Q52 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Mathematics

When $\mathbf{R}$ is the set of all real numbers, $\left\{x \in \mathbf{R}: \frac{\sqrt{12-x-x^2}}{x+10} \leq \frac{\sqrt{12-x-x^2}}{2 x+9}\right\}=$

  • A
    $(-4,1] \cup\{3\}$
  • B
    $[-4,1]$
  • Correct answer
    C
    $[-4,1] \cup\{3\}$
  • D
    $\phi$, the empty set

Solution

$\begin{array}{lc}& \text { We have, } \frac{\sqrt{12-x-x^2}}{x+10} \leq \frac{\sqrt{12-x-x^2}}{2 x+9} \\ \Rightarrow & \sqrt{12-x-x^2}(2 x+9-x-10) \leq 0 \\ \Rightarrow & \sqrt{12-x-x^2}(x-1) \leq 0 \\ \therefore & 12-x-x^2 \geq 0 \text { and } x \leq 1 \\ \Rightarrow & x^2+x-12 \leq 0 \text { and } x \leq 1 \\ \Rightarrow & (x+4)(x-3) \leq 0 \text { and } x \leq 1 \\ \Rightarrow & x \in[-4,3] \text { and } x \in(-\infty, 1] \\ \therefore & x \in[-4,1] \cup\{3\}\end{array}$

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

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