TS EAMCET · Maths · Quadratic Equation
The set of all real values of \(x\) satisfying the inequalíties \(x^2-4 x+3>0\) and \(x^2-2 x-8 \leq 0\), is
- A \([-2,1) \cup(3,4]\)
- B \([-1,2) \cup(3,4)\)
- C \([-2,2) \cup(2,4)\)
- D \([0,2) \cup(3,5)\)
Answer & Solution
Correct Answer
(A) \([-2,1) \cup(3,4]\)
Step-by-step Solution
Detailed explanation
We have, \[ \begin{aligned} & x^2-4 x+3>0 \text { and } x^2-2 x-8 \leq 0 \\ & (x-3)(x-1)>0 \text { and }(x-4)(x+2) \leq 0 \\ & x \in(-\infty, 1) \cup(3, \infty) \text { and } x \in[-2,4] \\ \therefore \quad & x \in[-2,1) \cup(3,4] \end{aligned} \]
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