If $\alpha$ and $\beta$ are the roots of the equation $2 x^2+6 x+k=0$, then the maximum value of…
If $\alpha$ and $\beta$ are the roots of the equation $2 x^2+6 x+k=0$, then the maximum value of $\left[\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right]$ when $k < 0$ is
$0$
$1$
$-1$
$-2$
Solution
Given, $\alpha$ and $\beta$ are the roots of $2 x^2+6 x+k=0$
$\Rightarrow \alpha+\beta=-\frac{6}{2}=-3$ and $\alpha \beta=\frac{k}{2}$
Now, $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha \beta}$
$=\frac{(\alpha+\beta)^2-2 \alpha \beta}{\alpha \beta}$
$\Rightarrow \frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{2(9-k)}{k}=\frac{18}{k}-2$
$\because \quad k < 0\left(\frac{18}{k}-2\right) < 0, \forall k < 0$
Thus, $\left[\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right]_{\max }=\left[\frac{18}{k}-2\right]=\left[\frac{18}{k}\right]-2=-2$
where, $\left[\frac{18}{k}\right]=0$ for $k < -18$