JEE MAIN - Mathematics Hindi (2023 - 8th April Evening Shift - No. 11)

यदि समीकरण $$a x^2+b x+1=0$$ के मूल $$\alpha > \beta > 0$$ हैं तथा $$\lim _\limits{x \rightarrow \frac{1}{\alpha}}\left(\frac{1-\cos \left(x^2+b x+a\right)}{2(1-\alpha x)^2}\right)^{\frac{1}{2}}=\frac{1}{k}\left(\frac{1}{\beta}-\frac{1}{\alpha}\right)$$ है, तो $$\mathrm{k}$$ बराबर है:
$$\beta$$
$$2\alpha$$
$$2\beta$$
$$\alpha$$

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