JEE Mathematics 2024 Question with Solution

Given: $\alpha, \beta \in \mathbb{N}$ are roots of $x^2 - 70x + \lambda = 0$ and $\lambda/2, \lambda/3 \notin \mathbb{N}$.

Find: The value of

when $\lambda$ is minimum.

Using Vieta's formulas for $x^2 - 70x + \lambda = 0$,

and

Since $\lambda/2 \notin \mathbb{N}$ and $\lambda/3 \notin \mathbb{N}$, the number $\lambda$ is neither divisible by $2$ nor by $3$. Also,

To get the minimum possible value, test the smallest natural values of $\alpha$.

For $\alpha = 1$, $\beta = 69$, so $\lambda = 69$, which is divisible by $3$.

For $\alpha = 2$, $\beta = 68$, so $\lambda = 136$, which is divisible by $2$.

For $\alpha = 3$, $\beta = 67$, so $\lambda = 201$, which is divisible by $3$.

For $\alpha = 4$, $\beta = 66$, so $\lambda = 264$, which is divisible by $2$.

For $\alpha = 5$, $\beta = 65$, so

This is not divisible by $2$ and not divisible by $3$. Hence the minimum possible value is $\lambda = 325$ with roots $\alpha = 5$ and $\beta = 65$.

Now evaluate the expression.

First,

Also,

and

Therefore,

So, the required value is $60$.

Given: $\alpha + \beta = 70$ and $\lambda = \alpha\beta$ with $\lambda$ not divisible by $2$ or $3$.

Find: The value of the required expression.

Start with the smallest possible natural value of one root. Since $\beta = 70-\alpha$,

Check sequentially:

• $\alpha=1 \Rightarrow \lambda=69$, divisible by $3$
• $\alpha=2 \Rightarrow \lambda=136$, divisible by $2$
• $\alpha=3 \Rightarrow \lambda=201$, divisible by $3$
• $\alpha=4 \Rightarrow \lambda=264$, divisible by $2$
• $\alpha=5 \Rightarrow \lambda=325$, allowed

So take $\alpha=5$, $\beta=65$, $\lambda=325$. Then

and

Hence,

Therefore, the required value is $60$.