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 $[x]$
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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}] $.
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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) $.
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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 $Similar Questions
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