Question:

$\color{{blue}}{{\text{{Step 1:}}}}$ To find the range of $f(x)$, we start by simplifying the given function. We have $f(x) = \cos^{-1}\left(\sqrt{\sin^{-1}\left(\sec\left(\ln\left(\frac{2x^2+3x-2}{x^2-3x+2}\right)\right)\right)}\right)$. First, let's simplify the expression inside the logarithm: $$\begin{aligned} \frac{2x^2+3x-2}{x^2-3x+2} &= \frac{(2x-1)(x+2)}{(x-1)(x-2)} \\ &= \frac{2x-1}{x-1} \cdot \frac{x+2}{x-2} \end{aligned}$$ $\color{{blue}}{{\text{{Step 2:}}}}$ Next, we consider the domain of the function to ensure that the expression inside the logarithm is positive. For the expression $\frac{2x-1}{x-1} \cdot \frac{x+2}{x-2}$ to be positive, either both factors must be positive or both must be negative. This leads to the conditions: - $x > 1$ and $x > 2$, or - $x < 1$ and $x < 2$. However, since $x$ cannot be simultaneously greater than 1 and 2, and less than 1 and 2, we refine these conditions: - $x > 2$, or - $x < 1$. $\color{{blue}}{{\text{{Step 3:}}}}$ Now, we analyze the behavior of the function $\sec\left(\ln\left(\frac{2x^2+3x-2}{x^2-3x+2}\right)\right)$. As $x$ approaches 2 from the right, $\frac{2x-1}{x-1} \cdot \frac{x+2}{x-2}$ approaches $+\infty$, and $\ln\left(\frac{2x^2+3x-2}{x^2-3x+2}\right)$ also approaches $+\infty$. Conversely, as $x$ approaches 1 from the left, $\frac{2x-1}{x-1} \cdot \frac{x+2}{x-2}$ approaches $-\infty$, and $\ln\left(\frac{2x^2+3x-2}{x^2-3x+2}\right)$ approaches $-\infty$. Since $\sec(\theta)$ is defined for $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$, the range of $\sec\left(\ln\left(\frac{2x^2+3x-2}{x^2-3x+2}\right)\right)$ will be $[-1, 1]$. $\color{{blue}}{{\text{{Step 4:}}}}$ Considering the next part of the function, $\sin^{-1}\left(\sec\left(\ln\left(\frac{2x^2+3x-2}{x^2-3x+2}\right)\right)\right)$, its range will also be $[-1, 1]$ because $\sin^{-1}$ is defined for inputs in $[-1, 1]$. $\color{{blue}}{{\text{{Step 5:}}}}$ Finally, for $\cos^{-1}\left(\sqrt{\sin^{-1}\left(\sec\left(\ln\left(\frac{2x^2+3x-2}{x^2-3x+2}\right)\right)\right)}\right)$, since $\cos^{-1}$ is defined for inputs in $[-1, 1]$, and the square root of a value in $[-1, 1]$ is in $[0, 1]$, the range of $f(x)$ will be $[0, \frac{\pi}{2}]$. $\color{{blue}}{{\text{{Step 6:}}}}$ The integers in the range of $f(x)$ are 0. Thus, $\sum \alpha_i^2 = 0^2 = 0$. Therefore, $1 + \left(\sum \alpha_i^2\right) = 1 + 0 = 1$. Therefore: $1$