If 0 are the roots of the equation a x 2 + b x + 1 = 0 , and lim x 1 ( 1 - cos ( x 2 + b x + a ) 2 ( 1 - x ) 2 ) 1 2 = 1 k ( 1 - 1 ) , then k is equal to [2023]

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Solution

Q.
$If α > β > 0 are the roots of the equation a x^{2} + b x + 1 = 0,$ and $\lim_{x \to \frac{1}{α}} {(\frac{1 - \cos (x^{2} + b x + a)}{2 {(1 - α x)}^{2}})}^{\frac{1}{2}} = \frac{1}{k} (\frac{1}{β} - \frac{1}{α}),$ then $k$ is equal to [2023]

1 $2 β$

2 $α$

3 $β$

4 $2 α$

Ans.
(4)

Given, $a x^{2} + b x + 1 = 0$ has roots $α, β$ , then $x^{2} + b x + a = 0$ has roots $\frac{1}{α}, \frac{1}{β}$ .

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

$= \lim_{x \to \frac{1}{α}} {(\frac{2 \sin^{2} (\frac{x^{2} + b x + a}{2})}{2 {(1 - α x)}^{2}})}^{1 / 2}$

$= \lim_{x \to \frac{1}{α}} {(\frac{2 \sin^{2} (\frac{(x - \frac{1}{α}) (x - \frac{1}{β})}{2})}{2 {(1 - α x)}^{2}})}^{1 / 2}$

$\lim_{x \to \frac{1}{α}} {(\frac{\frac{\sin^{2} (\frac{(1 - α x) (1 - β x)}{2 α β})}{{(\frac{(1 - α x) (1 - β x)}{2 α β})}^{2}} \times {(\frac{(1 - α x) (1 - β x)}{2 α β})}^{2}}{{(1 - α x)}^{2}})}^{\frac{1}{2}}$

$= \lim_{x \to \frac{1}{α}} (\frac{1 - β x}{2 α β}) = (\frac{α - β}{2 α^{2} β}) = \frac{1}{2 α} (\frac{1}{β} - \frac{1}{α})$

$= \frac{1}{k} (\frac{1}{β} - \frac{1}{α}) (\Rightarrow Given)$

$∴$ $k = 2 α$

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