All real values of x which satisfy x2 -3x-2 0 and x2 -3x -4 - 0 lie in the interval

X^2 3x+2 0 and x^2 3x 4 ≤ 0 x^2 3x+2 0 ⇒ (x 1)(x 2) 0 ⇒ x ∈ ( ∞, 1) ∪ (2, ∞) ⋯ (1) x^2 3x 4 ≤ 0 ⇒ (x +1)(x 4) ≤ 0 ⇒ x ∈ [ 1, 4] ⋯ (2) From (1) and ...

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Standard XIII

Mathematics

Method of Intervals

All real valu...

Question

All real values of $x$ which satisfy $x^{2} - 3 x + 2 > 0$ and $x^{2} - 3 x - 4 \leq 0$ lie in the interval

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

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[3, 6]

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(- 5, 0)

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[1, 6]

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Solution

The correct option is A $[- 1, 1) \cup (2, 4]$
$x^{2} - 3 x + 2 > 0$ and $x^{2} - 3 x - 4 \leq 0$
$x^{2} - 3 x + 2 > 0 \Rightarrow (x - 1) (x - 2) > 0 \Rightarrow x \in (- \infty, 1) \cup (2, \infty) \dots (1)$
$x^{2} - 3 x - 4 \leq 0 \Rightarrow (x + 1) (x - 4) \leq 0 \Rightarrow x \in [- 1, 4] \dots (2)$
From $(1)$ and $(2)$ ,
$x \in [- 1, 1) \cup (2, 4]$

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Method of Intervals

Standard XIII Mathematics

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All real values of x which satisfy x2−3x+2>0 and x2−3x−4≤0 lie in the interval

The correct option is A [−1,1)∪(2,4]x2−3x+2>0 and x2−3x−4≤0 x2−3x+2>0⇒(x−1)(x−2)>0⇒x∈(−∞,1)∪(2,∞)⋯(1) x2−3x−4≤0⇒(x+1)(x−4)≤0⇒x∈[−1,4]⋯(2) From (1) and (2), x∈[−1,1)∪(2,4]

All real values of $x$ which satisfy $x^{2} - 3 x + 2 > 0$ and $x^{2} - 3 x - 4 \leq 0$ lie in the interval