Q50 / 160 · AP EAMCET 2024 (20 May Shift 2) · Mathematics

Let $[r]$ denote the largest integer not exceeding $r$ and the roots of the equation $3 x^2+6 x+5+\alpha\left(x^2+2 x+2\right)=0$ are complex numbers whenever $\alpha\gt\mathrm{L}$ and $\alpha \lt \mathrm{M}$. If $(\mathrm{L}-\mathrm{M})$ is minimum, then greatest value of $[r]$ such that $\mathrm{L} y^2+\mathrm{M} y$ $+r \lt 0$ for all $y \in \mathbb{R}$ is,

  • Correct answer
    A
    L
  • B
    M
  • C
    $\mathrm{L}+\mathrm{M}$
  • D
    $\mathrm{M}-\mathrm{L}$

Solution

$3 x^2+6 x+5+\alpha\left(x^2+2 x+2\right)=0$ $\begin{aligned} & (\alpha+3) x^2+(2 \alpha+6) x+2 \alpha+5=0 \\ & \text { Roots are complex } \Rightarrow \mathrm{D} \lt 0 \\ & (2 \alpha+6)^2-4(\alpha+3)(2 \alpha+5) \lt 0 \\ & \Rightarrow-\alpha^2-5 \alpha-6 \lt 0 \\ & \alpha^2+5 \alpha+6\gt0 \Rightarrow(\alpha+3)(\alpha+2)\gt0 \\ & \alpha \lt -3 ; \alpha\gt-2 \\ & \mathrm{~L}=-2, \mathrm{M}=-3, \mathrm{~L}-\mathrm{M}=1 \\ & \mathrm{~L} y^2+\mathrm{M} y+r \lt 0 \\ & -2 y^2-3 y+r \lt 0 \forall y \in \mathrm{R} \\ & \mathrm{D} \lt 0 \\ & 9+8 r \lt 0 \Rightarrow r \lt -\frac{9}{8} \end{aligned}$
Maximum value of $[r]=-2=\mathrm{L}$

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

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