Mathematics · 2023
JEE Main · 8 April 2023, Shift 2 · Q9
If α>β>0 are the roots of the equation a x^2+b x+1=0, and lim_x → 1/(α)((1- cos (x^2+b x+a))/(2(1-α x)^2))^(1/2)=1/k(1/(β)-1/(α)), then k is equal to
If $\displaystyle \alpha>\beta>0$ are the roots of the equation $\displaystyle a x^2+b x+1=0$, and $\displaystyle \lim _{x \rightarrow \frac{1}{\alpha}}\left(\frac{1-\cos \left(x^2+b x+a\right)}{2(1-\alpha x)^2}\right)^{\frac{1}{2}}=\frac{1}{k}\left(\frac{1}{\beta}-\frac{1}{\alpha}\right)$, then k is equal to
Official answer
From NTA’s final answer key for this paper.
(3)
$\displaystyle 2 \alpha$
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JEE Main 2023 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.