The set of all real values of x which satisfy x-2x-3 - 2 - x-1x-2- is

(x 2)(x 3) ≤ 2 ≤ (x 1)(x 2) (i) (x 2)(x 3) ≤ 2 ⇒ x^2 5x+6 ≤ 2 ⇒ x^2 5x+4 ≤ 0 ⇒ (x 1)(x 4) ≤ 0 ⇒ x ∈ [1, 4] ⋯ (1) (ii) 2 ≤ (x 1)(x 2) ⇒ x^2 3x+2 ≥ 2 ⇒ ...

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The set of al...

Question

The set of all real values of $x$ which satisfy $(x - 2) (x - 3) \leq 2 \leq (x - 1) (x - 2),$ is

[3, 4]

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[1, 3]

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[1, 4]

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[0, 3]

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Solution

The correct option is A $[3, 4]$
$(x - 2) (x - 3) \leq 2 \leq (x - 1) (x - 2)$

$(i) (x - 2) (x - 3) \leq 2$
$\Rightarrow x^{2} - 5 x + 6 \leq 2 \Rightarrow x^{2} - 5 x + 4 \leq 0 \Rightarrow (x - 1) (x - 4) \leq 0 \Rightarrow x \in [1, 4] \dots (1)$

$(i i) 2 \leq (x - 1) (x - 2)$
$\Rightarrow x^{2} - 3 x + 2 \geq 2 \Rightarrow x^{2} - 3 x \geq 0 \Rightarrow x (x - 3) \geq 0 \Rightarrow x \in (- \infty, 0] \cup [3, \infty) \dots (2)$

From $(1)$ and $(2),$
$x \in [3, 4]$

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The set of all real values of x which satisfy (x−2)(x−3)≤2≤(x−1)(x−2), is

The correct option is A [3,4](x−2)(x−3)≤2≤(x−1)(x−2) (i) (x−2)(x−3)≤2 ⇒x2−5x+6≤2⇒x2−5x+4≤0⇒(x−1)(x−4)≤0⇒x∈[1,4] ⋯(1) (ii) 2≤(x−1)(x−2) ⇒x2−3x+2≥2⇒x2−3x≥0⇒x(x−3)≥0⇒x∈(−∞,0]∪[3,∞) ⋯(2) From (1) and (2), x∈[3,4]

The set of all real values of $x$ which satisfy $(x - 2) (x - 3) \leq 2 \leq (x - 1) (x - 2),$ is