Let a > 0 be a root of the equation 2x^2 + x - 2 = 0. If lim x → 1/a 16(1 - cos(2 +x - 2x^2)) / (1 - ax^2) =

Question

Let \(\mathrm{a}>0\) be a root of the equation \(2 \mathrm{x}^{2}+\mathrm{x}-2=0\).

If \(\lim \limits_{x \rightarrow \frac{1}{a}} \frac{16\left(1-\cos \left(2+x-2 x^{2}\right)\right)}{\left(1-a x^{2}\right)}=\alpha+\beta \sqrt{17}\), where \(\alpha, \beta \in \mathrm{Z}\) then \(\alpha+\beta\) is equal to ______.

Answer 1

Correct answer: 170

\(2 \mathrm{x}^{2}+\mathrm{x}-2=0^{\ \rightarrow \ a}_{\ \rightarrow\ b}\)

\(2 x^{2}-x-2=0^{\rightarrow\frac{1}{a}}_{\rightarrow \frac 1b}\)

\(\lim\limits _{x \rightarrow \frac{1}{a}} 16 \cdot \frac{\left(1-\cos 2\left(x-\frac{1}{a}\right)\left(x-\frac{1}{b}\right)\right)}{4\left(x-\frac{1}{b}\right)^{2}} \times \frac{4\left(x-\frac{1}{b}\right)^{2}}{a^{2}\left(x-\frac{1}{a}\right)^{2}}\)

\(=16 \times \frac{2}{\mathrm{a}^{2}}\left(\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{~b}}\right)^{2}\)

\(=\frac{32}{\mathrm{a}^{2}}\left(\frac{17}{4}\right)=\frac{17.8}{\mathrm{a}^{2}}=\frac{17 \times 8 \times 16}{(-1+\sqrt{117})^{2}}\)

\(=\frac{136.16}{18.2 \sqrt{7}} \times \frac{18+2 \sqrt{7}}{18+2 \sqrt{7}}\)

\(=\frac{136}{256}(18+2 \sqrt{7}) \cdot 16\)

\(=153+17 \sqrt{17}\)

\(=\alpha+\beta \sqrt{17}\)

\(\alpha+\beta=153+17=170\)