Question

If $e^{\lambda }$ and $e^{-\lambda }$ are the roots of $3x^2-(a+b)x+2a=0,a,b,\lambda ,ϵR,\lambda \ne 0$ then least integral value of b is

• A. $4$
• B. $5$
• C. $9$
• D. $10$

Answer 1

The correct option is B $5$

Sum of the roots in the given quadratic equation $=\frac{a+b}{3}$

and the product of the roots $=\frac{2a}{3}$

Roots of the equation are $e^{\lambda }$ and $\frac{1}{e^{\lambda }}$

$\therefore e^{\lambda }\times \frac{1}{e^{\lambda }}=1=\frac{2a}{3}⟹a=\frac{3}{2}$

$e^{\lambda }+\frac{1}{e^{\lambda }}=\frac{a+b}{3}=\frac{2b+3}{6}$

We know that the value of $e^{\lambda }+\frac{1}{e^{\lambda }}>2$, when $\lambda \ne 0$

$\therefore \frac{2b+3}{6}>2⟹b>4.5$

therefore the minimum integral value of $b=5$