Let a>0 be a root of the equation 2 x^2+x-2=0. If lim _x → 1 a 16 (1-cos (2+x-2 x^2 ) ) (1- a x)^2=α+β √17, where α, β ∈ Z, then α+β is equal to

$$\begin{aligned} & \lim _{x \rightarrow \frac{1}{a}} 16 \cdot \frac{\left(1-\cos 2\left(x-\frac{1}{a}\right)\left(x-\frac{1}{b}\right)\right)}{4\left(x-\frac{1}{b}\right)^2} \times \frac{4\left(x-\frac{1}{b}\right)^2}{a^2\left(x-\frac{1}{a}\right)^2} \\ & =16 \times \frac{2}{a^2}\left(\frac{1}{a}-\frac{1}{b}\right)^2 \\ & =\frac{32}{a^2}\left(\frac{17}{4}\right)=\frac{17.8}{a^2}=\frac{17 \times 8 \times 16}{(-1+\sqrt{117})^2} \\ & =\frac{136.16}{18.2 \sqrt{7}} \times \frac{18+2 \sqrt{7}}{18+2 \sqrt{7}} \\ & =\frac{136}{256}(18+2 \sqrt{7}) \cdot 16 \\ & =153+17 \sqrt{17}=\alpha+\beta \sqrt{17} \\ & \alpha+\beta=153+17=170\end{aligned}$$​x→a1​lim​16⋅4(x−b1​)2(1−cos2(x−a1​)(x−b1​))​×a2(x−a1​)24(x−b1​)2​=16×a22​(a1​−b1​)2=a232​(417​)=a217.8​=(−1+117​)217×8×16​=18.27​136.16​×18+27​18+27​​=256136​(18+27​)⋅16=153+1717​=α+β17​α+β=153+17=170​