Q54 / 160 · TS EAMCET 2023 (14 May Shift 1) · Mathematics

If $\mathbb{R}-(\alpha, \beta)$ is the range of $\frac{x+3}{(x-1)(x+2)}$, then the sum of the intercepts of the line $\alpha x+\beta y+1=0$ on the coordinate axes is

  • A
    $-8$
  • Correct answer
    B
    $10$
  • C
    $8$
  • D
    $9$

Solution

$y=\frac{(x+3)}{(x-1)(x+2)}$ $$ \begin{aligned} & (x-1)(x+2) y=x+3 \\ & x^2 y+(y-1) x-(2 y+3)=0 \\ & \because x \in \mathrm{R} \\ & \therefore \mathrm{D} \geq 0 \\ & (y-1)^2+4 y(2 y+3) \geq 0 \\ & (9 y+1)(y+1) \geq 0 \\ & y \in(-\infty,-1] \cup\left[-\frac{1}{9}, \infty\right) \Rightarrow y \in \mathrm{R}-\left(-1, \frac{-1}{9}\right) \\ & \alpha=-1 \\ & \beta=\frac{-1}{9} \\ & \alpha x+\beta y+1=0 \\ & -x-\frac{1}{9} y+1=0 \Rightarrow \frac{x}{1}+\frac{y}{9}=1 \end{aligned} $$ $\therefore$ Sum of intercepts $=1+9=10$

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

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