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
- Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
- Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
- If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
- Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1 + x)^n$, $n \in \mathbb{N}, 0 \leq r \leq n$. If $P_n = C_0 - C_1 + \frac{2^2}{3} C_2 - \frac{2^3}{4} C_3 + \dots + \frac{(-2)^n}{n+1} C_n$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2n}}$ equals.
- The number of elements in the relation $R = \{(x, y) : 4x^2 + y^2 < 52, x, y \in \mathbb{Z}\}$ is
- Let $S = \{z \in \mathbb{C} : 4z^2 + \bar{z} = 0\}$. Then $\sum_{z \in S} |z|^2$ is equal to :
Let f and g be functions satisfying $f(x+y) = f(x)f(y), f(1) = 7$ and $g(x+y) = g(xy), g(1) = 1$, for all $x, y \in \mathbb{N}$. If $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607$, then n is equal to :
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
- Let $S = \frac{1}{25!} + \frac{1}{3!23!} + \frac{1}{5!21!} + \dots$ up to 13 terms. If $13S = \frac{2^k}{n!}, k \in \mathbb{N}$, then $n + k$ is equal to
Important Questions from Algebra
- Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
- Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
- If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
- Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1 + x)^n$, $n \in \mathbb{N}, 0 \leq r \leq n$. If $P_n = C_0 - C_1 + \frac{2^2}{3} C_2 - \frac{2^3}{4} C_3 + \dots + \frac{(-2)^n}{n+1} C_n$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2n}}$ equals.
