Question:

Step 1: Determine the value of $b$. Let $\alpha$ and $\beta$ be the roots of the equation $x^2 - 5x + 5 = 0$. By Vieta's formulas, we have $\alpha + \beta = 5$ and $\alpha\beta = 5$. The expression for $b$ is given by: $$b = \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta}$$ Substitute $(\alpha^2 + \beta^2) = (\alpha+\beta)^2 - 2\alpha\beta$: $$b = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta} = \frac{(5)^2 - 2(5)}{5} = \frac{25 - 10}{5} = \frac{15}{5} = 3$$ Step 2: Simplify the expression for $t$. Substitute $b=3$ into the expression for $t$: $$t = x^2 - 4x + 3b - \frac{1}{5} + \frac{1}{x^2 - 4x + 9}$$ $$t = x^2 - 4x + 3(3) - \frac{1}{5} + \frac{1}{x^2 - 4x + 9}$$ $$t = x^2 - 4x + 9 - \frac{1}{5} + \frac{1}{x^2 - 4x + 9}$$ Let $u = x^2 - 4x + 9$. We can rewrite $u$ by completing the square: $$u = (x-2)^2 + 5$$ Since $(x-2)^2 \geq 0$ for all $x \in \mathbb{R}$, the minimum value of $u$ is $5$. Thus, $u \geq 5$. The expression for $t$ becomes: $$t = u - \frac{1}{5} + \frac{1}{u}$$ Step 3: Find the minimum value of $t$. Consider the function $f(u) = u + \frac{1}{u} - \frac{1}{5}$ for $u \geq 5$. To find the minimum value, we analyze the derivative of $f(u)$: $$f'(u) = 1 - \frac{1}{u^2}$$ For $u \geq 5$, $u^2 \geq 25$, so $\frac{1}{u^2} \leq \frac{1}{25}$. Therefore, $f'(u) = 1 - \frac{1}{u^2} \geq 1 - \frac{1}{25} = \frac{24}{25} > 0$. Since $f'(u) > 0$ for $u \geq 5$, the function $f(u)$ is strictly increasing on the interval $[5, \infty)$. The minimum value of $t$ occurs at the minimum value of $u$, which is $u=5$: $$t_{\min} = f(5) = 5 + \frac{1}{5} - \frac{1}{5} = 5$$ Thus, $t \geq 5$. Step 4: Evaluate the minimum value of $(b+t)$. The minimum value of $(b+t)$ is $b + t_{\min} = 3 + 5 = 8$. Step 5: Evaluate the maximum value of $\log_{1/5}(t)$. Since $t \geq 5$ and the base of the logarithm is $1/5 < 1$, the function $\log_{1/5}(t)$ is decreasing. Therefore, the maximum value of $\log_{1/5}(t)$ occurs at the minimum value of $t$: $$\max(\log_{1/5}(t)) = \log_{1/5}(t_{\min}) = \log_{1/5}(5) = -1$$ Step 6: Determine the range of $y = \cot^{-1}(\log_5 t)$. Since $t \geq 5$, we have $\log_5 t \geq \log_5 5 = 1$. The range of $\cot^{-1}(x)$ for $x \geq 1$ is $(0, \cot^{-1}(1)]$. Since $\cot^{-1}(1) = \frac{\pi}{4}$, the range of $y = \cot^{-1}(\log_5 t)$ is $\left(0, \frac{\pi}{4}\right]$. Step 7: Determine the range of $y = \cot^{-1}(\log_{1/5}(t))$. Since $t \geq 5$, we have $\log_{1/5}(t) \leq \log_{1/5}(5) = -1$. Let $z = \log_{1/5}(t)$. Then $z \leq -1$. The range of $\cot^{-1}(z)$ for $z \leq -1$ is $[\cot^{-1}(-1), \pi)$. Since $\cot^{-1}(-1) = \frac{3\pi}{4}$, the range of $y = \cot^{-1}(\log_{1/5}(t))$ is $\left[\frac{3\pi}{4}, \pi\right)$.