Let `f:R -> ( 0,(2pi)/3]` defined as `f(x) = cot^-1 (x^2-4x + alpha)` Then the smallest integral value of `alpha` such that, `f(x)` is into function is

Options

A. 2
B. 4
C. 6
D. 8

Explanation

To solve the problem, we need to determine the smallest integral value of \( \alpha \) such that the function \( f(x) = \cot^{-1}(x^2 - 4x + \alpha) \) is an into function. An into function is defined when the range of the function does not cover the entire codomain.

### Step-by-Step Solution:

1. **Understand the Function**:
We have \( f(x) = \cot^{-1}(x^2 - 4x + \alpha) \). The function \( \cot^{-1}(y) \) is defined for all real \( y \) and its range is \( (0, \pi) \).

2. **Identify the Codomain**:
The problem states that the codomain of \( f \) is \( (0, \frac{2\pi}{3}] \). For \( f(x) \) to be an into function, its range must not cover the entire codomain.

3. **Find the Minimum Value of the Quadratic**:
The expression \( x^2 - 4x + \alpha \) is a quadratic function. The minimum value of a quadratic \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \). Here, \( a = 1 \) and \( b = -4 \):
\[ x = -\frac{-4}{2 \cdot 1} = 2. \]
Now, substituting \( x = 2 \) into the quadratic:
\[ f(2) = 2^2 - 4 \cdot 2 + \alpha = 4 - 8 + \alpha = \alpha - 4. \]

4. **Determine the Condition for the Range**:
For \( f(x) \) to be an into function, the minimum value \( \alpha - 4 \) must be greater than the maximum value of \( \cot^{-1} \) that is equal to \( \cot^{-1}(-\frac{1}{\sqrt{3}}) \). Since \( \cot^{-1}(-\frac{1}{\sqrt{3}}) = \frac{2\pi}{3} \), we need:
\[ \alpha - 4 > -\frac{1}{\sqrt{3}}. \]

5. **Solve the Inequality**:
Rearranging gives:
\[ \alpha > 4 - \frac{1}{\sqrt{3}}. \]
Approximating \( \frac{1}{\sqrt{3}} \approx 0.577 \):
\[ \alpha > 4 - 0.577 \approx 3.423. \]

6. **Find the Smallest Integral Value**:
The smallest integer greater than \( 3.423 \) is \( 4 \).

### Conclusion:
Thus, the smallest integral value of \( \alpha \) such that \( f(x) \) is an into function is \( \alpha = 4 \).