JEE Main 202230 Jun, Morning shiftMathematicsHardComplex Numbers and Quadratic Equations264× since 2002Let S1={x∈R−{1,2}:(x+2)(x2+3x+5)−2+3x−x2≥0}{S_1} = \left\{ {x \in R - \{ 1,2\} :{{(x + 2)({x^2} + 3x + 5)} \over { - 2 + 3x - {x^2}}} \ge 0} \right\}S1={x∈R−{1,2}:−2+3x−x2(x+2)(x2+3x+5)≥0} and S2={x∈R:32x−3x+1−3x+2+27≤0}{S_2} = \left\{ {x \in R:{3^{2x}} - {3^{x + 1}} - {3^{x + 2}} + 27 \le 0} \right\}S2={x∈R:32x−3x+1−3x+2+27≤0}. Then, S1∪S2{S_1} \cup {S_2}S1∪S2 is equal to :AOption A. (−∞,−2]∪(1,2)( - \infty , - 2] \cup (1,2)(−∞,−2]∪(1,2)BOption B. (−∞,−2]∪[1,2]( - \infty , - 2] \cup [1,2](−∞,−2]∪[1,2]COption C. (−2,1]∪[2,∞)( - 2,1] \cup [2,\infty )(−2,1]∪[2,∞)DOption D. (−∞,2]( - \infty ,2](−∞,2]CheckSkip · show solution
