Question:

Step 1: Understand the condition for an into function. For $f(x) = \cot^{-1}(x^2 - 4x + \alpha)$ to be an into function (not onto), the range of $f$ must be a proper subset of the codomain $(0, \frac{2\pi}{3}]$. This means the range of $f$ should not cover the entire codomain. Step 2: Recall the range of the inverse cotangent function. The range of $\cot^{-1}$ is $(0, \pi)$. For $f$ to map into $(0, \frac{2\pi}{3}]$, we need to ensure that the output of $\cot^{-1}$ never exceeds $\frac{2\pi}{3}$. Step 3: Determine the constraint on the argument of $\cot^{-1}$. Since $\cot^{-1}$ is a decreasing function, the condition $\cot^{-1}(t) \leq \frac{2\pi}{3}$ is equivalent to: $$t \geq \cot\left(\frac{2\pi}{3}\right)$$ Computing $\cot\left(\frac{2\pi}{3}\right)$: $$\cot\left(\frac{2\pi}{3}\right) = \frac{\cos(2\pi/3)}{\sin(2\pi/3)} = \frac{-1/2}{\sqrt{3}/2} = -\frac{1}{\sqrt{3}}$$ Step 4: Find the minimum value of the quadratic expression. The argument of $\cot^{-1}$ is $g(x) = x^2 - 4x + \alpha$. This is a quadratic function with a minimum value occurring at $x = 2$ (the vertex). The minimum value is: $$g_{\min} = (2)^2 - 4(2) + \alpha = 4 - 8 + \alpha = \alpha - 4$$ Step 5: Apply the constraint to find the condition on $\alpha$. For $f(x)$ to be an into function, we need the minimum value of $g(x)$ to satisfy: $$\alpha - 4 \geq -\frac{1}{\sqrt{3}}$$ $$\alpha \geq 4 - \frac{1}{\sqrt{3}}$$ Step 6: Calculate the numerical value and find the smallest integer. Computing the right-hand side: $$4 - \frac{1}{\sqrt{3}} = 4 - \frac{\sqrt{3}}{3} \approx 4 - 0.577 \approx 3.42$$ Since $\alpha$ must be an integer and $\alpha \geq 3.42$, the smallest integral value of $\alpha$ is $\boxed{4}$. **Final Answer:** The smallest integral value of $\alpha$ is **4**, which corresponds to **Option 2**.