Let a0 be a root of the equation 2x^2+x-2=0.

Question

Let $a>0$ be a root of the equation $2{x}^{2}+x-2=0$. If $\underset{x	o \frac{1}{a}}{lim}\frac{16(1-\cos (2+x-2{x}^{2}))}{(1-ax{)}^{2}}=\alpha +\beta \sqrt{17}$, where $\alpha ,\beta \in Z$, then $\alpha +\beta $ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 170

Accepted Answer

$\because 2{x}^{2}+x-2=0$ has two roots where 
$a=\frac{\sqrt{17}-1}{4}$ and another root is $\frac{-\sqrt{17}-1}{4}$

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

Now $\underset{x	o \frac{1}{a}}{lim}\frac{16(1-\cos (2+x-2{x}^{2}))}{(1-ax{)}^{2}}$

$\begin{gathered} =\underset{x	o \frac{1}{a}}{lim}\frac{32{\sin }^{2}(\frac{2+x-2{x}^{2}}{2})}{{a}^{2}{(\frac{1}{a}-x)}^{2}} \ =\underset{x	o \frac{1}{a}}{lim}\frac{{(x+\frac{4}{\sqrt{17}+1})}^{2}32\cdot {(\sin (\frac{1}{2}\cdot (-2))(x-\frac{1}{a})(x+\frac{4}{\sqrt{17}+1}))}^{2}}{{a}^{2}{((x-\frac{1}{a})(x+\frac{4}{\sqrt{17}+1}))}^{2}} \ =2\cdot {(\frac{1}{a}+\frac{4}{\sqrt{17}+1})}^{2}\cdot {(\frac{4}{\sqrt{17}-1})}^{2} \ =2{(\frac{4}{\sqrt{17}-1}+\frac{4}{\sqrt{17}+1})}^{2}\cdot {(\frac{4}{\sqrt{17}-1})}^{2} \ =\frac{17	imes 4}{18-2\sqrt{17}}=\frac{68}{9-\sqrt{17}} \ =17(9+\sqrt{17}) \ \alpha +\beta =170 \end{gathered}$

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