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Quadratic Equation and Inequalities question

2022 · 30 Jun · Shift 1 · Q23
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  5. /2022 · 30 Jun · Shift 1 · Q23

Quadratic Equation and Inequalities question

2022 · 30 Jun · Shift 1 · Q23

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
Let 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 :
  1. A
    (−∞,−2]∪(1,2)( - \infty , - 2] \cup (1,2)(−∞,−2]∪(1,2)
  2. B
    (−∞,−2]∪[1,2]( - \infty , - 2] \cup [1,2](−∞,−2]∪[1,2]
  3. C
    (−2,1]∪[2,∞)( - 2,1] \cup [2,\infty )(−2,1]∪[2,∞)
  4. D
    (−∞,2]( - \infty ,2](−∞,2]
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