Q54 / 160 · AP EAMCET 2022 (08 Jul Shift 2) · Mathematics

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

  • A
    $0$
  • B
    $1$
  • C
    $-1$
  • Correct answer
    D
    $-2$

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 $k < -18$

Hence, the correct answer is option: 'D'.

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