-2 \le x < -1. If [x]=−3[x] = -3, then −3≤x<−2-3 \le x < -2. And so on. The union of all these intervals is x∈(−∞,0)x \in (-\infty, 0).

Case 2: 2[x]+1≥12[x] + 1 \ge 1 Subtract 1 from both sides: 2[x]≥02[x] \ge 0 Divide by 2: [x]≥0[x] \ge 0

This inequality means that the greatest integer less than or equal to xx must be greater than or equal to 00. This implies that [x][x] can be 0,1,2,…0, 1, 2, \dots. If [x]=0[x] = 0, then 0≤x<10 \le x < 1. If [x]=1[x] = 1, then 1≤x<21 \le x < 2. If [x]=2[x] = 2, then 2≤x<32 \le x < 3. And so on. The union of all these intervals is x∈[0,∞)x \in [0, \infty).

The domain of f(x)f(x) is the union of the solutions from Case 1 and Case 2: Domain =(−∞,0)∪[0,∞)=R= (-\infty, 0) \cup [0, \infty) = \mathbb{R}.

The domain of f(x)f(x) is all real numbers.