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(D) Given the quadratic equation $x^2-70x+\lambda=0$ with roots $\alpha, \beta \in \mathbb{N}$.By Vieta's formulas,$\alpha+\beta=70$ and $\alpha\beta=

Vedclass · Accepted Answer

$60$

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(D) Given the quadratic equation $x^2-70x+\lambda=0$ with roots $\alpha, \beta \in \mathbb{N}$.By Vieta's formulas,$\alpha+\beta=70$ and $\alpha\beta=\lambda$.We are given that $\frac{\lambda}{2} 
otin \mathbb{N}$ and $\frac{\lambda}{3} 
otin \mathbb{N}$,which implies $\lambda$ is not divisible by $2$ or $3$. Thus,$\lambda$ is not divisible by $6$.Since $\lambda = \alpha(70-\alpha)$,we test values of $\alpha$ such that $\lambda$ is not divisible by $2$ or $3$.If $\alpha=1$,$\lambda=69$ (divisible by $3$,reject).If $\alpha=2$,$\lambda=136$ (divisible by $2$,reject).If $\alpha=3$,$\lambda=201$ (divisible by $3$,reject).If $\alpha=4$,$\lambda=264$ (divisible by $2$ and $3$,reject).If $\alpha=5$,$\lambda=5 	imes 65 = 325$. $325$ is not divisible by $2$ or $3$. This is the minimum value.For $\alpha=5, \beta=65, \lambda=325$,we calculate the expression:$|\alpha-\beta| = |5-65| = 60$.$\sqrt{\alpha-1} = \sqrt{4} = 2$.$\sqrt{\beta-1} = \sqrt{64} = 8$.Expression $= \frac{(2+8)(325+35)}{60} = \frac{10 	imes 360}{60} = 10 	imes 6 = 60$.

List-$I$	List-$II$
$(A)$ The minimum value of $2x^2 + 4x + 5$	$(I)$ $-1$
$(B)$ The maximum value of $\frac{x^2 + 4x + 1}{x^2 + x + 1}$	$(II)$ $1$
$(C)$ If $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$, $\forall x \in [a, b]$ then $b =$	$(III)$ $2$
$(D)$ If $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$, $\forall x \in [a, b]$ then $a =$	$(IV)$ $3$
	$(V)$ $4$

Let $\alpha, \beta \in \mathbb{N}$ be roots of the equation $x^2-70x+\lambda=0$,where $\frac{\lambda}{2}, \frac{\lambda}{3} \notin \mathbb{N}$. 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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