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 = 17 , where, then is equal to _ _ _ . [2024]

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Solution

Q.
$Let a > 0 be a root of the equation 2 x^{2} + x - 2 = 0 . If \lim_{x \to \frac{1}{a}} \frac{16 (1 - \cos (2 + x - 2 x^{2}))}{{(1 - a x)}^{2}} = α + β \sqrt{17}, where α, β \in Z,$ $then α + β is equal to___.$ [2024]

Ans.
(170)

$We have, 2 x^{2} + x - 2 = 0$

$Given one root is a . Let another root be b .$

$Here, a = \frac{- 1 + \sqrt{17}}{4}, b = \frac{- 1 - \sqrt{17}}{4}$

$∴ 2 + x - 2 x^{2} = 0 has roots \frac{1}{a} and \frac{1}{b}$

$Now, \lim_{x \to \frac{1}{a}} \frac{16 (1 - \cos 2 (x - \frac{1}{a}) (x - \frac{1}{b}))}{a^{2} {(x - \frac{1}{a})}^{2}}$

$\lim_{x \to \frac{1}{a}} \frac{16 (1 - \cos 2 (x - \frac{1}{a}) (x - \frac{1}{b}))}{a^{2} {(x - \frac{1}{a})}^{2}} \times \frac{{(x - \frac{1}{b})}^{2}}{{(x - \frac{1}{b})}^{2}}$

$= \lim_{x \to \frac{1}{a}} \frac{16 \cdot 2 \sin^{2} (x - \frac{1}{a}) (x - \frac{1}{b})}{a^{2} {(x - \frac{1}{a})}^{2}} \times \frac{{(x - \frac{1}{b})}^{2}}{{(x - \frac{1}{b})}^{2}}$

$= 16 \times \frac{2}{a^{2}} {(\frac{1}{a} - \frac{1}{b})}^{2}$

$= \frac{32}{a^{2}} (\frac{17}{4}) = \frac{17 \times 8}{a^{2}} = \frac{17 \times 8 \times 16}{(- 1 + \sqrt{17})^{2}} = 153 + 17 \sqrt{17}$

$= α + β \sqrt{17}$

$\Rightarrow α = 153, β = 17$

$\Rightarrow α + β = 170$

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