The set of all real values of x satisfying the inequalities x2 - 4x + 3 > 0 and x2 - 2x - 8 ≤ 0 is

x2−4x+3>0 ⇒ x 3Now x2−2x−8≤0 ⇒ −2≤ x≤4∴ x∈[−2,1) ∪ (3,4]

The set of all real values of x satisfying the inequalities x2 - 4x + 3 > 0 and x2 - 2x - 8 ≤ 0 is

x2−4x+3>0 ⇒ x 3Now x2−2x−8≤0 ⇒ −2≤ x≤4∴ x∈[−2,1) ∪ (3,4]

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The set of all real values of x satisfying the inequalities x²- 4x + 3 > 0 and x²- 2x - 8 ≤ 0 is

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$[- 2, 2] \cup (2, 4)$

$[0, 2] \cup (3, 5)$

$[- 2, 1) \cup (3, 4]$

$(- 1, 2) \cup (3, 4)$

answer is C.

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Detailed Solution

$x^{2} - 4 x + 3 > 0 \Rightarrow x < 1 or x > 3$

Now $x^{2} - 2 x - 8 \leq 0 \Rightarrow - 2 \leq x \leq 4$

$∴ x \in [- 2, 1) \cup (3, 4]$

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The set of all real values of x satisfying the inequalities x2 - 4x + 3 > 0 and x2 - 2x - 8 ≤ 0 is

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The set of all real values of x satisfying the inequalities x²- 4x + 3 > 0 and x²- 2x - 8 ≤ 0 is