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JEE Main 30-Jan-2024 Paper - Shift 1 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 30-Jan-2024 Paper - Shift 14 Marks
Question
Let $\alpha, \beta \in N$ be roots of equation $x^2-70 x+\lambda=0,$ where $\frac{\lambda}{2}, \frac{\lambda}{3} \notin N$. If $\lambda$ assumes the minimum possible value, then $\frac{(\sqrt{\alpha-1}+\sqrt{\beta-1})(\lambda+35)}{|\alpha-\beta|}$ is equal to :
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Answer

$(a)\ (60)$
$x^2-70 x+\lambda=0$
$\alpha+\beta=70$
$\alpha \beta=\lambda$
$\therefore \alpha(70-\alpha)=\lambda$
Since $, 2$ and $3$ does not divide
$\lambda$$\therefore \alpha=5, \beta=65, \lambda=325$
By putting value of $\alpha, \beta, \lambda$ we get the required value $60 $.

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