JEE Main 2024

Question

Let $\alpha, \beta \in \mathbb{N}$ be roots of the equation $x^2-70x+\lambda=0$, where $\dfrac{\lambda}{2}, \dfrac{\lambda}{3}  
otin \mathbb{N}$. If $\lambda$ assumes the minimum possible value, then $\dfrac{(\sqrt{\alpha-1}+\sqrt{\beta-1})(\lambda+35)}{|\alpha-\beta|}$ is equal to ____. A). $60$ B). $62$ C). $65$ D). $70$

Edvaya · Accepted Answer

The correct answer is $60$. **Solution:** From Vieta's formulas, $\alpha+\beta=70$ and $\alpha\beta=\lambda$. Since $\alpha, \beta \in \mathbb{N}$, we can write $\lambda = \alpha(70-\alpha)$. We are given $\dfrac{\lambda}{2}, \dfrac{\lambda}{3} 
otin \mathbb{N}$, which means $\lambda$ is neither divisible by 2 nor 3. To minimize $\lambda = \alpha(70-\alpha)$, we must test the smallest possible natural numbers for $\alpha$: - $\alpha=1 \implies \lambda=69$ (Divisible by 3, rejected). - $\alpha=2 \implies \lambda=136$ (Even, rejected). - $\alpha=3 \implies \lambda=201$ (Divisible by 3, rejected). - $\alpha=4 \implies \lambda=264$ (Even, rejected). - $\alpha=5 \implies \lambda=325$ (Odd, and sum of digits $3+2+5=10$, not divisible by 3. Valid). Thus, $\alpha=5, \beta=65$, and $\lambda=325$. Evaluating the target expression: $\dfrac{(\sqrt{5-1}+\sqrt{65-1})(325+35)}{|5-65|} = \dfrac{(\sqrt{4}+\sqrt{64})(360)}{60} $ $ = \dfrac{(2+8)(360)}{60} = \dfrac{10 	imes 360}{60} = 60$

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