JEE Main 2024 (Online) 30th January Morning ShiftQuestion 1
Let $\alpha,\beta\in\mathbf{N}$ be roots of the equation $x^{2}-70x+\lambda=0$, where $\frac{\lambda}{2},\frac{\lambda}{3}\notin\mathbf{N}$. If $\lambda$ assumes the minimum possible value, then $\frac{\left(\sqrt{\alpha-1}+\sqrt{\beta-1}\right)(\lambda+35)}{|\alpha-\beta|}$ is equal to $\_\_\_\_\_\_$ .
Quadratic Equation · Relation Between Roots and Coefficients
Answer$60$
Solution
Concept. $\lambda=\alpha\beta$ with $\alpha+\beta=70$; list the smallest products that are not multiples of $2$ or $3$. Formulas used. (1) For $\alpha+\beta$ fixed, $\alpha\beta=\alpha(70-\alpha)$ increases as $\alpha$ goes from $1$ to $35$Step 1 - Odd $\lambda$. $\lambda$ odd needs both roots odd. Take $\alpha\le35$. Step 2 - Try in increasing order (1). $\alpha=1$: $\lambda=69=3\cdot23$. Rejected. $\alpha=3$: $\lambda=201=3\cdot67$. Rejected. $\alpha=5$: $\lambda=5\cdot65=325$, not divisible by $2$ or $3$. Accepted. So $\alpha=5$, $\beta=65$, $\lambda=325$. Step 3 - Value. $$\frac{(\sqrt4+\sqrt{64})(325+35)}{60}=\frac{10\cdot360}{60}=60.$$ Answer: 60