If $e^{\lambda}$ and $e^{-\lambda}$ are the roots of equation $3x^{2}-(a+b)x+2a=0,a,b,\lambda \epsilon R, \lambda \neq 0$ then least integral value of $b$ is
- $4$
- $5$
- $9$
- $10$
Given roots e^lambda and e^-lambda, their product is e^lambda * e^-lambda = 1. From the quadratic equation 3x^2 - (a+b)x + 2a = 0, the product of roots is 2a/3. Thus, 2a/3 = 1, so a = 1.5. The sum of roots is e^lambda + e^-lambda = (a+b)/3. Since e^lambda + e^-lambda >= 2 for real lambda, (1.5+b)/3 >= 2, so 1.5+b >= 6, b >= 4.5. The least integral value of b is 5.
Given e to the power of lambda and e to the power of negative lambda are the roots of 3x squared minus (a plus b)x plus 2a equals 0, the product of roots gives e to the power of lambda multiplied by e to the power of negative lambda equal to 2a divided by 3. This simplifies to 1 equals 2a divided by 3, meaning a equals 3 by 2. The sum of roots gives e to the power of lambda plus e to the power of negative lambda equal to (a plus b) divided by 3, which becomes (3 by 2 plus b) divided by 3, equaling b divided by 3 plus 1 by 2. Since e to the power of lambda plus e to the power of negative lambda is strictly greater than 2 for non-zero real lambda, we have b divided by 3 plus 1 by 2 greater than 2, meaning b divided by 3 is greater than 3 by 2, so b is greater than 4.5. The least integral value satisfying this is b equals 5.