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QJEEM24Ma48b5: JEE Main 2024 Mathematics - Class 11 JEEM24Ma48b5 Solution: Let be roots of the equation , where . If assumes the minimum…

Let α,β∈N\alpha, \beta \in \mathbf{N} be roots of the equation x2−70x+λ=0x^2-70 x+\lambda=0 , where λ2,λ3∉N\frac{\lambda}{2}, \frac{\lambda}{3} \notin \mathbf{N} . If λ\lambda assumes the minimum possible value, then (α−1+β−1)(λ+35)∣α−β∣\frac{(\sqrt{\alpha-1}+\sqrt{\beta-1})(\lambda+35)}{|\alpha-\beta|} is equal to :

Step-by-Step Verified Solution

amp;x2−70x+λ=0amp;α+β=70amp;αβ=λamp;∴α(70−α)=λ\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 λ\lambda

∴α=5,β=65,λ=325\therefore \alpha=5, \beta=65, \lambda=325

By putting value of α,β,λ\alpha, \beta, \lambda we get the required value 6060 .

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