Let (2,3) be the largest open interval in which the function f(x)=2 log _ e(x-2)-x^2+a x+1 is strictly increasing and (b, c) be the largest open…

$$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=\frac{2}{\mathrm{x}-2}-2 \mathrm{x}+\mathrm{a} \geq 0 \\ & \mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{-2}{(\mathrm{x}-2)^2}-2 < 0 \\ & \mathrm{f}^{\prime}(\mathrm{x}) \downarrow \\ & \mathrm{f}^{\prime}(3) \geq 0 \\ & 2-6+\mathrm{a} \geq 0 \\ & \mathrm{a} \geq 4 \\ & \mathrm{a}_{\min }=4 \\ & \mathrm{~g}(\mathrm{x})=(\mathrm{x}-1)^3(\mathrm{x}+2-\mathrm{a})^2 \\ & \mathrm{~g}(\mathrm{x})=(\mathrm{x}-1)^3(\mathrm{x}-2)^2 \\ & \mathrm{~g}{ }^{\prime}(\mathrm{x})=(\mathrm{x}-1)^3 2(\mathrm{x}-2)+(\mathrm{x}-2)^2 3(\mathrm{x}-1)^2 \\ & =(\mathrm{x}-1)^2(\mathrm{x}-2)(2 \mathrm{x}-2+3 \mathrm{x}-6) \\ & =(\mathrm{x}-1)^2(\mathrm{x}-2)(5 \mathrm{x}-8) < 0 \\ & \mathrm{x} \in\left(\frac{8}{5}, 2\right) \\ & 100(\mathrm{a}+\mathrm{b}-\mathrm{c})=100\left(4+\frac{8}{5}-2\right)=360\end{aligned}$$​f′(x)=x−22​−2x+a≥0f′′(x)=(x−2)2−2​−2<0f′(x)↓f′(3)≥02−6+a≥0a≥4amin​=4 g(x)=(x−1)3(x+2−a)2 g(x)=(x−1)3(x−2)2 g′(x)=(x−1)32(x−2)+(x−2)23(x−1)2=(x−1)2(x−2)(2x−2+3x−6)=(x−1)2(x−2)(5x−8)<0x∈(58​,2)100(a+b−c)=100(4+58​−2)=360​