Question

The domain of $f(x) = \cos^{-1} \left( \frac{4x + 2[x]}{3} \right)$ (where $[x]$ denotes greatest integer function) is

• A. $\left[ -\frac{1}{4}, \frac{3}{4} \right]$
• B. $\left[ -\frac{3}{4}, \frac{1}{4} \right]$
• C. $\left[ -\frac{3}{4}, -\frac{1}{4} \right]$
• D. $\left[ \frac{1}{4}, \frac{3}{4} \right]$

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For $\cos^{-1} u$ to be defined, $-1 \le u \le 1$.
Step 2: Detailed Explanation:
We need $-1 \le \frac{4x + 2[x]}{3} \le 1 \implies -3 \le 4x + 2[x] \le 3$.
Case 1: $x \in [0, 1) \implies [x] = 0$.
$-3 \le 4x \le 3 \implies x \in [0, 3/4]$.
Case 2: $x \in [-1, 0) \implies [x] = -1$.
$-3 \le 4x - 2 \le 3 \implies -1 \le 4x \le 5 \implies x \in [-1/4, 5/4]$.
Intersection with $[-1, 0)$ gives $x \in [-1/4, 0)$.
Case 3: $x \ge 1 \implies 4x + 2[x]>4+2=6>3$ (No solution).
Case 4: $x<-1 \implies 4x + 2[x]<-4 - 2 = -6<-3$ (No solution).
Combining the valid parts: $[-1/4, 0) \cup [0, 3/4] = [-1/4, 3/4]$.
Step 4: Final Answer:
Domain is $\left[ -\frac{1}{4}, \frac{3}{4} \right]$.