JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If $\alpha > \beta > 0$ are the roots of $a x^{2} + b x + 1 = 0$ , and $lim_{x \to \frac{1}{α}} (\frac{1 - c o s ( x ^{2} + b x + a )}{2 ( 1 - α x ) ^{2}})^{\frac{1}{2}} = \frac{1}{k} (\frac{1}{β} - \frac{1}{α})$ , then $k$ is equal to:

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Explanation

Roots of $a x^{2} + b x + 1 = 0$ are $α, β$ .
Replace $x \to \frac{1}{x} \Rightarrow x^{2} + b x + a = 0$ has roots $\frac{1}{α}, \frac{1}{β}$ .
$lim_{x \to \frac{1}{α}} [\frac{2 s i n ^{2} ( \frac{x ^{2} + b x + a}{2} )}{2 ( 1 - α x ) ^{2}}]^{\frac{1}{2}} = lim_{x \to \frac{1}{α}} \frac{∣ s i n ( \frac{( x - 1/ α ) ( x - 1/ β )}{2} ) ∣}{∣ - α ( x - 1/ α ) ∣}$
$= \frac{1}{α} \cdot \frac{1}{2} ∣ \frac{1}{α} - \frac{1}{β} ∣ = \frac{1}{2 α} ∣ \frac{1}{β} - \frac{1}{α} ∣$
Comparing with $\frac{1}{k} (\frac{1}{β} - \frac{1}{α}) \Rightarrow k = 2 α$ .

= \lim_{x \to \frac{1}{\alpha}} \left[ \frac{2\sin^2 \frac{\left(x - \frac{1}{\alpha}\right)\left(x - \frac{1}{\beta}\right)}{2}}{4 \times 2\alpha^2 \frac{\left(x - \frac{1}{\alpha}\right)^2 \left(x - \frac{1}{\beta}\right)^2}{4}} \left(x - \frac{1}{\beta}\right)^2 \right]^{\frac{1}{2}}

= \lim_{x \to \frac{1}{\alpha}} \left[ \pm \frac{1}{2} \frac{\sin \frac{\left(x - \frac{1}{\alpha}\right)\left(x - \frac{1}{\beta}\right)}{2}}{\alpha \frac{\left(x - \frac{1}{\alpha}\right)\left(x - \frac{1}{\beta}\right)}{2}} \left(x - \frac{1}{\beta}\right) \right]

= \lim_{x \to \frac{1}{\alpha}} \left[ \pm \frac{1}{2} \frac{\sin \frac{\left(x - \frac{1}{\alpha}\right)\left(x - \frac{1}{\beta}\right)}{2}}{\alpha \frac{\left(x - \frac{1}{\alpha}\right)\left(x - \frac{1}{\beta}\right)}{2}} \left(x - \frac{1}{\beta}\right) \right]

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