JEE MAIN - Mathematics (2024 - 30th January Morning Shift - No. 24)
Let $$\alpha, \beta \in \mathbf{N}$$ be roots of the equation $$x^2-70 x+\lambda=0$$, where $$\frac{\lambda}{2}, \frac{\lambda}{3} \notin \mathbf{N}$$. If $$\lambda$$ assumes the minimum possible value, then $$\frac{(\sqrt{\alpha-1}+\sqrt{\beta-1})(\lambda+35)}{|\alpha-\beta|}$$ is equal to :
Answer
60
Explanation
$$\begin{aligned} & x^2-70 x+\lambda=0 \\ & \alpha+\beta=70 \\ & \alpha \beta=\lambda \\ & \therefore \alpha(70-\alpha)=\lambda \end{aligned}$$
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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