\sqrt{17}-1}{4}$$ and another root is $$\frac{-\sqrt{17}-1}{4}$$

And $$2+x-2 x^2=-2\left(x-\frac{1}{a}\right)\left(x+\frac{4}{\sqrt{17}+1}\right)$$

Now $$\lim _\limits{x \rightarrow \frac{1}{a}} \frac{16\left(1-\cos \left(2+x-2 x^2\right)\right)}{(1-a x)^2}$$

$$\begin{aligned} & =\lim _{x \rightarrow \frac{1}{a}} \frac{32 \sin ^2\left(\frac{2+x-2 x^2}{2}\right)}{a^2\left(\frac{1}{a}-x\right)^2} \\ & =\lim _{x \rightarrow \frac{1}{a}} \frac{\left(x+\frac{4}{\sqrt{17}+1}\right)^2 32 \cdot\left(\sin \left(\frac{1}{2} \cdot(-2)\right)\left(x-\frac{1}{a}\right)\left(x+\frac{4}{\sqrt{17}+1}\right)\right)^2}{a^2\left(\left(x-\frac{1}{a}\right)\left(x+\frac{4}{\sqrt{17}+1}\right)\right)^2} \\ & =2 \cdot\left(\frac{1}{a}+\frac{4}{\sqrt{17}+1}\right)^2 \cdot\left(\frac{4}{\sqrt{17}-1}\right)^2 \\ & =2\left(\frac{4}{\sqrt{17}-1}+\frac{4}{\sqrt{17}+1}\right)^2 \cdot\left(\frac{4}{\sqrt{17}-1}\right)^2 \\ & =\frac{17 \times 4}{18-2 \sqrt{17}}=\frac{68}{9-\sqrt{17}} \\ & =17(9+\sqrt{17}) \\ & \alpha+\beta=170 \end{aligned}$$

19:T48b,

$$\because 2 x^2+x-2=0$$ के दो मूल हैं जहाँ $$a=\frac{\sqrt{17}-1}{4}$$ और दूसरा मूल है $$\frac{-\sqrt{17}-1}{4}$$

और $$2+x-2 x^2=-2\left(x-\frac{1}{a}\right)\left(x+\frac{4}{\sqrt{17}+1}\right)$$

अब $$\lim _\limits{x \rightarrow \frac{1}{a}} \frac{16\left(1-\cos \left(2+x-2 x^2\right)\right)}{(1-a x)^2}$$

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