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TS EAMCET · Maths · Quadratic Equation

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

  1. A \((-4,1] \cup\{3\}\)
  2. B \([-4,1]\)
  3. C \([-4,1] \cup\{3\}\)
  4. D \(\phi\), the empty set
Verified Solution

Answer & Solution

Correct Answer

(C) \([-4,1] \cup\{3\}\)

Step-by-step Solution

Detailed explanation

\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…