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

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

The correct answer is

Finding the Domain of $\cos^{-1}\left(\frac{4x+2[x]}{3}\right)$

The function is $f(x)=\cos^{-1}\left(\frac{4x+2[x]}{3}\right)$. The domain of the inverse cosine function, $\cos^{-1}(y)$, requires its argument $y$ to be in the interval $[-1, 1]$.

Domain Condition

For the given function $f(x)$, we must have:

$ -1 \le \frac{4x+2[x]}{3} \le 1 $

Multiplying the inequality by 3 gives:

$ -3 \le 4x+2[x] \le 3 $

We need to solve this compound inequality by considering different intervals for $x$ based on the greatest integer function, $[x]$.

Case Analysis for $[x]$

Case 1: $[x] = 0$
This occurs when $0 \le x < 1$. The inequality becomes:
$ -3 \le 4x + 2(0) \le 3 $ $ -3 \le 4x \le 3 $ $ -\frac{3}{4} \le x \le \frac{3}{4} $
Intersecting this with the condition $0 \le x < 1$, we get the interval $ [0, \frac{3}{4}] $.
Case 2: $[x] = -1$
This occurs when $-1 \le x < 0$. The inequality becomes:
$ -3 \le 4x + 2(-1) \le 3 $ $ -3 \le 4x - 2 \le 3 $
Adding 2 to all parts:
$ -1 \le 4x \le 5 $ $ -\frac{1}{4} \le x \le \frac{5}{4} $
Intersecting this with the condition $-1 \le x < 0$, we get the interval $ [-\frac{1}{4}, 0) $.
Other Cases
For $[x] = 1$ ($1 \le x < 2$), the inequality $-3 \le 4x+2 \le 3$ leads to $-1.25 \le x \le 0.25$, which has no intersection with $1 \le x < 2$. Similarly, for $[x] \ge 1$ or $[x] \le -2$, we find no valid $x$ values that satisfy both the inequality and the condition on $[x]$.

Combining Intervals for Domain

The total domain of the function $f(x)$ is the union of the intervals found in the valid cases:

$ \text{Domain} = \left[-\frac{1}{4}, 0\right) \cup \left[0, \frac{3}{4}\right] = \left[-\frac{1}{4}, \frac{3}{4}\right] $

The problem states the domain is $[\alpha, \beta]$. By comparing, we have:

$ \alpha = -\frac{1}{4} \quad \text{and} \quad \beta = \frac{3}{4} $

Final Calculation

We need to find the value of $12 (\alpha + \beta)$.

$ \alpha + \beta = -\frac{1}{4} + \frac{3}{4} = \frac{2}{4} = \frac{1}{2} $

Therefore,

$ 12 (\alpha + \beta) = 12 \times \frac{1}{2} = 6 $

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