Question

The domain of the function $\cos^{-1} \left\{ \frac{4x + 2[x]}{3} \right\}$, where $[\cdot]$ is greatest integer function, is

• A. $[0, 3/4]$
• B. $[-1/4, 1/4]$
• C. $[-1/4, 3/4]$
• D. $[-3/4, 1/4]$

Hint:

When solving inequalities involving both $x$ and $[x]$, splitting $x$ into its integer and fractional parts ($[x] + \{x\}$) is the most systematic way to find the solution set.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The domain of $\cos^{-1}(f(x))$ is defined by the inequality $-1 \le f(x) \le 1$.
Here, $f(x) = \frac{4x + 2[x]}{3}$.
Step 2: Key Formula or Approach:
We use the property $x = [x] + \{x\}$, where $0 \le \{x\}<1$.
The inequality is: $-3 \le 4x + 2[x] \le 3$.
Step 3: Detailed Explanation:
Substitute $x = [x] + \{x\}$:
\[ -3 \le 4([x] + \{x\}) + 2[x] \le 3 \]
\[ -3 \le 6[x] + 4\{x\} \le 3 \]
We test possible integer values for $[x]$:

Case 1: If $[x] = 0$
\[ -3 \le 0 + 4\{x\} \le 3 \implies -3/4 \le \{x\} \le 3/4 \]
Since $0 \le \{x\}<1$, we get $0 \le \{x\} \le 3/4$.
Thus, $x \in [0, 3/4]$.

Case 2: If $[x] = -1$
\[ -3 \le -6 + 4\{x\} \le 3 \implies 3 \le 4\{x\} \le 9 \implies 3/4 \le \{x\} \le 9/4 \]
Since $0 \le \{x\}<1$, the valid interval is $3/4 \le \{x\}<1$.
$x = [x] + \{x\} \implies x \in [-1 + 3/4, -1 + 1) \implies x \in [-1/4, 0)$.

Case 3: If $[x] = 1$
\[ -3 \le 6 + 4\{x\} \le 3 \implies -9 \le 4\{x\} \le -3 \implies -9/4 \le \{x\} \le -3/4 \]
Not possible as $\{x\} \ge 0$.

Combining valid intervals from Case 1 and Case 2:
\[ x \in [-1/4, 0) \cup [0, 3/4] = [-1/4, 3/4] \]
Step 4: Final Answer:
The domain is $[-1/4, 3/4]$.