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JEE Math Practice Question

Let $${S_1} = \left\{ {x \in R - \{ 1,2\} :{{(x + 2)({x^2} + 3x + 5)} \over { - 2 + 3x - {x^2}}} \ge 0} \right\}$$ and $${S_2} = \left\{ {x \in R:{3^{2x}} - {3^{x + 1}} - {3^{x + 2}} + 27 \le 0} \right\}$$. Then, $${S_1} \cup {S_2}$$ is equal to :

  1. A.$$( - \infty , - 2] \cup (1,2)$$
  2. B.$$( - \infty , - 2] \cup [1,2]$$
  3. C.$$( - 2,1] \cup [2,\infty )$$
  4. D.$$( - \infty ,2]$$

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