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(D) Given,$\alpha$ and $\beta$ are the roots of $2x^2 + 6x + k = 0$. From the relation between roots and coefficients,$\alpha + \beta = -\frac{6}{2} =

Vedclass · Accepted Answer

$-2$

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(D) Given,$\alpha$ and $\beta$ are the roots of $2x^2 + 6x + k = 0$. From the relation between roots and coefficients,$\alpha + \beta = -\frac{6}{2} = -3$ and $\alpha\beta = \frac{k}{2}$. Now,consider the expression $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}$. Substituting the values,we get $\frac{(-3)^2 - 2(k/2)}{k/2} = \frac{9 - k}{k/2} = \frac{18 - 2k}{k} = \frac{18}{k} - 2$. Since $k As $k$ becomes more negative (i.e.,$k 	o -\infty$),$\frac{18}{k} 	o 0$,so $f(k) 	o -2$. For $k Specifically,for $k However,for $-18 \le k The maximum value of the greatest integer function $\left[\frac{18}{k} - 2ight]$ for $k < 0$ is $-2$.

If $\alpha$ and $\beta$ are the roots of the equation $2x^2 + 6x + k = 0$,then the maximum value of $\left[\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\right]$ when $k < 0$ is (where $[\cdot]$ denotes the greatest integer function)

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