Question

Let [·] denote the greatest integer function. If the domain of the function \[ f(x) = \cos^{-1}\left(\frac{4x + 2\lfloor x \rfloor}{3}\right) \] is \([\alpha, \beta]\), then \(12(\alpha + \beta)\) is equal to:

• A. 6
• B. 8
• C. 9
• D. 4

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The domain of the inverse cosine function \(\cos^{-1}(y)\) is \([-1, 1]\).
Therefore, for the function \(f(x)\) to be defined, the argument must satisfy \(-1 \le \frac{4x+2[x]}{3} \le 1\).
Step 2: Key Formula or Approach:
We can express any real number \(x\) as \(x = I + f\), where \(I = [x]\) is the integer part and \(0 \le f<1\) is the fractional part.
Setting up the inequality:
\[ -1 \le \frac{4x+2[x]}{3} \le 1 \implies -3 \le 4x + 2[x] \le 3 \] Substitute \(x = I + f\) and \([x] = I\):
\[ -3 \le 4(I + f) + 2I \le 3 \implies -3 \le 6I + 4f \le 3 \] Step 3: Detailed Explanation:
We analyze the possible integer values for \(I\). Since \(0 \le 4f<4\), we can substitute different values of \(I\):
Case 1: Let \(I = 0\).
\[ -3 \le 6(0) + 4f \le 3 \implies -3 \le 4f \le 3 \] Since \(f \ge 0\), we have \(0 \le 4f \le 3 \implies 0 \le f \le \frac{3}{4}\).
For \(I=0\), \(x = 0 + f\), so \(x \in \left[0, \frac{3}{4}\right]\).
Case 2: Let \(I = -1\).
\[ -3 \le 6(-1) + 4f \le 3 \implies -3 \le -6 + 4f \le 3 \implies 3 \le 4f \le 9 \] This gives \(\frac{3}{4} \le f \le \frac{9}{4}\). Since \(f<1\), the valid range is \(\frac{3}{4} \le f<1\).
For \(I=-1\), \(x = -1 + f\), so \(x \in \left[-1 + \frac{3}{4}, -1 + 1\right) = \left[-\frac{1}{4}, 0\right)\).
Case 3: Let \(I = 1\).
\[ -3 \le 6(1) + 4f \le 3 \implies -9 \le 4f \le -3 \] This implies \(f \le -\frac{3}{4}\), which is not possible since \(f \ge 0\).
Case 4: Let \(I = -2\).
\[ -3 \le 6(-2) + 4f \le 3 \implies 9 \le 4f \] This implies \(f \ge \frac{9}{4}>1\), which is not possible.
Step 4: Final Answer:
The domain is the union of the valid intervals for \(x\):
\[ x \in \left[-\frac{1}{4}, 0\right) \cup \left[0, \frac{3}{4}\right] = \left[-\frac{1}{4}, \frac{3}{4}\right] \] Comparing this with \([\alpha, \beta]\), we get \(\alpha = -\frac{1}{4}\) and \(\beta = \frac{3}{4}\).
\[ \alpha + \beta = -\frac{1}{4} + \frac{3}{4} = \frac{1}{2} \] We need to find \(12(\alpha + \beta)\):
\[ 12 \left(\frac{1}{2}\right) = 6 \]